Physiology MCQs ANZCA Flashcards

Question

BP12 [Feb12] Which of the following is not true regarding intracellular organelles: A. The Endoplasmic reticulum is involved in protein synthesis B. The Golgi apparatus ..?.. C. ..Maybe an option about gene transcription..?.. D. All cells contain a nucleus E. ..something true about peroxisome?

Answer 1

A. True Rough ER is involved in protein synthesis | D. False Anucleate cells include RBCs, platelets, outer layer of epidermis.

Answer 2

``` B. ECG changes with hypokalaemia prolongation of the PR interval ST segment depression T wave: decreased T wave amplitude, late inversion prominent U waves If the T and U waves merge, the apparent QT interval is prolonged, but, if they are separated, the QT interval is seen to be normal. In addition, hypokalaemia: hyperpolarise the membrane causes ventricular extrasystoles Hypokalaemia does: NOT cause q waves NOT shorten QRS NOT prolong QT Does that mean both hypo/hyperkalemia prolong PR segment? As every other source I've read says increased K increases PR segment, and then Ganong says low K increases PR ```

Answer 3

Firstly the idea of a semipermeable membrane is that the solvent water can cross the membrane but the solutes cannot. A. - All the other options are wrong (see below) so by excluson this option is the correct one. However, the remembered wording is very poor and is wrong. What the option should say is something like: "A suitable hydrostatic pressure applied to A will stop the osmotic movement of water from B into A". This afterall is essentially the definition of "osmotic pressure". However the wording of "stop osmotic pressure" is wrong and this is probably because the idea of the option has been remembered rather than the exact wording. B. - WRONG. Net water movement is by diffusion from B->A as determined by the osmotic gradient. This is not bulk flow but due to diffusion of water molecules. C. - WRONG. Water moves from B->A (down its osmotic gradient) so the fluid level in B will drop. D. - WRONG but "technically correct". The [Na+] in A will decrease as water moves into the solution. However, strictly speaking, NaCl is TOTALLY ionised in water so there is in reality NO undissociated NaCl present in the solution, only Na+ and Cl- ions. So the [NaCl] = 0 throughout - it "remains the same". However the examiners are unlikwely to be asking for your knowledge of this situation. E. - WRONG but "technically correct". Water moves BOTH ways across a semi-permeable membrane (by diffusion) but where there is an osmotic gradient (ie difference in water concentrations across the membrane) there will be a NET movement of water in one direction, in this case from B->A. So presumingtb this option really refers to net movement then it is incorrect. Comment Depends on what you think semi-permeable means- permeable to water, permeable to small cations? If you think the membrane will be permeable to Na/Cl - then i think B will be right. Comment Bulk flow refers to movement of water and solutes together down a pressure gradient. This is seen in the circulation of CSF. As solute movement does not occur in this scenario, bulk flow is not present. Diffusion refers to the net movement of water down a concentration gradient due to the random movement of individual molecules. If the examiners are referring to net movement (in this admittedly poorly phrased question), then this option is potentially correct. The osmotic pressure refers to the tendancy of water to diffuse across the membrane. It is defined by the hydrostatic pressure that must be applied to the recipient solution in order to prevent said movement. For a solution with osmolality of 287mOsmol/L, this is 5547mmHg. If the examiners are indeed referring to the hydrostatic pressure required to prevent the movement of water from B to A, this option is the most correct. All we can be sure of is that the question will be better written on the day!

Answer 4

N/saline is distributed throughout the ECF only as its [Na+] restricts its distribution. In contrast, note that 'pure water' is distributed throughout the total body water so each compartment would obtain an amount of the infused fluid in proportion to its % contribution to total body water. (You would not want to infuse "pure water" as it could cause haemolysis but you could drink a couple of litres of pure water). Note that giving IV 5% dextrose solution is the equivalent of infusing 'pure water' as the glucose is rapidly taken up by cells. Infusing N/saline would not cause osmotic shifts of water as its osmolality is the same as that of body fluids. (This may not be true if the person is either hyponatraemic or alternatively hypertonic for some reason but you can assume it is true for the normal situation). So ECF [Na+] would not change. The distribution of the infused N/saline would be totally to the ECF. The ICF volume would not change. So we would get increased ECF, unchanged ICF, unchanged [Na+] which is the correct option. Comment Technically you should get a slight rise in ECF [Na+] as [Na+] in N/Saline is 150 mmol/L, compared with 140 mmol/L in plasma Question says 'rapid'. In other words, kidneys have no time to compensate for the additional [Na+] load. Thus expected increase in [Na+] = 20mmol/14L = 1.43 mmol/L Further Comment: Ref: The Australian and New Zealand Primary Exam by Faunce pg 119 Rapid infusion of 2L NS causes inc ECF, unchanged ICF, and UNCHANGED [Na+] due to plasma solids effect Alternative view As per Brandis (2nd edition), the plasma consists of 93% plasma water and 7% plasma solids (protein and lipid). However, measurement of Na+ by flame emission spectrophotmetry or ISE measure it as a whole. Hence, the measured plasma Na+ appears less than the actual amount of Na+ contained in the plasma. Again as per Brandis "...the decrease in measured plasma Na+ due to the plasma solids effect is about the same as the increase in Na+ that occurs due to the Gibbs-Donnan effect...", with the result being that measured plasma Na+ is about the same as measured ISF Na+. Hence, the postulated reduction in plasma [Na+] that is present as a result of the plasma solids effect is negated, and we can effectively treat the system (for measurement purposes only) as though the two do not exist. Consequently, infusing a solution that contains 150mmol/L of NaCl into a compartment that contains 140mmol/L of NaCl will result in: Expansion of ECF No change in ICF (despite a measured osmolality of 300mOsmol/L, this is not true in-vivo) Increase in measured [Na+] The above relies on the fact that NaCl dissociates completely in solution and in vivo. However, this may not be the case as: Serum osmolality is 285 mOsmol/L, and whilst 0.9% saline has an osmolality of 300mOsmol/L it is still considered isotonic; The solution contains 154 mmol/L of Na+ and 154 mmol/L of Cl-, yet has an osmolality of 300mOsmol (should be 308mOsmol if complete dissolution were to occur). Can anyone reconcile the above? Preferably within the next 7 days?

Answer 5

B (?before E, also correct) Not A/C Not D - may be a complication of digoxin toxicity, but hypokalemia potentiates digoxin toxicity

Answer 6

B (although I have also read A in a few areas too -- for e.g. Brandis page 15 under "Potassium" ECG changes with hyperkalaemia (in order that they occur) Tall peaked T waves loss of the P wave prolongation of the QRS complex ventricular arrhythmias - tachycardias and fibrillation asystole

Answer 7

A. Clotting factors - Yes. All Lymph contains clotting factors (just low fibrinogen) B. Higher protein content than plasma – WRONG: Always lower even in hepatic lymph C. Similar composition to ISF – WRONG: Thoracic duct lymph contains more protein than usual ISF because of the contribution of hepatic lymph which has a relatively high protein concentration (as compared to lymph from the rest of the body). D. Rarely contains fat – WRONG: Lymphatics from small bowel contains fat (chylomicrons) after meals – important in fat absorption (approx 90% fat absorbed this way) E. ? Thoracic duct lymph contains protein about 50g/L (hepatic lymph 60g/L), whereas lymph from muscle/skin contains 20g/L Plasma contains about 70g/L protein. Albumin 40g/L Globulins 25g/L Fibrinogen 5g/L

Answer 8

Gibbs-Donnan leads to : E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood. Gibbs-Donnan effect The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules. Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. "About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone." (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno). I thought that 'oncotic pressure' a.k.a. 'colloid osmotic pressure' was osmotic pressure due to proteins alone, whereas 'osmotic pressure' refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me. Re: Aug 2011 version: A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium B. (seems like an incomplete option) C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells. D. ? E. ? Best answer: 'A' Alternate view - Why couldn't B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?

Answer 9

Gibbs-Donnan leads to : E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood. Gibbs-Donnan effect The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules. Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. "About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone." (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno). I thought that 'oncotic pressure' a.k.a. 'colloid osmotic pressure' was osmotic pressure due to proteins alone, whereas 'osmotic pressure' refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me. Re: Aug 2011 version: A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium B. (seems like an incomplete option) C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells. D. ? E. ? Best answer: 'A' Alternate view - Why couldn't B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?

Answer 10

Gibbs-Donnan leads to : E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood. Gibbs-Donnan effect The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules. Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. "About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone." (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno). I thought that 'oncotic pressure' a.k.a. 'colloid osmotic pressure' was osmotic pressure due to proteins alone, whereas 'osmotic pressure' refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me. Re: Aug 2011 version: A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium B. (seems like an incomplete option) C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells. D. ? E. ? Best answer: 'A' Alternate view - Why couldn't B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?

Answer 11

Gibbs-Donnan leads to : E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood. Gibbs-Donnan effect The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules. Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. "About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone." (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno). I thought that 'oncotic pressure' a.k.a. 'colloid osmotic pressure' was osmotic pressure due to proteins alone, whereas 'osmotic pressure' refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me. Re: Aug 2011 version: A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium B. (seems like an incomplete option) C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells. D. ? E. ? Best answer: 'A' Alternate view - Why couldn't B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?

Answer 12

Decreased urine output due to increased ADH secretion. Osmoreceptors are very sensitive to a decrease in osmolality and the decreased osmolality will result in a decreased ADH secretion. However, the low pressure baroreceptors (volume receptors) in the right atrium and great veins though being less sensitive than the osmoreceptors respond far more powerfully. The peak [ADH] resulting from hypovolaemia is much higher than that produced by maximal response to osmoreceptor input. So the knowledge this question is testing is whether you know that hypovolaemia over-rides the inhibition from the hypo-osmolality. Volume is maintained at the expense of a decreased osmolality. But either way it is true that permeability of collecting duct to water is increased, therefore correct answer D.

Answer 13

Answers 1st wording - none correct Edit - dont think so, Hartmanns is actually hypotonic relative to plasma, I would go with B Hartmann's solution contains Ca++ but no Mg++. The Ca++ content of Hartmanns is important as it can result in clotting in the tubing if blood is given slowly before or after it in the same line. Because of this problem the traditional nursing practice in the past was to give a 100mls of normal saline before and after each bag of blood. This clotting problem is less of a concern during anaesthesia if we fluids are given quickly (and since we don't give whole blood much anymore). Some people use ‘Plasmalyte 148’ which contains Mg++ but no Ca++. The other difference between these 2 solutions is the slightly lower [Na+] in Hartmann’s solution: 130mmol/l compared to 140mmol/l in Plasmalyte. Some neuroanaesthetists prefer not to use Hartmanns because of this. Hartmann's Solution Solute Concentration Na+ 129 mmol/L K+ 5 mmol/L Ca2+ 2 mmol/L Cl- 109 mmol/L Lactate 29 mmol/L (converted into HCO3- via oxidation and gluconeogenesis by the liver) pH 6.5 Osmolarity 279 mosm/L Comment: Most sources (inc. Sasada & Smith) say Na-131mM, Cl-111mM.

Answer 14

Answers 2nd wording - no magnesium Edit - dont think so, Hartmanns is actually hypotonic relative to plasma, I would go with B Hartmann's solution contains Ca++ but no Mg++. The Ca++ content of Hartmanns is important as it can result in clotting in the tubing if blood is given slowly before or after it in the same line. Because of this problem the traditional nursing practice in the past was to give a 100mls of normal saline before and after each bag of blood. This clotting problem is less of a concern during anaesthesia if we fluids are given quickly (and since we don't give whole blood much anymore). Some people use ‘Plasmalyte 148’ which contains Mg++ but no Ca++. The other difference between these 2 solutions is the slightly lower [Na+] in Hartmann’s solution: 130mmol/l compared to 140mmol/l in Plasmalyte. Some neuroanaesthetists prefer not to use Hartmanns because of this. Hartmann's Solution Solute Concentration Na+ 129 mmol/L K+ 5 mmol/L Ca2+ 2 mmol/L Cl- 109 mmol/L Lactate 29 mmol/L (converted into HCO3- via oxidation and gluconeogenesis by the liver) pH 6.5 Osmolarity 279 mosm/L Comment: Most sources (inc. Sasada & Smith) say Na-131mM, Cl-111mM.

Answer 15

Answers 3rd wording - both A and B are correct Edit - dont think so, Hartmanns is actually hypotonic relative to plasma, I would go with B Hartmann's solution contains Ca++ but no Mg++. The Ca++ content of Hartmanns is important as it can result in clotting in the tubing if blood is given slowly before or after it in the same line. Because of this problem the traditional nursing practice in the past was to give a 100mls of normal saline before and after each bag of blood. This clotting problem is less of a concern during anaesthesia if we fluids are given quickly (and since we don't give whole blood much anymore). Some people use ‘Plasmalyte 148’ which contains Mg++ but no Ca++. The other difference between these 2 solutions is the slightly lower [Na+] in Hartmann’s solution: 130mmol/l compared to 140mmol/l in Plasmalyte. Some neuroanaesthetists prefer not to use Hartmanns because of this. Hartmann's Solution Solute Concentration Na+ 129 mmol/L K+ 5 mmol/L Ca2+ 2 mmol/L Cl- 109 mmol/L Lactate 29 mmol/L (converted into HCO3- via oxidation and gluconeogenesis by the liver) pH 6.5 Osmolarity 279 mosm/L Comment: Most sources (inc. Sasada & Smith) say Na-131mM, Cl-111mM.

Answer 16

One source gives a total plasma osmotic pressure of plasma is about 5562mmHg or 7.31 atmospheres of 733kPa. This can be calculated using the van't Hoff equation and the actual value in a person depends on the plasma osmolality. See the Fluid Physiology tutorial on this site (section 2.4.2) where a value of 5409 mmHg is calculated for a plasma osmolality of 280 mOsm/kg. Using this equation, the value of 5562mmHg quoted above is the value obtained when the osmolality is 288mOsm/kg. The oncotic P is the component due to colloids is 25 to 28mmHg, or 0.5% of the total (ie 19.3 x plasma osmolality due to colloids) Using Van't Hoff in KB's book a total plasma osmotic pressure of 5900mmHg would mean that the plasma osmolality would have to be 305 mOsm/kg. This sounds high but is actually the upper end of the physiological range quoted in Yentis A-Z (280-305 mOsm/kg). This could mean that answer C is close enough to be true. NB:More likely is that the actual value on the paper was lower than the remembered 5,900mmHg value quoted in this reconstructured version of the MCQ. So option C is correct but may have been somrthing like 5,500mmHg on the paper. Surely given the figures above assuming that they are remembered correctly then E is the correct answer as working backwards 5900 mmHg gives a 305mOsm/kg which is above the normal or 288-293 mOsm/kg depending on the text you read. Obviously if option C had read 5500mmHg then you could say that is was correct, but at 5900mmHg its a bit high

Answer 17

Normal saline pH range 4.0 - 7.0 Osmolality 300mOsm

Answer 18

Comment: All figures seemed slightly off from standard text Response: Thats because 4 of the options have to be wrong. The actual question will expect an answer consistent with values in the ANZCA 'Recommended texts'. Don't grieve too much over the values in this remembered version; just use this MCQ as a reminder to learn the standard values so you can answer the question if it appears. Daily Water Loss (Power & Kam) 900ml insensible loss from skin and lungs 50ml as sweat in normal climates 100ml in faeces 430ml urine - minimum volume to excrete the daily solute load (600mOsmol / 1400mosmol/kg) It seems that 500ml for urine loss is probably the closest answer! (Comment - 500mls is the minimum daily urine, not normal daily loss, which is about 1.5L/day. Food for thought.) Fluid Output (for a normal individual): (Guyton & Hall) insensible-skin 350ml insensible-lung 350ml sweat 100ml faeces 100ml urine 1400ml I think that 700ml from the lung and skin is therefore the best answer. I think the wording of the 1st question as "obligatory" water loss would suggest that the answer is urine, rather than insensible losses from skin/lung (see link to fluid physiology notes below) "An obligatory urine loss occurs because of the need to remove various solutes from the body" Comment: totally agree. Obligatory refers to urine excretion which is obligatory to prevent toxic buildup of certain byproducts of metabolism. Maximum human urine osmolality is ~ 1400mOsm/L, and at that concentration we would have to produce 400-500ml urine at least to remove adequate waste products.

Answer 19

Comment: All figures seemed slightly off from standard text Response: Thats because 4 of the options have to be wrong. The actual question will expect an answer consistent with values in the ANZCA 'Recommended texts'. Don't grieve too much over the values in this remembered version; just use this MCQ as a reminder to learn the standard values so you can answer the question if it appears. Daily Water Loss (Power & Kam) 900ml insensible loss from skin and lungs 50ml as sweat in normal climates 100ml in faeces 430ml urine - minimum volume to excrete the daily solute load (600mOsmol / 1400mosmol/kg) It seems that 500ml for urine loss is probably the closest answer! (Comment - 500mls is the minimum daily urine, not normal daily loss, which is about 1.5L/day. Food for thought.) Fluid Output (for a normal individual): (Guyton & Hall) insensible-skin 350ml insensible-lung 350ml sweat 100ml faeces 100ml urine 1400ml I think that 700ml from the lung and skin is therefore the best answer. I think the wording of the 1st question as "obligatory" water loss would suggest that the answer is urine, rather than insensible losses from skin/lung (see link to fluid physiology notes below) "An obligatory urine loss occurs because of the need to remove various solutes from the body" Comment: totally agree. Obligatory refers to urine excretion which is obligatory to prevent toxic buildup of certain byproducts of metabolism. Maximum human urine osmolality is ~ 1400mOsm/L, and at that concentration we would have to produce 400-500ml urine at least to remove adequate waste products.

Answer 20

A. False B. False - osmolality of Hartmann's is 274mOsm C. False - 0.9% saline pH range 4.0 - 7.0 D. True - 0.9% sailne osmolality 300mOsm

Answer 21

``` Based on Ganong: GFR = 180 litres/day Urine = 1.0 litres/day Percent reabsorption = 99.4% Other figures calculated: 99.7% if assume 500ml obligate urine output 99.2% if assume 1500ml urine output) The 99.9% option is incorrect as the urine volume is below that required for the minimum urine volume to excrete the daily solute load (typically 600-700 mosmoles/day) ```

Answer 22

``` Intracellular Concentration Na+ = 10mmol/L HCO3- = 10mmol/L Ca+2 = 100nmol/L (note: nanomoles/l) Mg+2 = 10mmol/L K+ = 150mmol/l Answer is C - Calcium has the lowest intracellular concentration ```

Answer 23

Total Plasma Osmolarity can be calculated by several formulae, one of which is: Osmolarity (mOsmol/kg) = [2 x (Na + K)] + urea + glucose according to fluid website ( link below): Each mOsm/kg of solute contributes about 19.32mmHg to the osmotic pressure (derived from van't Hoff eqn) So maybe close enough for 19.20 to be correct? Addition: the wording in the question is a bit confusing. If the answer is D the question must have been asking about total plasma osmotic pressure. Osmotic pressure can be measured (eg oncometer) or calculated using the Vant Hoff equation (ref Physiology viva, KB). Osmolarity can be measured (eg freezing point depression) or calculated: Osmolarity (mOsmol/kg) = [2 x (Na + K)] + urea + glucose Answer: Depends on the wording of the question If question says: Osmolality: E - none of the above - osmolality can never be calculated Osmolarity: E - none of the above - an appropraite formula may be = 2(Na+K) + glucose + urea, for example Osmotic pressure: D - mOsm x 19.3 = osmotic pressure (in mmHg). Or could use Hoff's Law: π = cRT JB2012

Answer 24

Reduce plasma potassium Beta adrenergic receptor AGONIST Insulin Aldosterone Hyperglycaemia Carbonic anhydrase inhibitors Norepinephrine and epinephrine cause an initial rise via liver release and then a prolonged fall in plasma potassium due to skeletal muscle (via B2 adrenergic receptors) Ganong pg 360 Beta receptor agonism causes increased movement of potassium intracellularly, presumably due to increased Na/K ATPase activity, as well as increasing insulin release from the pancreas (which has a similar effect). Think salbutamol in hyperkalaemia. Increase plasma potassium Metabolic acidosis http://www.aic.cuhk.edu.hk/web8/Potassium%20metabolism.htm and haemolysis - think haemolysed sample K levels in U+E's.

Answer 25

Reduce plasma potassium Beta adrenergic receptor AGONIST Insulin Aldosterone Hyperglycaemia Carbonic anhydrase inhibitors Norepinephrine and epinephrine cause an initial rise via liver release and then a prolonged fall in plasma potassium due to skeletal muscle (via B2 adrenergic receptors) Ganong pg 360 Beta receptor agonism causes increased movement of potassium intracellularly, presumably due to increased Na/K ATPase activity, as well as increasing insulin release from the pancreas (which has a similar effect). Think salbutamol in hyperkalaemia. Increase plasma potassium Metabolic acidosis http://www.aic.cuhk.edu.hk/web8/Potassium%20metabolism.htm and haemolysis - think haemolysed sample K levels in U+E's.

Answer 26

This CANNOT be "osmotic pressure" as ISF and plasma have the same osmolality so must have the same osmotic pressure. The stem must be about "oncotic pressure". Additionally, osmotic pressure has pressure units (eg mmHg) and NOT concentration units as here. Disagree, the above statement is wrong. Guyton and Hall "textbook of medical physiology" (11th ed 2006)- sites the difference between as being about 1mosmol/L and that plasma proteins is the cause of this. States that it is responsible for about 20mmHg greater pressure in the capillaries than in the interstitial fluid.

Answer 27

Clearly these options have been fairly poorly recalled. I think the answers should be: 1st version: A) wrong (anterior hypothalamus not posterior) B) true C) false (baroreceptor afferents go to medullary vasomotor centre) D) & E) false as ADH and aldosterone do not directly affect thirst although these are both TRUE if the question really is "response to osmolarity" as they are both responses to this Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst: Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes Increases in ADH Dry mucous membranes Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly "prandial drinking" (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above. It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT. The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia. The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region. ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei. The key parts of the hypothalamus involved in water balance are: Osmoreceptors Thirst centre OVLT & SFO (respond to angiotensin II) Supraoptic & paraventricular nuclei (for ADH synthesis)

Answer 28

Alt version 03A A) false - ADH produced from hypothal, secreted from post pituitary B) see above C) ?? true depending on wording. Thirst is in part regulated by angiotensin Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst: Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes Increases in ADH Dry mucous membranes Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly "prandial drinking" (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above. It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT. The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia. The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region. ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei. The key parts of the hypothalamus involved in water balance are: Osmoreceptors Thirst centre OVLT & SFO (respond to angiotensin II) Supraoptic & paraventricular nuclei (for ADH synthesis)

Answer 29

Alt version: A) true (subfornical organ & OVLT are cirumventricular organs) Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst: Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes Increases in ADH Dry mucous membranes Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly "prandial drinking" (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above. It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT. The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia. The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region. ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei. The key parts of the hypothalamus involved in water balance are: Osmoreceptors Thirst centre OVLT & SFO (respond to angiotensin II) Supraoptic & paraventricular nuclei (for ADH synthesis)

Answer 30

Re 2004B version: A) false B) true C) false D) false (see above for all) Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst: Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes Increases in ADH Dry mucous membranes Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly "prandial drinking" (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above. It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT. The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia. The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region. ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei. The key parts of the hypothalamus involved in water balance are: Osmoreceptors Thirst centre OVLT & SFO (respond to angiotensin II) Supraoptic & paraventricular nuclei (for ADH synthesis)

Answer 31

Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst: Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes Increases in ADH Dry mucous membranes Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly "prandial drinking" (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above. It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT. The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia. The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region. ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei. The key parts of the hypothalamus involved in water balance are: Osmoreceptors Thirst centre OVLT & SFO (respond to angiotensin II) Supraoptic & paraventricular nuclei (for ADH synthesis)

Answer 32

``` Re Aug14 version: A - T B - F - C - F - sensed by hypothalamus D - F - ATII stimulate SFO and OVLT ``` Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst: Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes Increases in ADH Dry mucous membranes Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly "prandial drinking" (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above. It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT. The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia. The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region. ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei. The key parts of the hypothalamus involved in water balance are: Osmoreceptors Thirst centre OVLT & SFO (respond to angiotensin II) Supraoptic & paraventricular nuclei (for ADH synthesis) Regarding whether angiotensin II affects SFO or OVLT osmoreceptors or BOTH to stimulate thirst: Ganong says there is evidence that OVLT, not just SFO, are stimulated by ATII to produce thirst But this study [1] suggests it's only SFO that stimulates thirst However Ganong is a prescribed text and therefore correct in exam parlance Also, regarding the primary thirst-stimulating mechanism in hyperosmolar states Ganong states that thirst is due to (a) osmoreceptors and (b) ATII The response due to osmolarity appears to be the primary response The OVLT and SFO appear to be in the anterior and lateral but not posterior hypothalamus Referenced from Ganong 21st ed p244

Answer 33

FE20 From 'The Physiology Viva' by Kerry Brandis (p17): Mg++ is the major intracellular divalent cation. Intracellular function Catalysing Mg++ dependent enzymes all enzymes for phosphate transfer all enzymes requiring thiamine phosphorylase as a co-factor (therefore Na+/K+ pump, oxidative phosphorylation, all reactions involving ATP) acts as 'plug' in NMDA receptors ->voltage-dependent block of channel. Extracellular function Reduces nerve and membrane excitability (similar to but

Answer 34

FE20b | B. Correct - cofactor in Na/K/ATPase

Answer 35

B: Gibbs Donnan effect: 'because of proteins in cells, there are more osmotically active particles in cells than in interstitial fluid, osmosis would make them swell and rupture if not for Na/K/ATPase pumping ions back out' (p6, Ganong 20thed)

Answer 36

As far as I can find: At rest loss of [Na] is approximately 11 mmols/l by sweating (physiology for the anaesthetist). However this can vary from 30 to 65 mmols/l depending on the level of acclimatisation. Aldosterone decreases the [Na+] of sweat. This concentration is less than plasma [Na] concentration of 140 mmols/l Therefore A is correct

Answer 37

"For the DI question there were a set of normal electrolytes as an option too - which is missing from the bank thus far (the problem with the discussion previously on the Bulletin Board ie the most 'normal' set of electrolytes were hyponatraemic/hypoosmotic - if your thirst mechanism is intact and you have access to water you have normal electrolytes but tend towards hypernatraemia/hyperosmolality. The actual numbers in the MCQ were Na 140 K 3.5 Osm 300. (One of the options also had a K of 6.0!)" "I must admit I still answered normal electrolytes in spite of the 'untreated' bit in the Q since drinking is part of the normal physiology of DI (most DI's are not the head injured ventilated pt's we tend to see in ICU but rather the compensating DI's in renal clinics)... ie I regarded drinking lots as normal physiology for DI, not a treatment for it." Comment Ganong 22nd ed pg 247 states 30% are neoplastic, 30% post-traumatic, 30% idiopathic, remainder vascular lesions, infections, systemic diseases eg sarcoidosis. Surgical removal of post hypothesis does not guarantee permanent DI as suprooptic and paraventricualr fibres can regrow. Contrary View: Plasma osmolality is high, Urine osmolality is low,Plasma sodium may be high, Dehydrated. Case report notes K high: http://adc.bmj.com/cgi/content/full/80/6/548 Perhaps D? - I agree that the blood results would likely be abnormal. In a normal physiological state, there is ADH, thirst, ATII, etc, etc, so I think even if you were able to drink water, without one of the other fine-controlling mechanisms you would have at least a partially abnormal end result Oh´s Intensive Care Manual: "The plasma osmolality measures in central DI is usually in the higher regions of the normal range or very slighltly supranormal" "Hyperosmolality or hypernatraemia suggest impaired sensation of thirst or inabilty to access water"--Kajuprim 20:52, 26 Feb 2010 (EST)

Answer 38

The Colligative properties of a solution are those properties that depend ONLY on the particle concentration (number of solute particles per unit volume) - and NOT on the chemical properties of the substance or size of the particles. (as concentration of particles increases) These properties are: Freezing Point Depression Boiling Point Elevation Vapour Pressure Depression – reduction of the solvent molecules ability to leave the solution Osmotic pressure

Answer 39

Mg++ is an essential co-factor for the Na+-K+ ATPase (cell membrane sodium pump). Perhaps options A or B were an ion of ATP?

Answer 40

Firstly: The MW of NaHCO3 is 84 so a one molar solution would contain 84 grams in a liter (ie 8.4 grams in a 100mls which is an 8.4% solution). Secondly: Sodium bicarbonate splits in two ions on dissolving: Na+ and HCO3-. So an 8.4% solution of sodium bicarbonate contains 1,000mmols of Na+ and 1,000mmoles of HCO3-. The osmolality is this 2 Osm/litre (or 2,000 mOsm/litre). A 100mls of this solution is an osmotic load of 200 Mosmoles.

Answer 41

FE27 Answer = B. (don't think this is right- hyperkalaemia doesn't cause alkalosis- can't remember where I read from but it says something along the lines that in an attempt to correct acidosis, the cells will take H+ in in exchange for a K+ ion- hence acidosis is a/w high K+.) Metabolic acidosis leads to hyperkalaemia due to compartment shifts. K+ shifts out of cells causing [K+] increase by 0.6mmol/L per each pH decrease by 0.1 (effect is greater with non-organic acids eg KCl - Cl is obligatory extracellular anion so as H+ taken up by cells, K+ must move out to maintain neutrality) K+ depletion stimulates H+ secretion however hypokalaemia decreases aldosterone release which decreases H+ secretion. Combination of hypokalaemia and secondary hyperaldosteronism synergistically to stimulate H+ secretion causing metabolic alkalosis. Ref Mark Finnis notes Interrelation of potassium and hydrogen ion gradients in metabolic alkalosis Jared J. Grantham 1 and Paul R. Schloerb 1 To study the interrelationships of potassium and hydrogen ion intra- and extracellularly, as an extension of previous investigations on gastric alkalosis, metabolic alkalosis was produced in dogs by dietary K depletion, NaHCO3 loading, and deoxycorticosterone administration, with and without chloride depletion. Analysis of skeletal muscle and measurement of intracellular pH confirmed the development of cellular K depletion, with increases of intracellular sodium and hydrogen ion associated with extracellular alkalosis and hypokalemia. These changes were not influenced by concomitant dietary chloride restriction. The intracellular K concentration was linearly and inversely related to the intra-extracellular hydrogen ion gradient. The intra-extracellular K gradient was about 10 times greater than the corresponding hydrogen ion gradient. These respective gradients were linearly related, in this proportion, over the range of K depletion studied in this type of metabolic alkalosis as well as that resulting from gastric juice loss. Deviation from this relationship occurred only with the most severe K depletion. These experimental observations demonstrate in vivo that the potassium and hydrogen ion gradients change in the same direction and are nearly constantly related in at least two types of metabolic alkalosis. Kerry says Acidosis is commonly said to cause hyperkalaemia by a shift of potassium out of cells. The effect on potassium levels is extremely variable and indirect effects due to the type of acidosis present are much more important. For example hyperkalaemia is due to renal failure in uraemic acidosis rather than the acidosis. Significant potassium loss due to osmotic diuresis occurs during diabetic ketoacidosis and the potassium level at presentation is variable (though total body potassium stores are invariably depleted).

Answer 42

FE27b A. Partly correct - plasma H+ and K+ are associated through an unknown mechanism, possibly modulating the cellular Na-K-ATPase, but this is not a direct linear correlation as implied in the option B. Partly correct - acidosis increases plasma concentration and renal filtration of potassium, but low intracellular pH inhibits Principal cell capacity to secret K+ (paradoxical potassium retention) "These topics are fraught with difficulty because the effects are not consistently seen." C. Wrong - insulin increases cellular uptake of potassium (important for plasma potassium regulation with meal boluses), but does not have a direct affect on the interaction between potassium and H+ D. Wrong - hypokalaemia stimulates H+ excretion and HCO3- generation in tubular cells, itself generating a metabolic alkalosis E. ? This is a problematic topic, but my "most correct" answer would be: A

Answer 43

A is correct, B is false (aldosterone causes Na retention and K excretion) From Vander p147: Low plasma pH is usually associated with hyperkalaemia. There are 2 known reasons for the effects of acid base status on potassium. changes in extracellular [H+] cause exchange of H+ with cellular cations (mainly K+). Acidosis results in H+ being taken up by cells which is balanced by efflux of K+ from the cell. intracellular pH inhibits Na-K-ATPase contributing to K+ loss from cells and an inability of cells to uptake extracellular K+. Principal cells have their Na-K-ATPase pumps and luminal membrane K+ channels inhibited which results in paradoxical K+ retention From Vander p141: Aldosterone stimulates K+ secretion by the principal cells by 3 mechanisms: activation of apical K+ channels (ROMK). stimulation of basolateral membrane Na-K-ATPase pumps which increases intracellular K+ and the gradient driving K= secretion into the lumen. increases the activity or number of luminal membrane Na channels which allows more Na to enter the principal cell to then be pumped out by the Na-K-ATPase basolaterally (K+ secretion is dependent on Na+ being pumped out so more Na+ entering the cell allows more K+ to leave)

Answer 44

Answer = A Rate of diffusion = K. A. (P2-P1)/D Fick's law states that the rate of diffusion of a gas across a membrane is: Constant for a given gas at a given temperature by an experimentally determined factor, K Proportional to the surface area over which diffusion is taking place, A Proportional to the difference in partial pressures of the gas across the membrane, P2 − P1 Inversely proportional to the distance over which diffusion must take place, or in other words the thickness of the membrane, D. The exchange rate of a gas across a fluid membrane can be determined by using this law together with Graham's law. I looked at the formula and this is not what I got used to from J West book: Diffusion through a tissue sheet= A x D x ( P1-P2)/ T where A - area D - diffusion constant ( P1- P2) the difference in partial pressure T - thickness Fick's Law of diffusion: - rate of diffusion of a gas through a tissue slice is proportional to the area but inversely proportional to the thikness. - diffusion rate is proportional to the partial pressure difference - diffusion rate is proportional to the solubility of the gas in the tissue but inversely proportional to the square root of the molecular weight. So... A is correct, B is incorrect - diffusion rate inversely proportional to the square root of the MW

Answer 45

c N/saline infusion causes a hyperchloraemic metabolic acidosis.

Answer 46

Answer: ?A Ganong states "assisted when muscles contract" (pg 593) B: This is approximately the interstitial pressure and a higher pressure would reduce lymph formation by decreasing the hydrostatic pressure gradient across the capillary membrane. C: Flow 100mls/hr (in ref) 120 ml/hr total lymph flow at rest includes 100 ml/hr of thoracic duct lymph Interstitial hydrostatic pressure usually b/w 0 and -3mmHg - Guyton

Answer 47

E. Free water excess Both excess lipids and excess glucose can cause a pseudohyponatraemia but the most common cause is probably water excess.

Answer 48

A - wrong - not great reference but http://lifeinthefastlane.com/ecg-library/basics/hypokalaemia/ B - seems most correct of these options C - wrong - this happens in hyperkalemia - Ganong pp566 Ed 21 D - probably wrong as chronic electrolyte changes are usually better tolerated than acute E - wrong - "RMP in hyperkalemia decreases" Ganong Generally, these questions are difficult to reference to source texts...

Answer 49

FE36 A: not sure.. hyperchloremia usually ends up with a decreased serum bicarb - normal anion gap metabolic acidosis, but is it causal? Under the Stewart acid-base system, hyperchloraemia ought to decrease the Strong Ion Difference of plasma (usually about 40 mEq/L), reducing bicarbonate concentrations (a weak ion), and so cause a metabolic acidosis. The problem is that this assumes an isolated hyperchloraemia, which is a rare event, and we don't have complete blood chemistry to really comment. Hard call. B: is true, value is 9 mM from Ganong Ed 21 pp 8 This is true, but higher in erythrocytes (~77 mM). Assuming this question was correctly recalled, B seems the "most correct" answer.

Answer 50

``` FE36b A. Partly correct - per A above B. Wrong - Cl- is an anion (but is the major anion extracellularly) C. Wrong - subject to the acid-base definition being employed, it is a strong base (taken to be "H+ acceptor" under the Brønsted-Lowry definition) D. ? E. Partly correct - per B above "Most correct" answer: E References ```

Answer 51

Not clear whether this MCQ had example values, or was just high,low, normal options This is a table straight from Ganong: In adrenal insufficiency Na low Cl low (goes with Na presumably) K high (hypoaldosteronism so Na wasted, K retained) HCO3 unchanged Ganong pp380, Ed 21

Answer 52

Dextran 40 has been shown to improve microcirculation presumably by decreasing blood viscosity thereby improving laminar flow in the microcirculatory beds. Both Dextran 40 and Dextran 70 possess antiplatelet effects and may interfere with blood typing. (Nagelhout pg. 411, M&M pg. 694) References

Answer 53

Dextran 40 has been shown to improve microcirculation presumably by decreasing blood viscosity thereby improving laminar flow in the microcirculatory beds. Both Dextran 40 and Dextran 70 possess antiplatelet effects and may interfere with blood typing. (Nagelhout pg. 411, M&M pg. 694) References

Answer 54

A - False - Normally 0.7-1.1mmol/L B - True - Slow sinoatrial conduction/increases SA node recovery time [1] C - False - Mg2+ decreases uterine tone D - False - No diuretic effects E - False - Respiratory depression at toxic doses (>3.0mmol/L) due to diaphragm paralysis from neuromuscular blockade (not direct)

Answer 55

Ans D. I think answer D is incorrect. PO2 of 40mmHg correlates with oxygen saturations of 75%. While this fits some patients who need oxygen therapy, it does not fit with the vast majority of COPD patients. I think the most correct answer is B (morphine overdose). Nope, not B, there is a compensated respiratory acidosis, has to be longer term than morphine OD. Ans IS D I agree that the answer is D, and I imagine that the HCO3 was probably included in the question. The fact that the pO2 is so low really rules out an acute morphine overdose(particularly if in conjunction with metabolic compensation HCO3 34). Using the "4 for 10" rule for Chronic Resp Acidosis and the presumptive HCO3 of 34. Expected HCO3 = 24 + 4[(60-40/10)] = 32. Close to what we have. I dont think B is correct. Using alveolar equation with Pco2 of 60, will give much higher Po2. Assuming otherwise normal lung function, morphine OD alone will nit cause this picture

Answer 56

I think the correct answer is A. Morphine overdose would cause acute hypoventilation with hypoxia (paO2 60) with an acute respiratory acidosis (pH7.29, paCO2 55). Renal compensation will take some time with increases in plasma HCO3. (BE within +/-1). B is incorrect because it shows a fully corrected respiratory acidosis with a positive base excess which usually occurs in alkalosis not acidosis. This does follow the 1 in 10 rule i.e for each 10mmHg increase in CO2 there will be 1mmol increase in HCO3...but is probably still wrong because the base excess should be negative. I disagree - both show a primary respiratory acidosis for every 10mmHg of CO2, the HCO3 will compensate by 1 acutely... thus in A, it has actually compesated by 8 which doesn't make sense in B, the CO2 is 20mmHg above 40, so you'd expect the bicarb to rise by 2, which it has. Therefore either the person in A already has some primary metabolic alkalosis before the event, or the numbers got remembered wrong (which is more likely) Both of these stems seem incorrect to me. The base excess clearly has been remembered incorrectly as it should reflect the bicarb so for option A), the base excess should be about +7 and for option B) the base excess should be about 0 +/- 1. Option A) is incorrect as usually morphine overdose is acute so there should not be this much metabolic compensation by the kidneys as yet. Option B) is incorrect because although the patient should have an acute respiratory acidosis, we would expect the pH to be lower than this for a CO2 of 60 and the base excess doesn't make any sense. Answer should be: hypoxic with acute, uncompensated respiratory acidosis and a base excess that reflects the bicarb - so neither!! Addit comment: If you calculate the Standardised Base Excess (SBE) from the data in option B it is: SBE = 0.93*[26]+13.77*[7.37]-124.58 (wikpedia in mEq/L) = +1.08 Which is correct for the situation of acute respiratory acidosis. Option B therefore fits a whole lot better (for what its worth). response to the 2nd comment above - a compensatory process does not move the pH into the normal range - in this case there must be a second acid-base disorder going on, and given the BE of +5, it's likely that this person would also have an underlying metabolic alkalosis, allowing the pH to be 7.37 in the face of a respiratory acidosis. Agree neither answer seems to fit. This should be an acute respiratory acidosis. Power & Kam re: in vivo titration curves - for every increase pCO2 by 10mmHg, corresponding decrease in pH by 0.07 (gives 7.29 for CO2 55, 7.26 for CO2 60), increase in HCO3- by 0.08 (gives 25 for CO2 55, 26 for CO2 60). BE should be negative.

Answer 57

Please Note: I took this question to mean how a bicarbonate infusion would buffer a metabolic acidosis when given for that clinical indication. Apologies if I've misinterpreted it. Notes from the Acid-Base Physiology section of Kerry Brandis' Physiology Book, also on-line at: http://www.anaesthesiamcq.com/AcidBaseBook/ab2_2.php ``` Answer A: NO Metabolic acidosis: 57% of buffering occurs intracellularly and 43% occurs extracellularly (Of the 43% EC component, 80% is performed by the bicarbonate buffer system). Metabolic alkalosis: 30% IC, 70% EC Respiratory acidosis: 99% IC, 1% EC Respiratory alkalosis: 97% IC, 3% EC ``` Answer B: YES Accounts for about 6% of buffering in metabolic acidosis "Protein buffers in blood include haemoglobin (150g/l) and plasma proteins (70g/l). Buffering is by the imidazole group of the histidine residues which has a pKa of about 6.8. This is suitable for effective buffering at physiological pH. Haemoglobin is quantitatively about 6 times more important then the plasma proteins as it is present in about twice the concentration and contains about three times the number of histidine residues per molecule." (sic) Answer C: YES The body's ability to change alveolar ventilation ("physiological buffering") rapidly is what makes the system such a good buffer despite the pKa being low at 6.1. Answer D: "YES" ICF buffering (57%) occurs by protein phosphate and bicarbonate buffers due to entry of H+ by: Na+-H+ exchange 36% K+-H+ exchange 15% Other 6% (mostly Cl-) Comment. An infusion of HCO3- will cause a metabolic alkalosis; there will be renal and respiratory compensation, as well as buffering. Compensation will take the form of renal excretion of HCO3- and a respiratory acidosis. Thus RR will decrease. Therefore C is incorrect. In addition, B is less correct than D, leaving D as the correct answer assuming that the question is remembered correctly. Reference: Brandis Alternate View If interpreted as the buffering for a bicarbonate load, i.e. metabolic alkalosis, then compensation is CO2 retention: Hence, (A) ~70% will occur in ECF (B) Enters RBC through Cl- exchange and forms CO2 and H2O (C) Decreased respiratory rate and (D) Is incorrect. If above is true then correct answer is B. Note that respiratory compensation is not the same as buffering to produce CO2. Addit notes: Bicarbonate formed by CO2 passage into the RBC is 70% exchanged for Cl- with plasma. The buffering of Bicarbonate therefore happens intracellularly through the catalyst carbonic anyhydrase. Excess H is buffered by proteins and Hb and there is a significant pH change between venous and arterial blood, a process with is reversed when in the lungs and CO2 is lost to atm.

Answer 58

For a buffer to be effective in a given location you need: A high enough concentration of it must be present the buffer's pKa to be within 1 pH unit of the local pH The acid to be buffered cannot be itself A high enough concentration of it must be present. The phosphate concentration is high enough in urine and ICF to be an important buffer there. In comparison, in ECF, the phosphate concentration is too low for it to provide much buffering (despite meeting the other requirement for buffer effectiveness ie pKa within +/- 1 pH unit of prevailing pH) The buffer's pKa to be within 1 pH unit of the local pH Actually this rule is for 'closed' buffer systems - which is the common situation. If there is an 'open' buffer system then the buffer can be effective wuith a larger ph-pKa difference - as occurs with the bicarbonate buffer which buffers in plasma with a pH of 7.4 despite having a pKa of 6.1. For phosphate buffer system, the relevant reaction is the conversion between monohydrogen phosphate and dihydrogen phosphate. This reaction has a pKa of 6.8 and so is within the required range (of 7.4 +/- 1). In plasma and ISF however the phosphate concentration is just too low to actually contribute much buffering. The acid to be buffered cannot be itself For example, the bicarbonate buffer cannot buffer a respiratory acidosis because it cannot buffer itself. Consequently most of the buffering of respiratory acid-base disorders (97-99%) occurs intracellularly on protein and phosphate buffers.

Answer 59

For a buffer to be effective in a given location you need: A high enough concentration of it must be present the buffer's pKa to be within 1 pH unit of the local pH The acid to be buffered cannot be itself A high enough concentration of it must be present. The phosphate concentration is high enough in urine and ICF to be an important buffer there. In comparison, in ECF, the phosphate concentration is too low for it to provide much buffering (despite meeting the other requirement for buffer effectiveness ie pKa within +/- 1 pH unit of prevailing pH) The buffer's pKa to be within 1 pH unit of the local pH Actually this rule is for 'closed' buffer systems - which is the common situation. If there is an 'open' buffer system then the buffer can be effective wuith a larger ph-pKa difference - as occurs with the bicarbonate buffer which buffers in plasma with a pH of 7.4 despite having a pKa of 6.1. For phosphate buffer system, the relevant reaction is the conversion between monohydrogen phosphate and dihydrogen phosphate. This reaction has a pKa of 6.8 and so is within the required range (of 7.4 +/- 1). In plasma and ISF however the phosphate concentration is just too low to actually contribute much buffering. The acid to be buffered cannot be itself For example, the bicarbonate buffer cannot buffer a respiratory acidosis because it cannot buffer itself. Consequently most of the buffering of respiratory acid-base disorders (97-99%) occurs intracellularly on protein and phosphate buffers.

Answer 60

Correct Answer is A. Metabolic alkalosis as pH is higher than 7.4 and HCO3 is high. There is some respiratory compensation as pCO2 > than 40mmHg. OR Alkalaemia as pH is greater than 7.4 Metabolic as HCO3 is greater than normal. Rule of thumb 6. (0.7 x HCO3) + 20 = pCO2 (+/- 5mmHg) Here = 42 Within range of appropriate respiratory compensation. Therefore most likely = primary metabolic alkalosis with respiratory compensation. The changed stem is more difficult. "Partial compensation" of A is as close as "none of the above". Alternative: Boston Rules: pH 7.46 pCO2 46mmHg HCO3 31 mM/L Step 1 pH>7.44, so alkalosis. Step 2 Both pCO2 (35-45) and HCO3 (22-28) are high: This suggests metabolic alkalosis OR respiratory acidosis But, we already know it is an alkalosis. Step 3 Look for clues - none provided (eg: AG, Crea, Glucose, etc.) Step 4 Assess compensatory response using The Boston Rules: Rule 6: The Point Seven Plus 20 Rule Expect pCO2 = 0.7 [HCO3] + 20 (range: +/- 5) = 0.7 x 31 + 20 = 41.7 +/- 5 = TRUE! Thus, respiratory compensation plausible. Step 5 Metabolic alkalosis.

Answer 61

The answer is A Metabolic acidosis -> exchange of H+ for K+ across the cell membrane -> increased intracellular H+. Bicarbonate production is INCREASED also: hypoalbuminaemia can cause a metabolic alkalosis UNLESS, of course, the acidosis is caused by a decreased HCO3 production in the first place.

Answer 62

For a buffer system [1] to be useful it ordinarily must possess two properties: Its components must be present in a sufficient concentration It must have a pKa in the range +/-1 of the prevailing pH For example, if we assess the phosphate buffer system in the ECF against these 2 criteria: Concentration is about 1 mmol/l pKa is 6.8 (& ECF pH is 7.4) So, the phosphate bufffer has a suitable pKa BUT its ECF concentration is much too low to be useful here. [2] Now, consider the bicarbonate buffer: Concentration is 24 mmol/l pKa is 6.1 (& ECF pH is 7.4) So, the concentration is OK (plenty of it here) BUT its pKa is more than i pH unit away from 7.4. We would predict that it is not a useful buffer in these circumstances. This is correct if we assess its usefulness as a simple closed buffer system (that is as a simple chemical buffer system like the phosphate system). However, the bicarbonate buffer system is not closed but open. Indeed it is said to be "open at both ends". This means that both components of this buffer system are regulated by physiological mechanisms: bicarbonate is maintained by the kidneys, and CO2 is excreted via the lungs. The effect of these physiological mechanisms is to overcome the limitation of having a pKa more than 1 pH unit away from ECF pH. The main reasons why the bicarbonate system is the most important ECF buffer are: Bicarbonate is present in significant (ie sufficient) concentrations in the ECF The buffer system is open at both ends and this overcomes the limitation of its low pKa There is a lack of other useful buffer systems in the ECF. This final point is important as to be the "most important" it helps if there is little competition. For example: The phosphate buffer is not important (concentration too low in ECF) Plasma proteins provide the only other buffer system in the ECF but the concentration is lower than that of bicarbonate AND the protein concentration in the interstitial fluyid is even lower then in plasma However, it is wise not to dismiss the "proteins" too lightly. The reasons for this are: Firstly, the plasma proteins can buffer the H+ produced when bicarbonate is formed (Obviously, this cannot be buffered by the bicarbonate system as a buffer system cannot buffer itself). Secondly, another important protein contributes very significantly to buffering of the H+ produced by bicarbonate formation. This is haemoglobin in the red cells. Now because of its intracellular location it is NOT really an extracellular buffer. However, as the bicarbonate (& its H+) are produced in the red cell before diffusing out into the plasma, the haemoglobin really does a significant amount of buffering of this H+. Thus the haemoglobin is sometimes said to be an important extracellular buffer EVEN THOUGH it is actually intracellular. [3] Comment I think the answer is E - due to the open-ended nature of the system, and ability to regulate CO2 via the lungs, HCO3 via the kidneys. quote from Ganong the system is one of the most effective buffer systems in the body because the amount of dissolved CO2 is controlled by respiration. In addition, the plasma concentration of HCO3– is regulated by the kidneys B is not quite right, because it is not so much HCO3 excretion by the kidneys, as regulation (it mainly reabsorbs HCO3, not excretes) Hmmmmm If the kidney doesn't reabsorb a filtered load is it excreted? I thought so??. thats why it handles an alkilosis so easily? I think B is correct, but so is E and prob safer if they both pop up- surely they didn't have both stems so similar? It is the use of the word regulate: It is the lungs that exchange CO2, they do not regulate it, regulation is part of the control system mediated by the respiratory centre, chemoreceptors, efferent nerves ect. Though the excretion of HCO3?? I'd favour B. Re: above Yes, the respiratory centre regulates the lung, but the actions/abilities/healthiness of the lung regulates CO2 levels. Does the speed of the motor regulate the wind or is it the speed of the fan blades? Much of a muchness.

Answer 63

For a buffer system [1] to be useful it ordinarily must possess two properties: Its components must be present in a sufficient concentration It must have a pKa in the range +/-1 of the prevailing pH For example, if we assess the phosphate buffer system in the ECF against these 2 criteria: Concentration is about 1 mmol/l pKa is 6.8 (& ECF pH is 7.4) So, the phosphate bufffer has a suitable pKa BUT its ECF concentration is much too low to be useful here. [2] Now, consider the bicarbonate buffer: Concentration is 24 mmol/l pKa is 6.1 (& ECF pH is 7.4) So, the concentration is OK (plenty of it here) BUT its pKa is more than i pH unit away from 7.4. We would predict that it is not a useful buffer in these circumstances. This is correct if we assess its usefulness as a simple closed buffer system (that is as a simple chemical buffer system like the phosphate system). However, the bicarbonate buffer system is not closed but open. Indeed it is said to be "open at both ends". This means that both components of this buffer system are regulated by physiological mechanisms: bicarbonate is maintained by the kidneys, and CO2 is excreted via the lungs. The effect of these physiological mechanisms is to overcome the limitation of having a pKa more than 1 pH unit away from ECF pH. The main reasons why the bicarbonate system is the most important ECF buffer are: Bicarbonate is present in significant (ie sufficient) concentrations in the ECF The buffer system is open at both ends and this overcomes the limitation of its low pKa There is a lack of other useful buffer systems in the ECF. This final point is important as to be the "most important" it helps if there is little competition. For example: The phosphate buffer is not important (concentration too low in ECF) Plasma proteins provide the only other buffer system in the ECF but the concentration is lower than that of bicarbonate AND the protein concentration in the interstitial fluyid is even lower then in plasma However, it is wise not to dismiss the "proteins" too lightly. The reasons for this are: Firstly, the plasma proteins can buffer the H+ produced when bicarbonate is formed (Obviously, this cannot be buffered by the bicarbonate system as a buffer system cannot buffer itself). Secondly, another important protein contributes very significantly to buffering of the H+ produced by bicarbonate formation. This is haemoglobin in the red cells. Now because of its intracellular location it is NOT really an extracellular buffer. However, as the bicarbonate (& its H+) are produced in the red cell before diffusing out into the plasma, the haemoglobin really does a significant amount of buffering of this H+. Thus the haemoglobin is sometimes said to be an important extracellular buffer EVEN THOUGH it is actually intracellular. [3] Comment I think the answer is E - due to the open-ended nature of the system, and ability to regulate CO2 via the lungs, HCO3 via the kidneys. quote from Ganong the system is one of the most effective buffer systems in the body because the amount of dissolved CO2 is controlled by respiration. In addition, the plasma concentration of HCO3– is regulated by the kidneys B is not quite right, because it is not so much HCO3 excretion by the kidneys, as regulation (it mainly reabsorbs HCO3, not excretes) Hmmmmm If the kidney doesn't reabsorb a filtered load is it excreted? I thought so??. thats why it handles an alkilosis so easily? I think B is correct, but so is E and prob safer if they both pop up- surely they didn't have both stems so similar? It is the use of the word regulate: It is the lungs that exchange CO2, they do not regulate it, regulation is part of the control system mediated by the respiratory centre, chemoreceptors, efferent nerves ect. Though the excretion of HCO3?? I'd favour B. Re: above Yes, the respiratory centre regulates the lung, but the actions/abilities/healthiness of the lung regulates CO2 levels. Does the speed of the motor regulate the wind or is it the speed of the fan blades? Much of a muchness.

Answer 64

Bicarbonate is the most important ECF buffer against metabolic acids (including hydrochloric acid). I disagree... See http://www.anaesthesiamcq.com/AcidBaseBook/ab2_2b.php Experiments in metabolic acidosis have shown that 57% of buffering occurs intracellularly and 43% occurs extracellularly. The processes involved in this buffering are: ECF 43% (by bicarbonate & protein buffers) ICF 57% (by protein phosphate and bicarbonate buffers) due to entry of H+ by: Na+-H+ exchange 36%, K+-H+ exchange 15%, Other 6% [yet another disagreement] I think the [sneaky] answer is proteins, because collectively intracellular proteins and Hb (a protein) BOTH contribute synergistically to the buffering effect. [yet another disagreement with the disagreement] Which finally brings us back to the original answer: that bicarbonate is the most important buffer because it is in BOTH ECF and ICF and is open ended - which makes it more effective than proteins in buffering metabolic acids!! Consider the maths: (a) intra cellular buffer most important (57%) BUT most of the extracellular buffering (43%) is by bicarbonate (b) Bicarbonate "less" intracellularly but still important e.g. in RBC, bicarbonate accounts for 18% buffering whilst Hb 35% (power and cam p.225), and so % buffering by bicarbonate can't be that much less in other cells - the only question is whether the "other" cells have as much protein as Hb is in RBCs... probably not - so I still maintain bicarbonate as the most important... but I can see why protein looks like the right answer. From Ganong 22nd ed, (where many of the MCQs have been lifted, from a single sentence), it says "in metabolic acidosis, only 15-20% of the acid load is buffered by the H2CO2-HCO3 system in the ECF, and most of the remainder is buffered in cells. In metabolic alkalosis, about 30-35% of the OH load is is buffered in cell, whereas in respiratory acidosis and alkalosis, almost all the buffering is intracellular" (page 733) ..... SInce infusion of acid is a metabolic acidosis, it would follow that the answer to the question is C, intracellular buffers. I like C too. As stated above, "57% occurs intracellularly". It doesn't say by which buffers, so it includes HCO3, proteins/Hb, phosphate. Fantastic. Brandis Page 36 suggests answer is B - but he also says "assuming all the buffering is extracellular and is by bicarbonate. These assumptions are not really correct..." The other question is how long has this infusion been going on for? if it is a slow infusion over long period of time... probably C is most likely. but if the duration is short. the infused HCl will probably be buffered by ECF HCO3 and breathed out before it is shifted into the cells! It is not exactly the same situation as endogenous metabolic acidosis where acid production is intracellular... I would probably still go for B! The rate of infusion is not important. Even if it is dumped into the blood the actual buffering in the blood occurs either in proteins or Hb where Hb is 6 times more important (due to quantity of Hb and number of histidine residues - 38, ganong 22nd ed pg 732). To buffer with Hb the acid load dealt with in the ICF due to the presence of carbonic anhydrase in the (RBC). Another view As per Power & Kam (2nd Ed, Pg. 256), "...approximately 70% of non-carbonic acids produced by the body are buffered by plasma bicarbonate...", whilst "...more than 90% of the capacity of the blood to buffer carbonic acid is contributed by haemoglobin...". Metabolic acidosis assumes that the H+ is produced intracellularly, and so the intracellular buffers will have first contact. As HCl is infused, rather than produced intracellularly, the correct answer is likely B - bicarbonate buffer.

Answer 65

Re MCQ AD09: Diabetic ketoacidosis is a metabolic acidosis. Acid Base Physiology Case 2:A Sick Diabetic Patient A. ? B. Respiratory compensation (Kussmaul breathing) is hyperventilation which causes a decreased arterial pCO2 C. The keto-acids are produced in the liver and released into the ECF so only only the liver cells have a low pH (increased [H+]) due to intracellular acid production. However CO2 is very lipid soluble and crosses cell membranes easily. The low arterial pCO2 results in low intracellular CO2 -> decreased intracellular [H+]) D. The kidney excretion of titratable acids increases due partly to the fact that when the urine pH is low, keto-acids themselves contribute to "titratable acidity". Comment: I think that C is true - compensation for metabolic acidosis involves exchange of intracellular K+ for ECF H+ and so a rise in H+ in intracellular fluid. This is a more general effect in metabolic acidosis than ketone synthesis in the liver. From KB Acid Base eBook: In diabetic ketoacidosis (DKA), the ketoacids are produced in the liver and not in every cell in the body. The intracellular alkalinising effect of the compensatory hypocapnia that occurs will however affect every cell and not just the hepatocytes. Does this mean that DKA produces an extracellular rise in [H+] but the opposite change in most tissues (excluding the liver) where the net effect is a fall in intracellular [H+] due to the compensatory hypocapnia? Ketoacids can enter most cells and be used as an energy substrate and this would initially cause a fall in intracellular [H+]. Intracellular pH may not be altered much once maximal respiratory compensation has been achieved because of these opposing effects. It is possible that though the maximal respiratory compensation does not fully correct the extracellular acidaemia, it may be sufficient to prevent much change in intracellular pH. This discussion is speculative and has not been fully investigated. The purpose here is merely to show that looking at acid-base disorders from the intracellular viewpoint can lead to ideas which are different from those of the conventional extracellular viewpoint.

Answer 66

In the Mar 2005 version, I think D is correct, which suggests C is the answer for the 2000-01 version.

Answer 67

Again, the only answer that makes sense is that H+ is increased intracellularly Alternative: Unlike previous years, the 2012 question as recalled here specifically asks for a compensatory mechanism. Increase intracellular pH is only compensatory if you consider intracellular buffering to be a 'compensation', which is not a common usage of 'compensation'. Assuming that option A was correctly recalled as, "Increased ammonium production", it would be the best answer available.

Answer 68

Loss of a sufficient amount of bicarbonate rich pancreatic or biliary secretions will cause a hyperchloraermic (or normal anion gap) metabolic acidosis. This is the 'standard' acid-base disorder associated with a pancreatic or biliary fistula. A correct answer

Answer 69

A healthy young male so assume lungs & heart otherwise noromal. Pneumothorax-> sensation of dyspnoea -> hyperventilation -> decreased arterial pCO2 (resp alkalosis). Now assume the arterial pCO2 is 25mmHg (as in the given options), what then would the arterial pO2 be? Using the Alveolar gas equation for alveolar pO2: Alveolar pO2 = inspired pO2 - (pCO2)/0.8 (as it is reasonable to assume respiratory exchange ratio = normal (0.8) and that alveolar pCO2 = arterial pCO2). Substituting: Alveolar pO2 = 149 - (25/0.8) = 149 -31 = 118 mmHg. Now of course the arterial pO2 MUST be lower than this because of the shunt and V/Q mismatch. However, assuming one lung is totally collapsed-> assume minimal pulmonary blood flow here (thus shunt this blood to the ventilated lung), so pO2 would be lower than 118mmHg but not really low. The option of pO2 = 90mmHg would be a reasonable expectation. So the option with pO2 =90 & pCO2 =25mmHg would seem realistic. If pO2 is 90, pCO2 is 25, then there is neither hypoxic or hypercapnic resp drive -TRUE, neither hypoxaemia or hypercapnia will occur so NEITHER is important here.

Answer 70

This is a Compensated Respiratory Alkalosis. (NB: This comment and analysis refers to the 1st version above) A CO2 this low (24) would be expected to produce a pH of about 7.53. (0.1 change for every 12mmHg of CO2 away from 40mmHg). Therefore there has been some metabolic 'compensation' to reduce the pH to 7.48, which is consistent with a low HCO3 (19). (Ref: [1]) In an Acute Respiratory Alkalosis there you would expect drop in HCO3- by 2 mmol/l (from 24 mmol/l)for every 10mmHg decrease in pCO2 from the reference value of 40mmHg. The lower limit of 'compensation' for this process is 18mmol/l - so bicarbonate levels below that in an acute respiratory alkalosis indicate a co-existing metabolic acidosis. This would give us an expected HCO3- = 21 mmol/l (Ref: Acid-Base on-line tutorial at [2]) In a Chronic Respiratory Alkalosis an average 5 mmol/l decrease in HCO3- (from 24 mmol/l) per 10mmHg decrease in pCO2 from the reference value of 40mmHg would be expected. This takes 2 to 3 days. The limit of compensation is a HCO3- of 12-15 mmol/l. This would give us an expected HCO3- = 16 (+/- 2) mmol/l (Ref: Acid-Base on-line tutorial at [3]) If it was a metabolic acidosis with respiratory compensation the expected CO2 would be about 36-37, from the formula: Expected pCO2 = 1.5 (Actual HCO3 ) + 8 mmHg (Ref: Acid-Base on-line tutorial at [4]) Can someone please explain why the BE would be positive then? Unless the number is remembered wrongly. Respiratory alkalosis or metabolic acidosis are associated with a negative base excess wherelese respiratory acidosis and metabolic alkalosis are associated with positive base excess which doesn't fit if 15 is the right number. I am happy to accept the above answers without the BE. Good point. I struggled with this when I researched the answer, and came to the conclusion that the number was probably wrong, as a BE of +15 would indicate a mixed respiratory and metabolic alkalosis which is not in the list of possible answers. Perhaps the answers are wrong. Kerry Brandis' acid-base tutorial has a good argument for not using BE which includes the point: "It does not distinguish compensation for a respiratory disorder from the presence of a primary metabolic disorder." For further info via acid-base calculators using BE see: [5] and [6] Calculations: pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15 Step 1 Alkalosis as pH > 7.44 Step 2 Both pCO2 and HCO3 trending down - either respiratory alkalosis OR metabolic acidosis Step 3 No clues Step 4 Assess compensations: If acute respiratory alkalosis, Boston Rule 3 predicts: Expected [HCO3] = 24 - 2 [(40 - actual pCO2) / 10]= 24 - 2 [(40-24)/10] = 20.8 mmHg If chronic respiratory alkalosis, Boston Rule 4 predicts: Expected [HCO3] = 24 - 5 [(40 - actual pCO2) / 10] = 24 - 5 [(40-24)/10] = 16.0 mmHg Step 5 Formulate diagnosis: Appears to be a partially compensated acute respiratory alkalosis. Q. how long does it take for full renal compensation? See Ganong Table 36-3 Values for respiratory alkalosis: After 3 wks at 4000m altitude pH 7.48, HCO3- 18.7, pCO2 26 In voluntary hyperventilation pH 7.53, HCO3- 22, pCO2 27 Whilst not exactly correct CO2 values, I think this table makes a good argument for E in the first version and G in the second version.

Answer 71

This is a Compensated Respiratory Alkalosis. (NB: This comment and analysis refers to the 1st version above) A CO2 this low (24) would be expected to produce a pH of about 7.53. (0.1 change for every 12mmHg of CO2 away from 40mmHg). Therefore there has been some metabolic 'compensation' to reduce the pH to 7.48, which is consistent with a low HCO3 (19). (Ref: [1]) In an Acute Respiratory Alkalosis there you would expect drop in HCO3- by 2 mmol/l (from 24 mmol/l)for every 10mmHg decrease in pCO2 from the reference value of 40mmHg. The lower limit of 'compensation' for this process is 18mmol/l - so bicarbonate levels below that in an acute respiratory alkalosis indicate a co-existing metabolic acidosis. This would give us an expected HCO3- = 21 mmol/l (Ref: Acid-Base on-line tutorial at [2]) In a Chronic Respiratory Alkalosis an average 5 mmol/l decrease in HCO3- (from 24 mmol/l) per 10mmHg decrease in pCO2 from the reference value of 40mmHg would be expected. This takes 2 to 3 days. The limit of compensation is a HCO3- of 12-15 mmol/l. This would give us an expected HCO3- = 16 (+/- 2) mmol/l (Ref: Acid-Base on-line tutorial at [3]) If it was a metabolic acidosis with respiratory compensation the expected CO2 would be about 36-37, from the formula: Expected pCO2 = 1.5 (Actual HCO3 ) + 8 mmHg (Ref: Acid-Base on-line tutorial at [4]) Can someone please explain why the BE would be positive then? Unless the number is remembered wrongly. Respiratory alkalosis or metabolic acidosis are associated with a negative base excess wherelese respiratory acidosis and metabolic alkalosis are associated with positive base excess which doesn't fit if 15 is the right number. I am happy to accept the above answers without the BE. Good point. I struggled with this when I researched the answer, and came to the conclusion that the number was probably wrong, as a BE of +15 would indicate a mixed respiratory and metabolic alkalosis which is not in the list of possible answers. Perhaps the answers are wrong. Kerry Brandis' acid-base tutorial has a good argument for not using BE which includes the point: "It does not distinguish compensation for a respiratory disorder from the presence of a primary metabolic disorder." For further info via acid-base calculators using BE see: [5] and [6] Calculations: pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15 Step 1 Alkalosis as pH > 7.44 Step 2 Both pCO2 and HCO3 trending down - either respiratory alkalosis OR metabolic acidosis Step 3 No clues Step 4 Assess compensations: If acute respiratory alkalosis, Boston Rule 3 predicts: Expected [HCO3] = 24 - 2 [(40 - actual pCO2) / 10]= 24 - 2 [(40-24)/10] = 20.8 mmHg If chronic respiratory alkalosis, Boston Rule 4 predicts: Expected [HCO3] = 24 - 5 [(40 - actual pCO2) / 10] = 24 - 5 [(40-24)/10] = 16.0 mmHg Step 5 Formulate diagnosis: Appears to be a partially compensated acute respiratory alkalosis. Q. how long does it take for full renal compensation? See Ganong Table 36-3 Values for respiratory alkalosis: After 3 wks at 4000m altitude pH 7.48, HCO3- 18.7, pCO2 26 In voluntary hyperventilation pH 7.53, HCO3- 22, pCO2 27 Whilst not exactly correct CO2 values, I think this table makes a good argument for E in the first version and G in the second version.

Answer 72

This is a Compensated Respiratory Alkalosis. (NB: This comment and analysis refers to the 1st version above) A CO2 this low (24) would be expected to produce a pH of about 7.53. (0.1 change for every 12mmHg of CO2 away from 40mmHg). Therefore there has been some metabolic 'compensation' to reduce the pH to 7.48, which is consistent with a low HCO3 (19). (Ref: [1]) In an Acute Respiratory Alkalosis there you would expect drop in HCO3- by 2 mmol/l (from 24 mmol/l)for every 10mmHg decrease in pCO2 from the reference value of 40mmHg. The lower limit of 'compensation' for this process is 18mmol/l - so bicarbonate levels below that in an acute respiratory alkalosis indicate a co-existing metabolic acidosis. This would give us an expected HCO3- = 21 mmol/l (Ref: Acid-Base on-line tutorial at [2]) In a Chronic Respiratory Alkalosis an average 5 mmol/l decrease in HCO3- (from 24 mmol/l) per 10mmHg decrease in pCO2 from the reference value of 40mmHg would be expected. This takes 2 to 3 days. The limit of compensation is a HCO3- of 12-15 mmol/l. This would give us an expected HCO3- = 16 (+/- 2) mmol/l (Ref: Acid-Base on-line tutorial at [3]) If it was a metabolic acidosis with respiratory compensation the expected CO2 would be about 36-37, from the formula: Expected pCO2 = 1.5 (Actual HCO3 ) + 8 mmHg (Ref: Acid-Base on-line tutorial at [4]) Can someone please explain why the BE would be positive then? Unless the number is remembered wrongly. Respiratory alkalosis or metabolic acidosis are associated with a negative base excess wherelese respiratory acidosis and metabolic alkalosis are associated with positive base excess which doesn't fit if 15 is the right number. I am happy to accept the above answers without the BE. Good point. I struggled with this when I researched the answer, and came to the conclusion that the number was probably wrong, as a BE of +15 would indicate a mixed respiratory and metabolic alkalosis which is not in the list of possible answers. Perhaps the answers are wrong. Kerry Brandis' acid-base tutorial has a good argument for not using BE which includes the point: "It does not distinguish compensation for a respiratory disorder from the presence of a primary metabolic disorder." For further info via acid-base calculators using BE see: [5] and [6] Calculations: pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15 Step 1 Alkalosis as pH > 7.44 Step 2 Both pCO2 and HCO3 trending down - either respiratory alkalosis OR metabolic acidosis Step 3 No clues Step 4 Assess compensations: If acute respiratory alkalosis, Boston Rule 3 predicts: Expected [HCO3] = 24 - 2 [(40 - actual pCO2) / 10]= 24 - 2 [(40-24)/10] = 20.8 mmHg If chronic respiratory alkalosis, Boston Rule 4 predicts: Expected [HCO3] = 24 - 5 [(40 - actual pCO2) / 10] = 24 - 5 [(40-24)/10] = 16.0 mmHg Step 5 Formulate diagnosis: Appears to be a partially compensated acute respiratory alkalosis. Q. how long does it take for full renal compensation? See Ganong Table 36-3 Values for respiratory alkalosis: After 3 wks at 4000m altitude pH 7.48, HCO3- 18.7, pCO2 26 In voluntary hyperventilation pH 7.53, HCO3- 22, pCO2 27 Whilst not exactly correct CO2 values, I think this table makes a good argument for E in the first version and G in the second version.

Answer 73

The story and gases are typical of acute diabetic ketoacidosis. The low pH despite the very low pCO2 indicates a very low bicarbonate. This is a severe metabolic acidosis with respiratory compensation. Additional Comment pH = 6.1 + log ( [HCO3] / (0.03 x pCO2) ), Hence log ( [HCO3] / (0.03 x pCO2) ) must = 1, and ( [HCO3] / (0.03 x pCO2) ) = 10, As such, [HCO3] would = 3.3. Given pCO2 = 1.5 x [HCO3] + 8 +/- 2, which is 12.95 +/- 2, this is consistent with metabolic acidosis at maximal respiratory compensation, not complicated by other acid-base disorders. Also, PaO2 = FiO2 x (760-47) - ([CO2] / 0.8) = 135.98 mmHg. There is a slight A-a gradient (135.98 - 110) and this patient is unlikely to be receiving supplemental oxygen. LOVE mathematics today said, likely answer is D. 1. pH = 6.1 + log ( [HCO3] / (0.03 x pCO2), PH=7.1, pCO2=11, aHCO3=3.3. 2. expected HCO3 in acute=24-2((40-pCO2)/10), pCO2=11, eHCO3=18.2 expected HCO3 in chronic=24-5((40-pCO2)/10), pCO2=11, eHCO3=9.5 3.aHCO3

Answer 74

We are unable to reliably interpret a blood gas without having a history. We can however say whether something is consistent or not. In this case, option B (acute respiratory acidosis) as a single disorder (ie a simple acid base disorder) is not possible as the bicarbonate (see calculation below) is too high. The 1 for 10 rule (Rule 1) would predict a HCO3 of 25. Option C (fully compensated metabolic acidosis) is ridiculous. The compensation for a metabolic acidosis is hyperventilation to reduce the arterial pCO2 (below 40, to a value predicted by Rule 5 ie pCO2=1.5 Bicarb + 8). In any case the elevated bicarbonate excludes metabolic acidosis. Option E (as remembered here) is also impossible. A mixed acidosis can NEVER result in a normal pH of 7.4. OPtion A may be true but there is no sure way of knowing on the information given - so this CANNOT be the correct answer. If the pO2 was high (eg >140 mmHg say) then we could say that the person must be breathing a high pO2 but this is not the case here. (Technically though this need not be a high FIO2 as the FIO2 for example could be low but the inspired pO2 high if the pertson was in a hyperbaric chamber.) The only option left is option D. Now this does not depend on knowing any clinical histoiry so does not violate the essential history requirement prior to interpreting a blood gas. This is not an interpretation, just a simple application of the Henderson-Hasselbalch equation. THUS: Application of Henderson-Hasselbach indicates that there MUST be an increase in bicarbonate in order for a normal pH to be maintained: 7.4 = 6.1 + log ([HCO3]/0.03 pCO2) So: log ([HCO3]/1.5) = 1.3 And as: log 2 = 0.3 and log 10 = 1 (thus log 20 = 1.3) So: [HCO3] = 20 x 1.5 = 30 mmol/l D is correct: Bicarbonate is elevated. Comment It is possible to roughly work out what the FiO2 is, using the alveolar gas equation (and assuming pAO2 is very close to paO2) - I think it roughly works out to room air FIO2, so A is wrong anyway as for anything else, the pH is normal - so it doesn't matter what the primary acid-base disorder is, if we know that paCO2 is raised, then HCO3 would be raised also (either as the primary or compensating mechanism)

Answer 75

Option E - CO2 decrease and O2 increase - is correct. Traditional (and simple answer) is that ↑ in minute vol. in response to hypoxia at altitude is limited by the resulting respiratory alkalosis which tends to ↓ MV via central chemoreceptors. Then the renal compensation to respiratory alkalosis (HCO3 excretion - maximal at 3-4 days) results in HCO3 movement out of CSF and thus correcting CSF pH toward normal - 'releasing the brakes' on respiratory centre and allowing further increase in minute ventilation and further decrease in pCO2, and further improvement of pO2. Real situation appears to be more complex (isn't it always?). Has now been shown that changes in MV don't actually parallel the compenstaion of respiratory alkalosis. Not well understood but observed changes as above probably due to a change in the 'set-point' of the carotid bodies repsonse to hypoxia, or changes in the hypoxic repsonse of the respiratory centre itself.

Answer 76

Correct Answer is E. Gas shows a metabolic alkalosis with respiratory compensation. A. Altitude causes hyperventilation because of hypoxia leading to a respiratory alkolosis which can last days (4 days?) with metabolic compensation. Even chronically may have high respiratory rate. B. COAD would cause a respiratory acidosis with metabolic compensation. C. Metabolic acidosis is wrong. pH is >7.4 D. ? E. Prolonged Vomiting. Loss of acid+ will cause a metabolic alkalosis which will be compensated for by hypoventilation and rise in pCO2. Boston Rules Strategy: pH 7.48 PO2 70 pCO2 48 HCO3 35 Step 1 pH>7.44 = alkalosis Step 2 pCO2 and HCO3 raised = metabolic alkalosis OR respiratory acidosis Step 3 No clues Step 4 Assess respiratory compensation- One and a Half Plus Eight Rule Expected pCO2 = 1.5[HCO3] + 8 = 1.5[35]+8 = 60.5 >> 48 (but 58 about right) Step 5 Metabolic alkalosis with appropriate respiratory compensation Only plausible answer is vomiting (loss of acid), although note can lose bicard dependent on where vomiting came from (see brandis). I think the wrong Bedside rule has been applied. It should be the Point Seven + Twenty Rule instead Hence Expected pCO2 = 0.7(35)+20 = 44.5(+/-5) Food for thought: Assuming this person is breathing room air & at sea level, there is an A-a gradient i.e. Expected pO2 = 150 - 48/0.8 = 90 Why??? --cos the formula you used was wrong! The A-a gradient in this case is [0.21(760-47) - 48/0.8] - 70 = 19.73mmHg (which isn't that nasty) You used exactly the same formula! And yes the A:a gradient is raised.. we know nothing about the patient's age, which will impact on it..

Answer 77

Base excess or deficit is the amount of acid or base required to titrate whole blood at 37 degrees celcius and PaCO2 of 40 mmHg to a pH of 7.4 Edit - yes, but that's not the same as saying assumes a CO of 40 is it? Doesn't the base excess adjust the values to what they would be IF PCO2 were 40mmHg? (1) Base excess assesses the metabolic component by applying PCO2 40 mmHg during lab measurement. (2) As the measurement occurs "as if" ABG PCO2 was 40 mmHg, 'A' will be the "most correct" of the available answers.

Answer 78

Base excess or deficit is the amount of acid or base required to titrate whole blood at 37 degrees celcius and PaCO2 of 40 mmHg to a pH of 7.4 Edit - yes, but that's not the same as saying assumes a CO of 40 is it? Doesn't the base excess adjust the values to what they would be IF PCO2 were 40mmHg? (1) Base excess assesses the metabolic component by applying PCO2 40 mmHg during lab measurement. (2) As the measurement occurs "as if" ABG PCO2 was 40 mmHg, 'A' will be the "most correct" of the available answers.

Answer 79

Stewart (in his quantitative acid-base analysis) said there were 3 independent variables that determined pH (and the other dependent variables required for acid-base analysis) in each body fluid compartment, namely: pCO2 Strong ion difference (SID) Total concentration of non-volatile weak acids (Atot) will determine the status of body pH, HCO3, & other dependent variables in body fluid Plasma SID = (Sum of concentration of all the strong cations) - (Sum of the concentration of all the strong anions) = about 40-42 mEq/l under normal conditions Common Strong Cations: Na+,K+,Mg++,Ca++ Common Strong Anions: Cl-,Lactate-, Ketone bodies Atot is the total concentration of non-weak volatile acid. (i.e. not CO2) in that compartment. In plasma this mainly consists of: Inorganic phosphate, serum proteins, and albumin From Acid-base text: In particular, the substances which affect acid-base balance in body fluids can all be classified into 3 groups based on their degree of dissociation. This allows certain generalisations and simplifications which are useful in understanding complex solutions. Body fluids can be considered as aqueous solutions that contain: * strong ions * weak ions * non-electrolytes Strong ions in solution are those ions which are always fully dissociated. They exist only in the charged form. Weak ions in solution are only partially dissociated Strong ions are mostly inorganic (eg Na+, Cl-, K+) but some are organic (e.g. lactate anion). In general, any substance which has a dissociation constant greater then 10-4 Eq/l is considered as a strong ion because in the body it exists ONLY in the charged form, and so does not participate in proton transfers (this last thing being the essential difference from the weak ions and the non-electrolytes.) The actual answer to this MCQ depends on the options on the actual MCQ and thge wording of the stem (which may be different from what has been remembered above). In this remembered version, there are 2 correct answers as both the chloride ion and the sulphate ion are 'strong ions'. However sulphate concentration ([SO4--in plasma is low and it is not normally measured so does not figure much in calculations (though Peter Lloyd - see comments below - does use it in his calculator). A significant problem in determining derived parameters for a Stewart analysis is that they involve mathematical manipulations of multiple ions, the determination of each of which has a measurement error. Consequently the total error is the sum of the errors of the individual measurements and so the range within which the correct values lies is large.

Answer 80

Answer is A Arterial blood gas machines measure PaO2, PaCO2 and pH. HCO3 is determined by substitution of the known values into the Henderson Hasselbach equation

Answer 81

B - chronic pulmonary disease. pH is near normal, but mildly acidaemic. pCO2 is raised, so the primary disorder is a respiratory acidosis. Using the 1 for 10 rule (acute RA) = HCO3 = 26 Using the 4 for 10 rule (chronic RA) = HCO3 = 32 I think that's the best answer, so chronic respiratory acidosis caused by pulonary disease.

Answer 82

Fits with an uncompensated acute respiratory alkalosis, so "Hyperventilation" it is.

Answer 83

NB: These 2 versions differ in that the first one is looking for a reason why Hb is better than PPs, whereas the second is just about Hb as a buffer. The answer (as always) depends on the wording of the actual question, not on what the remembered version says. The options here are so close these remembered versions are certainly the same question. Hb has 38 carboxyl residues: Incorrect Hb has got 38 Histidine (imidazole) residues which are important in buffering (i.e. not carboxyl residues) Great amount: Correct "Hb is about six times more important quantitatively as a buffer compared to the plasma proteins because : Its concentration is about twice as much[Hb] 150 gr/L versus [plasma proteins] 70 gr/L Each haemoglobin molecule contains about three times more histidine residues than the average plasma protein." Ref: from p25 physiology viva Kerry Brandis Plasma protein pKa near pH of Blood - The relevant group for buffering by proteins is the imidazole group of histidine residues. This has a pKa about 6.8 so as a 'closed' buffer system this provides effective buffering for a pH of 7.4 (i.e. by the pKa +/- 1 rule). However this applies equally to both plasma proteins and Hb. In summary: A. Correct - present in large concentrations B. Wrong - has 38 HISTIDINE residues, and that in total rather than per globin molecule C. Correct - pKa of ~6.8 is close to physiological pH 7.4 (note that Feb 2011 recalls 'closer', which would make this option more tenuous as the reason for a difference between Hb and plasma proteins) D. Wrong - deoxyHb is a better H+ acceptor (base) than oxyHb E. Not entirely correct - Proteins are zwitterions with multiple pKas. As physiological pH, only those proteins with imidazole groups have the a pKa 6.8 With two correct choices, one wonders whether the Feb 2011 option C is in fact the correctly recalled option with 'closer' pKa. If we were talking about the relative buffering power of Hb compared to plasma proteins, 'A' would be correct.

Answer 84

f pH is 7, then [H+] is 10-7 = 100 nmol/L, therefore option "D". Note: 40 nmol/L for pH 7.4, 36 nmol/L for pH 7.44 46 nmol/L for pH 7.36

Answer 85

f pH is 7, then [H+] is 10-7 = 100 nmol/L, therefore option "D". Note: 40 nmol/L for pH 7.4, 36 nmol/L for pH 7.44 46 nmol/L for pH 7.36

Answer 86

This is 1 MAC of CO2. Henderson hasselbach equation suggests the increased dissociation of carbonic acid will raise H+ concentration (ICF where carbonic anhydrase is) leading to exchange across the cellular membrane and consequent decrease of ICF K+ and therefor increase in ECF K+ to maintain electroneutrality. No idea regarding the other answers. Yep so A correct.

Answer 87

Answer = D | By definition, a strong ion is essentially completely dissociated at physiological pH. JB2012

Answer 88

There is a significant alkalaemia present (pH > 7.44), so there must be an underlying alkalosis (unless there is a measurement error). The pattern of a high HCO3 and a high pCO2 indicate a metabolic alkalkosis. The commonest causes of metabolic alkalosis (accounting for 90% of cases) are: loss of gastric acid (prolonged vomiting, or prolonged NG suction, in the absence proton pump inhibitors of course), and use of certain diuretics (thiazides, frusemide). On this basis, a quick check of the options indicates option E as the correct option. A is wrong: Hyperventilation occurs at (significant) altitude (due to low arterial pO2 stimulating the peripheral chemoreceptors), so arterial pCO2 is decreased. This increase in ventilation is initially limited by central chemoreceptor mediated inhibition of the respiratory centre (due to the low pCO2). With time at altitude, there is equilibration across the blood brain barrier and the pH around the central chemoreceptor returns towards normal (despite continuing low pCO2) so ventilation can increase further. This is a respiratory alkalosis. This helps maintain a higher arterial pO2 at altitude. The high pCO2 in this blood gas excludes this option. B is wrong. The acid base disorder associated with chronic pulmonary disease is a respiratory one, not metabolic alkalosis. C is wrong. "Diabetic coma" is due to diabetic ketoacidosis (though actual "coma" is uncommon). This is a metabolic acidosis. A hyperosmolar nonketotic state can occur in diabetics and this can also result in impaired consciousness ('coma') but without an acid base disorder; indirectly any coma could result in airway obstruction with a respiratory acidosis. D is wrong. Five minutes of hyperventilation causes an acute respiratory alkalosis, that is the arterial pCO2 is low.

Answer 89

A change in pH from 7.4 to 7.1 is a change of 0.3 and the solution becomes more acidic. The [H+] doubles, so this is an increase of 100% (in the terms of the options). This is easy to work out if you are familiar with a small trick. pH is a log to base 10 scale, and the key fact to know is that log102 = 0.3. So a change in pH of 0.3 represents a change in [H+] by a factor of 2, either increasing or decreasing depending on the direction of the change: more acidic is a doubling, and less acidic is a halving). Thus: If the pH is decreased by 0.3 (from any starting pH), then this is becoming more acidic (lower pH) so the [H+] increases by a factor of 2 (i.e. it doubles). If the pH is increased by 0.3 (from any starting pH), then this is becoming less acidic (higher pH) so the [H+] decreases by a factor of 2 (i.e. it halves). You should know that pH = 7.4 (the typical value for arterial pH) is a [H+] = 40 nmoles/l. So when the pH is 7.1, the [H+] = 80 nmoles/l, and a further 0.3 pH drop to 6.8 is a further doubling to 160 nmoles/l). Going ther other way, then a pH of 7.7 is a [H+] = 20 nmoles/l. Another easy [H+] to remember is 100 nmoles/l at pH = 7.0 and so the [H+] at pH = 7.3 is 50 nmoles/l (0.3 pH units above 7.0), and [H+] at pH = 7.6 is 25 nmoles/l (and so on). The reference range for pH in arterial plasma is 7.36 to 7.44 which is almost exactly a [H+] range of 44 to 36 nmoles/l. (That is, pH=7.44 is [H+]=36nmoles/l; and pH=7.36 is [H+]=44nmoles/l.) So now its easy to work out in your head what the [H+] is at pH = 7.06, and at pH = 7.14 (and so on) as these are 0.3 pH units away from pH values where you know the [H+]. Knowing this (log2 = 0.3) fact makes this question extremely easy to answer, and certainly a lot easier than than doing conversions from pH to [H+] by calculating it from first principles. . Another way this knowledge is useful is when considering the normal values in the Henderson-Hasselbalch equation for the bicarbonate buffer, and substituting normal values for arterial pCO2(=40) and bicarbonate(=24), and for pKa(=6.1),we get: pH = 6.1 + log[24 / (0.03 x 40)] pH = 6.1 + log[24/1.2] pH = 6.1 + log(20) Now 20 = 2 x 10 so log(20) log(2 x 10) = log(2) + log(10) [i.e. the log of a product is the sum of the numbers as you will remember from high school maths] and we know log2 = 0.3, and log10 = 1, so log(20) = 0.3 + 1.0 = 1.3. So: pH = 6.1 + 1.3 = 7.4 (as expected)

Answer 90

A change in pH from 7.4 to 7.1 is a change of 0.3 and the solution becomes more acidic. The [H+] doubles, so this is an increase of 100% (in the terms of the options). This is easy to work out if you are familiar with a small trick. pH is a log to base 10 scale, and the key fact to know is that log102 = 0.3. So a change in pH of 0.3 represents a change in [H+] by a factor of 2, either increasing or decreasing depending on the direction of the change: more acidic is a doubling, and less acidic is a halving). Thus: If the pH is decreased by 0.3 (from any starting pH), then this is becoming more acidic (lower pH) so the [H+] increases by a factor of 2 (i.e. it doubles). If the pH is increased by 0.3 (from any starting pH), then this is becoming less acidic (higher pH) so the [H+] decreases by a factor of 2 (i.e. it halves). You should know that pH = 7.4 (the typical value for arterial pH) is a [H+] = 40 nmoles/l. So when the pH is 7.1, the [H+] = 80 nmoles/l, and a further 0.3 pH drop to 6.8 is a further doubling to 160 nmoles/l). Going ther other way, then a pH of 7.7 is a [H+] = 20 nmoles/l. Another easy [H+] to remember is 100 nmoles/l at pH = 7.0 and so the [H+] at pH = 7.3 is 50 nmoles/l (0.3 pH units above 7.0), and [H+] at pH = 7.6 is 25 nmoles/l (and so on). The reference range for pH in arterial plasma is 7.36 to 7.44 which is almost exactly a [H+] range of 44 to 36 nmoles/l. (That is, pH=7.44 is [H+]=36nmoles/l; and pH=7.36 is [H+]=44nmoles/l.) So now its easy to work out in your head what the [H+] is at pH = 7.06, and at pH = 7.14 (and so on) as these are 0.3 pH units away from pH values where you know the [H+]. Knowing this (log2 = 0.3) fact makes this question extremely easy to answer, and certainly a lot easier than than doing conversions from pH to [H+] by calculating it from first principles. . Another way this knowledge is useful is when considering the normal values in the Henderson-Hasselbalch equation for the bicarbonate buffer, and substituting normal values for arterial pCO2(=40) and bicarbonate(=24), and for pKa(=6.1),we get: pH = 6.1 + log[24 / (0.03 x 40)] pH = 6.1 + log[24/1.2] pH = 6.1 + log(20) Now 20 = 2 x 10 so log(20) log(2 x 10) = log(2) + log(10) [i.e. the log of a product is the sum of the numbers as you will remember from high school maths] and we know log2 = 0.3, and log10 = 1, so log(20) = 0.3 + 1.0 = 1.3. So: pH = 6.1 + 1.3 = 7.4 (as expected)

Answer 91

A change in pH from 7.4 to 7.1 is a change of 0.3 and the solution becomes more acidic. The [H+] doubles, so this is an increase of 100% (in the terms of the options). This is easy to work out if you are familiar with a small trick. pH is a log to base 10 scale, and the key fact to know is that log102 = 0.3. So a change in pH of 0.3 represents a change in [H+] by a factor of 2, either increasing or decreasing depending on the direction of the change: more acidic is a doubling, and less acidic is a halving). Thus: If the pH is decreased by 0.3 (from any starting pH), then this is becoming more acidic (lower pH) so the [H+] increases by a factor of 2 (i.e. it doubles). If the pH is increased by 0.3 (from any starting pH), then this is becoming less acidic (higher pH) so the [H+] decreases by a factor of 2 (i.e. it halves). You should know that pH = 7.4 (the typical value for arterial pH) is a [H+] = 40 nmoles/l. So when the pH is 7.1, the [H+] = 80 nmoles/l, and a further 0.3 pH drop to 6.8 is a further doubling to 160 nmoles/l). Going ther other way, then a pH of 7.7 is a [H+] = 20 nmoles/l. Another easy [H+] to remember is 100 nmoles/l at pH = 7.0 and so the [H+] at pH = 7.3 is 50 nmoles/l (0.3 pH units above 7.0), and [H+] at pH = 7.6 is 25 nmoles/l (and so on). The reference range for pH in arterial plasma is 7.36 to 7.44 which is almost exactly a [H+] range of 44 to 36 nmoles/l. (That is, pH=7.44 is [H+]=36nmoles/l; and pH=7.36 is [H+]=44nmoles/l.) So now its easy to work out in your head what the [H+] is at pH = 7.06, and at pH = 7.14 (and so on) as these are 0.3 pH units away from pH values where you know the [H+]. Knowing this (log2 = 0.3) fact makes this question extremely easy to answer, and certainly a lot easier than than doing conversions from pH to [H+] by calculating it from first principles. . Another way this knowledge is useful is when considering the normal values in the Henderson-Hasselbalch equation for the bicarbonate buffer, and substituting normal values for arterial pCO2(=40) and bicarbonate(=24), and for pKa(=6.1),we get: pH = 6.1 + log[24 / (0.03 x 40)] pH = 6.1 + log[24/1.2] pH = 6.1 + log(20) Now 20 = 2 x 10 so log(20) log(2 x 10) = log(2) + log(10) [i.e. the log of a product is the sum of the numbers as you will remember from high school maths] and we know log2 = 0.3, and log10 = 1, so log(20) = 0.3 + 1.0 = 1.3. So: pH = 6.1 + 1.3 = 7.4 (as expected)

Answer 92

Body metabolism every day produces a net excess of acid that in order to maintain acid base homeostasis has to be excreted by the body. The largest amount of net 'acid' production due to metabolism is of carbon dioxide (or equivalently, carbonic acid). This is excreted by the lungs. As it evident here, it is CO2 that is excreted, not hydrogen ion. However it can be seen that the CO2 excretion is equivalent to the elimination of the same amount of carbonic acid. The question asks about excretion by the kidneys, so CO2 is not relevant. The net "fixed acid" production by the body per day is 1.0-1.5 mmols/kg/day which for the standard 70kg person is 70-100 mmol/day. Strictly it is not the hydrogen ion that is excreted (so in that sense the question is wrongly worded) but in a less strict sense this excretion of fixed acid (by the kidney) is commonly spoken of as acid excretion (and thus moire loosely, as "hydrogen ion excretion"). Obviously the question expects you to understand this common use. Of the options available (assuming they are correctly remembered), thern the correct answer os E. 68 mmol. Technical Note Assuming the remembered version are correct, then this question is VERY poorly structured. Why? Because the correct answer can be determined by a pattern analysis of the answers without any knowledge of acid-bases physiology at all. Firstly, note that 4 options have mmol (millimol) and only one answer has nmol (nanomol), so eliminate A. Secondly, 3 answers have 68 in them, so eliminate A and B (which have 32) Thirdly, note that 3 of the options have larger numbers in them (32, 32 & 68) without a decimal point, so eliminate C & D. The remaining option then is E, which is correct. Quite a few years ago now the college appointed a Director of Education (now long gone) who was an educationalist, not medical at all. He reviewed the MCQs and noted that a significant number had internal structural clues to the correct answer, which would enable a person to guess the correct answer without any knowledge of the subject at all. Writing MCQs with this defect is a very common thing to do and most people are not aware they are doing it. Anyway, as a consequence the college then reviewed ALL the MCQs and removed some and reworded many others. This was of course a few years ago, and with change in staff then there is a risk of forgetting this lesson in MCQ writing - maybe this is what happened here, but of course this is just a 'remembered' version and it could be that the person remembering the question and knowing the correct answer was the person who unknowingly caused the pattern. If interested, see for flaws in MCQs: [1] The answer of 68 mmol is of course not the same as 70-100 mmol as I stated above (or the same as 70 mmol as given in some texts) but is very close to the answer I state. But then why is it different? The answer is probably related to the rule that the examiners use when writing primary MCQs, namely that each question and answer must be referenced to one (or more) of the recommended books in the reference text list. My informed guess is that one or more of these sources uses the 68 mmol figure.

Answer 93

Base excess is not measured, it is calculated according to a formula programmed into the machine. To see the formula used just ask your pathology department to let you view the manual of the machine. It is used to assess metabolic acid base status It is NOT the same as bicarbonate Answer:B

Answer 94

"The mean total number of alveoli has recently been established as 480 million" - Nunn's applied respiratory physiology 2005. Not 3 million, so A. is incorrect. Each has a diameter of 1/3mm not 3 mm so B. is incorrect(The size of the alveoli is proportional to lung volume but owing to gravity they are normally larger in the upper part of the lung, except at maximal inflation, when the vertical gradient in size disappears. At FRC the mean diameter is 0.2mm- (Ref: Nunn's Applied Respiratory Physiology' p19). Alveolar surface area is 50-100m2 External surface area is much less than 10m2 Answer is E. None of the above

Answer 95

Collapse of one lung will result in shunt. The paO2 will decrease as decreased O2 content in blood from the collapsed lung causes significant decline in pO2 (shifts point down on to steep part of Oxygen Hb Dissociaton curve).Increasing inhaled O2 concentration to 100% does not correct pO2 ,as the addition of a small amount of shunted blood with its low O2 concentration greatly reduces the p02 of arterial blood. (West-p61) Hypoxic pulmonary vasoconstriction in the collapsed lung, and the increased pulmonary vascular resistance as a consequence of the loss of radial stretch on the vessels will both result in a decrease in the amount of blood shunting through the collapsed lung. The collapsed lung will result in a sensation of increased difficulty with breathing (dyspnoea) AND impair pulmonary gas exchange. BOTH the dyspnoea, and any tendency for arterial pCO2 to rise will stimulate ventilation. This would tend to maintain at least a normal paCO2 (or even result in a lowered paCO2) as the decreased CO2 in blood from the normal lung will tend to counteract the increase in pCO2 in blood from the collapsed lung. This ability to "counteract" fairly well is due to both the linear shape of the CO2 dissociation curve and the ability of paCO2 to stimulate ventilation. In summary, the shunt will impair both oxygenation and CO2 excretion, but the increased ventilation in this situation will overcome the CO2 excretion difficulty (linear dissociation curve) BUT will not be able to overcome the oxygenation problem. So, the answer will be the one with a low pO2 and a normal or low paCO2. The logic is that: PaO2 will be impaired by the shunt (

Answer 96

Part B A - unlikely to be correct pCO2 very low B - definitely incorrect; pO2 too high C - most likely D - possible Application of the Alveolar Gas Equation/A-a gradient Assessing the proposed values above withe the above in light of a significant shunt may help to narrow down a more likley set of values. In Part A, both A and C are the most probable answers on face value. A results in a small A-a gradient (8mmHg)which seems unlikely, C results in an A-a gradient of 37.5mmHg which would more accurately reflect the situation. In Part B, a similar application of the A-a gradient can narrow down the field by allowing you to exclude those answers with a normal A-a gradient. Most likely answers are C or D For those like me that have forgotten so close to the exam! A-a (O2) = (FiO2%/100) * (Patm - 47 mmHg) - (PaCO2/0.8) - PaO2 Thus, for pO2 50 and pCO2 50 A-a (O2) = 0.21(760-47) - (50/0.8) - 50 = 37.2mmHg

Answer 97

Pulmonary vascular resistance is minimal at FRC, and increases with a change in lung volume in either direction, ie with increased or decreased lung volume. The increase in PVR with decreased lung volume is due to changes in the extra alveolar vessels, which are pulled open by radial traction at higher volumes, or close at lower volumes due to their smooth muscle tone. The increase in PVR at high lung volumes is due to the alveolar vessels, which are distorted and their calibre decreased. PVR decreases with an increase in pulmonary arterial pressure, firstly due to recruitment of previously closed vessels (increasing the number of open capillaries as much as threefold), and then at higher pressures due to distension of vessels (by as much as twofold). PVR also decreases with increased pulmonary venous pressure, although not as much. Therefore an increased left atrial pressure should lead to a decrease in PVR. It seems to me that probably an elevation in CVP would exert effects on PVR indirectly through effects on pulmonary artery pressure, although I haven't yet found anything that specifically addresses this question. "Accumulation of CO2 leads to a drop in pH in the area, and a decline in pH also produces vasoconstriction in the lungs, as opposed to the vasodilation it produces in other tissues" (Ganong p641). Therefore option B - hypocarbia and C - alkalosis are BOTH wrong. Refs Guyton 11th ed pgs 483- 487, West's Pulmonary Physiology and Pathophysiology pgs 86-92, Brandis p156-157 Above comments noted. ABOVE SAYS PVR dcrease as pulmonary venous pressure increase,it should be otherway around as the pulmonary venous pressure or the lt atrial pressure increase this will increase PVR as PVR = PAP-PVP/ CO so i think the answer is clearly PVR INCREASE IF PV PRESSURE OR THE LT ATRIAL PRESSURE INCREASE I don't think this above comment is correct. If you increase Pulmonary Venous Pressure, this will make the numerator in the above formula smaller and thus DECREASE pulmonary vascular resistance. The textbooks say that increasing pulmonary venous pressure will recruit non-patent pulmonary vessels and distend those that are already patent - thereby increasing their total surface area and decreasing resistance to flow. The trouble with using this equation is that if PVP goes up, PAP and CO would change with it. PVP goes up, CO would probably come down a bit. So you can't just apply this equation with an isolated change in PVP. In any case, yeah, it was applied incorrectly to start with. This whole concept of decreasing PVR by recruitment and distension is dodgy. If the PVR truly decreases when PAP goes up, which is what the books are implying is the cause of the r&d, then PAP wouldn't have gone up in the first place. The fact that PAP went up, meant that the decrease in PVR was inadequate to compensate for whatever is putting up the pressure, so even though the total PVR decreases, the PVR per unit of CO actually increased. Of course West doesn't mention this to make it sound neat. Or was it because he didn't really understand it? Perhaps nobody does. And now you have a bunch of confused people trying to make sense of it all for an exam. --- Hypercapnia has a slight pressor effect on PVR Acidosis (metabolic or respiratory) augments HPV Alkalosis (metabolic or respiratory) causes pulmonary vasodilation Alkalosis (metabolic or respiratory) can reduce or even abolish HPV Page 102 (Ch 7) Nunns 6th ed. If CVP were to affect pul vascular resistance via pulm art pressure, a rise in JVP would likely cause rise in PA pressure, causing recruitment of pulm vasculature, thus reduction in overall resistance to flow. To clarify the comment on change in PVR with change in lung volume: PVR is sum of intraalveolar vessel resistance and extra-alveolar resistance. As lung volume increases intraalveolar resistance increases and extraalveoar resistance falls, and visa-versa for a decrease in lung volume, (for the reasons prevously mentioned). The sum of the two is minimal at lung volume = FRC. Graph with three lines, for intra-alveolar, extra-alveolar, and total resistance, can demonstrate this. Summary: PVR increases with increased H+, pCO2, lung volume (> FRC), pulmonary pressure, arterial pressure. (Note fig 4-4 West: demonstrates decreasing PVR with increasing pulmonary aterial and venous pressures - due to distension and recruitment) PVR decreases with decreasing lung volume (

Answer 98

Well I think that increasing left Atrial pressure will end up increasing pulmonary vascular resistance. As has been pionted out pressure gradient between PA pressure and LA pressure deterimines flow. with increase LA pressure there would be smaller pressure gradient. If flow has decreased then the resistance to the flow would have been increased. thus PVR must be increased since its one of the factors that resists blood flow. my reasoning may not be acceptable to some but you will find that most descriptions of pathological conditions that increase LAP eg mitral stenosis state that pulmonary hypertension is present. if indeed there were to be a decrease in PVR then pulm HPT would not be a problem in these groups of patients. [2] emedicine mitral stenosis [3] wikipedia Mitral stenosis Comment: This reasoning is incorrect with regards to PVR. It is true to say that increasing LA pressure reduces lung perfusion pressure. That would reduce flow only if resistance is unchanged. However, recruitment and distension of pulmonary capillaries will occur with increase of either pulmonary artery OR venous pressure. This is because the recruitment is acting against the alveolar gas pressure: even venous blood pressure will improve capillary engorgement. In this way, increased LA pressure will reduce PVR. Increased LA pressure also leads to pulmonary hypertension. However hypertension does not equal resistance. The implication is that the is low pulmonary vascular resistance due to recruitment and distension, but a pulmonary hypertension due to cardiac valvular disease. The pulmonary capillary bed is thus compensating for the increased resistance within the left heart: low pulmonary vascular resistance compensating for high intra-cardiac resistance. SO for the Feb 12 version, hypoxic pulmonary vasoconstriction as the best answer? Yes, 'A'.

Answer 99

RE03c [Aug 11] Which of the following increase pulmonary vascular resistance: A. Wrong - Hypocarbia, with respiratory alkalosis, causes pulmonary vasodilation. Hypercarbia has a pressor effect. B. Wrong - Alkalosis causes pulmonary vasodilation C. Wrong - Raised pulmonary artery pressure decreases PVR through recruitment and distension of capillaries D. Wrong - Raised left atrial pressure decreases PVR through recruitment and distension of capillaries (even though perfusion pressure may have decreased) E. None of the above Most correct answer: E

Answer 100

Physiological dead space is the part of tidal volume that does not take part in gas exchange. It comprises of both anatomical and alveolar dead space. Anatomical dead space is the volume of the conducting airways (some sources list it as the total lung volume minus the volume of alveoli). Alveolar dead space is where alveoli are ventilated but not (adequately) perfused. - - Pulmonary embolism leads to an increase of alveolar dead space (before any atelectasis) as previously perfused areas of the lung are now no longer perfused. Atelectasis/collapse will mean the involved lung isn't ventilated; it doesn't receive any of the tidal volume. A pneumothorax will decrease lung volume, and decrease how much it is ventilated. Bronchoconstriction reduces ventilation, not perfusion. (If you constrict your bronchioles you decrease anatomical dead space) Obesity shouldn't cause a decrease in perfusion??? (although may cause atelectasis, see above) - Agreed - Pressure of abdominal contents, especially if supine, should decrease basal ventilation, potentially causing atelectasis. However, there would be Alveolar hypoxia causing increased vascular resistance in that lung region, with decreased perfusion to return V/Q towards 1. So the best answer is A.

Answer 101

A1. Correct - Water vapour pressure will remain constant throughout the lung. B1. False - Apex 553mmHg, Base 582 mmHg - West says variation in effect default due to variations in others. C1. False - pCO2 lower at apex D1. True - pO2 higher at apex (130mmHg vs 88 Kam p 92) E1. False - V/Q ratio is much greater at apex (>3.0) F1. False - greater ventilation achieved at base because greater compliance at base G1. True - compliance greater at base V/Q is higher at the apex than the base. Both ventilation and perfusion decrease from base to apex, but perfusion decreases more. Compliance is greater at the base than the apex, because of the position of the lung on the pressure-volume curve. Therefore ventilation, when seen as a change in volume per unit of resting volume, is greater at the base than the apex (West Pulmonary physiology and pathophysiology, p 41-42). It is the change in V/Q ratio which affects regional pCO2 and pO2 differences. If the alveoli were ventilated but not at all perfused, the alveolar gas would approach inspired gas composition (pO2 of 149mmHg and pCO2 close to 0). If the alveoli were perfused but not at all ventilated, the alveolar pO2 and pCO2 would approach venous values (pO2 of roughly 40mmHg and pCO2 of roughly 45mmHg). So therefore pO2 must be higher at the apex than the base, and pCO2 higher at the base than at the apex. pN2 is lower at the apex (553mmHg at apex compared to 582mmHg at base). The variation is by default as the total alveolar gas pressure is constant throughout the lung. [The Riley approach to V/Q; Nunn p.116] Incidentally, i dimly remember something about TB granulomas being generally in the upper regions of the lungs because of the higher pO2, although I can't remember where I read this(West p66) West p41-42, Guyton 11th ed p500. West pg 66 A- Water Vapour pressure is constant throughout the lung- TRUE, as temperature and relative humidity are the factors that alter water vapour, both these are not variable conditions in the lung. The saturated water vapour is higher in warm than cool air ( 20C it is 17.5mmHg. at body temp 37C it is 47mmHg) Decreases with increasing humidity, at 40% humidity at 20C it is 7mmHg. C- False ,apex - high V/Q ration of 3, leading to high Po2 130mmHg, low Pco2 28mmHg. D -false, Base- low V/Q ratio of 0.6 , with low pO2 88mmHg, high pCO2 42mmHg F- false, ventilation is highest in the lower zones (West Fig 2-7, p 21)

Answer 102

REFERENCES in West: Fig 5-8, p.65: V, Q, V/Q vs base-apex lung. Fig 7-8, p.102: Explanation of regional differences in ventilation. Illustrates: 1) Why ventilation is greater at base than apex, and 2) why compliance (dV / dP) is greater at base than apex. Fig 4-7, p.43: Explanation of regional differences in perfusion (ie: gravity). Fig 5-10, p.66, Regional difference in gas exchange down lung with numbers for Vol, VA, Q, pN2, etc. physi16.jpg E1: If Ventilation much greater as stated in comment above, then 'V/Q is higher at base than at apex' should be true (see RE06) E1 is talking about the V/Q ratio, not "ventilation". The base is relatively more perfused than ventilated (low V/Q) and the apex is relatively more ventilated than perfused (high V/Q). It's the ratio not the actual amount of "ventilation" or gas exchange that goes on so E1 is INCORRECT. Compliance is generally greater at the bottom of the lung compared with the top. However, compliance at the bottom of the lung can be smaller at low lung volumes. Therefore option G is not strictly correct. In alternative version: Isn't PaO2 referring to arterial pO2 and will reflect venous admixture in left atrium and therefore area of lung become irrelevant to Q?. Is PaO2 and PaCO2 referring to pulmonary end capillary tension or is it just a typo and meant to be PAO2 & PACO2? http://www.kerrybrandis.com/wiki/mcqwiki/index.php?title=RE05

Answer 103

E1: If Ventilation much greater as stated in comment above, then 'V/Q is higher at base than at apex' should be true (see RE06) E1 is talking about the V/Q ratio, not "ventilation". The base is relatively more perfused than ventilated (low V/Q) and the apex is relatively more ventilated than perfused (high V/Q). It's the ratio not the actual amount of "ventilation" or gas exchange that goes on so E1 is INCORRECT. Compliance is generally greater at the bottom of the lung compared with the top. However, compliance at the bottom of the lung can be smaller at low lung volumes. Therefore option G is not strictly correct. In alternative version: Isn't PaO2 referring to arterial pO2 and will reflect venous admixture in left atrium and therefore area of lung become irrelevant to Q?. Is PaO2 and PaCO2 referring to pulmonary end capillary tension or is it just a typo and meant to be PAO2 & PACO2? http://www.kerrybrandis.com/wiki/mcqwiki/index.php?title=RE05

Answer 104

For 1996,1999,2001 version: (A to E below refer to first version above): A. WRONG - Perfusion gradient is steeper B. WRONG - Ventilation is higher at the base C. WRONG - Perfusion is higher at the base D. TRUE -V/Q ratio is higher (3.3) than at the base (0.63).Ventilation is less at the top than the bottom,but the differences in blood flow are more marked.Consequently,the ventilation-perfusion ratio decreases down the lung,and all the differences in gas exchange follow from this(West p 66). (These values for V/Q ratio are the standard ones from West "Respiratory Physiology") E. WRONG - Because D. is correct. D is obviously correct. But whoever wrote this question has no strict mathematical understanding of the word "gradient". The gradient changes depending on which direction you are plotting the graph. There is no reason why the graph must be plotted as shown in West. Replace the word "greater" with "steeper" and this would make more mathematical sense.

Answer 105

``` For Feb 12 version A is wrong, V/Q higher at apex B is wrong, less perfusion at apex C is correct, alveolar pressure is the same throughout the lung and intrapleural pressure is more negative at apex, hence higher transmural (probably meant to be transpulmonary) D is wrong ```

Answer 106

A right shift of the haemoglobin oxygen dissociation curve (ODC) indicates a decreased oxygen affinity, that is increased O2 unloading in the capillaries as indicated by a lower saturation at any particular pO2 value. The haemoglobin ODC is shifted to the right by: increased PaCO2 increased temperature increased H+ concentration (ie decreased pH) increased 2,3 DPG. The only correct option is A: Increases with increased paCO2 [NB: This MCQ has not resurfaced since asked in 1996; this is presumably because it was so easy, it had a poor discrimination index and was eliminated from the ANZCA Question Bank]

Answer 107

A: Correct. Alveolar dead space is a component (subset) of physiological dead space B: Incorrect. Mechanical ventilation typically creates/increases alveolar dead space C: Correct. (Pulmonary) hypotension can increase alveolar dead space by increasing zone 1 of the lung D: Fowler's method measures Anatomical dead space. Physiological dead space is the part of tidal volume which does not take part in gas exchange. It consist of anatomical and alveolar dead space. Alveolar dead space is where alveoli is ventilated but not perfused. Pleural effusion, pneumothorax and CCF leading to pulmonary oedema all compromise ventilation more than perfusion and decrease alveolar dead space. So for RE08b, A,B & C are FALSE Hypotension and alveolar dead space is a reference to "zone 1" of the lung from p43 of West "...zone 1 does not occur under normal conditions...but may be present if the arterial pressure is reduced (eg following severe hemorrhage) or if alveolar pressure is raised (during positive pressure ventilation). This ventilated but unperfused lung is useless for gas exchange and is called alveolar dead space" Therefore alveolar dead space is increased with hypotension so for RE08b, D is correct Further discussed in Nunn's: "Low cardiac output (?relate to hypotension -ed), regardless of the cause, results in pulmonary hypotension and failure of the uppermost parts of the lungs(Zone 1). During anaesthesia with controlled ventilation, sudden changes in end-expiratory CO2 therefore usually indicate changing alveolar dead space secondary to abrupt variations in cardiac output." - Nunn's

Answer 108

A: Correct. Alveolar dead space is a component (subset) of physiological dead space B: Incorrect. Mechanical ventilation typically creates/increases alveolar dead space C: Correct. (Pulmonary) hypotension can increase alveolar dead space by increasing zone 1 of the lung D: Fowler's method measures Anatomical dead space. Physiological dead space is the part of tidal volume which does not take part in gas exchange. It consist of anatomical and alveolar dead space. Alveolar dead space is where alveoli is ventilated but not perfused. Pleural effusion, pneumothorax and CCF leading to pulmonary oedema all compromise ventilation more than perfusion and decrease alveolar dead space. So for RE08b, A,B & C are FALSE Hypotension and alveolar dead space is a reference to "zone 1" of the lung from p43 of West "...zone 1 does not occur under normal conditions...but may be present if the arterial pressure is reduced (eg following severe hemorrhage) or if alveolar pressure is raised (during positive pressure ventilation). This ventilated but unperfused lung is useless for gas exchange and is called alveolar dead space" Therefore alveolar dead space is increased with hypotension so for RE08b, D is correct Further discussed in Nunn's: "Low cardiac output (?relate to hypotension -ed), regardless of the cause, results in pulmonary hypotension and failure of the uppermost parts of the lungs(Zone 1). During anaesthesia with controlled ventilation, sudden changes in end-expiratory CO2 therefore usually indicate changing alveolar dead space secondary to abrupt variations in cardiac output." - Nunn's

Answer 109

``` The above equation appears not to be working so the answer is: Bohr Equation Vd/Vt = (PaCO2-PeCO2)/PaCO2 ∴ 1/3 = (45-PeCO2)/45 ∴ 15 = 45 - PeCO2 ∴ PeCO2 = 30 Answer is C ```

Answer 110

RE10 The below answers all have an effect on the oxygen dissociation curve... Question possibly remembered incorrectly too! A - ? correct; hypothermia would decrease metabolic requirements (although VO2 is the same ruling this out); hypothermia will cause a left shift in the haemoglobin desaturation curve meaning that less oxygen has been liberated at the tissues?? B - increased paCO2 will cause a right shift and a decrease in SvO2 C - decreased 2-3 DPG will cause a left shift D - Alkalosis will also cause a left shift E - July 03 B - anaemia; decreased oxygen carrying capacity in the blood will mean that greater extraction has to take place from the blood delivered to the tissues and hence a lower SvO2; also remember O2 content = 1.34*Sat%*Hb + 0.003*pO2, so with lower Hb, there will be lower O2 content (but stem now says TENSION which is partial pressure not CONTENT.) - comment - a lower mixed venous content will also result in a lower mixed venous pO2/saturation The relevant formula is Fick's law as applied to oxygen uptake in the lungs. When re-arranged this results in: Mixed venous O2 content = Arterial O2 content - (VO2 /CO) If VO2 & CO are constant, then a consideration of this equation shows that the ONLY way for mixed venous O2 content to increase is if there is an increase in arterial O2 content. --- Now, considering the oxygen flux equation, the ONLY ways to increase arterial oxygen content (if CO constant) are: an increase in [Hb], an increase in saturation, or an increase in dissolved O2. The only way to increase dissolved O2 content is to increase arterial pO2. Now given that FIO2 is fixed this is very unlikely to occur. Also if FIO2 is constant than it is unlikely that arterial O2 saturation will change. So the only possible option then is to increase [Hb]. If this is an option in the "actual" version of the question then it would be correct. If it were not an option, then the answer would be the "none of the above" option. Consideration of factors which affect the position of the ODC are TOTALLY IRRELEVANT. The ODC is essentially a graphical way to convert a pO2 value into a Saturation value. This question talks about O2 CONTENT, not pO2, so there is absolutely no sense in referring to the ODC. As mentioned above, the only way that pO2 comes into it is because "dissolved O2 content" is proportional to pO2 (as per Henry's Law), so it is possible that a large increase in arterial pO2 can significantly increase arterial O2 content. As per the Fick's Law equation then this could result in an increase in mixed venous pO2. However, the MCQ says "fixed FIO2" thus effectively excluding this possibility. COMMENT re: above. I may be wrong but I think possibly an increased CO2 may be the answer in the first stem. CvO2 = CaO2 - VO2/CO and we are holding VO2, CO and FiO2 all constant Therefore the only thing that can change is CaO2 which equals Hb x sats x 1.34 + PaO2 x 0.003 Sats is unlikely to change because FiO2 is constant. Therefore an increase in CO2 will shift the ODC to the right. If we hold sats constant on that new curve, the new PaO2 will be slightly higher. This means that CaO2 will be a bit higher (although not much as have to multiply by 0.003) So although it's a silly question and I think maybe remembered incorrectly, perhaps increased CO2 is the answer?? Further comments: With a fall of temp you get increased gas solubility as per Henry's law - shouldn't this mean you get an increase content of oxygen - (Probably a small increase but an increase none the less)? It will be a fall in PO2, but an increase in dissolved O2. Reply: Yes this is correct. BUT the temp difference would be small (a few degrees maybe) and the increased amount of dissolved O2 would be very small indeed. Probably too little to even be able to measure. Thus, this would not be a meaningful option in my opinion. ALSO, with hypothermia, VO2 would decrease but the question (in the top version anyway) says constant VO2. The idea with this question is to understand and be able to apply Fick's Law. The correct answer will depend on the actual wording, and the options on the actual MCQ on the day. Comment I think Jul 03 - answer is anaemia. need greater oxygen extraction per unit volume blood due to decreased oxygen carrying capacity, therefore decreasing pvO2 I think the first question would be "none of the above", as most of them need the position of the ODC to change or you would not be able to have a change in oxygen extraction if you could not change position of ODC for a given CO, oxygen uptake, etc For the first question, the only thing that could change the mixed venous O2 content given the constants - would be arterial oxygen content (if oxygen uptake, and CO are constant, then by the fick principle, mixed venous content would only change for arterial content) if the question uses the term "tension", then B would increase the pvO2 for a given venous content - as it R shifts the curve (as with all the constants - venous content does not change) Comment re Jul 03 Anaemia would not be correct because O2 tension is not affected by Hb. A person can have normal O2 tension with Hb of 5. However Anaemia would lead to a decrease O2 CONTENT (not partial pressure). yes you are correct, anaemia does not alter partial pressure. But this only applies to arterial side. As oxygen content is decreased that is supplying the tissue, with no change in oxygen consumption, there will be much less content left in venous blood, which will result in decreased partial pressure in venous blood. The PO2 is related to the CONTENT of O2 through the ODC which is unchanged (as given in the question). Lower CONTENT equals lower PO2 because the other unchanged variables means the Hb oxygen complex will have a lower %O2 and if you follow the ODC at a lower saturation: a lower PO2. Which makes sense really. Comment re Jul 03 I am thinking maybe hypercarbia could cause it perhaps if due to increased metabolism (eg Malignant hyperthermia) then it stands to reason that there would be increased in Co2 production / o2 consumption and then this would indeed reduce mixed venous o2 sats. but i guess on the D day with all options available and correctly worded it would be hopefully easier to chose the one best answer! As it stands its probably B or D Hypercarbia without change in VO2 or Hb would lead to an increase in the PO2 from the right shift of the ODC. Without changing the ODC (!) yeah right I'd like to try controlling for that in that lab. IMHO these are not the same question and should be dissociated. From Peter Kam's NSW primary course the answer to the first question with oxygen content is E but it is annotated that the question was abandoned due to an unclear best answer.

Answer 111

Not sure how acute this question is. Initially at altitude, acute hypoxaemia leads to increased ventilation (acting via peripheral chemoreceptors in aortic and carotid bodies) resulting in decreased arterial pCO2 secondary to increased exhalation; this inhibits the increase in ventilation (acting via the central chemoreceptors). This means that initial Hypoxic Ventilatory Response (HVR) is blunted but increases over the next 2-3 days as the pH of arterial blood and CSF is normalized (see below). In native lowlanders, the carotid bodies increase in size and weight due to increasing vascularity. In native highlanders, they increase through hyperplasia. They respond to the lower arterial oxygen tension, not the lower oxygen content. Over the next 2 to 3 days, the respiratory centre of the brain stem loses around four fifths of its sensitivity to pCO2 and H+ concentrations, thereby removing much of the hypocapnoeic inhibition on the ventilation increase. This increased ventilation then limits the amount of hypoxaemia (at the expense of a very low arterial pCO2). This is achieved firstly by "normalising" the pH of CSF by increased transport of HCO3- across the blood brain barrier (out of the CSF), and secondly (slower) by renal compensation for low pCO2 by increased excretion of bicarbonate. So by the third day or so, someone at altitude should be ventilating much greater volumes than they were on the first day. The action of acetazolamide to stimulate respiratory drive is by inhibiting carbonic anhydrase, so that H+ ions accumulate in the CSF, lowering its pH and stopping the alkalotic blunting of the HVR. As part of acclimatisation to altitude, the concentration of red cell 2,3 DPG increases. This is caused by hypoxic induced alkalosis, which inhibits the enzyme 2,3-DPG Diphosphatase that is responsible for the breakdown of 2,3-DPG. Slight elevations of the 2,3-DPG concentration are seen after the first few hours of exposure to altitude. This leads to a right shift of the oxygen-haemoglobin dissocation curve, therefore increased P50 and increased oxygen unloading in the peripheries. This right shift is often ablated by the respiratory alkalosis which shifts the ODC to the left. ACIDaemia reduces the affinity of haemoglobin for oxygen, ie also shifts the curve right, therefore increasing P50 with resulting increased oxygen unloading in the peripheries. Tissue hypoxia causes the JuxtaGlomerular Apparatus (JGA) to secrete Erythropoietin which stimulates erythropoiesis (increased by about 30%) in the bone marrow, increasing the haemoglobin concentration. EPO secretion begins to rise in the first few hours and is maximal at about 48 hours of exposure to hypoxia. Increase of Hb concentration is one of the slowest parts of acclimatization to altitude. The red blood cells maintain their normal lifespan of 120 days. In brief: Ventilation increases at altitude. Cardiac output and heart rate will increase (& this increases tissue O2 delivery) Plasma and red cell volume both increase Coronary blood flow decreases and myocardial oxygen extraction increases, both by about 30% Hypoxic Pulmonary Vasoconstriction increases the Pulmonary Vascular Resistance Polycythaemia will develop Red cell 2,3 DPG will increase due to decreased breakdown Pulmonary and cerebral oedema are known complications of ascent to altitude. However, at extreme altitude, a left shift of the oxygen-haemoglobin dissociation curve is seen, secondary to extreme respiratory alkalosis. this leads to reduced P50 and improved O2 uptake in the lungs. High altitude pulmonary oedema occurs acutely in unacclimatised people who ascend to altitude. They experience the usual symptoms of pulmonary oedema, and the condition may be fatal. Treatment is descent. High altitude cerebral oedema mirrors high altitude pulmonary oedema. Features are confusion, ataxia, hallucinations, and eventually loss of consciousness and death; again, treatment is descent. Notes: It would be helpful to know how rapidly the onset of polycythaemia occurs - this paper (http://jap.physiology.org/cgi/content/abstract/81/2/636) notes that there is no change within 13 days - so NOT an acute effect by any means... References

Answer 112

Not sure how acute this question is. Initially at altitude, acute hypoxaemia leads to increased ventilation (acting via peripheral chemoreceptors in aortic and carotid bodies) resulting in decreased arterial pCO2 secondary to increased exhalation; this inhibits the increase in ventilation (acting via the central chemoreceptors). This means that initial Hypoxic Ventilatory Response (HVR) is blunted but increases over the next 2-3 days as the pH of arterial blood and CSF is normalized (see below). In native lowlanders, the carotid bodies increase in size and weight due to increasing vascularity. In native highlanders, they increase through hyperplasia. They respond to the lower arterial oxygen tension, not the lower oxygen content. Over the next 2 to 3 days, the respiratory centre of the brain stem loses around four fifths of its sensitivity to pCO2 and H+ concentrations, thereby removing much of the hypocapnoeic inhibition on the ventilation increase. This increased ventilation then limits the amount of hypoxaemia (at the expense of a very low arterial pCO2). This is achieved firstly by "normalising" the pH of CSF by increased transport of HCO3- across the blood brain barrier (out of the CSF), and secondly (slower) by renal compensation for low pCO2 by increased excretion of bicarbonate. So by the third day or so, someone at altitude should be ventilating much greater volumes than they were on the first day. The action of acetazolamide to stimulate respiratory drive is by inhibiting carbonic anhydrase, so that H+ ions accumulate in the CSF, lowering its pH and stopping the alkalotic blunting of the HVR. As part of acclimatisation to altitude, the concentration of red cell 2,3 DPG increases. This is caused by hypoxic induced alkalosis, which inhibits the enzyme 2,3-DPG Diphosphatase that is responsible for the breakdown of 2,3-DPG. Slight elevations of the 2,3-DPG concentration are seen after the first few hours of exposure to altitude. This leads to a right shift of the oxygen-haemoglobin dissocation curve, therefore increased P50 and increased oxygen unloading in the peripheries. This right shift is often ablated by the respiratory alkalosis which shifts the ODC to the left. ACIDaemia reduces the affinity of haemoglobin for oxygen, ie also shifts the curve right, therefore increasing P50 with resulting increased oxygen unloading in the peripheries. Tissue hypoxia causes the JuxtaGlomerular Apparatus (JGA) to secrete Erythropoietin which stimulates erythropoiesis (increased by about 30%) in the bone marrow, increasing the haemoglobin concentration. EPO secretion begins to rise in the first few hours and is maximal at about 48 hours of exposure to hypoxia. Increase of Hb concentration is one of the slowest parts of acclimatization to altitude. The red blood cells maintain their normal lifespan of 120 days. In brief: Ventilation increases at altitude. Cardiac output and heart rate will increase (& this increases tissue O2 delivery) Plasma and red cell volume both increase Coronary blood flow decreases and myocardial oxygen extraction increases, both by about 30% Hypoxic Pulmonary Vasoconstriction increases the Pulmonary Vascular Resistance Polycythaemia will develop Red cell 2,3 DPG will increase due to decreased breakdown Pulmonary and cerebral oedema are known complications of ascent to altitude. However, at extreme altitude, a left shift of the oxygen-haemoglobin dissociation curve is seen, secondary to extreme respiratory alkalosis. this leads to reduced P50 and improved O2 uptake in the lungs. High altitude pulmonary oedema occurs acutely in unacclimatised people who ascend to altitude. They experience the usual symptoms of pulmonary oedema, and the condition may be fatal. Treatment is descent. High altitude cerebral oedema mirrors high altitude pulmonary oedema. Features are confusion, ataxia, hallucinations, and eventually loss of consciousness and death; again, treatment is descent. Notes: It would be helpful to know how rapidly the onset of polycythaemia occurs - this paper (http://jap.physiology.org/cgi/content/abstract/81/2/636) notes that there is no change within 13 days - so NOT an acute effect by any means... References

Answer 113

Not sure how acute this question is. Initially at altitude, acute hypoxaemia leads to increased ventilation (acting via peripheral chemoreceptors in aortic and carotid bodies) resulting in decreased arterial pCO2 secondary to increased exhalation; this inhibits the increase in ventilation (acting via the central chemoreceptors). This means that initial Hypoxic Ventilatory Response (HVR) is blunted but increases over the next 2-3 days as the pH of arterial blood and CSF is normalized (see below). In native lowlanders, the carotid bodies increase in size and weight due to increasing vascularity. In native highlanders, they increase through hyperplasia. They respond to the lower arterial oxygen tension, not the lower oxygen content. Over the next 2 to 3 days, the respiratory centre of the brain stem loses around four fifths of its sensitivity to pCO2 and H+ concentrations, thereby removing much of the hypocapnoeic inhibition on the ventilation increase. This increased ventilation then limits the amount of hypoxaemia (at the expense of a very low arterial pCO2). This is achieved firstly by "normalising" the pH of CSF by increased transport of HCO3- across the blood brain barrier (out of the CSF), and secondly (slower) by renal compensation for low pCO2 by increased excretion of bicarbonate. So by the third day or so, someone at altitude should be ventilating much greater volumes than they were on the first day. The action of acetazolamide to stimulate respiratory drive is by inhibiting carbonic anhydrase, so that H+ ions accumulate in the CSF, lowering its pH and stopping the alkalotic blunting of the HVR. As part of acclimatisation to altitude, the concentration of red cell 2,3 DPG increases. This is caused by hypoxic induced alkalosis, which inhibits the enzyme 2,3-DPG Diphosphatase that is responsible for the breakdown of 2,3-DPG. Slight elevations of the 2,3-DPG concentration are seen after the first few hours of exposure to altitude. This leads to a right shift of the oxygen-haemoglobin dissocation curve, therefore increased P50 and increased oxygen unloading in the peripheries. This right shift is often ablated by the respiratory alkalosis which shifts the ODC to the left. ACIDaemia reduces the affinity of haemoglobin for oxygen, ie also shifts the curve right, therefore increasing P50 with resulting increased oxygen unloading in the peripheries. Tissue hypoxia causes the JuxtaGlomerular Apparatus (JGA) to secrete Erythropoietin which stimulates erythropoiesis (increased by about 30%) in the bone marrow, increasing the haemoglobin concentration. EPO secretion begins to rise in the first few hours and is maximal at about 48 hours of exposure to hypoxia. Increase of Hb concentration is one of the slowest parts of acclimatization to altitude. The red blood cells maintain their normal lifespan of 120 days. In brief: Ventilation increases at altitude. Cardiac output and heart rate will increase (& this increases tissue O2 delivery) Plasma and red cell volume both increase Coronary blood flow decreases and myocardial oxygen extraction increases, both by about 30% Hypoxic Pulmonary Vasoconstriction increases the Pulmonary Vascular Resistance Polycythaemia will develop Red cell 2,3 DPG will increase due to decreased breakdown Pulmonary and cerebral oedema are known complications of ascent to altitude. However, at extreme altitude, a left shift of the oxygen-haemoglobin dissociation curve is seen, secondary to extreme respiratory alkalosis. this leads to reduced P50 and improved O2 uptake in the lungs. High altitude pulmonary oedema occurs acutely in unacclimatised people who ascend to altitude. They experience the usual symptoms of pulmonary oedema, and the condition may be fatal. Treatment is descent. High altitude cerebral oedema mirrors high altitude pulmonary oedema. Features are confusion, ataxia, hallucinations, and eventually loss of consciousness and death; again, treatment is descent. Notes: It would be helpful to know how rapidly the onset of polycythaemia occurs - this paper (http://jap.physiology.org/cgi/content/abstract/81/2/636) notes that there is no change within 13 days - so NOT an acute effect by any means... References

Answer 114

Central chemoreceptors were thought to be in the medullary respiratory centre, but are now recognised as separate. They: lie on the surface of the anterior medulla, a short distance away from the respiratory neurones of the medulla (respiratory centre). are bathed in ECF, the composition of which is determined by CSF, local blood flow and metabolism. respond to increased hydrogen ion concentration, although the precise mechanism by which pH change causes stimulation of these chemoreceptors is still unknown. are mediated by muscarinic ACh receptors. Some light seems to have been thrown on to the mechanism by which pH canges cause stimulation of chemoreceptors: (See: Am J Physiol Cell Physiol 287: C1493–C1526, 2004.) which suggests an interplay of intracellular Ca, gap junctions, oxidative stress, and neurotransmitters regulate the action of K+ and Ca2+ channels. Answers: A - false, bathed in ECF not CSF B - false, respond to ECF pH not CSF pH C - true D - false, close to (but not in) the respiratory centre Comment This is a terrible question and I hope it's been abandoned. True, the chemoreceptors are bathed in ECF. However this ECF is right next to CSF and according to West, CSF is "apparently the most important" factor in determining the ECF pH. Furthermore at the top of p107 he summarises the preceding section with three points, one of which is" "(Central Chemoreceptors:) respond to the change in pH of the ECF/CSF". This is one of those anal questions that has no application in clinical practice and is really testing us on whether or not we have read a prescribed text in detail. In conclusion, B is true, however I agree that C is "more" true.

Answer 115

A. they do have a nonlinear response to paO2 changes- in fact they register changes as high as pO2 500mmHg, but the response really gears up at around 70mmHg, and maximum response is below 50mmHg. B. their response is eliminated by as little as 0.1 MAC- bad news for CO2 retainers C. they respond to a rise in paCO2 D. they respond around five times more rapidly than the central chemoreceptors to an increase in paCO2 E. they respond to acidaemia (their response is to pO2, pH and pCO2). NB:They also do respond to reduction in their perfusion rate (Nunn 63) F. therefore they do not respond only to H+ G. or only to arterial hypoxaemia H. they are innervated by Hering's nerves -> Glossopharyngeal n -> dorsal respiratory centre of the medulla I. they have a high metabolic rate J. Probably not. Nunn (6th p65) states Cyanide and Carbon Monoxide are chemical stimulants due to interference with the cellular cytochrome system (Nicotine and acetylcholine are also stimulants, via sympathetic ganglia). However, while Ganong (19th p644) agrees CO is toxic to the tissue cytochromes, this is at an amount 1000x the lethal dose, so plays no role in clinical poisoning. Ganong goes on to say CO is not a stimulant of the peripheral chemoreceptors as the PaO2 remains normal. Also Kam (2nd p109): "The peripheral chemoreceptors are stimulated by low oxygen tension (PaO2), not low oxygen content (CaO2) of blood. Conditions with low blood oxygen content but relatively normal oxygen tension (anaemia, carboxyhaemoglobin) do not stimulate ventilation via the peripheral chemoreceptors. K. Cyanide is a chemical stimulants due to interference with the cellular cytochrome system, decreasing ATP synthesis. Nunn (6th p65) L. blood flow is 20 times their weight per minute- 20mL/g/min or 2,000mL/100g/min M. the aortic bodies are innervated by the vagus -> dorsal medullary respiratory centre N. No. See J. O. depite their high metabolic rate, their small size and the enormous amount of blood flow results in very little oxygen being removed; therefore they have a very low A-V O2 difference, and respond to arterial rather than venous pO2 P. yes they have Type I and II glomus cells

Answer 116

``` Feb 04 Version Peripheral chemoreceptors: A. In the carotid sinus B. Have glomus cells C. Low A-V difference D. Innervated by glossopharyngeal nerve E. Blood flow of 200mls/g/min ``` A. the carotid bodies are at the carotid bifurcation, near the carotid sinus (which contains the baroreceptors). B. yes, contain glomus cells C. yes, they have a low A-V difference D. yes, they are innervated by the glossopharyngeal nerve (via Hering's nerves) E. see L above RE13b [Feb04] Carotid bodies (Similar to RE13) A. Have glomus cells B. Innervated by vagus C. Blood flow of 200mls/g/min D. High A-V difference E. Afferents via CN IX (See also:RE36) A. true B. false. the aortic bodies are innervated by the vagus, the carotid bodies by the glossopharyngeal nerve C. true. See L of the first question (L states 20mls/g/min, C is false.) D. false. they have a low A-V difference, despite high metabolic rate, because they have a very high blood flow E. true. Ganong (18th p629) Afferents from the carotid body ascend to the medulla via the carotid sinus and IX nerves, fibres from the aortic bodies ascend in the vagi.

Answer 117

Re RE14: A. not sure- but the saline-filled lung pressure-volume curve doesn't show hysteresis Surfactant is a radially- dependent surface tension reducing agent - providing more reduction in surface tension with smaller radius ie it counteracts Laplace's Law. Also Fig 7.7B West shows the lung extract (surfactant) change in surface tension with area follows the hysteresis curve. Can these be correlated?? My conclusion is that surfactant is at least involved with hysteresis. B. produced by type II pneumocytes C. is commonly deficient in premmies D. Yes. Detergent molecules (hydrophobic) repel each other, reducing the attractive forces between water moleculres and thus surface tension. Surfactant does the same. (NO! Look at Fig 7-7, p.101, West , 7th Ed. Detergent does reduce surface tension, but it is independent of area, whereas dipalmitoyl phosphatidylcholine (DPPC) curve is different [hysteresis]. Would appreciate comments!!) E. Yes. Reduction of surface tension reduces static lung recoil pressure, therefore reduces the amount of negative IPP F. production is fast, but so is turnover G. true, it increases pumonary compliance

Answer 118

Re: RE14b: A. true, surface tension is decreased with increasing surfactant concentration B. lung compliance increases with surfactant C. produced by type II cells D. stabilises alveoli to prevent them emptying into bigger ones E. decreases surface tension in smaller alveoli to promote stability Hysteresis is shown by all other elastic bodies. It is due to intrisic elastic forces within the lung tissues. Nunn Page 30)

Answer 119

Work of breathing = Work to overcome elastic recoil + Work to overcome resistance forces 2 factors are responsible for the lung's elastic recoil - surface tension + elastic tissues During quiet breathing, expiration is passive in that no active muscular contraction occurs, the energy used for expiration is the stored elastic energy achieved by inspiration. Thus the answer should be B. Passive due to surface tension in the alveoli and elastic tissue recoil.

Answer 120

``` Correct answer : B Venous CO2 : 53.5ml/100m blood Arterial CO2: 49ml/100ml blood A-V difference : 4.5ml/100ml blood Refer arterial and mixed venous CO2 dissociation curves ``` Ganong (23rd ed, pg 611, table 36-1) gives slightly different values, but with a similar outcome: arterial CO2 = 46.4 mL/dL venous CO2 = 49.7 mL/dL difference = 3.3 (close to 4)

Answer 121

Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs. Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung. Up to 80% of bradykinin is inactivated (by ACE). Angiotensin I is converted to angiotensin II (by ACE). aldosterone is unaffected. Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO. Vasopressin is unaffected. ADH is unaffected. The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites. Prostaglandins E2 and F2alpha are removed. Leukotrienes are removed. prostaglandin A2 and prostacyclin are not affected. Histamine is unaffected. Surfactant has a fast turnover, but I'm not sure whether it's actually degraded. Guyton: 'surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells

Answer 122

Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs. Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung. Up to 80% of bradykinin is inactivated (by ACE). Angiotensin I is converted to angiotensin II (by ACE). aldosterone is unaffected. Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO. Vasopressin is unaffected. ADH is unaffected. The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites. Prostaglandins E2 and F2alpha are removed. Leukotrienes are removed. prostaglandin A2 and prostacyclin are not affected. Histamine is unaffected. Surfactant has a fast turnover, but I'm not sure whether it's actually degraded. Guyton: 'surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells' References

Answer 123

Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs. Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung. Up to 80% of bradykinin is inactivated (by ACE). Angiotensin I is converted to angiotensin II (by ACE). aldosterone is unaffected. Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO. Vasopressin is unaffected. ADH is unaffected. The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites. Prostaglandins E2 and F2alpha are removed. Leukotrienes are removed. prostaglandin A2 and prostacyclin are not affected. Histamine is unaffected. Surfactant has a fast turnover, but I'm not sure whether it's actually degraded. Guyton: 'surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells' References

Answer 124

Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs. Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung. Up to 80% of bradykinin is inactivated (by ACE). Angiotensin I is converted to angiotensin II (by ACE). aldosterone is unaffected. Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO. Vasopressin is unaffected. ADH is unaffected. The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites. Prostaglandins E2 and F2alpha are removed. Leukotrienes are removed. prostaglandin A2 and prostacyclin are not affected. Histamine is unaffected. Surfactant has a fast turnover, but I'm not sure whether it's actually degraded. Guyton: 'surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells' References

Answer 125

Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs. Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung. Up to 80% of bradykinin is inactivated (by ACE). Angiotensin I is converted to angiotensin II (by ACE). aldosterone is unaffected. Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO. Vasopressin is unaffected. ADH is unaffected. The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites. Prostaglandins E2 and F2alpha are removed. Leukotrienes are removed. prostaglandin A2 and prostacyclin are not affected. Histamine is unaffected. Surfactant has a fast turnover, but I'm not sure whether it's actually degraded. Guyton: 'surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells' References

Answer 126

Breathing 100% Oxygen can cause absorption atelectasis therefore reduces vital capacity. Question not specific enough in this form - I breathe oxygen everyday without any problems. me too. Count me in too. I believe A is the correct answer. Nunn's 5th Ed, pg 564. " Voluntary maximal inspirations have been effective in clearing areas of absorption atelectasis in subjects who had been breathing oxygen near residual volume. The process imparted a distinctive tearing sensation in the chest, but rapidly restored the PO2 and CXR to Normal."

Answer 127

CO2 is carried in the blood in three forms: Bicarbonate Carbamino compounds Dissoved CO2 In arterial blood, the proportions of CO2 present in these 3 forms is as in option A in the question. The majority of CO2 (90%) is present as bicarbonate. However, the question is asking about the percent contribution of these three forms of carriage to the additional CO2 picked up in systemic capillaries. The 'standard' textbook values are the figures in option B above. Note the increased importance of carbamino compounds to the carriage of this extra 4mls CO2/dl of blood.

Answer 128

Physiological dead space is "the volume of gas not equilibrating with blood (wasted ventilation)" and comprises of the anatomical dead space plus the alveolar dead space. Anything that increases anatomical dead space (the volume of the conducting airways) or alveolar dead space (area of the lung where the alveoli are ventilated but not perfused) will increase physiological dead space. Both B and G are true! A. False: Physiological dead space increases by approx 1ml/yr from early adulthood. B. Physiological dead space increases with anaesthesia - due to increased alveolar dead space the exact cause of which is unclear - it's not a Zone 1. (Nunn's 6th Ed. p310) C. Supine position makes West Zone 1 unlikely (well, West Zones are in the upright lung, but the same principle applies - the effect due to gravity is less). [Clarification:"In the supine position differences in blood flow between apices and bases are replaced by differences between posterior and anterior regions". That is, the V-Q relationship is much the same. And whilst "the VD/VT ratio decreases from a mean value of 34% in the upright position to 30% in the supine position", this is actually "largely explained by differences in anatomical dead space". [Nunn 6th Ed p114 and 121]] (Physiologic dead space consists of anatomic + alveolar dead space. Anatomic reduces from 150mL to 100mL when moving from sitting to lying - Nunn 6th Ed Pg 119) D. The Bohr equation uses the pCO2 of mixed expired gas, not end-tidal pCO2. E. See D. The Enghoff modification to the Bohr equation used arterial pCO2 as an estimate of average (ideal) alveolar pCO2. F. False, see the definition above. G. Yes, but only a slight increase in the short term in normal lungs. Long term PEEP can cause a greater increase in physiological dead space. (Nunn's Respiratory Physiology 6th Ed. p 433). See below quote and also answer to B in RE20b below. "The acute application of moderate amounts of PEEP causes only a slight increase in VD/VT ratio. The alveolar component of physiological dead space may be increased by ventilation in patients with lung injury or when mean intrathoracic pressure is high, such as with significant amounts of PEEP. Under the latter conditions, lung volume is increased to such an extent that not only does cardiac output fall but pulmonary vascular resistance rises as well. Perfusion to overexpanded alveoli is reduced and areas of lung with high V/Q ratios develop, which constitute alveolar dead space. In healthy lungs this is not seen until PEEP levels exceed 10-15cm H2O. However with IPPV in lung injury there is now good evidence that overdistension occurs in the relatively small number of functional alveoli and local perfusion to these lung units is likely to be impeded. There is indirect evidence that long term application of PEEP may cause a very large increase in the dead space, probably because of bronchiolar dilation." - Nunn's Respiratory Physiology p433

Answer 129

Here the best answer is B: Hypotension A. Pulmonary hypertension: Pulmonary vascular obstruction can occur late in pulmonary hypertension, which would lead to an increase in alveolar dead space. (Decreased perfusion also increases this, so perhaps A is a possible answer) B. Hypotension: This can lead to the development of West Zone 1 - where pA > pa > pv (i.e. alveolar pressure is greater than pulmonary arterial pressure, therefore no perfusion occurs) leading to an increase in alveolar dead space, (which by definition increases physiological dead space). C. Atelectasis: This causes a region of lung to not be ventilated - therefore it can't contribute to dead space. D. Pleural effusion: occurs outside the lung, so won't increase dead space directly.

Answer 130

To calculate shunt, you need all of the above. shunt blood flow/total blood flow = (oxygen content in end capillary blood - oxygen content in arterial blood)/(oxygen content in end capillary blood - oxygen content in mixed venous blood) note that oxygen CONTENT is in mL/L (not pO2 - but can figure it out from pO2 by using ODC) and flows are in L/min total blood flow = cardiac output (L/min) CcO2 - pulmonary end capillary oxygen content is assumed to be that of oxygen content that you would have with PAO2 as the blood at the end of the pulmonary capillaries has equilibrated with the alveoli. summarised a little more succintly as Qs/Qt = (CcO2 - CaO2)/(CcO2 - CvO2) which i like to think of as proportion of shunt flow to total flow = proportion of impairment of oxygenation to oxygenation that should be achieved which makes it a little easier for me to remember

Answer 131

During normal quiet breathing alveolar pressure oscillates between -1 to 1 cmH2O, and is zero during pauses between inspiration and expiration. So B and D are correct. In the second version, D is correct.

Answer 132

During normal quiet breathing alveolar pressure oscillates between -1 to 1 cmH2O, and is zero during pauses between inspiration and expiration. So B and D are correct. In the second version, D is correct.

Answer 133

FIRST VERSION Best answer is B A: INCORRECT: Shallower gradient in effort dependent part of the loop C: INCORRECT: increased volume means that the lung compliance curve is at the top shallower (less compliant) end. Nunn's (5th ed p48) says that 'emphysema is unique in that static pulmonary compliance is increased' - does this mean that C could also be true, or is it going to far to assume that they mean COPD/emphysema when they say chronic airflow limitation? Comment: I believe that chronic airflow limitation (CAL) is an old acronym for what is now called COPD.

Answer 134

2015 VERSION A: CORRECT: secondary to hyperinflation from gas trapping B:INCORRECT: shallower slope of effort dependent portion C: INCORRECT: although total lung capacity increases, much of this is gas trapping so is part of RV rather than VC D:INCORRECT: increased compliance

Answer 135

One lung anaesthesia creates a large shunt, hence A and B are incorrect. Lying ventilated lung side down would theoretically give better V/Q matching of ventilated lung - helping unventilated lung become West's Zone 1. D would be correct, hyperventilation of the one lung may not completely abolish the hypercarbia. PEEP - Ventilated lung should have mild PEEP to avoid atelectasis and maximise ventilation of this lung. However, PEEP to ventilated lung may exacerbate hypoxia by reducing cardiac output and forcing more blood through the uppermost, non-ventilated lung. It may be tried to remedy hypoxia or with a 'sinking' mediastinum, but the results are unpredictable. I think in the context of the question the answer is false to these parts. (A-Z under 'one-lung anaesthesia', Oxford Handbook ref. from Miller) Initial version a - high FiO2 will completely correct PaO2 incorrect one lung ventilation creates an obligatory right to left transpulmonary shunt through the non-ventilated non-dependent lung. For the same FiO2 (as two lung ventilation), one lung ventilation results in a much larger PAO2-PaO2 gradient and lower PaO2 than two lung ventilation. -- Miller's 6th ed., One Lung Ventilation i.e. Large shunt from non-ventilated lung means that PaO2 cannot be *completely* corrected by increased PAO2 in ventilated lung. b - CPAP will completely correct PaO2 incorrect applying PEEP to the ventilated lung has the risk of PEEP induced increase in lung volume which can cause compression of small dependent lung intra-alveolar vessels and increase this lungs PVR and therefore resulting in a shunt to non-ventilated lung --> increasing shunt and decreasing PaO2. PEEP is a trade off between positive effect of increased FRC and negative effect of increased PVR and shunt Millers 6th One lung ventilation section However, mild CPAP to non-ventilated lung can be helpful in decreasing the shunt, but won't completely correct it. i.e. Large shunt from non-ventilated lung means that PEEP to ventilated lung cannot *completely* correct PaO2. c - supine position will give better V/Q matching incorrect ventilated (dependent) lung has better V/Q matching in a lateral position, compared with supine, due to gravity effect of improving perfusion d - associated with hypercarbia correct during one-lung ventilation, ventilated lung can eliminate enough carbon dioxide to compensate for non-ventilated lung. Overtime however retention of CO2 from blood traversing non-ventilated lung usually slightly exceeds increased elimination of CO2 from blood traversing ventilated lung, PaCO2 will usually slowly increase. -- Miller's 6th ed., One Lung Ventilation

Answer 136

July 2010 version a - oxygenation better in supine false oxygenation better with ventilated lung as the dependent lung in LDP (lateral decubitus position) b - should have 20 cmH2O PEEP to lower lung false PEEP should be cautiously applied as above c - usually/can be associated with hypercarbia true can be associated with hypercarbia as above Just ventilate more? The patient is anaesthetised and likely ventilated. Whether hypercarbic or not is completely dependent on how much you ventilate him. It's more likely that you would over ventilate than under ventilate, to keep up PO2. So unless you are slack, shouldn't this be usually associated with low PCO2? In the normal surgical patient, this is essentially correct. However, in one-lung ventilation, there is a dramatic increase in shunt. There is a limit as to how much you can ventilate a patient, especially when you have only one lung available. Oxygenation is worse (but with SpO2 typically >90%), but increased ventilation is limited. Hence, since PaCO2 ~ PACO2 ~ VCO2/VA (i.e. CO2 arterial tension is proportional to production / alveolar ventilation), the decrease in ventilation will tend to result in hypercarbia. Mild CPAP to the non-ventilated lung can also be helpful at maintaining oxygenation. (I think this is why there is an (incorrect) answer regarding CPAP to the ventilated lung.) Ref: One-lung ventilation and arterial oxygenation. Current Opinion in Anaesthesiology. 24(1):24-31, February 2011.

Answer 137

``` This can be calculated as follows: Assuming: barometric pressure (PB) = 760mmHg inspired O2 concentration (FIO2) = 0.21 dry air so pH2O = 0 mmHg then: pIO2 = (0.21)(760 - 0) = 159.6 mmHg ```

Answer 138

None of the above. None of them become deranged enough to exert an effect. Something else causes an increase in ventilation before they can be affected. Maybe a CNS effect. West discusses A. (page 133) but goes on to say that no current theory adequately explains observed increase in ventilation. Agreed None of the above is the most suitable answer. Muscle mechanoreceptors & raised body temperature (via higher centres) thought to contribute. Faunce mentions a possible role of increased plasma K+ (via peripheral chemoreceptors) & of PCO2 fluctuations. Johnson BD - Chest - 01-AUG-1999; 116(2): 488-503 does not directly answer the question however does provide an insight into minute ventilation changes during exercise. However, Power&Kam say on pages 97 and 326. The rapid response in minute ventilation at the onset of exercise result from "neural inputs to the respiratory centre from the motor cortex and proprioceptive receptors in the exercising muscle". The gradual response in minute ventilation that comes later may be related to fluctuations in arterial oxygen tension and hence the "oscillatory discharge of central chemoreceptors provides a potent respiratory stimulus in exercise". Nunn 6thEd' (p244) says that "peripheral chemoreceptors do contribute, in a small way, to exercise-induces hyperpnoea". Other factors are "the release of epinephrine and norepinephrine, and a rise in blood temperature". So "A" is possible? The sentence after the above Nunn quote is "In spite of this caveat, it is difficult to avoid the conclusion that arterial blood gas tensions acting on the chemoreceptors cannot be the main factor in the increase of ventilation during exercise. Option E probably remains the best. West 8th Ed P135: "One hypothesis is that oscillations in arterial PO2 and PCO2 may stimulate the peripheral chemoreceptors, even though the mean level remains unaltered." So A is definitely possible. Nunns applied physiology- 7th edition-page-266 Peripheral chemo receptors are responsible for exercise induced hyperopnoea- during the non steady state. This may not result from changes in Po2, but from oscillations in arterial Pco2. Humoral factors are important in heavy and severe exercise. Lactic acidocis contributes to excess ventilation during heavy and severe exercise. Slight additional respiratory drive may result from hyperthermia. I agree with the answer A since it provides a possible mechanism as West suggested which makes sense. We know that hypoxaemia/hypercarbia/acidosis do not occur during exercise. Comment: perhaps the answer lies in the difference between the words "cause" and "hypothesis"?

Answer 139

As worded, the question is not asking about efficiency, but about how much of basal energy expenditure is devoted to respiration - "...as % of total VO2..." Work of breathing is ~3mL O2/min, or 1% of basal O2 consumption, as per Brandis page 148,

Answer 140

PEEP: Increases FRC Increases lung compliance in lower dependant parts of the lung and reduces compliance in non-dependant parts (Nunn 6th Ed, pg 434) therefore also improving VQ matching Does not reduce (total) lung water, but may redistribue extravascular lung water. Reduces airway resistance So answer should be D. PEEP also improves V/Q inequalities (Morgan & Murray) I agree with above statements, agree answer D and also add: significant PEEP (raised intrathoracic pressure) can increase alveolar component of physiological dead space, can also occur with ventilation in lung injury (high V/Q ratios, which constitute alveolar dead space, don't develop in healthy lungs until PEEP > 10-15cmH20) Nunn 6th Ed, p 432 PEEP doesn't change total amount of lung water but greater proportion in extra-alveolar interstitial space and increased lymphatic drainage (in haemodynamic pulmonary oedema in dogs) Nunn p 392 animal models of pulmonary oedema indicate that increased lung volume increases capacity of interstitium to hold liquid Nunn p 392 increasing lung volume decreases airway resistance and helps to prevent gas trapping (achieved with CPAP in spontaneously breathing or PEEP in ventilated patient) Nunn p 44

Answer 141

Step 1: Assuming by "moist" inspired pO2 it means saturated air at 37C in the airways (so pH20 = 47mmHg), then: pO2 = (247 -47) x 0.21 Step 2: Now approximating 0.21 to 0.2 (ie a one fifth so divide by 5 which is easy) then: pO2 = 200/5 = 40mmHg. Query: The MCQ is asking about moist ispired pO2. I think we don`t need to subtract 47 mmHg from 247 mmHg. So answer is D 50 mmHg? Response: This is of course a 'remembered' version of the actual MCQ so the wording is not perfect. I think that 47 has to be subtracted for the following reasons: If you think the question means the air is already 'moist', then that will only be for the ambient temperature (which is not stated) and in any case it doesn't say 'saturated' so some value for pH2O will have to be subtracted but you have no way of knowing what. (SVP is temp dependent). So the assumption you make (if you followed it correctly) just renders the MCQ unanswerable. Being 'moist' before inspiration still dilutes the ambient pO2 in any case but as you do NOT know the ambient temperature then again you do NOT know what value of pH2O to use to correct for this. The MCQ on your assumption is unanswerable. The given value of 247mmHg suggests that the MCQ is trying to make the calculation easy as "(247-47)/5" is easy mental arithmetic. The exam afterall is not a test of mathematical skill and MCQs that require calculation are set up so the calculation is easy to do. The problem is knowing the correct formula - in this case: pO2 = (pB - pIO2) x FIO2 where pB is barometric pressure. So, once you know that calculation MCQs have to: have easy calculations (Exception if you are supposed to 'know' the answer as a standard value) AND that MCQs have to have an answer (Exception if the examiners actually make a mistake -which happens) THEN the confusion is overcome. The expectation is that the candidates who do not understand the topic very well will get confused about whether to subtract 47 or not, and where to subtract it from, or whether to use 0.2 or not and the distractors will be designed to exploit this confusion, for example: some might think it is: (0.2 x 247)-47 = 2 (which is option B); some will forget that air is only 21% O2 and just use 247-47=200 (option A) some may try (247-47)-(40/0.8) = 150mmHg (possibly the missing option E), and -- incorrect alveolar gas equation yeap, would be (247-47)x0.21 - (40/0.8) = an impossible answer some will try 247 * 0.2 = about 50mmHg (option D).

Answer 142

Type I pneumocytes are flat with minimal organelles. They are easily damaged, and then replaced by proliferating type II pneumocytes which then differentiate into type I pneumocytes. Neither are macrophages. Type II have microvilli whereas Type I are a flat thin sheet. Surfactant works by sitting on top of the water layer anyway, so would be no good bound to a brush border. Type II pneumocytes produce surfactant. ``` neumocito.gif Type II pneumocyte secreting surfactant Surfactant is cleared from alveolar fluid by alveolar macrophages and type II cell reuptake (Faunce p21). Therefore, answer RE30 is D. II is bigger than I, therefore type II pneumocytes are fatter than type I pneumocytes, and have more equipment to do more stuff. Additional information: from Nunn 6th edition p 21,22,26,27 type 2 (alveolar epithelial cells) are stem cells and don't function as gas exchange membranes type 1 cells develop from type 2 type 2 cells contain stored surfactant in striated osmiophiluc organelles (surfactant formed and released from type 2 cells) type 2 are involved in pulmonary defence mechanisms, secrete cytokines and contribute to pulmonary inflammation type 2 are resistant to oxygen toxicity and replace type 1 cells after prolonged exposure to high oxygen concentrations (type 1 are very sensitive to damage from high O2 concentrations) type 2 have SP-A receptors, stimulation of these receptors produces negative feedback on surfactant secretion from type 2 cells and increases the reuptake of surfactant components into the type 2 cells ``` "Type II pneumocytes are plump, cuboidal cells that make up about 16% of the cell population of the alveolar region. They have microvilli on their outer edges (seen in the scanning EM on the right). They cover a small amount of alveolar surface area - only ~5% (compared with 95% of surface area covered by Type I cells)." clip_image002_000.gif "Under normal conditions, Type II cells synthesize surfactant, store it within cytoplasmic vacuoles (electron dense structures called lamellar bodies) and then release that surfactant onto the alveolar surface where it forms a thin film. Surfactant serves to lower surface tension and therefore it helps alveoli remain expanded, especially during the expiratory phase of breathing. Following injuries, Type II cells proliferate and replace damaged Type I cells." -from [1] dustcells.jpg Pneumocytes are naughty... when no one is looking '2 become 1'

Answer 143

Type I pneumocytes are flat with minimal organelles. They are easily damaged, and then replaced by proliferating type II pneumocytes which then differentiate into type I pneumocytes. Neither are macrophages. Type II have microvilli whereas Type I are a flat thin sheet. Surfactant works by sitting on top of the water layer anyway, so would be no good bound to a brush border. Type II pneumocytes produce surfactant. ``` neumocito.gif Type II pneumocyte secreting surfactant Surfactant is cleared from alveolar fluid by alveolar macrophages and type II cell reuptake (Faunce p21). Therefore, answer RE30 is D. II is bigger than I, therefore type II pneumocytes are fatter than type I pneumocytes, and have more equipment to do more stuff. Additional information: from Nunn 6th edition p 21,22,26,27 type 2 (alveolar epithelial cells) are stem cells and don't function as gas exchange membranes type 1 cells develop from type 2 type 2 cells contain stored surfactant in striated osmiophiluc organelles (surfactant formed and released from type 2 cells) type 2 are involved in pulmonary defence mechanisms, secrete cytokines and contribute to pulmonary inflammation type 2 are resistant to oxygen toxicity and replace type 1 cells after prolonged exposure to high oxygen concentrations (type 1 are very sensitive to damage from high O2 concentrations) type 2 have SP-A receptors, stimulation of these receptors produces negative feedback on surfactant secretion from type 2 cells and increases the reuptake of surfactant components into the type 2 cells ``` "Type II pneumocytes are plump, cuboidal cells that make up about 16% of the cell population of the alveolar region. They have microvilli on their outer edges (seen in the scanning EM on the right). They cover a small amount of alveolar surface area - only ~5% (compared with 95% of surface area covered by Type I cells)." clip_image002_000.gif "Under normal conditions, Type II cells synthesize surfactant, store it within cytoplasmic vacuoles (electron dense structures called lamellar bodies) and then release that surfactant onto the alveolar surface where it forms a thin film. Surfactant serves to lower surface tension and therefore it helps alveoli remain expanded, especially during the expiratory phase of breathing. Following injuries, Type II cells proliferate and replace damaged Type I cells." -from [1] dustcells.jpg Pneumocytes are naughty... when no one is looking '2 become 1'

Answer 144

A. Pneumotaxic centre in upper pons inhibits the inspiratory ramp, therefore regulates inspiration volume and time. B. Apneustic centre in lower pons has an excitatory effect on inspiratory area of medulla tending to prolong ramp action potentials (?function in humans) C. Inspiratory control originates from the dorsal medullary respiratory centre. D. Ventral medullary respiratory centre is associated with expiration ie C is correct

Answer 145

D. Pa 60mmHg, SaO2 91% ("ICU point")

Answer 146

Fowler's method measures anatomical dead space. The Bohr equation calculates physiological dead space. Alveolar dead space is physiological dead space minus anatomical dead space.

Answer 147

A: False. Superoxide dismutase is an antioxidant that protects against oxygen toxicity B: False C: False. Causes absorption atelectasis D: True E: False F: True Addit: Pulmonary oxygen toxicity occurs after 16 hours at levels of 50kPa or more (~375 mmHg or 50% oxygen) So E is true West p146 says CNS toxicity at PO2 exceeding 760mmHg. doesn't it mean B is correct? Central nervous system toxicity is caused by short exposure to high concentrations of oxygen at greater than atmospheric pressure. Pulmonary and ocular toxicity result from longer exposure to elevated oxygen levels at normal pressure. Pulmonary toxicity occurs with prolonged exposure of 16–24 hours or more to elevated concentrations of oxygen greater than 50% at normal atmospheric pressure. O2 toxicity causes oxidative damage to cell membranes, the collapse of the alveoli in the lungs, retinal detachment, and seizures. Additional comment: According to Wikipedia, need 160kPa for CNS toxicity. Atmospheric pressure is about 100kPa. Also says need more than 50kPa for O2 toxicity. http://en.wikipedia.org/wiki/Oxygen_toxicity. Agree with either D or F.

Answer 148

C= True hmmm...... Pulmonary stretch receptors are aka slowly adapting pulmonary stretch receptors. Lie within airway smooth muscle. They discharge in response to distension of the lung, activity is sustained with lung inflation therefore showing little adaptation. Impulses travel via Vagus nerve. Reflex is to slow respiratory frequency due to an increase in in expiratory time = the Hering-Breuer inflation reflex. Inflation of the lungs tend to inhibit further insp muscle activity. opposite is also seen = the deflation reflex. These reflexes are largely inactive in humans unless tidal vol exceeds 1 litre

Answer 149

B= True The peripheral chemoreceptors are located in the carotid bodies and the aortic bodies. There's a good outline of peripheral chemoreceptors at question RE13.

Answer 150

A and F true Option A: For each CO2 molecule which diffuses into a RBC either an HCO3 or chloride atom appears inside cell (the latter due to chloride shift when some HCO3- (out)exchanges for a Cl- (in). This results in the presence of one osmotically active particle for each CO2, which attracts H2O and causes the RBC to swell slightly. This together with a very small amount of fluid returning via lymphatics means that the haematocrit of venous blood is normally about 3% greater than arterial blood. Saturation is 75% so option B is wrong. Higher pCO2 = 46mmHg means lower pH so option C is wrong Option D is incorrect because true mixed blood incorporates venous blood from all venous beds, therefore cannot be sampled from the right atrium as blood not adequately mixed here. can only be sampled from pulmonary artery. Option E: pO2 is 40mmHg (when breathing room air) whereas coronary sinus blood has a pO2 of 20mmHg. Re option F: Just wondering if there is a reference for 'F'? Brandis says pO2 = 20mmHg - and I'm not sure where this correlates on the ODC Calculating it using O2 content = (1.39)(Hb)(SO2) + (0.003)(pO2) assuming Hb=15; SO2=97; and ignoring the dissolved O2 portion... And AV O2 difference for heart muscle being 114ml/L O2 content in venous blood is 86ml/L or 8.6ml/L which working backwards gives SO2 of 40% (ish) Thoughts? Alternatively: Oxygen extraction ratio across the coronary circulation is typically 55-65% (i.e. high compared to systemic circulation as a whole) so starting from a nominal 97% this would predict a mived venous saturation of 32-42%. Extraction ration can get as high as 70% which would result in 27%. So the 30% in option E is a tad low but not unreasonable, though 35-40% would be a more comfortable answer in this option. Of course there are 6 remembered options (instead of 5) so this question is wrong. We can assume the "real" question would be more clearcut. Another way of calculating sats from pO2 of 20mmHg for coronary sinus blood is as follows: We know CaO2 (arterial content of blood) is approx 20mls O2 / 100mls (Hb x sats x 1.34 = 15 x 1 x 1.34) We know that coronary blood flow is about 250mls / min. Therefore the coronaries are getting about 50mls O2 / min They extract about 55%. Meaning 45% is returned to the coronary sinus (ie 45% of 50 is 22.5 mls O2 in 250mls) 22.5 in 250 is equal to 9mls O2 / 100mls = CvO2 Working back from CvO2 = Hb x sats x 1.34 where we assume Hb is about 15, we get sats = 45% PS i doubt we would have to do all this on a single MCQ!! A simpler way would be that we know p50 on the ODC is about 26. So 20mmHg must be a bit less than sats 50%! Re F: Has a nice interactive ODC curve where you set pH/temp/PCO2 and then can enter Sat/PO2 to calculate the other. With CO2 46 and pO2 20--> gives Sat 33% addit: based on ODC then the real sats would be even lower than 33% since the lower pH causes a right shift

Answer 151

The 'standard' textbook values for CO2 carriage are: Bicarbonate - 90% Carbamino compounds - 5% Dissolved CO2 - 5% Also the degree of oxygenation of haemoglobin has an important affect on the carriage of CO2 (due to the Haldane effect) As written above, none of the options are correct but 85% bicarbonate is the closest and may well be the figure used in some sources. The 60,30,10 values refer to proportions for carriage of the AV difference for CO2. See also the comments for RE19

Answer 152

B. True Deoxyhaemoglobin is 3.5 times more able to form carbamino compounds (as compared to oxyhaemoglobin). This is important in CO2 transport and accounts for 70% of the magnitude of the Haldane effect. Carbonic anhydrase catalyses the reaction CO2 + H2O - H2CO3, not formation of carbamino acid. Decrease in O2 tension favours formation of carbamino-Hb through the Haldane effect. Decrease in pH (increase H+) would decrease formation of carbamino-Hb also decreased pH means the HbO2 curve is shifted to the right and therefore mixed venous PO2 is higher, therefore less reduced Hb and less H+ buffering for formation of HCO3 and less CO2 carried as carbamino A reduction in pH causes decrease formation of Carbamino-Hb due to the Left shift of the following chemical equation: Hb.NH2 + CO2 Hb.NHCOOH Hb.NHCOO- + H+ I agree with above, but perhaps here is a spanner in the works... The decrease in oxygen tension, would have to be significant enough to cause de-saturation of Hb, for the top statement to be correct. If the paO2 drops from 400mmhg, to 200mmhg for eg, then Hb saturation would be unchanged at 100% (meaning E could be the "trick" answer). (However, I hope they wouldn't be that nasty, and that I have just got too much time on my hands to think about things like this) Not sure how well this question was memorised, but it appears tricky to me. I would have thought a decrease in pH shift the ODC to right, i.e. promoting offloading of O2 from Hb. More DeoxyHb = increased ability to carry CO2 via H+ liberated when carbonic acid dissociates and via formation of carbamino-Hb. Disagree with the above comment regarding decrease in oxygen tension being a 'trick' answer because even if PaO2 reaches 760mmhg, once the Hb is saturated the rest are just dissolved O2, and PvO2 is not going to be too much higher than having a PaO2 of 100mg. But I guess it also really depends on what 'decrease in oxygen tension' really means here -- does it mean by decrease in O2 saturation with Hb?? Additional: The interaction of oxygen with Hb is at best described not as an absolute but as a reaction constant or probability if you like. Therefore the increase of PO2 does actually increase the saturation of Hb but to an increasingly small amount (likelihood).

Answer 153

B. True Deoxyhaemoglobin is 3.5 times more able to form carbamino compounds (as compared to oxyhaemoglobin). This is important in CO2 transport and accounts for 70% of the magnitude of the Haldane effect. Carbonic anhydrase catalyses the reaction CO2 + H2O - H2CO3, not formation of carbamino acid. Decrease in O2 tension favours formation of carbamino-Hb through the Haldane effect. Decrease in pH (increase H+) would decrease formation of carbamino-Hb also decreased pH means the HbO2 curve is shifted to the right and therefore mixed venous PO2 is higher, therefore less reduced Hb and less H+ buffering for formation of HCO3 and less CO2 carried as carbamino A reduction in pH causes decrease formation of Carbamino-Hb due to the Left shift of the following chemical equation: Hb.NH2 + CO2 Hb.NHCOOH Hb.NHCOO- + H+ I agree with above, but perhaps here is a spanner in the works... The decrease in oxygen tension, would have to be significant enough to cause de-saturation of Hb, for the top statement to be correct. If the paO2 drops from 400mmhg, to 200mmhg for eg, then Hb saturation would be unchanged at 100% (meaning E could be the "trick" answer). (However, I hope they wouldn't be that nasty, and that I have just got too much time on my hands to think about things like this) Not sure how well this question was memorised, but it appears tricky to me. I would have thought a decrease in pH shift the ODC to right, i.e. promoting offloading of O2 from Hb. More DeoxyHb = increased ability to carry CO2 via H+ liberated when carbonic acid dissociates and via formation of carbamino-Hb. Disagree with the above comment regarding decrease in oxygen tension being a 'trick' answer because even if PaO2 reaches 760mmhg, once the Hb is saturated the rest are just dissolved O2, and PvO2 is not going to be too much higher than having a PaO2 of 100mg. But I guess it also really depends on what 'decrease in oxygen tension' really means here -- does it mean by decrease in O2 saturation with Hb?? Additional: The interaction of oxygen with Hb is at best described not as an absolute but as a reaction constant or probability if you like. Therefore the increase of PO2 does actually increase the saturation of Hb but to an increasingly small amount (likelihood).

Answer 154

CO forms a very tight bond with Hb, therefore a large amount can be taken up by the red cell with almost no increase in partial pressure. The amount of CO which diffuses into blood in the pulmonary capillaries is NOT limited by the amount of blood available (therefore A correct).

Answer 155

A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher. B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty. C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure. E - Correct As above. Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.

Answer 156

A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher. B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty. C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure. E - Correct As above. Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.

Answer 157

A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher. B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty. C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure. E - Correct As above. Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.

Answer 158

A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher. B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty. C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure. E - Correct As above. Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.

Answer 159

Correct answer is E- no change. 0.04% is what we actually breathe. see question 45. see http://en.wikipedia.org/wiki/Carbon_dioxide I don't know, I often get tachypnoea when breathing 0.04% CO2... although I admit this is usually when I'm thinking about the exam

Answer 160

A: False. Mainly in the medium airays (up to 7th generation) B: True. Airway resistance decreases with increased lung volume due to radial traction increasing calibre of airways. C: False. Decreased resistance with stimulation of beta2 adrenergic receptors. D: False. Measured by pressure difference divided by flow rate. E. False. Decreased by breathing helium-oxygen due to decreased density.

Answer 161

According to the equation R= (8nl)/(pi* r^4), with each unit decrease in radius, there would be a corresponding 4 time increase in airway resistance It should be 16 times E = true I think its probably improperly worded, people state 'a unit' decrease will lead to 16 times inc in resistance, but from what i understand decreasing radius by a certain ratio will cause a increase in resistance by that ratio to the power 4. IE halving the radius will inc resistance 16 times. Another view: Nunn p42 states that laminar flow does not happen until the 11th airway generation. Predominantly turbulent flow occurs in the conducting airways. We know that resistance = deltaP/q and Davis and Kenny page 93 say that pressure is proportional to flow in laminar flow and pressure is proportional to the square of the flow in turbulent flow so assuming turbulent flow then the answer would be C. Comments please. - a note about the above statement, you are right about the pressure drop being proportional to the square of the flow in turbulent flow, however the question is asking about a change in airway diameter. It is my understanding, that in turbulent flow, the flow is proportional to the 5th power of the radius. (see Examiner's Report on question about Factors affecting airway resistance from 1998 on this site)[[1]]. Therefore, it would be a 32-fold change, which is not mentioned as an answer, so E sounds most correct. This question doesn't make sense as written. It only changes the resistance by 16 times if the radius is halved or doubled. For instance, changing the radius from 5 units to 4 units will change the resistance by 2.43 times. The correct answer should be "altered by power of 4 of the radius" or something like that. This is obviously an incompletely remembered question. The answer is obvious but the above discussion is lacking, hence my comment. The answer lies in the Hagen-Poiseuille equation: Q=(P1-P2) x pi x r^4 divided by 8 x n x L Ohm's Law states that V=IR i.e. P1-P2=QxR Rearranging the two formulae, R = (P1-P2)/Q = 8 x n x L divided by pi x r^4 as someone correctly stated above. ∴ If radius doubles, resistance decreases by a factor of 16 If radius is halved, resistance increases by a factor of 16.

Answer 162

According to Guyton 11th ed p493, atmospheric air is : PO2 159mmHg (20.84%) PN2 597mmHg (78.62%) PCO2 0.3mmHg (0.04%) PH2O 3.7mmHg (0.50%) Note that there is lots of room for PH2O to change with humidity, eg in a tropical summer with temperature of 37 degrees and 80% humidity. Guyton's doesn't give temperature or humidity with these figures. Nunn 6th Ed, Page 4 (Table 1.1), indicates that atmospheric air is composed of: O2 20.95% N2 78.08% CO2 0.037% Argon 0.93% The closest answer that approximates these figures is E. ``` A-Z Anaesthesia: N2 78.03 O2 20.00 CO2 0.03 Argon 0.93 Answer: E ``` I agree with E. This is from Ganong: "The composition of dry air is 20.98% O2, 0.04% CO2, 78.06% N2, and 0.92% other inert constituents such as argon and helium" I guess if you can't be bother to remember argon, just remember that O2 21% + N2 78% = 99%, knowing CO2 is 0.04%, the rest is going to be near 0.9%!

Answer 163

A. False. Increases static compliance (but decreases dynamic compliance) [Do you have a reference??!] Static compliance is increased in emphysema but unchanged in chronic bronchitis B. True C. False. Increases airway resistance D. False. Decreases FEV1 Definitions The static compliance of the lung is the change in volume for a given change in transpulmonary pressure with zero gas flow. Dynamic compliance measurements are made by monitoring the tidal volume used, while intra thoracic pressure measurements are taken during the instance of zero air flow that occur at the end inspiritory and expiratory levels with each breath.

Answer 164

his remembered version does not indicate that the MCQ is talking about ARTERIAL blood, but we will assume its arterial blood (or arterial plasma) that was in the real MCQ A is incorrect - plasma contains 0.3mLO2/100ml at PaO2 100mmHg B is correct -> see calculation below C is incorrect -> paO2 will not be high enough. D is incorrect - as it says per litre of blood (Would be 3 mlsO2/litre of blood) E is incorrect - see B (West's Pulmonary Physiology and Pathophysiology, p134) Calculations: Dissolved O2 is 0.003 ml O2/mmHg pO2/100ml blood. So for PaO2 of 100mHg -> 0.3 mlO2/100ml blood NOW for a person breathing 100% at 3 atmospheres, the ALVEOLAR pO2 can be calculated using the alveolar gas equation: PAO2 = PIO2 - (PACO2/RR) = [(760x3)-47] - (40/0.8) = 2233 - 50 = 2183 mmHg Now the arterial pO2 will be lower than this but cannot be calculated. So just to get a ball-park figure, lets use a guesstimate of 2100 mmHg Amount of O2 dissolved in ARTERIAL blood at this pO2 is: PaO2 = 0.003 x 2100 = 6.3 mlO2/100ml blood This is pretty close to the 6 value given in the MCQ. (Actually a value of 6 is correct for an arterial pO2 = 2,000mmHg). Nunn (5th ed p262): Normal arterial content O2 = 0.25-0.3 ml/100ml (NB: The exact value depends on the arterial pO2 Plasma dissolves 0.003ml O2 per mmHg pO2 per 100ml blood 100% O2 @ 1 atm = 2mL/100mL 100% O2 @ 3 atm = 6mL/100mL If there is 6 mls O2 dissolved in arterial blood (and assuming CO & regional flows are the same and arterial-venous difference in O2 is 5mL/100mL, then breathing 100% O2 at 3 atmospheres means haemoglobin in mixed venous blood will be 100% saturated as all the body's oxygen required can be supplied by dissolved O2. However breathing 3 ATM of O2 results in convulsions so this is only a very short-term situation. Of course, Hb will desaturate in areas with high O2 extraction ratios eg the coronary circulation. Here at 3 ATMs, a coronary flow of 250 mls/min & extraction of 8 mls O2/100 mls of blood (as MVO2 = 7-9 mlsO2/min), the 6 mls dissolved will not be enough.

Answer 165

A. False - Decreases in parallel with FRC (Accord. to Nunn's 6th Ed p304) B. False - closing volume is about 10% of vital capacity in young normal subjects (West 7th ed p. 163) C. False - increases with age. CC=FRC at mean age 44 supine, CC=FRC at mean age 66 erect D. I'd say false, it is more the reduction in FRC and its reaching CC under anaesthesia that causes the atelectsis, not the CC itself. Although this may be the only right answer ??? E. False - unless 44 is considered "elderly" I presumed that if it is equal to FRC supine at 44yo, then it will be at least equal to FRC supine in elderly, so E would be correct. Although not perfectly correct. Thoughts??] -- in elderly, say 66 for arguments sake, CC is the same as FRC in erect position, but CC exceeds FRC in supine, so I'll stick with False for E p118 Brandis - Closing Capacity is the lung volume at which the small airways in the lung first start to close. This closure occurs first in the dependent parts of the lung. So in the erect person, the small airways in the base are the first to close. This airway closure impairs gas exchange as any blood flow through the non-ventilated areas is shunted blood. The closing capacity is the sum of the residual volume and closing volume CC = RV + CV Firstly, quoting Kerry above does not answer the question as to which is the best option. Finally, "The role of airway closure in causing relative hypoxaemia during anaesthesia has been confirmed and accounts for approximately 35% of the increase in venous admixture."(Wahba, p. 1147). Hence answer is D Comment: I interpret the above statement as airway closure as a result of reduced FRC and the resulting shunts rather than 'increased closing capacity/volume' as such. My understanding of closing capacity has always been pathological or due to aging secondary to small airways disease. Appreciate comments.

Answer 166

A is true B - can't think of a reason for this to be true C Helium is used because very little is lost into blood as poorly soluble. D is false Just my 5c: For comment re:B is this because the ratio of O2 used to CO2 produced is not 1:1. Therefore, would expect that the calculation becomes skewed especially at high FiO2s another 2c: ?absorption atelectasis - therefore decreases your FRC, and end up with a reading that does not correlate to your original FRC.... Having said that, I like A as the answer addit: 1. not sure how FiO2 would affect the result given that the test is measuring pressure and volume changes 2. like the idea of absorption atelectasis -- but I guess it probably only matter in people with actual diseased, obstructed airway. probably not significant in younger person, yes? I support answer A 4th 2c: mcq technique -- having the ability to ignore the unusual answers when there is a most correct answer screaming at you. now, move on please

Answer 167

Absolute humidity = actual Water Vapour content When air is saturated at 37C, the absolute humidity is 44 mgs/litre = 44 g/m3. (Note that the units are equivalent because 1,000 litres in a m3 and 1,000 mg in a gram). 44mg/L is equal to 44g/m3 (not 44mg/m3). Answer is D Absolute humidity is the mass of dissolved water vapour per unit volume of total moist air. Absolute humidity of air saturated at 37°C is 44mg/L = 44g/m3 Absolute humidity of air saturated at 20°C is 17mg/L = 17g/m3 i.e. 27mg/L water vapour is added on inspiration from "room temperature" to body temperature.

Answer 168

RE14 includes the above options, and more. This question (RE51) is probably mis-remembered, imo. Comments SURFACTANT stabilises small alveoli to prevent them empty into the larger ones. SURFACE TENSION causes small alveoli to collapse into the larger ones. I still think both A and B are true, since the question is about surface tension rather than surfactant.

Answer 169

Shunt & V/Q mismatch

Answer 170

Its about 48 to 50 mmHg (making all the "usual assumptions" ie CO is constant, no changes in organ blood flow distribution, normal O2 consumption, so a-v O2 content difference stays at 5 ml O2/dL blood). "Mixed venous blood" is a mixture of blood from SVC, IVC and coronary sinus - this can only be obtained from one site in the body: the pulmonary artery. The pO2 of mixed venous sample in a normal healthy person is about 40mmHg (and 75% oxygen saturation of Hb). My boss today told me I was wrong when I spouted this fact in a tutorial. She insisted PvO2 would be 90% if breathing 100% oxygen. So, to prove the point: For FiO2 of 21%: PAO2 = 149 - 40/.08 + 2 = 101mmHg [let us assume A-a gradient of 4mmHg] PaO2 = 97mmHg [let us assume SpO2 = 98%; and Hb = 120g/L, or 12g/dL] CaO2 = 1.39 x 12 x .98 + 0.003 x 97 = 16.28 ml/100ml [let us assume 4.8ml/100ml extraction] CvO2 = 16.28 - 4.8 = 11.48 ml/100ml Now, this part is a bit dodgy mathematically, but: Because CvO2 = 1.39 x 12 x SvO2 + 0.003 x PvO2, for ease of math I'm going to assume 0.003 x PvO2 is insignificant; Therefore SvO2 = 11.48/(1.39x120) = 68.8% And if we refer to the O2 dissociation curve... PvO2 =~ 35mmHg For FiO2 = 100% PAO2 = 713 - 40/.08 + 2 = 663mmHg PaO2 = 659mmHg [let us assume SpO2 = 100% with the higher FiO2, just for the benefit of my consultant]' CaO2 = 1.39 x 12 x 1 + 0.003 x 663 = 18.67 ml/100ml [let us assume 4.8ml/100ml extraction] CvO2 = 18.67 - 4.8 = 13.87 ml/100ml Therefore SvO2 = 13.87/(1.39x120) = 83.2% And if we refer to the O2 dissociation curve... PvO2 =~ 48mmHg Presumably the difference between my results and the stated normal value for PvO2 is the rightward shift of the curve in venous blood, but this does prove the point that PvO2 will be only about 10mmHg higher in venous blood when breathing 100% oxygen. So, if you were in any doubt, there you go. Now, back to actual study...

Answer 171

B is true. WHY????? paO2 is decreased in V/Q mismatch - why? Say there are 3 alvoeli - Alveolus A has low V/Q, B has normal V/Q, C has high V/Q. The total O2 content would be the addition of O2 content in the end capillaries of A, B and C. Alveolus A has low V/Q, low pO2, high pCO2. The low pO2 lies on steeper part of the ODC, therefore O2 content in end capillary A is significantly decreased (eg 16mL/dL as in the example in West) Alveolus B has normal V/Q, O2 content is normal (19.5mL/dL) Alveolus C has high V/Q, high pO2, low pCO2. The high pO2 means that it lies on the flattened part of the ODC, therefore O2 content in end capillary C is not significantly greater than that of normal (20mL/dL in West) So when we add all 3 end capillary blood together to give us the final pulmonary venous blood, the low O2 content from the low V/Q alveoli depresses the total O2 content much greater than the high V/Q alveoli would increase total O2. Therefore the effect of V/Q mismatch on PaO2 is because of the shape of the dissociation curve. For CO2, the dissociation curve is almost linear in the physiological range, so the high V/Q and low V/Q should balance each other out and PaCO2 should not be affected. Note: If there is severe hypoxaemia from V/Q mismatch eg in a large shunt, hyperventilation may occur due to peripheral chemoreceptors. In this situation the pCO2 will then by affected. However, in this question the best answer is still B. _____________________________________________________________________________ The dissociation curve for CO2 is linear in the physiological range however the contribution from areas of low VQ ratios is going to be more because of the higher blood flow in these units. I agree that for O2 the shape of the HbO2 dissociation curve produces the greatest effect, but in terms of CO2 I would argue that hyperventilation is more important. Can anyone comment on this? The point is that the reason for the differing effects of hyperventilation on PaCO2 and PaO2 is the difference in their respective disocciation curves. The way I understood what West said was that you are able to excessively ventilate the high VQ ratio units to remove the CO2 reducing content here (not altering O2 content) by an amount that compensates for the low VQ ratio units (which are not really able to be ventilated much more than they are already. The linear disscoiation curve of CO2 means you can do this. Hyperventilation itself doesn't to me seem like a good answer as it is not the reason for the lower O2 is not able to be improved. Does that help at all? nr The other clue in the question could be the use of "hyperventilation". I would argue that if the ventilation needs to be increased to eliminate CO2 - to keep paCO2 in the normal range - then this is merely "adequate" ventilation, not hyperventilation.

Answer 172

From The Physiology Viva (Kerry Brandis 2007 ed) FRC is about 30ml/kg in adults and children. It is almost fully established in the neonate at 60mins after birth. Increased with: increased height changing from supine to erect (30% higher) less elastic recoil (ie, emphysema) Decreased with obesity pregnancy supine positioning anaesthesia (with or without paralysis) pulmonary disease causing increased elastic recoil (pulmonary fibrosis) It does not really change with age - however, the relationship between closing capacity and FRC does. Functions of the FRC Oxygen Store Arterial pO2 buffer minimizes airway resistance, PVR, V/Q mismatch, work of breathing prevents atelectasis (also part of minimizing WOB) improves positioning on compliance curve (also part of minimizing WOB) FRC measurement Body plethysmography Gas dilution (nitrogen washout, Helium wash-in) As PEEP improves FRC, you can see all the reasons above why a bit of PEEP is good here and there.

Answer 173

I think A? If hypoxaemia is related to atelectasis then, yes, increasing airway pressures between ventilation (PEEP) would correct this. (Brings FRC up). Isn't this then V/Q matching also? It is but the initiating problem is atelectasis treatable with PEEP. Hypoxaemia during anaesthesia occurs due to decreased FRC, below closing capacity, and atelectasis with consequent shunting. Hypoxic pulmonary vasoconstriction is impaired by volatile agents and shunting is more of a problem than in the awake patient with similar degree of atelectasis. Unfortunately there are not good interventions to improve HPV but PEEP (or recruitment manoeuvres) will reduce the atelectasis which is the original cause of the problem. For this reason I would choose A. Addition: On the other hand... PEEP can create more West zone 1 thus increase alveolar dead space so i thinks it B or C Add: But a shunt eg atelectasis is more likely to cause hypoxaemia than a small increase in dead space? I would go for A, would always reach for the peep knob instead of the adrenaline or fluid in a mild dropping sats in OT.. It's a funny way to say PEEP, why don't they just say 'PEEP"? (NB: They may do so - this is just the partially remembered version which is both incomplete and possibly a bit misleading - KB) A is the best answer because that's what you do in real life - increase FiO2, ensure adequate minute ventilation, then put on some PEEP. (while at the same time exploring sinister causes)Duncan 18:20, 5 Dec 2006 (EST) if the patient hypotensive , then PEEP will kill him , B looks better Applying some PEEP increases FRC or reduces atelectasis and hence improve V/Q matching, for patient in supine position, in the posterior part of the lungs. Intra-abdominal surgery/obesity/pregnancy/muscle paralysis are going to worsen V/Q matching by reduciing FRC, so I think applying PEEP is a sensible and obviously a common move to improve hypoxaemia. Anything related to increase West zone 1 and dead space is clearly less significant in contributing to hypoxaemia -- that's why a big PE doesn't cause hypoxaemia via dead space but mainly via creating significant shunt, and hence giving O2 alone produces only small improvement. As for the comment regarding hypotension -- I can only hope that we all watch at the monitor closely and not to dial the PEEP up too quickly... you may be surprised how much difference a PEEP of 2 cmH2O can make, and I don't think a PEEP of 2 would result in so much haemodynamic instability that would stop you from using it completely. And I guess if the patient was hypotensive at the first instance with hypoxaemia, restore circulation first! Answer from me is A

Answer 174

E Surface Tension- seems to the the closest, as in it has the greatest influence on Compliance, the others look wrong. You need a respiratory lab when measuring static compliance as uses oesophageal balloon. Compliance (both static & dynamic) is measured at points of "no flow" so airway resistance does NOT come into it. Yes, but dynamic compliance is "related to airway resistance during the equilibration of gases throughout the lung at end-inspiration or expiration" (Yentis p130) Add: One can measure static (and dynamic) compliance in paralysed patient on appropriate ventilators eg in ICU, but then you are measuring static compliance of whole system including abdo and chest wall.. Add: surface tension causes alveoli to collapse - surfactant increases compliance ie surface tension decreases compliance so E is wrong I think what you mean is that Surface Tension contributes significantly to the Elastic Recoil of the lung. Elastance is the reciprocal of Compliance. The Compliance of the Lung is very much related to Surfactant. The clinical correlation is a surfactant-depleted lung (such as lung disease of prematurity or even ARDS), where the lung without Surfactant becomes "stiff" and poorly Compliant. The Compliance of the normal lung is partially related to the presence of Surfactant, but also to the inherent elasticity of the lung parenchyma. E is a possible answer. [To my mind, B and C and D make little sense]. [Nunn Ch3] So what's the actual answer???? Sounds like it is resistance. The actual answer is D. Difference between static and dynamic due to airflow resistance Static Compliance Measured after lung volume has been held at a fixed volume for as long as is practicable (in conscious subjects, obviously much easier in paralyzed victims). This means sufficient time should be given for functional respiratory units to equilibrate and contribute equally in the calculation of compliance. Dynamic Compliance Sacrifices this luxury for the sake of simplicity in measurement, whereby compliance is (hastily) assessed at points of no-flow (at the mouth -- not within the lung) at end-inspiration and end-expiration. Therefore not allowing complete equilibration in the case of diseased lung where airway resistance in differing functional units will yield different time constants. HOWEVER, doesn't the answer A correct as well? I can't see how this could be done outside the respiratory lab -- moving the spirometry and oesophageal barometry devices to some other places doesn't count!! ----> we measure compliance on the anaesthetic machine every day... But I do agree that answer D is correct too. And, static compliance IS higher than dynamic compliance. Edit: I don't think D is right - a major component of the difference between static and dynamic compliance is stress relaxation of elastic bodies, which has nothing to do with AWR. The AWR contributes to part of the difference, i.e. heterogeneity of time constants, which itself is a function of both AWR and compliance. Think A has to be right, you can't do lung compliance on its own without oesophageal monometry, an anaesthetic machine is measuring the respiratory system, careful.. not the lung. Comment: I agree with the 'edit' above. 'A' is the "most correct" answer. A - Correct. TOTAL respiratory compliance can be measured on a ventilator, and many can measure both dynamic and static compliance. However, LUNG compliance requires oesophageal manometry. B - Wrong. Dynamic pressure > Static pressure for given volume, therefore Dynamic compliance

Answer 175

At 18,000ft, or 5486m, pressure is half that of sea level (Yentis) Barometic pressure decreases roughly in a exponential manner. Highest permanent human habitation is 4900-5000m in the Andes. The summit of Mt Everest is at 8848m (around 29,000 ft) where PiO2 43mmHg, and Base Camp sits at 5400m, which is half of the barometric pressure at sea level. At 20,000m (60,000ft) barometric pressure is around 47mmHg, so FIO2 is zero. When going underwater, barometric pressure increases by 1 atm roughly every 10m depth (or every 33 feet).

Answer 176

C- correct, HbS where valine replaces glutamic acid on beta-chain causes critical loss of solubility if reduced Hb leading to polymerisation and "sickle" shape of Hb at PaO2 less than 40mmHg. (Nunn resp phys) Methaemoglobin Haemoglobin in which the iron has been oxidised (trivalent ferric form). metHb is unable to combine with oxygen but is slowly reconverted to haemoglobin (can be treated with ascorbic acid or methylene blue). May arise from OxyHb scavenges nitric oxide (physiological process to limit endogenous NO activity) Drugs (prilocaine, benzocaine, nitrites, dapsone) metHb ---> Hb via four mechanisms NADH-metHb reductase enzymes in erythrocytes (most important mechanism) Ascorbic acid via direct chemical reaction (accounts for 16%) Glutathione-based reductive enzymes NADPH-dehydrogenase in erythrocytes (considered 'reserve' metHb reductase)

Answer 177

Venous admixture is the adition of mixed venous blood to end alveolar capillary blood. It is the entrance of blood to the arterial circulation which has not been exposed to ventilated alveoli (ie shunted blood). It accounts for the PaO2 being less than the PAO2 in healthy individuals at rest, and is due to blood from the bronchial veins and thebesian vein returning to the left side of the heart without undergoing gas exchange in the lungs. Venous admixture can be calculated by the shunt equation as detailed in West; derived from the fact that total oxygen must equal the sum of end capillary oxygen and shunt oxygen (providing the shunted blood is mixed venous blood). The shunted blood reduces the PaO2, even if the shunt is small. A small drop in oxygen content leads to a large drop in PaO2 due to the almost flat ODC. Increasing Fi02 to correct hypoxaemia cannot compensate for a shunt of more than 30% of cardiac output. 100% oxygen has a less than expected impact on PaO2. A shunt usually has no impact on PaCO2, as chemoreceptors detect an inrcease in CO2 content and respond by increasing ventilation. Shunt fraction --> arterial content on top, venous content on bottom: Qs/Qt = CcO2-CaO2/CcO2-CvO2.

Answer 178

Is typically 200ml/cmH2O (1mmHg=1.36cmH2O).Is measured in vivo with a conscious,relaxed,upright subject,at different degrees of deflation from near TLC when no air is flowing. Factors affecting static compliance: lung vol (normally measured at FRC. If only 1 lung dV is half that for a given dP) FRC variation due to body size (specific compliance 0.05cmH2O relates changes in dV & dP to FRC (dV/dP)/FRC ) pulmonary blood vol (comp v with congestion) alveolar collapse lung disease Quite rhetorical, because if there were no surface tension, then surfactant would do nothing. Physiologically speaking, the determinants of lung compliance are its intrinsic elasticity, and the surface tension (about 50% each).

Answer 179

oth O2 and CO2 - increased solubility with hypothermia Since partial pressure is measuring dissolve gases (Brandis p135), then both PaCO2 and PaO2 should increase. I think you might have this backwards. The solubilty increases, which means the amount dissoved increases. If the system is closed then this means that the partial pressure must drop. Henry's law: gas content dissolved in soln is directly proportional to the sol coefficent x partial pressure in the gas phase.The solubility coefficent increases as temp deceases. pp = conc dissolved gas / sol coefficent shows the inverse relationship btw partial pressure and solubility coefficent when temp changing. Therefore if temp v, sol coeff ^,and pp v. In the case of a closed system the partial pressures falls with v temp as the solubilty increases.Note that this law applies only when there is no reaction with the solvent.(Mnemonic Henry Lawson was bit of a piss pot and the amount of the intoxicant he could dissolve was directly proportional to the pressures on him and his attempts to maintain a constant of proportion) ref: faunce http://gasboys.net It depends on how the question is asked - if the patient is hypothermic then the A-v difference in O2 would drop as well right? So it's important whether that's asking A (as in alveoli) or a (as in arterial) Comment Now let me get this right: Don't matter what the temperature is, atmospheric pressure should be 760mmHg (PV = nRT only applies to a set volume....) Now does that mean that inspired, saturated alveolar pO2 will still be (760-47) x 0.21 - 40/0.8 = 100? --> nope, the point here is, 5% of CO2 carried in blood is dissolved normally, but now more can be dissolved for LESS overall CO2!! --> there is less VO2 hence less VCO2 when you are hypothermic (Brain VO2 definitely down 7% per defgree, but ?14% decrease in BMR for each decrease temperature! p.281 Ganong - says about inc temp, not other way) So PCO2 has to decrease. So if CO2 decreases, that means O2 has to increase because pCO2 is which means if PAO2 increases, PaO2 should increase too... (now this assumes that R is constant) So my conclusion? decreased temperature --> (a) increased solubility (b) decreased CO2 production --> decreased arterial PCO2 partial pressure --> increased alveolar O2 partial pressure --> increased arterial O2 partial pressure. (Correct me if I'm wrong please!!!!!!) Now, having said all that above.... I believe the above conclusion drawn does not answer this MCQ. My conclusion here basically means that there will be MORE OXYGEN CONTENT taken up into the body, i.e. increased arterial oxygen partial pressure due to increased availability of alveolar oxygen partial pressure (not to mention left shift of the ODC). However, if the question asks about the PaO2 for a GIVEN OXYGEN CONTENT when temperature decreases, then partial pressure of oxygen will DECREASE, as quoted from Kerry's site (see below). adding to the confusiong -- hypothermia itself reduces VO2 as well, it sort of complicates the whole scenario of how the question was asked then because that could affect the PaO2 and PvO2. But I guess that's just me being silly because when we put ppl under anaesthesia they are having surgery, that means surgical stress is applied which results in increased VO2... the conclusion is, this is a s*** question. Or a decent question being remembered badly. Quote from http://www.anaesthesiamcq.com/EduResources/Gases/Gas005.php "pO2 decreases with a decrease in temperature for reasons as for pCO2. There is an additional factor here as the haemoglobin oxygen dissociation curve shifts to the left with a decrease in temperature - this means Hb has a higher oxygen affinity at a lower temperature. These 2 factors (increased solubility & increased Hb affinity) make for a more complex situation then with CO2. If pO2 is high the Hb is already fully saturated and the change in affinity won't have much effect. At lower pO2 levels, the increased affinity means some of the dissolved oxygen will bind to Hb causing a decrease in pO2. This is additional to the decrease in pO2 that occurs with the increased solubility of O2 at the lower temperature. Total oxygen content does not change."

Answer 180

Anatomical dead space is the volume of the conducting airways which is around 150ml or 2.2ml/kg for the average adult. It is influenced by size (17mls for every 10cm) age posture (sitting 157ml supine 101ml) the posistion of the head and neck,lung volume and intubation (increases apperatus dead space decreases anatomic). Fowler's method is used to calculate anatomic dead space.(single insp of 100% O2 followed by measurement of expired N2. Increasing conc of expired N2 is plotted against expired volume). The Bohr equation (nmemonic 'DAETA')is used to calculate physiologic dead space. Anatomic dead sapce is mostly atmospheric air with a little CO2

Answer 181

Fowlers method is used to calculate anatomic dead space (single insp of 100% O2 followed by measurement of expired N2.The increasing conc of expired N2 is plotted against expired volume). The Bohr equation (mnemonic 'DAETA')is used to calculate physiologic dead space. A- false, the Bohr equation is used to measure physiological dead space B- false,nitrogen washout, used in the Fowler method is used to measure anatomical dead space C- Calculation of physiological/total dead Space(Vt) requires end-tidal pCO2 and alveolar pCO2. {EDIT} Its not caculated from Endi-tidal, its Mixed Expired C02; they are all wrong. By utilising and modifying Bohr equation, anatomical dead space can be calculated-by replacing alveolar with end-expiratory gas in the Bohr equation. Hence, Bohr equation can theoretically be used to calculate anatomical dead space. Ref: Nunn 6th ed., p121. Physiological dead space = anatomical dead space + alveolar dead space One can use Bohr's equation and substitute mixed expired Pco2 with end-tidal Pco2 and that gives you the alveolar dead space. By calculating the physiological dead space using mixed expired Pco2 as well, using the above formula anatomical dead space is derived. By this mean you ARE using Bohr's equation to measure anatomical dead space. Unless the answers was remembered wrongly, e.g. D is Alveolar dead space can be calculated from end-tidal CO2 and alveolar CO2, or Physiological dead space can be calculated from mixed expired CO2 and alveolar CO2

Answer 182

On a plot of respiratory rate vs work of breathing, as respiratory rate increases work to overcome elastic forces decreases and work to overcome airway resistance increases. The total work is minimal when both elastic work and airway resistance work contribute 50%. The total work vs respiratory rate curve is U-shaped and is minimal at rest respiratory rates. However this minimum work point occurs at higher rates in pathologies that increase elastic work (such as pulmonary fibrosis) and at lower rates in pathologies that increase airway resistance work (such as COPD). work due to airway resistance depends on the respiratory rate. change in work due to increasing respiratory rate depends on what the respiratory rate was initially. E is definately true and is the best answer Components that make up the work of breathing during quiet inspiration: Nonelastic work (Viscous resistance (7%)Airway resistance (28%)) Elastic work (65%) The higher the breathing rate the faster the flow velocity (which by using Reynold's number - is more likely to cause turbulent flow, and require increasing pressure changes for the same volume) and the larger the work done in overcoming non elastic work. The work done in overcoming the elastic forces is decreased as respiratory rate increases. Therefore, people with highly compliant lungs and high airways resistance - breath at higher volumes and mroe slowly (ie, emphysema) people with less compliant lungs breath at faster rates, and more shallow volumes (ie, pulmonary fibrosis) _______________________________________________________________________________________________________________ I do not believe E to be the correct answer. Assuming the question is correctly remembered, C is the best option. The integral of the pressure volume LOOP only encompasses the area within the loop itself ie work done to overcome non elastic forces of inhalation (active) and non elastic forces of exhalation (supplied by potential energy stored by inspiratory work done against compliance). The true amount of total work encompasses the integral of the entire loop, as well as the area between the y axis and the loop itself. Also, I disagree with another line above, stating that "The work done in overcoming the elastic forces is decreased as respiratory rate increases." Elastic forces are not related to resp rate, but to compliance ie surface tension and intrinsic lung or chest wall elasticity. Therefore, work done against elastic forces are increased with increasing tidal volumes, and work done against non-elastic forces increases with increasing respiratory rate. I agree with pretty much everything else above however, especially this bit: Therefore, people with highly compliant lungs and high airways resistance - breath at higher volumes and mroe slowly (ie, emphysema) people with less compliant lungs breath at faster rates, and more shallow volumes (ie, pulmonary fibrosis)

Answer 183

A-a gradient = PAO2 - PaO2 where: PaO2 is partial pressure of O2 in the artery -obtained from the arterial blood gases. PAO2 is partial pressure of O2 in the alveoli - obtained from the Alveolar Gas equation: PAO2 = ( FiO2 * (PB - pH2O)) - (PaCO2 / R) Alveolar gas equation where: PB is atmospheric pressure (eg 760mmHg at sea level) pH2O is SVP of water at body temp (eg 47mmHg at 37C) R is the respiratory exchange ratio (In steady state or over any period of time, this equals RQ; a typical value on a typical diet is 0.8) Thus (if breathing room air at sea level and if arterial pCO2 = 40): PAO2 = (FiO2 * (PB - PH2O)) - (PaCO2 / R) = (0.21 * (760-47)) - (40/0.8) = 150 - 50 = 100 mmHg The Aa gradient provides an assessment of gas exhange between alveolar and capillaries, and is used as an index of V/Q mismatch in the lung. Aa gradients can be increased due to an increase in FiO2, respiratory disease/venous admixture. In room air the normal Aa gradient is

Answer 184

As: Pulmonary capillary blood volume = 80 mls Total blood volume = 5,000mls So: % = 80 x 100 / 5000 = 1.6% Did it really say pulmonary capillaries? Or pulmonary circulation? - Because the pulmonary circulation has the capacity to hold 400-500mL blood as a reservoir. Which would make it 8-10%. Additional comment: P+K (p121 2nd ed) states that 6% of total blood volume is in the pulmonary capillaries. -- Actually, on page 120 rather than 121, its says 6% is in the systemic capillary exchange vessels, and only 3% is in the pulmonary capillaries

Answer 185

Answer is B Blood draining from an unventilated part of the lung will have no opportunity to gain O2 and should not lose any either. Therefore it should have a composition identical to the blood supplying that part of the lung, i.e. the pulmonary artery. Consideration of the other options reveals a. coronary sinus - Incorrect. The primary venous drainage of the myocardium, has a high O2 extraction and pO2 of ~20mmHg, far lower than mixed venous O2 b. pulmonary artery - correct answer - see above c. bronchial artery - Incorrect. These arise from origins in the aorta and therefore have a normal systemic PaO2. (Typically ~100mmHg, allowing for a PAO2 of 105 - physiological Aa gradient.) d. Alveolar Gas - Incorrect. PAO2 is described by the alveolar gas equation as (760-47 * 0.21) - pCO2 / respiratory quotient = ~150 - 45 = 105mmHg. Clearly impossibly high as higher than arterial O2

Answer 186

A: Could be, I think most likely , from Miller (e-version no page sorry - chapter on PACU), "Atelectasis and alveolar hypoventilation are the most common causes of transient postoperative arterial hypoxemia in the immediate postoperative period." So in other words, could be A or C B - no C - could be right see above, not sure we are supposed to definitively answer this, anyone?

Answer 187

A. Intubation - decrease (anatomical dead space as distinct from apparatus dead space) B. Chin tuck position - decrease C. Moving from supine to erect - increase - the correct answer D. Moving from sitting to semi-recumbent - decrease E. Bronchospasm - decrease Supine 101mls, sitting 147mls, semi recumbent 124mls NUNN Nunn, factors affecting anatomical dead space: size of subject age (decrease birth to 6yr; increases early adulthood onwards) posture (supine = 2/3 sitting) position neck and jaw (increases from neck flexion to extension) lung volume end inspiration (increases with increasing volume) intubation/LMA (apparatus dead space bypassing 1/2 anatomical dead space) drugs (bronchodilators increase) Vtidal + RR (reducing Vt reduces dead space as measured by Fowler's method; due to air flow changes) Answer: C

Answer 188

A. Intubation - decrease (anatomical dead space as distinct from apparatus dead space) B. Chin tuck position - decrease C. Moving from supine to erect - increase - the correct answer D. Moving from sitting to semi-recumbent - decrease E. Bronchospasm - decrease Supine 101mls, sitting 147mls, semi recumbent 124mls NUNN Nunn, factors affecting anatomical dead space: size of subject age (decrease birth to 6yr; increases early adulthood onwards) posture (supine = 2/3 sitting) position neck and jaw (increases from neck flexion to extension) lung volume end inspiration (increases with increasing volume) intubation/LMA (apparatus dead space bypassing 1/2 anatomical dead space) drugs (bronchodilators increase) Vtidal + RR (reducing Vt reduces dead space as measured by Fowler's method; due to air flow changes) Answer: C

Answer 189

Assuming: 70kg Fit young adult = 3 litres/min 70year = 2 litres/min Sedentary = reduced by 50% Middle age = ?? half way between 2 and 3 = 2.5 and sedentary = 1.25L/minute = 17ml/kg/minute is there such thing as sedentary VO2max? I think answer should be 40ml/kg/min. I also agree with C. The "most correct" answer given above would give a V02 MAX of 770ml/min - a MAXIMUM just over three times basal rate. I would be surprised if he could do up his shoelaces at this rate! Nunn clearly states that VO2 MAX is the oxygen consumption of a subject when exercising as hard as possible for that subject which means that this sedentary persons VO2 MAX may be reduced compared to an active person but that is still the MAX for him (p335 5th Ed). So if we use Nunn's definition VO2 MAX for this pork chop = 1.5L/min = 20ml.kg.min - which leaves me with no answer... dunno I took this is meaning the VO2-max of an untrained 40 yr old as opposed to a marathron-running 40 year old. Both would have the same VO2-couch, but very different VO2-max. 40mL/kg/min was then an easy answer to circle. I suggest, based upon this study, the ball park answer is not 11 but 40 ml/kg/min. Aerobic work capacity in sedentary men and active athletes in Israel Agreed that this question is likely incorrectly remembered. There clearly is no such thing as sedentary VO2 max. The options would likely be: what is VO2 max in a 40yr old male? what is VO2 in a sedentary 40yr old male? what is VO2 max in an untrained 40 yr old male: option 1: VO2max in a 40yo male (with some degree of fitness) is likely to be approximately 40mls/kg/min option 2: VO2 in a sedentary 40yo male (Basal metabolic rate ) 200-250mls /min or 2/8-3/8mls/kg/min for a 70kg individual option 3: VO2 max in an untrained 40yo male ? between 28-34mls/kg/ min (number from the Israeli study above) Disagree that this is incorrectly recalled. It is not the VO2max that is sedentary, it is the male subject! Measuring VO2max will excercise the slovenly, sedentary subject as hard as his jelly legs will take him. (As a sedentary 39 year-old, I'm allowed to write that.) Nunn, "Maximal oxygen uptake refers to the oxygen consumption of a subject when exercising as hard as possible for that subject." Per the assumptions made earlier, his VO2max can be no better than a fit young adult (~42 mL/kg/min) and no worse than a sedentary 70 year-old (~17 mL/kg/min). My guess is that the "most correct" answer is 'C'. I think the above discussions are making it a bit more complicated than it needs to be. Nunn says, a fit young person should be able to attain a V02max of 3L/min. Athletes should be able to get up to 5L/min. This "sedentary" 40 year old is thus not an athlete and would have the usual V02 max of 3L/min. Assuming he's 70kg, that would mean 3000L/min = 42ml/kg/min therefore C is the correct answer.

Answer 190

[Comments for A-D in the 2nd version above A. - Incorrect => "Not to be confused" B - Incorrect - This is metabolic rate - R can be measured at any instant in time and does not require equilibrium to have been reached C - Incorrect - Increases during severe exercise as CO2 increases - can rise to 2 D - Increases - Decreases whilst repaying oxygen debt - can fall to 0.5 Hence E must have been something correct or others remembered differently

Answer 191

[Comments for A-D in the 2nd version above A. - Incorrect => "Not to be confused" B - Incorrect - This is metabolic rate - R can be measured at any instant in time and does not require equilibrium to have been reached C - Incorrect - Increases during severe exercise as CO2 increases - can rise to 2 D - Increases - Decreases whilst repaying oxygen debt - can fall to 0.5 Hence E must have been something correct or others remembered differently

Answer 192

a) correct p109 west, but isn't tracheal flow sinusoidal. Peak flow is 0.5L/s. and technically alveolar pressures are between those values (-1 to +1). intrapleural pressure curve is not sinusoidal.

Answer 193

A. Correct - intrapleural pressure varies between -5 cmH2O to -8 cmH2O per (West 9th Ed, Fig 7-13 p 112) B. Partly correct - Alveolar pressure during tidal breathing varies from +1 cmH2O to -1 cmH2O, which is within the range but a significantly different number. Making this clear is probably why the question was changed from +/-2 cmH2O to +/-5 cmH2O. C. Wrong - Vtidal ~ 7 mL/kg which is about 525 mL in a healthy 75 kg person D. Wrong - RR 12/min and I:E 4 is pathological "Most correct" answer is 'A'.

Answer 194

A - no, is effort independent - "It has the theoretical advantage of avoiding measurement during the more effort-dependent first quarter of the FVC." Gibson, Clinical Tests of Respiratory Function, ED 2, pp46 B no - is the middle half of expiration of the FVC C - D - ? not sure.. E - should be decreased in airway obstruction

Answer 195

Alternative: A. Wrong - specifically designed to exclude effort dependent expiration B. Wrong - measured in middle half C. Partly correct - FEV1 and FEF both vary with obstruction/ restrictive disease D. Wrong - time is not a consideration in a flow-volume relationship E. Wrong - COPD will typically have some early dynamic airway closure, reducing FEF (which is also why they have a longer expiratory phase). Most correct answer: 'C'

Answer 196

Don't understand this question - sampling from a pulmonary artery catheter is how you obtain mixed venous blood values isn't it? So how could anything in it be different from mixed venous blood? So A and C would both be correct. My reading is that this is about sampling blood distal to the wedge. After aspirating an intermediate 5 mL sample, blood aspirated distal to the wedge should be similar to an ABG. This technique is used to confirm correct placement, hence validate PCWP measurements. Hence "most correct" answer: 'D'

Answer 197

C may be the most correct - http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3919066/

Answer 198

Residual volume - and any capacity which includes RV eg FRC, TLC

Answer 199

Bicarbonate accounts for 90% of CO2 carriage in arterial blood, and for 60% of the additional 4mls/dl picked up after passage through the systemic capillaries. Bicarbonate is formed inside the red cell (due presence of carbonic anhydrase) but then exchanges for Cl- across the cell membrane to enter the plasma.

Answer 200

Yentis states that hypercapnia is associated with increased sympathetic activity, thus increasing circulating catecholamine levels. These catecholamines offset the direct myocardial depressant effect of CO2 itself.

Answer 201

A seems to be correct.

Answer 202

ANSWER A and C A: CORRECT: Resistance = (8Lη)/(πr4), therefore decreased viscosity -> decreased resistance B: INCORRECT: decreases with increasing lung volume C: CORRECT: Resistance = pressure/flow D: ? E: INCORRECT: β2 Dr mitta 06:34, 3 December 2015 (CST) C may be more correct than A as the Hagen–Poiseuille equation is based on laminar flow. Most of the flow in lower airways is transitional.

Answer 203

``` A - T (See below quote from Ganong) B - T C - F D - T E - F ``` Ganong (21 ed) p568 - "RA systole precedes LA systole and contraction of the RV starts after that of the LV. However, since pulmonary arterial pressure is lower than aortic pressure, RV ejection begins before LV ejection. During expiration, the pulmonary and aortic valves close at the same time; but during inspiration, the aortic valve closes slightly before the pulmonary." Physiological systole lasts from the start of isovolumic contraction to the peak of the ejection phase, so that physiological diastole commences as the LV pressure starts to fall (Braunwald Heart disease 6th 463-4) Further comment: Ganong 21ed pg 567. Figure 29-3. Defines atrial systole from the mid point of the p-wave to the closure of the atrioventricular valves. Ventricular systole is from the closure of the atrioventricular valve to the closure of the semilunar valves. Diastole is from the closure of the semilunar valvue until the onset of atrial systole. When conflicts arise between texts I think the examiners would give more priority to texts such as Ganong so I would use this definition. And further... S2 splitting attributed to "lower impedance" of pulmonary circuit by Ganong (p.566, 22nd Ed) - whatever that means. Other texts suggest that it is due to increased right ventricular return and therefore delay in right pulmonary valve closing when inspiring as decreased intrathoracic pressure. splitting can be attributed to the fact that the semilunar valves only close once the pressure in the respective ventricles fall below the pressures in the circuits they are perfusing. logically since the pressure in the pumonary circuit is lower than the systemic circulation, it would close at a lower pressure, causing splitting. - aortic valve closes earlier due to the higher impedance or resistance of the systemic circulation compared to the pulmonary circulation and its effects on the pulmonary valve. Valve closure requires a reversal of pressure gradient, and amount of back flow I suppose, and this pressure depends on the resistance/impedance of the circulations. In expiration, there is slightly higher intrathoracic pressure which probably adds to the pulmonary circulation's impedance - meaning PV closes earlier (same time as AV). In summary: (pg 566 Ganong 22nd Edn Chp 29) RA contract before LA RV contract after LV but RV ejection before LV ejection In inspiration Aortic valve closes before pulmonary valve In expiration Aortic valve and Pulmonary valve close at the same time

Answer 204

``` Alt Ver 1 A - T (See below quote from Ganong) B - F C - F D - F E - F ``` In summary: (pg 566 Ganong 22nd Edn Chp 29) RA contract before LA RV contract after LV but RV ejection before LV ejection In inspiration Aortic valve closes before pulmonary valve In expiration Aortic valve and Pulmonary valve close at the same time

Answer 205

``` Alt Ver 2 A - T (See below quote from Ganong) B - F C - F D - F (coincides with systole but not peak flow) E - T ``` In summary: (pg 566 Ganong 22nd Edn Chp 29) RA contract before LA RV contract after LV but RV ejection before LV ejection In inspiration Aortic valve closes before pulmonary valve In expiration Aortic valve and Pulmonary valve close at the same time

Answer 206

CV01B C - T In summary: (pg 566 Ganong 22nd Edn Chp 29) RA contract before LA RV contract after LV but RV ejection before LV ejection In inspiration Aortic valve closes before pulmonary valve In expiration Aortic valve and Pulmonary valve close at the same time

Answer 207

C From Ganong 21st Ed (p 571) "The c wave is the transmitted manifestation of the rise in atrial pressure produced by the bulging of the tricuspid valve into the atria during isovolumetric ventricular contraction." A wave coincides with atrial contraction C wave coincides with isovolumetric contraction - bulging of the tricuspid valve. V wave mirrors the rise in atrial pressure before the tricuspid valve opens in diastole. It reflects peripheral venous return filling the atrium. X decent occurs after the C wave because contraction of the right ventricle pulls the fibrous A-V ring that results increased compliance of the atrium (the atrium does not enlarge as written in text but capacity to expand for a given pressure improves). Y decent after the tricuspid valve opens and reflects the decreasing compliance of the right ventricle allowing blood to spill over into this cavity from the right atrium. COMMENT in Berne + Levy's Wigger's diagram, the c wave of the JVP corresponds with LV ejection, due to the time taken for the wave to travel there I presume I think the C wave differs depending on whether you talk about atrial pressure waves or JVP waves atrial pressure wave - caused by the bulging of the atroventricular valve into the atrium during isovolumetric contraction JVP - coincides with rapid ejection phase due to time taken. Caused by above, but also (According to Levy & Pappano) by the pulsation in the carotid

Answer 208

B From Ganong 21st Ed (p 571) "The c wave is the transmitted manifestation of the rise in atrial pressure produced by the bulging of the tricuspid valve into the atria during isovolumetric ventricular contraction." A wave coincides with atrial contraction C wave coincides with isovolumetric contraction - bulging of the tricuspid valve. V wave mirrors the rise in atrial pressure before the tricuspid valve opens in diastole. It reflects peripheral venous return filling the atrium. X decent occurs after the C wave because contraction of the right ventricle pulls the fibrous A-V ring that results increased compliance of the atrium (the atrium does not enlarge as written in text but capacity to expand for a given pressure improves). Y decent after the tricuspid valve opens and reflects the decreasing compliance of the right ventricle allowing blood to spill over into this cavity from the right atrium. COMMENT in Berne + Levy's Wigger's diagram, the c wave of the JVP corresponds with LV ejection, due to the time taken for the wave to travel there I presume I think the C wave differs depending on whether you talk about atrial pressure waves or JVP waves atrial pressure wave - caused by the bulging of the atroventricular valve into the atrium during isovolumetric contraction JVP - coincides with rapid ejection phase due to time taken. Caused by above, but also (According to Levy & Pappano) by the pulsation in the carotid

Answer 209

B According to Ganong (p 568 21st Ed)(p 565 22nd Ed): the end-diastolic volume of each ventricle is about 130ml and end-systolic volume of each is about 50ml (SV 70-90ml at rest, EF about 65%).

Answer 210

E According to Ganong (p 568 21st Ed)(p 565 22nd Ed): the end-diastolic volume of each ventricle is about 130ml and end-systolic volume of each is about 50ml (SV 70-90ml at rest, EF about 65%).

Answer 211

In moderate exercise LV end systolic volume stays the same (Berne and Levy p274) Power and Kam seems to disagree. I'd probably go with Power and Kam as they sit on the examination board. There is a non-linear increase in SV during excercise. This increase in SV occurs mainly in light to moderate excerice, with changes at maximal excercise contributing little. Most of the increase in SV is from an increase in end-diastolic volume (from increased VR) and from a decrease in end-systolic volume (caused by increased emptying due to increased sympathetic activity). Power and Kam pg 163 Answer will be B or C depending on which text you believe! Question doesn't state whether isotonic or isometric exercise... New version of B&L: "Cardiovascular Physiology", M. N. Levy & A. J. Pappano, 9th ed, Mosby, 2007. page 243 states SV increases 10-35% in mild to moderate exercise - so if 70ml is the upper limit of normal stroke volume (50-70 ml) I think 30ml is probably the best answer. "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 633 states with isometric exercise little change in SV, but with isotonic "marked increase in SV". Take your pick! Gray Guyton p103 claims the LVESV can fall as low as 10-20mls with increased contraction - doesn't classify the degree of exertion. Also states the LVEDV can increase to 150-180mls in the normal heart and the stroke volume output can double! B&L states the SV increases by 10-35%. That give a SV ranges of 45-63mls if normal SV is 70 mls or 32.5-45 if normal SV is 50mls. Doesn't comment on range of increase in LVEDV hence difficult to subtract down to LVESV. I think I'd pick B (30 mls) as despite increase in LVEDP associated with moderate exertion I would anticipate an increased EF thus a less than normal LVESV. I think in moderate exercise you would have a 10-30 % increase in SV, most of the increase in CO comes from increasing HR (like in L & P where they talk about exercising dogs - denervated hearts have increased SV but normally innervated hearts have more of an increase in HR). I'd pick B too. I disagree: There are some misleading statements here. LVESV = LVEDV - SV. LVEDV and SV both increase in moderate exercise, whereas LVESV only decreases mildly if at all - not a massive drop to 30ml. That might occur but only in strenuous exercise.

Answer 212

The fall in venous return to the right heart, and the decrease in hydrostatic pressure relative to the heart (i.e above the heart due to gravity), causes an immediate fall in arterial pressure at the carotid and arch barroreceptors. This decreases their level of firing. This is projected to the NTS which decreases the firing of the neurons projecting from the NTS to the RVLM, (the main efferent vasomotor/cardiac centre). The normal signal from the NTS to the RVLM is inhibitory, so decreasing it's firing results in decreased inhibition of the RVLM. Thus there is greater SNS efferent signal to the heart(increasing heart rate and contractility and thus cardiac output),and the resistance vessels,(increasing vasomotor tone and thus PVR). To a lesser degree there is also an increase in venomotor tone increasing venous return. Overall the increase in VR adds to the rise in CO, and this coupled with the rise in PVR, tends to restore the BP measured at the barroreceptors. The primary physiological aim is to maintain CPP. The increase in SNS output/vasomotor tone will decrease skin blood flow and this effect occurs very quickly, (effectively immediately) as the barroreceptor are responsible for response to rapid transient changes.

Answer 213

Answer: C The pulmonary artery is only site for sampling for measurement of mixed venous blood. Right atrial blood is not mixed sufficiently for consistent and reliable results. Mixed venous oxygen saturation is a direct measurement of the blended blood in the right ventricle, a mix of blood from the inferior vena cava, the superior vena cava and the coronary circulation. Mixed venous oxygen saturation is the percentage of reduced hemoglobin left after tissue oxygen extraction. The normal value for SvO2 is 60 - 80%. The measurement can be made from a blood sample drawn from the ventricle using the pulmonary artery port of a Swan-Ganz catheter. Reference: MIXED VENOUS OXYGEN SATURATION by David Kissin, BS, RRT Berne & Levy - CVS Physiology book Same answer in The Physiology Viva Q&A by KB p90 Mixed venous blood can only be obtained from the pulmonary artery (Providing that no shunts have been contaminated with blood). This is the only site in which adequate mixing of the 3 venous streams can be said to have occurred (venous blood from SVC, IVC and coronary sinus).

Answer 214

Answer: A "When ICP is elevated to more than 33mmHg over a short period, cerebral blood flow is significantly reduced. The resultant ischaemia stimulates the vasomotor area and systemic blood pressure rises. Stimulation of vagal outflow produces bradycardia, and respiration is slowed. The BP rise, which was described by Cushing and is sometimes called the Cushing reflex, helps to maintain the cerebral blood flow." - Ganong 20th ed p595 "When intracranial pressure is increased, the blood supply to the vaso motor area is compromised, and the local hypoxia and hypercapnia increase its discharge... ...and over a considerable range the (resultant) blood pressure rise is proportional to the increase in ICP. The rise in BP decreases heart rate via the arterial baroreceptor reflex". (Respiratory rate not mentioned). "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 609. ``` Mneumonic for causes of raised ICP: Trauma Hydrocephalus Infection Neoplasia Bleeding Benign Intracranial Hypertension (THIN BB) Template:Www.anaesthesiauk.com Initially described in patients with severely raised intracranial pressure, the phenomenon consisted of bradycardia, hypertension and cessation of respiration. Cushing H: Concerning a definite regulatory mechanism of the vasomotor centre which controls blood pressure during cerebral compression. Bull Johns Hopkins Hosp 12:290-292, 1901 Mohran CV Physiology describe bradycardia, hypertension and respiratory abnormality with increased ICP but not specific as to the respiratory effects ```

Answer 215

A - Correct (Ganong 22nd ed, p575) B - Correct (Power and Kam, p110, Ganong (22nd ed) p566) C - Incorrect D - Incorrect (Shortens - Levy & Pappano p31) E - ? (The diastolic filling TIME certainly decreases, and at high heart rates filling can be compromised, however tachycardia may be associated with increase diastolic filling in exercise) F - Incorrect - coronary perfusion increases (Brandis Revised ed, pg 71 - shorter diastole means less time for coronary perfusion however with a normal coronary circulation autoregulation of coronary blood flow will fully compenstate for this and will meet the increased myocardial oxygen demand due to the higher heart rate - ie. coronary perfusion must increase due to the increased myocardial oxygen demand caused by higher HR) Comment: do they really mean "vascular filling" or could it mean "ventricular filling"?? Ganong 22e says that up to 180bpm filling is adequate providing venous return is normal. So C could also be correct. Thoughts?

Answer 216

Best answer - B A. The blood flow of resting muscle is known to double after sympathectomy, suggesting that some decrease in tonic vasoconstrictor discharge may contribute to any increase in blood flow. Exercising muscle however, can increase its blood flow as much as x30, hence sympathetic vasodilation is unlikely to be the major cause of increase in blood flow. B. Local mechanisms maintaining a high blood flow in exercising muscle include: Fall in tissue pO2 Rise in tissue pCO2 Accumulation of K+ and other vasodilator metabolites (?role of lactate) Rise in temperature This is the correct answer. C. When a muscle contracts or "pumps", blood flow actually ceases during the contraction. It is only between contractions that flow is increased and this is maintained by local mechanisms (refer B above). Although metabolic vasodilation is the MAJOR source of increased blood flow in the excercins muscle, SYMPATHETIC VASODILATION may contribute to blood flow in the excercising muscle. "In addition to their vasoconstrictor innervation, resistance vessels in skeletal muscles are innervated by vasodilator fibres, which although they travel with the sympathetic nerves, are cholinergic (Sympathetic Cholinergic Vasodilator System). There is no tonic activity in the vasodilator fibres, but the vasoconstrictor fibres to most vascular beds have some tonic activity. When the sympathetic nerves are cut (sympathectomy), the blood vessels dilate. In most tissues, vasodilation is produced by decreasing the rate of tonic discharge in the vasoconstrictor nerves, although in SKELETAL MUSCLE it can be produced by activating the SYMPATHETIC CHOLINERGIC VASODILATOR SYSTEM.

Answer 217

CV10b Skin blood flow is regulated mainly by the sympathetic nervous system (Berne and Levy Cardiovascular Physiology p241-2) In lung hypoxia has the most important influence in pulmonary vasomotor tone (Berne and Levy Cardiovasc Phys p254) In skeletal muscle local factors predominate in exercise, whilst at rest neural control predominates (Berne and Levy Cardiovasc Phys p247) Renal blood flow is autoregulated via the myogenic mechanism in glomerular arterioles and tubuloglomerular feedbeck (Brandis p74) In the liver the portal venous system (providing 60-80% or total hepatic blood flow) does not autoregulate (Berne and Levy Cardiovasc Phys p264) Assuming the question refers to exercising skeletal muscle then the answer is C Could the other way of thinking about it be - which of each organs autoregulatory mechanisms would "win" if pitted against each other? eg, SNS regulation due to temperature regulation would "win" against local metabolic factors in skeletal muscle, local autoregulation tends to predominate (during exercise) against the vasoconstrictive effects of the SNS CV10 Autoregulation in intestinal circulation is not as well developed as other vascular beds, eg brain and kidney. The principal mechanism for autoregulation is metabolic although a myogenic mechanism probably also participates (Berne and Levy Cardiovascular Physiology p261) I think the best answer may be D. Splanchic In Ganong (p623, 22nd edition), it states "The blood flow to the mucosa...responds to chnages in metabolic activity", but it also states "The intestinal circulation is capable of extensive autoregulation" If metabolic autoregulation refers to things such as, adenosine, K, decreased pH, increased CO2 - then couldn't lung also be seen as metabolically regulated? (just mainly opposite to the rest of the circulation - ie, these things cause vasoconstriction, at least hypoxia, hypercapnoea, and acidosis do anyway according to Ganong)... so maybe the best answer is the lungs (as this may be a greater effect than SNS stimulation, compared with the other answers) Infact, Levy & Pappano says that "Hypoxia has the most important influence on pulmonary vasomotor tone", which would tend to mean that this form of "metabolic autoregulation" outweights the importance of the autonomic nervous sytem and humoral factors. But the hypoxia that controls pulmonary blood flow relates to inspired (alveolar) gas, and isn't related to the metabolic rate of the lung itself - I'd go with splanchnic as the best of the poor choices here but hope that E was something more obvious.

Answer 218

(Alt wording: Which tissues autoregulate blood flow prominently: ) (Alt wording in July 03: Metabolic regulation of arteriolar resistance is most important in) (Comment received Jul03: "Options D & E were Lungs & Kidneys - two clearly incorrect answers since their flow is varied according to factors important to the body as a whole rather than for their own benefit ie oxygenation & filtration (& temperature regulation, in the case of skin). " Circulation in skeletal muscles match tightly with the metabolic demand. Studies reviewed that capillaries capable to co-ordination of muscle blood flow responses to changed muscle metabolism. Further research are done to showed there are links to direct arterials to communicate with up branching vessels. So not just appply the condition to exercising muscle but skeletal muscle as a whole would match its metabolic demand

Answer 219

``` Answer C Shock can be divided into: Distributive Hypovolaemic Cardiogenic Obstructive ``` General terms such as shock require general answer. Therefore the obvious must be stated – ischaemia is the relative decrease in oxygen supply to demand. So it can be caused by: Coronary blood flow or oxygen carrying capacity declines relative to demand. Demand for O2 increases without the corresponding increase in oxygen supply. Ischaemia therefore results. The best answer is then relative. Comment: If the precipitating cause is shock....a low output/supply state then the cause is a supply issue. If someone has a heightened metabolic state then maybe the demand may be the primary issue. Sure, someone in shock becomes tachycardic and has adrenaline etc which increases myocardial o2 requirements, but it is the initial low supply (shock) that started it all. So C I would argue that by definition, shock means lower blood pressure, therefore lower coronary perfusion pressure (being the cause of decreased oxygen supply). This is a quote from the AnaesthesiaUK site - "A vicious circle of poor coronary perfusion with worsening cardiac output occurs" Having said that, here's something to muddy the waters from Oh's Intensive Care Manual - "decreased coronary blood flow is not the usual cause of myocardial dysfunction in severe sepsis as coronary sinus catheter studies showed normal coronary blood flow" (I guess this was talking about myocardial dysfunction in sepsis, as opposed to ischaemia) Sepsis is a special form of shock - the cardiac output is high. Therefore the coronary flow was normal. Why not B? considering in shock your blood volume is preserved for vital organs like brain and heart in the expense of periphery... so your coronary blood flow should be only slightly reduced but the demand is largely increased all because of the significantly increased workload due to the increased SNS activity. relatively speaking the decreased supply probably less than the increased demand...

Answer 220

C. 30% -most textbooks No simple answer really. Increases with age. One paper (Am J Cardiol. 1987 May 1;59(12):1174-8) suggested that 20yo may be 12% while 80yo with out heart disease may be 45%. Increases with heart rate. Suggestions of 10% at rest up to 40% at maximum HR Increases with most forms of heart disease.

Answer 221

Standing causes blood to pool in the legs which results in a sudden decrease in venous return/cardiac output. The baroreceptor reflex (predominantly carotid baroreceptor) causes simultaneous sympathetic activation and vagal inhibition. This causes an increase in the tone of resistance vessels resulting in decreased perfusion to areas such as the skin and will also increase heart rate. Ganong 21st ed. p633

Answer 222

Go for E My assumption for this question is that the man is standing quietly, and that the question is referring to the pressure in a blood vessel in the dorsum of the foot. They haven't specified artery or vein, but according to Ganong, in either case the answer is (E). p588 states that in an upright adult human with MAP of 100 mmHg, the pressure in a large artery in the foot is 180 mm Hg. p595 states that during quiet standing, venous pressure at the ankle is 85-90 mm Hg. (NB: If the man is contracting his leg muscles, the venous pressure can be reduced to less than 30 mm Hg)

Answer 223

Moving from supine to erect causes hypotension, increased TPR and increased ADH. (Ganong p608) See excellent Fig 33-1, p. 557 "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. 23Ed; (631 21Ed) Additional to this: Cardiac output drops secondary to reduced venous return and subsequent stroke volume reduction, although this is partially compensated for by an increased heart rate. Minimal venoconstriction. Rapid increase in circulating renin and aldosterone (Ganong 21Ed p633; 23Ed p557).

Answer 224

Moving from supine to erect causes hypotension, increased TPR and increased ADH. (Ganong p608) See excellent Fig 33-1, p. 557 "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. 23Ed; (631 21Ed) Additional to this: Cardiac output drops secondary to reduced venous return and subsequent stroke volume reduction, although this is partially compensated for by an increased heart rate. Minimal venoconstriction. Rapid increase in circulating renin and aldosterone (Ganong 21Ed p633; 23Ed p557).

Answer 225

The myocardial fibres in the ventricle have the lowest intrinsic discharge rate, which is zero. These fibres have a stable phase 4 and are not pacemaker cells. Hence as Ganong (20th ed p530, 22nd ed p548) says, "atrial and ventricular muscles do not have pre-potentials, and they discharge spontaneously only when injured or abnormal." The bundle branches contain "latent pacemakers" and only take over setting the heart rate when the SA & AV nodes are depressed or conduction from them is blocked. (does no intrinsic discharge mean lowest? or should we choose the one that actually has an intrinsic activity?)

Answer 226

``` The slowest conduction velocity is in the SA & AV nodes. According to the table in Ganong (20th ed p530, 22nd ed p549): SA node: 0.05 m/s Atrial pathways: 1 m/s AV node: 0.05 m/s Bundle of His 1 m/s Purkinje system: 4 m/s Ventricular muscle: 1 m/s ```

Answer 227

Answer is E - see below. Also note that hepatic artery flow less than portal vein hence B and C automatically wrong Different texts give different numbers. Power & Kam p. 152 indicate that the ratio is 1:3. "Portal vein normally accounts for three quarters of the blood supply" Figure 32-15 on page 624 of Ganong 22nd Edition shows hepatic artery average blood flow of 500mL/min and portal blood flow of 1000mL/min. This would give a ratio of 1 : 2. SAQ examiner comment: 25% and 75% resp.

Answer 228

Obviously without the diagram this question isn't much use. The diagram in Ganong doesn't outline the changes with temperature, hypothermia and acidosis. Those factors all modify cerebral blood flow, which according to Ganong, hovers at 11.2 mmCSF. Between 68mmCSF and 200mmCSF, CSF absorption increases linearly, while production remains the same. Increased cerebral blood flow will increase intracranial pressure and thus increase aborption of CSF slightly. Comments "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. 23Ed; page 572 figure 34-3 is a diagram or CSF pressure versus flow showing absorption and formation. The units of the X-axis is mmCSF. I don't know if this is the same diagram but it would fit.

Answer 229

VIP causes vasodilatation. Neuropeptide Y causes vasoconstriction. Vasoconstrictors: TXA2 - promotes platelet aggregation and vasoconstriction Endothelin - three types produced by endothelial cells, endothelin-1 is one of most potent vasoconstrictors Angiotensin II - increased with elevated renin secretion when BP falls or ECF volume drops, works to maintain BP Serotonin - has effects on CNS, GIT and vascular systems Neuropeptide Y - augments vasoconstrictive effects of noradrenergic neurons, found in brain and autonomic nervous system Vasodilators: Prostacyclin - inhibits platelet aggregation and promotes vasodilation VIP - found in nerves in GIT and circulating blood ACh, Histamine via H1Rs, Bradykinin, Substance P and VIP act on endothelial system Adenosin, Histamine via H2Rs produced relation of vascular smooth muscle independent of endothelium Nitric Oxide Reference "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 603 Table 31-2 Ganong 20th Edition page 575 cGRP - calcitonin gene related peptide is also a vasodilator

Answer 230

VIP causes vasodilatation. Neuropeptide Y causes vasoconstriction. Vasoconstrictors: TXA2 - promotes platelet aggregation and vasoconstriction Endothelin - three types produced by endothelial cells, endothelin-1 is one of most potent vasoconstrictors Angiotensin II - increased with elevated renin secretion when BP falls or ECF volume drops, works to maintain BP Serotonin - has effects on CNS, GIT and vascular systems Neuropeptide Y - augments vasoconstrictive effects of noradrenergic neurons, found in brain and autonomic nervous system Vasodilators: Prostacyclin - inhibits platelet aggregation and promotes vasodilation VIP - found in nerves in GIT and circulating blood ACh, Histamine via H1Rs, Bradykinin, Substance P and VIP act on endothelial system Adenosin, Histamine via H2Rs produced relation of vascular smooth muscle independent of endothelium Nitric Oxide Reference "Review of Medical Physiology", W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 603 Table 31-2 Ganong 20th Edition page 575 cGRP - calcitonin gene related peptide is also a vasodilator

Answer 231

Locally released platelet serotonin and neuropeptide Y both cause vasoconstriction. Ref: Ganong 22nd Edition page 603 table 31-2 Note: Serotonin vasoconstricts in most tissue beds but vasodilates in skeletal muscle and heart. (Faunce Pg 75 and also Goodman's and Gilman's on ANZCA website) Oxytocin can also cause vasoconstriction and hypertension [1]

Answer 232

Neuropeptide Y is found in postganglionic sympathetic nerves. CGRP is found in sensory nerves near blood vessels. --> incomplete. Ganong: Two forms in rats and thus probably humans: CGRPα and CGRPβ CGRPα found in GI tract CGRPβ found in: (a) primary afferent neurons (b) neurons projecting taste into thalamus (c) neurons in medial forebrain bundle (d) sensory nerves near blood vessels Causes weak vasodilation and has other weak effects Ref: Ganong 22nd Edition page 602, 114

Answer 233

Does lying down really cause vasoconstriction??? Comment: cerebral autoregulation - Yes it does then I guess. Comment: Except if you consider supine to standing in which case it causes vasoconstriction. Bradykinin causes vasodilatation. Ganong 22nd Edition page 603 table 31-2

Answer 234

Answer - A: heart acts as a demand pump for metabolic requirements of body. Increased requirement = increased CO Anticipation of exercise -> stimulates cerebrocortical activation of SympNS -> Inc HR, Inc Contractile force, Inc SVR -> Inc CO and BP Berne &Levy Chap 12 In a long distance race, metabolism is aerobic and the increased O2 requirements are met by an increase in cardiac output, particularly an increase in blood flow to skeletal muscles. A 100m is a sprint so there is an increase in cardiac output supplying more oxygen to the exercising muscle but this is insufficient to met energy requirements and anaerobic metabolism ensues. But anaerobic metabolism does not involve oxygen which is what the question asks. The oxygen flux equation says that the only way that oxygen delivery can be substantially increased is by an increase in cardiac output. The other factors in the equation ([Hb}, SaO2, pO2) CANNOT be much increased. A factor that is relevant are an increased oxygen extraction due to a right shift of the oxygen dissociation curve consequent upon the altered microcirculatory conditions in hot, acidotic exercising muscle.

Answer 235

Bainbridge reflex: opposite to baroreceptor reflex. Infusion of volume tends to increase heart rate when heart rate is slow/blood volume is high (Ref: Berne+Levy 8th ed p91, "Cardiovascular Physiology", M. N. Levy & A. J. Pappano, 9th ed, Mosby, 2007. page 86). The opposite may occur if initial HR is higher - however, according to Ganong this may be "competition" with the baroreceptor reflex Bezold-Jarisch reflex: coronary chemoreflex. Serotonin /capsaicin/veratridine/phenyldiguinide and some other drugs injected into coronary vessels supplying left ventricle stimulates C fibre endings - afferent vagal response casues apneoa, followed by rapid breathing, hypotension and bradycardia. (Ref: Ganong 20th ed p585, 22nd ed - p608). Cushing reflex: caused by increased intracranial pressure which compromises blood supply to the vasomotor area causing local hypoxia and hypercapnia which in turn increases its neuronal discharge resulting in an increase in blood pressure. This increased blood pressure stimulates the baroreceptors leading to a decrease in heart rate. Pulmonary chemoreflex: similar to the coronary chemoreflex (Bezold-Jarisch reflex-see above) except that the drugs are injected into the pulmonary arteries. Causes a decreased heart rate. Stimulation of the atrial stretch receptors: An increase in the rate of discharge of the atrial stretch receptors results in vasodilation and an increase in heart rate. Stretching the atrium: This will result in an increase in the rate of discharge of type B atrial stretch receptors, so it will increase the heart rate. L ventricular stretch: Similar to arterial baroreceptor reflex, will cause a decrease in HR Carotid Chemoreflex: Primarily involved in regulation of respiration, however, hypoxic stimulation of chemoreceptors causes vasoconstriction and bradycardia - this may be overridden by the hypoxic stimulation of adrenaline secretion which causes tachycardia (Ganong) Hering-Breuer Reflex: Primarily regulates lung inflation/deflation, no effect on HR as far as I can see - therefore, stretching the atrial wall / stimulating atrial stretch receptors is the only answer that causes an increased HR (ie, Bainbridge may not cause increased HR if higher HR to start with) - isn't Bainbridge reflex = stretching of atrial receptors???? at least thats how it seems to be on Power & Kam Fill in the table: Reflex: HR, RR, BP, Role, Vagotomy inhibits?

Answer 236

Anterior tibial artery (mean BP 90mmHg when supine) and vein (few mmHg) is largest pressure difference Pulmonary artery 25mmHg (systolic) - Pulmonary vein 5 mmHg Femoral vein 6 mmHg - Right atrium 2 mmHg Renal afferent arteriole 60 mmHg - renal vein 4 mmHg Pressure of 60mmHg at arterial end of glomerular capillaries (Guyton & Hall 9th ed p317) "The pressure in the renal vein is about 4mmHg" (Ganong 20th ed p680) There is a high hydrostatic pressure in the glomerular capillaries and this (along with the very high filtration coefficient of the glomerular capillaries) aids filtration. The pressure drop along the short length of the glomerular capillaries is small (1-3mmHg); the main pressure drop occurs across the peritubular capillaries. This second capillary bed is responsible for the low pressure in the renal veins.

Answer 237

Answer - A: think about it... you pump blood before the measured point, the pressure is going to go up. If you measure the aortic pressure after contraction of muscle (heart), it goes up. If you are measuring in the ventricle, the pressure goes down after contraction. Taking a step forward would cause the muscular pump to contract and push venous blood up towards the right atrium. The mechanism of the muscle pump is important in decreasing venous pressure from >80 to less than 30mmHg in the lower limbs. The valves in the veins helps to move blood unidirectionally. Useful information (also from Ganong) is that the magnitude of pressure change on standing is 0.77 mmHg/cm at normal blood density (this is just the conversion factor for mmHg to cm in any fluid column) , either increasing as you go down or decreasing as you go up. The muscle contraction on stepping forward will increase venous pressure initially via vein compression. As muscle relaxes the pressure will drop, and valves effectively break the fluid column, limiting its hydrostatic effects. So pressure will initially increase, then drop, then rise again unless the muscle pump reactivates.

Answer 238

CV25 - B highest oxygen extraction is in the heart Myocardial oxygen extraction ratio is 55-65%(Kerry Brandis the physiology Viva book) This extraction ratio is much higher than the average oxygen extraction ratio 25% for the body as a whole. Two coronary arteries that supply the myocardium arise from the sinuses of valsalva at the root of aorta,resting coronary blood flow is 225 to 250 mls/min.,or 4 to 5 %of the cardiac output=75 ml/100g/min Resting myocardial oxygen consumption is 8 to 10 mls/100g/min=10%total body oxygen consumption Oxygen consumption of the arrested ,nondistended and normothermic heart is 1ml/100g/min - compared with 5 ml/100g/min in both the arrested distended and beating but empty heart. Lowering myocardial temperature from 37C to 11C produces only a modest 5% further decrease in myocardial oxygen consumption compared with the arrested and nondistended heart EVEN in normal resting state,oxygen extraction by cardiac cells is nearly maximal (approaches 70%) This means increases in myocardial oxygen consumption must be met by commensurate increases in coronary blood flow,because there is little additional oxygen extraction that can occur. Heart (114ml/L) > Brain (62) > Skeletal Muscle (60) > Liver (34) > Skin (25) > kidney (14) Ganong 23rd ed Table 34.1 Pneumonic: Heavy Breathing May Leave Skin (K)clammy Not easy to find a number for the carotid body oxygen extraction - does anyone have a number?

Answer 239

Re CV25b -> A Ganong 21st ed p705: Heart (114ml/L O2 diff) > Brain (62ml/L) > Kidney (14ml/L) Renal cortical blood flow is high with a low O2 extraction, the medulla has a high oxygen extraction but a very low blood flow (0.6ml/g/min compared with 2.5ml/g/min in the cortex). Heart (114ml/L) > Brain (62) > Skeletal Muscle (60) > Liver (34) > Skin (25) > kidney (14) Ganong 23rd ed Table 34.1 Pneumonic: Heavy Breathing May Leave Skin (K)clammy Not easy to find a number for the carotid body oxygen extraction - does anyone have a number?

Answer 240

``` CV25 c Answer is E heart 114ml/L, skeletal muscle 60ml/L, liver 34 ml/L, skin 25ml/L, kidneys 14ml/L see Ganong table 32.1 Mnemonic: Heart > Brain >= Muscle > liver > skin > kidneys so: Heavy Breathing May Leave Skin (K)lammy. ``` Heart (114ml/L) > Brain (62) > Skeletal Muscle (60) > Liver (34) > Skin (25) > kidney (14) Ganong 23rd ed Table 34.1 Pneumonic: Heavy Breathing May Leave Skin (K)clammy Not easy to find a number for the carotid body oxygen extraction - does anyone have a number? The oxygen extraction ratio (i.e. same as A-V O2 difference) for the body as a whole is 25% but that figure is not actually typical of much of the body. One of the main reasons for this is that the kidney receives 25% of the cardiac output (= O2 delivery of 250mls/min) as that huge percent flow for such a small (by weight) organ is because its needed to do what kidneys do (i.e. to perform its function). Renal blood flow is not determined by the need for oxygen. After taking say 18mls oxygen out, it returns this large relatively oxygen rich venous blood to the vena cava. Because this volume of blood is so large it in effect causes the average venous oxygen content for the body as a whole to be higher than it would otherwise be. In contrast the coronary circulation is subject to a significant Starling resistor effect. Blood flow to the left ventricle (the biggest consumer of oxygen in the heart) during systole is low because the arteries are compressed. The net result of various factors is that the myocardial oxygen extraction is 55-65%. It can go a little higher (say to 75% or so) but not much. A consequence ofthis is that if the heart needs more oxygen then it needs more coronary blood flow. It cannot use the other way that tissues respond to an acute demand for more oxygen, that is, it cannot extract more as extraction is already very high. Anyway the simple fact is that the HIGHEST O2 extraction is in the body is in the heart. The brain takes 15% of the cardiac output at rest, so 15% of the 1,000mls oxygen flux, which is an oxygen supply of 150mls/min. The brain's oxygen consumption - known as CMRO2 for "cerebral metabolic rate of oxygen consumption" - is 3-3.5 mlsO2/100G/min, which works out to about 45-50mls O2/min. Thus O2 extraction ratio = (approx) 50/150 or 33%. This is higher then the body as a whole (25%). Skin receives a fairly low basal blood flow (say 300mls/min) appropriate to its oxygen needs. Skin is usually at a temperature lower than core temperature so has a lower metabolic rate (= less O2 consumption/G) because of this. The key thing with the skin is that the blood flow can increase by a large amount when there is a heat stress. Much excess heat is lost from the skin. A skin blood flow of 3,000mls/min is a typical value at maximal heat stress. Clearly under these conditions then the oxygen extraction ratio will be very low. The liver is metabolically very active, and consumes about 50mls O2/min from its blood supply (= portal flow + hepatic artery flow) of 1,500mls/min. Roughly speaking, it gets twice the blood supply as the brain but consumes the same amount of oxygen as the brain . However the 30% of cardiac output it receives does not mean 30% of the oxygen flux as 1,200 mls of the supply is low pO2 portal venous blood. The blood flow to the carotid body is remarkable, and also illustrates that the blood flow that some organs receive is related to the function they perform. For many tissues blood flow is matched to tissue oxygen demand (aka 'flow-metabolism coupling'), as this oxygen (actually the energy from aerobic metabolism) is what they need to perform their function. But sometimes performing the function needs more blood than oxygen so there can be a disconnect. The kidney (as noted above) is an important exception to the flow-metabolism coupling rule as there is a major disconnect between renal oxygen demand and the flow needed to perform its functions (filtration and reabsorption in the tubules). The carotid body is the most extreme example in the body of a disconnect between flow required for function and the flow required for its aerobic metabolism (this being the major consumer of O2 in every cell). The carotid bodies are chemoreceptors, and what they essentially need to do their function is access in their cappillaries to arterial blood. The chemoreceptor cells get access to capillary blood and the flow to the carotid body has to be high enough so that the composition of oxygen, carbon dioxide (and pH) in the capillary blood in the carotid bodies is not altered by the organs metabolic need for oxygen, or delivery of CO2 into the capillaries. Consider this: if the carotid body blood flow was low (on a weight basis that is) then uptake of O2 in the capillaries, and output of CO2 to the capillaries would cause the capillary blood to have a lower pO2 and a higher pCO2 than the arterial blood supplying it. So a chemoreceptor cell sampling capillary blood would not be measure arterial values. BUT if the flow was very high (on a weight basis that is) then the carotid body capillaries are being flushed so rapidly that the pO2 and pCO2 would be the same as in arterial blood. So the receptor cells effectively sample arterial blood in their capillaries. The venous blood draining from the carotid bodies has the same concentration of everything as arterial blood; it is for all intents and purposes 'arterial blood' (in composition). And of course 'arterial blood' in the systemic circulation (excluding of course pulmonary artery blood here) has a particular property: "All (systemic) artery blood has the same composition". So the chemoreceptors in the carotid body provide information about pO2, pCO2, and pH in the arterial blood supply to every tissue and organ in the body. For the purposes of this MCQ though, the point is that the O2 extraction ratio (or equivalently, the A-V difference) is EXTREMELY LOW. So whatever tissues/organs that any version of this MCQ could have, then the heart will be in the leftmost position (highest O2 extraction) and the carotid body will be in the rightmost position (lowest O2 extraction), and the kidney (under normal resting conditions) will be placed just before the carotid body. And if there is great heat stress then the skin will have high blood supply per unit wt but low O2 extraction. A final point about the carotid bodies: they get a huge blood supply if considered on a weight basis (2,000 mls/100G/min) but their weight is EXTREMELY small (something like 20-30mg in total), so the actual total arterial flow involved is VERY SMALL indeed. A quick calculation now gives 0.3ml/min!

Answer 241

CV26a - B and D correct CV26a The CV26a version says the "initial phase' so this means phase 1. Phase 1 is brief and reflex changes to increase heart rate and SVR have not yet had time to occur. However, the BP does increase transiently during phase 1 so option D is correct for this version. Option B is also correct for this version because transient increase in BP is caused by increased cardiac output from increased venous return to left heart from lungs as increased intra-pulmonary pressure squeezed on pulmonary vessels. CV26a - Answer B and D are correct: Initial pressure rise: On application of expiratory force, pressure rises inside the chest forcing blood out of the pulmonary circulation into the left atrium. This causes a mild rise in blood pressure (associated with increase in cardiac output) The Valsalva manoeuvre is sometimes defined as forced expiration against a closed glottis after a full inspiration, but this is wrong because a closed glottis is NOT a necessary component. According to Brandis (p53): "The essential or defining feature of the Valsalva manoeuvre is the increase in intrathoracic pressure". As originally described by Valsalva, the manoeuvre was performed by closing off the mouth and nose and raising intrathoracic pressure by contracting the expiratory muscles. When doing this, the glottis is OPEN and the raised pressure is transmitted to the cavities of the mouth, nose (and the middle ear provided the Eustachisn tubes are patent). Valsalva described this as a technique for expelling pus from the middle ear via a perforated ear drum, but in current times, this technique is more commonly used to equalise pressure across the ear drum (eg when travelling by air). Phase I: Blood is expelled from the thoracic vessels by the increase in intrathoracic pressure. This causes a transient increase in blood pressure. Phase II: The increase in intrathoracic pressure causes a reduction of venous return, lowering the preload and BP. The baroreceptor reflex is activated, causing vasoconstriction and a tachycardia, raising the BP towards normal. Phase III: As intrathoracic pressure suddenly drops there is a pooling of blood in the pulmonary vessels, causing a further drop in BP. Phase IV: With venous return restored there is an overshoot as compensatory mechanisms contineu to operate. The increased BP causes a baroreceptor mediated bradycardia.

Answer 242

CV26b - A and C correct (but poorly worded) The Valsalva manoeuvre is sometimes defined as forced expiration against a closed glottis after a full inspiration, but this is wrong because a closed glottis is NOT a necessary component. According to Brandis (p53): "The essential or defining feature of the Valsalva manoeuvre is the increase in intrathoracic pressure". As originally described by Valsalva, the manoeuvre was performed by closing off the mouth and nose and raising intrathoracic pressure by contracting the expiratory muscles. When doing this, the glottis is OPEN and the raised pressure is transmitted to the cavities of the mouth, nose (and the middle ear provided the Eustachisn tubes are patent). Valsalva described this as a technique for expelling pus from the middle ear via a perforated ear drum, but in current times, this technique is more commonly used to equalise pressure across the ear drum (eg when travelling by air). Phase I: Blood is expelled from the thoracic vessels by the increase in intrathoracic pressure. This causes a transient increase in blood pressure. Phase II: The increase in intrathoracic pressure causes a reduction of venous return, lowering the preload and BP. The baroreceptor reflex is activated, causing vasoconstriction and a tachycardia, raising the BP towards normal. Phase III: As intrathoracic pressure suddenly drops there is a pooling of blood in the pulmonary vessels, causing a further drop in BP. Phase IV: With venous return restored there is an overshoot as compensatory mechanisms contineu to operate. The increased BP causes a baroreceptor mediated bradycardia.

Answer 243

CV26c - A correct CV26c The wording in the version labelled CV26c is used as this version is an accurate transcription by a single candidate. This reveals that the MCQ refers not to phase I (as in the CV26a version) but to the whole raised intrathoracic pressure (ITP) phase. This is phase 1 (but only briefly) and phase 2 (mostly). As regards the options: During increased intrathoracic pressure of a Valsalva manoeuvre: A. Diastolic filling of the right ventricle is decreased - Correct, as the raised ITP tends to decrease venous return. Additionally the increased heart rate decreases time for ventricular filling. B. Arterial baroreceptor activation produces bradycardia -Wrong, the BP drop results in baroreceptor-induced tachycardia C. Increased venous pressure augments cardiac output - Incorrect - The increased venous pressure will 'tend' to overcome the block to venous return caused by the typically high (eg 40mmHg) ITP. However, this is generally not sufficient to make much difference to cardiac output. Looked at a slightly different way, the increased intrathoracic pressure impairs venous return (& cardiac output) is decreased during the manoeuvre. D. Total peripheral resistance is decreased - Incorrect - TPR is reflexly increased. This helps maintain arterial BP and thus cerebral perfusion pressure. Actually disagree because decreased CO means decreased TPR to maintain flow which is the important thing. the heart is a demand pump. Ie Correct. > "decreased CO means decreased TPR" - Huh?? This makes no sense. E. Arterial blood pressure initially decreases - Incorrect - BP transiently increases during phase 1 The Valsalva manoeuvre is sometimes defined as forced expiration against a closed glottis after a full inspiration, but this is wrong because a closed glottis is NOT a necessary component. According to Brandis (p53): "The essential or defining feature of the Valsalva manoeuvre is the increase in intrathoracic pressure". As originally described by Valsalva, the manoeuvre was performed by closing off the mouth and nose and raising intrathoracic pressure by contracting the expiratory muscles. When doing this, the glottis is OPEN and the raised pressure is transmitted to the cavities of the mouth, nose (and the middle ear provided the Eustachisn tubes are patent). Valsalva described this as a technique for expelling pus from the middle ear via a perforated ear drum, but in current times, this technique is more commonly used to equalise pressure across the ear drum (eg when travelling by air). Phase I: Blood is expelled from the thoracic vessels by the increase in intrathoracic pressure. This causes a transient increase in blood pressure. Phase II: The increase in intrathoracic pressure causes a reduction of venous return, lowering the preload and BP. The baroreceptor reflex is activated, causing vasoconstriction and a tachycardia, raising the BP towards normal. Phase III: As intrathoracic pressure suddenly drops there is a pooling of blood in the pulmonary vessels, causing a further drop in BP. Phase IV: With venous return restored there is an overshoot as compensatory mechanisms contineu to operate. The increased BP causes a baroreceptor mediated bradycardia.

Answer 244

The Valsalva manoeuvre is sometimes defined as forced expiration against a closed glottis after a full inspiration, but this is wrong because a closed glottis is NOT a necessary component. According to Brandis (p53): "The essential or defining feature of the Valsalva manoeuvre is the increase in intrathoracic pressure". As originally described by Valsalva, the manoeuvre was performed by closing off the mouth and nose and raising intrathoracic pressure by contracting the expiratory muscles. When doing this, the glottis is OPEN and the raised pressure is transmitted to the cavities of the mouth, nose (and the middle ear provided the Eustachisn tubes are patent). Valsalva described this as a technique for expelling pus from the middle ear via a perforated ear drum, but in current times, this technique is more commonly used to equalise pressure across the ear drum (eg when travelling by air). Phase I: Blood is expelled from the thoracic vessels by the increase in intrathoracic pressure. This causes a transient increase in blood pressure. Phase II: The increase in intrathoracic pressure causes a reduction of venous return, lowering the preload and BP. The baroreceptor reflex is activated, causing vasoconstriction and a tachycardia, raising the BP towards normal. Phase III: As intrathoracic pressure suddenly drops there is a pooling of blood in the pulmonary vessels, causing a further drop in BP. Phase IV: With venous return restored there is an overshoot as compensatory mechanisms contineu to operate. The increased BP causes a baroreceptor mediated bradycardia.

Answer 245

The last part of the heart to be depolarised is the epicardial surface of the left ventricular wall at the base of the heart. So, for CV27, (A) is correct.

Answer 246

The last part of the heart to be depolarised is the epicardial surface of the left ventricular wall at the base of the heart. So, for CV27, (A) is correct.

Answer 247

Ref: Ganong 22nd edition page 549 Table 28.1 SA node and AV node: 0.05 m/sec Atrial pathways, Bundle of His, Ventricular muscle: 1 m/sec Purkinjie system 4m/sec All the textbooks agree that the fastest conduction is in the Purkinje system, but there is some variation in the actual figures given (which is probably why the questions don't mention numbers, just which is fastest). As well as the Ganong value of 4m/sec above, Berne & Levy, 8th Ed, p38 states that "the conduction of the action potential over the Purkinje fiber system is faster than in any other...tissue within the heart; estimates of conduction velocity vary from 1 to 4 m/s Rang et al, 5th Ed, p265 gives a value of ~200 cm/s

Answer 248

Ref: Ganong 22nd edition page 549 Table 28.1 SA node and AV node: 0.05 m/sec Atrial pathways, Bundle of His, Ventricular muscle: 1 m/sec Purkinjie system 4m/sec All the textbooks agree that the fastest conduction is in the Purkinje system, but there is some variation in the actual figures given (which is probably why the questions don't mention numbers, just which is fastest). As well as the Ganong value of 4m/sec above, Berne & Levy, 8th Ed, p38 states that "the conduction of the action potential over the Purkinje fiber system is faster than in any other...tissue within the heart; estimates of conduction velocity vary from 1 to 4 m/s Rang et al, 5th Ed, p265 gives a value of ~200 cm/s

Answer 249

Isovolumetric contraction: Occurs as the first part of ventricular systole Mitral and Tricuspid valves close as ventricles contract and intra-ventricular pressures begin to rise from zero Ventricular pressures rise, but during this phase: RV pressure

Answer 250

Answer - B Fick Principle: uptake or output of a substance by a tissue must be equal to the difference between the amount entering (flow x arterial concentration) and the amount leaving the tissue (flow x venous concentration). This is true because of the conservation of mass. Using the Fick Principle, we can determine cardiac output from a sample of mixed venous blood, arterial blood and a measurement of O2 uptake. Cardiac Output = O2 uptake/ (arterial O2 content - mixed venous O2 content For example, substituting some typical values: O2 uptake = 250 mls O2/min arterial O2 content = 200mlsO2/l of blood mixed venous O2 content: 150mlsO2/l of blood thus: CO = 250/(200-150) = 5 l/min. If you use O2 content in units of mlsO2/dl then the answer will be in decilitres. Though often overlooked, this use of the Fick equation for measuring cardiac output includes the assumption that O2 consumption of the lung from the ventilated gas is negligible. If this was not true, the measured O2 consumption would be higher and the calculated CO higher. Therefore B is true.

Answer 251

From Fick Principle: Cardiac output = O2 uptake / Arterio-venous O2 difference = 280 / (150-110) = 280 / 40 Cardiac output = 7 L/min Hence, answer is C.

Answer 252

There is an excellent table in Ganong: 32-1. The organ with the greatest whole organ blood flow at rest is the liver: 1500 mL/min The organ with the greatest blood flow per 100g mass is the kidney: 420 mL/100g/min Assuming the question refers to whole organ blood flow answer B liver is correct. Doesn't the carotid body have a flow rate of 2000ml.100g.min? Re above: yes, has a very high blood flow Re above: yep - and that infinitely thin ring of tissue of almost negligible mass around the aorta right after the heart has a flow rate of a bazillion ml/100g/min

Answer 253

CV34 ”NB – the stem seemed to imply total organ blood flow as opposed to per unit weight Excellent table in Ganong 22nd edition page 612 Table 32.1/23rd ed p570 Table 34-1: Oxygen consumption in mL/100g/min: ``` Heart 9.7 Kidneys 6.0 Brain 3.3 Liver 2.0 Skin 0.3 Skeletal Muscle 0.2 Oxgen consumption for whole organs (ml/min): ``` ``` Liver 51.0 Skeletal Muscle 50.0 Brain 46.0 Heart 29.0 Kidneys 18.0 Skin 12.0 ```

Answer 254

Excellent table in Ganong 22nd edition page 612 Table 32.1/23rd ed p570 Table 34-1: Oxygen consumption in mL/100g/min: ``` Heart 9.7 Kidneys 6.0 Brain 3.3 Liver 2.0 Skin 0.3 Skeletal Muscle 0.2 Oxgen consumption for whole organs (ml/min): ``` ``` Liver 51.0 Skeletal Muscle 50.0 Brain 46.0 Heart 29.0 Kidneys 18.0 Skin 12.0 ```

Answer 255

CV34c On Apr 2001 paper there were 2 questions on oxygen consumption, one at rest]] [CV34] [Jul96] and one during exercis] [Mar98]] [CV34] [Mar97]. Neither question specified whether absolute consumption or ml/100g/min which is a significant oversight. Options were said to be the same for both. KB 26-May-01 Excellent table in Ganong 22nd edition page 612 Table 32.1/23rd ed p570 Table 34-1: Oxygen consumption in mL/100g/min: ``` Heart 9.7 Kidneys 6.0 Brain 3.3 Liver 2.0 Skin 0.3 Skeletal Muscle 0.2 Oxgen consumption for whole organs (ml/min): ``` ``` Liver 51.0 Skeletal Muscle 50.0 Brain 46.0 Heart 29.0 Kidneys 18.0 Skin 12.0 ``` ``` Answer to CV34c [Apr01] Reference from www.cvphysiology.com Oxygen consumption of the heart (mL/100g/min): Arrested: 2 Resting: 8 Heavy Exercise: 70 Oxygen consumption of skeletal muscle (mL/100g/min): Resting muscle: 1 Contracting muscle: 50 Mass of Heart = 0.3 kg; Mass of Skeletal Muscle = 31.0 kg (Ref: Table 32-1 p615 Ganong 21st Ed) Hence: B correct for O2 consuption per 100g C correct for whole organ O2 consumption ```

Answer 256

Plasma volume should return to normal in: Brandis page 58: 12-72 hrs. Power and Kam pg 156: 12-24. Ganong 22nd ed pg 638: 12-72 hrs.

Answer 257

Plasma volume should return to normal in: Brandis page 58: 12-72 hrs. Power and Kam pg 156: 12-24. Ganong 22nd ed pg 638: 12-72 hrs.

Answer 258

As worded answer is A and C A. Lying down → return of venous blood pooled in legs → ↑ venous retrun to RA → ↑ cardiac output → ↑ BP → ↑ carotid sinus stimulation → venodilation. B. Valsalva → ↓ venous retrun to RA → ↓ cardiac output → ↓ BP → ↓ carotid sinus stimulation → venoconstriction. C. Carotid sinus compression → ↑ carotid sinus stimulation → venodilation. Stimulation of the carotid sinus (e.g by increased BP or physical compression) leads to increased firing of the baroreceptor (via the glossopharyngeal nerve and the nucleus tractus solitarius) which causes inhibition of SNS outflow and stimulation of the vagus nerve. This leads to vasodilation, venodilation, bradycardia and reduced cardiac output.

Answer 259

C is true, especially in LV diastole. A: wrong - Right in 50%, equal in 30% B: wrong - diastole C: correct - greater blood flow to endocardium during diastole D: wrong - diastole E: wrong - R>L because of lower pressure developed F: wrong There is a superb diagram in Brandis P. 69. main points from his book: L ventricular supply occurs during diastole, and **subendocardial flow to L ventricle ceases in systole.** Difference still present, but less pronounced in R ventricle (as above comment; due to lower pressure differential) Also, Kam says (P103) that 50%of people Rt side is dominant, 20% left side is dominant. Either way, answer A is wrong. Coronary circulation nicely covered in detail in "Cardiovascular Physiology", M. N. Levy & A. J. Pappano, 9th ed, Mosby, 2007. chapter 11 page 203.

Answer 260

C is true, especially in LV diastole. A: wrong - Right in 50%, equal in 30% B: wrong - diastole C: correct - greater blood flow to endocardium during diastole D: wrong - diastole E: wrong - R>L because of lower pressure developed F: wrong There is a superb diagram in Brandis P. 69. main points from his book: L ventricular supply occurs during diastole, and **subendocardial flow to L ventricle ceases in systole.** Difference still present, but less pronounced in R ventricle (as above comment; due to lower pressure differential) Also, Kam says (P103) that 50%of people Rt side is dominant, 20% left side is dominant. Either way, answer A is wrong. Coronary circulation nicely covered in detail in "Cardiovascular Physiology", M. N. Levy & A. J. Pappano, 9th ed, Mosby, 2007. chapter 11 page 203.

Answer 261

While adenosine does block AV conduction, this question seems to be more about the receptor itself and its second messenger (given IP3 as one of the options) rather than the overall result of the drug. The absence of some of the options makes it difficult to be certain. art_adenosine.gif In terms of adenosine's effect on the AV node, Rang et al, 5th ed, p277 says that the A1 adenosine receptor is linked to the KACh potassium channel. On p194-195 (and in Ganong) they suggest that the adenosine receptors inhibit or stimulate adenylate cyclase. "Stimulation of the A1 receptor has a myocardial depressant effect by decreasing the conduction of electrical impulses and suppressing pacemaker cell function, resulting in a decrease in heart rate. This makes adenosine a useful medication for treating and diagnosing tachyarrhythmias, or excessively fast heart rates. This effect on the A1 receptor also explains why there is a brief moment of cardiac standstill when adenosine is administered as a rapid IV push during cardiac resuscitation". (from Wikipedia ref)

Answer 262

CV39 - C When a constrictor is placed on the aorta above the renal arteries, the blood pressure in both kidneys falls, renin secreted, angiotensin formed causing vasoconstriction. Within a few days, salt and H2O retention occurs so that the arterial pressure in the lower body at the level of the kidneys rises approx. to normal wheras high pressure persists in the upper body. Now the kidneys are no longer ischaemic so secretion of renin will decrease. Arterial pressure in the lower body is usually almost normal whereas the pressure in the upper body is far higher than normal (55% higher). Blood flow in arms is normal (despite higher BP) because of compenstation to elevated pressure by local autoregulation. Blood flow in legs where the pressure is not elevated is almost exactly normal. More Comments BP in upper body stays elevated for the life of the patient. So then, doesn't the baroreceptor sensitivity decrease over time due to chronic HT? Hence reduced function? What you say sounds entirely reasonable. But in the above MCQs there is no option which reads "decreased sensitivity", only "inactivated", which I would say is too absolute and not correct. References Guyton 9th edn, pg232 "Textbook of Medical Physiology", A. C. Guyton & J. E. Hall, 11th ed, W. B. Saunders, 2006. page 227.

Answer 263

July 99 - C. When a constrictor is placed on the aorta above the renal arteries, the blood pressure in both kidneys falls, renin secreted, angiotensin formed causing vasoconstriction. Within a few days, salt and H2O retention occurs so that the arterial pressure in the lower body at the level of the kidneys rises approx. to normal wheras high pressure persists in the upper body. Now the kidneys are no longer ischaemic so secretion of renin will decrease. Arterial pressure in the lower body is usually almost normal whereas the pressure in the upper body is far higher than normal (55% higher). Blood flow in arms is normal (despite higher BP) because of compenstation to elevated pressure by local autoregulation. Blood flow in legs where the pressure is not elevated is almost exactly normal. More Comments BP in upper body stays elevated for the life of the patient. So then, doesn't the baroreceptor sensitivity decrease over time due to chronic HT? Hence reduced function? What you say sounds entirely reasonable. But in the above MCQs there is no option which reads "decreased sensitivity", only "inactivated", which I would say is too absolute and not correct. References Guyton 9th edn, pg232 "Textbook of Medical Physiology", A. C. Guyton & J. E. Hall, 11th ed, W. B. Saunders, 2006. page 227.

Answer 264

? When a constrictor is placed on the aorta above the renal arteries, the blood pressure in both kidneys falls, renin secreted, angiotensin formed causing vasoconstriction. Within a few days, salt and H2O retention occurs so that the arterial pressure in the lower body at the level of the kidneys rises approx. to normal wheras high pressure persists in the upper body. Now the kidneys are no longer ischaemic so secretion of renin will decrease. Arterial pressure in the lower body is usually almost normal whereas the pressure in the upper body is far higher than normal (55% higher). Blood flow in arms is normal (despite higher BP) because of compenstation to elevated pressure by local autoregulation. Blood flow in legs where the pressure is not elevated is almost exactly normal. More Comments BP in upper body stays elevated for the life of the patient. So then, doesn't the baroreceptor sensitivity decrease over time due to chronic HT? Hence reduced function? What you say sounds entirely reasonable. But in the above MCQs there is no option which reads "decreased sensitivity", only "inactivated", which I would say is too absolute and not correct. References Guyton 9th edn, pg232 "Textbook of Medical Physiology", A. C. Guyton & J. E. Hall, 11th ed, W. B. Saunders, 2006. page 227.

Answer 265

Answer is C "The first portions of the ventricles to be excited are the interventricular septum (except at its basal portion) and the papillary muscles." "Cardiovascular Physiology", M. N. Levy & A. J. Pappano, 9th ed, Mosby, 2007. page 41. Order of excitation and start of contraction is RA-LA-LV-RV, however RV ejection begins before LV due to PA pressure beings so much lower than aortic (Ganong 22ed page 566). However Ganong also states that the last part of the heart to be depolarised is the posterobasal portion of the LV, the pulmonary conus and uppermost (basal) portion of the septum (Ganong 22ed page 549). Presumably the larger mass of the LV means that despite its excitation begining before the RV, it finishes after. Comments?

Answer 266

Option A refers to Ahlqvist who first suggested the presence of alpha & beta adrenergic receptors. -KB March 2003: "This was actually on the Pharmacology MCQ paper"] B is true

Answer 267

B true: see figure 29-3 Ganong p569 21st Ed – the Cardiac Cycle Aortic valve closes at the end of ventricular ejection, when LV pressure = 80 mmHg and RV pressure = 15 mmHg. This is followed by the period of isovolumetric ventricular relaxation when RV and LV pressures drop to zero. In expiration, AV and PV close at the same time; in inspiration PV closes after AV.

Answer 268

Answer is either B or C Velocity proportional to flow, such that V=Q/A where a is the cross sectional area Ganong Fig 30-14 clearly shows that aortic flow velocity is highest Pulmonary vein receives the same flow as pulmonary circulation, and during diastole there should be flow due to stored energy. Small arteries has the highest velocity of blood flow This is because: Flow = pressure / resistance Systemic blood pressure decreases as blood travels from the aorta to large veins(Guyton, Hall) Resistance in the small arteries increases causing the mean arterial pressure to decrease to about 85mmHg at the beginning of the arterioles. (Stoelting Hillier) The arterioles has the highest resistance to flow, 50% of the entire systemic circulation.(Stoelting Hillier) Pressure is about 30mmHg when blood reaches the capiparies.(Stoelting Hillier) Pressure within veins decreases from 10mmHg to nearly 0mmHg indicates that the resistance to flow is higher than expected, caused by compression of the veins by external forces that keep many of them, especially the venae cavae, partially collapsed. (Stoelting Hillier) Velocity = Flow / cross-sectional area Could this be the pulmonary vein? I see from the graph in Ganong (18th ed, p544) that blood flow is highest in aorta; pulm v flow must be the same , so unless the pulm v is much bigger in cross-sectional area, then velocity of flow must be equal?, and I guess that pulm v flow is higher in diastole as this is when there is more filling of the heart and so less resistance , what do you lot think re this? Comment: I think the correct answer is the pulmonary vein: V = Q/CSA Excludes Capillaries and small arteries (CSA > ICV and PV) PV has additional flow above that of the IVC because of the contribution of the bronchial circulation (not to mention the SVC) Published Values: Hard to find data: Aorta and Pulm Artery around 70 cm/sec [1] Peak velocities during ventricular systole ranged from 30 to 45 cm/sec in the inferior, and from 10 to 35 cm/sec in the superior, vena cava. [2] Peak systolic velocity in the right lower pulmonary vein was 47 +/- 11 cm/s, peak diastolic velocity was 40 +/- 9 cm/s, and peak backflow velocity was 14 +/- 3 cm/s. [3] I'm going for the pulmonary vein, but the error in these values would be larger than the difference. [4] - a little pic that still does nothing to end this debate NB there is no "pulmonary vein" as such; rather 2 PAIRS of pulmonary veins from either lung -- Oh Please... Comment (11/2/2011) -- I suspect the best answer here is pulmonary veins. As the above has pointed out -- the same cardiac output is going through the pulmonary circulation, so for that amount of blood to be drained from the capillaries into the L side of the heart the velocity must be quite high to travel the relatively short distance. -- SYL

Answer 269

A. Cardiac output is 7 litres per minute In a resting supine man of average size, cardiac output averages out at 5.0 L/min (72bpm x 70ml) In a trained athlete, SV increases and HR decreases, but CO at rest is similar B. Left ventricular end-systolic volume is 60mls (?? OR: end-diastolic volume is ?60mls ?100mls) In an average person, the end-diastolic ventricular volume is about 130ml. Stroke volume is about 70-90mls, leaving an end-systolic volume of about 50mls. In a trained athlete, HR is decreased, increasing the time for diastolic filling. Thus the end diastolic volume will be greater and stroke volume increases through the Frank-Starling mechanism. The end-systolic volume will be unchanged at about 50 mls. C. Stroke volume is 70mls 70mls is the stroke volume of an average 70kg male at rest in the supine position. A trained athlete has a larger stroke volume and proportionately lower HR. D. Oxygen consumption is 350mls/min Oxygen consumption at rest is normally 250mls/minute E. A-v O2 extraction is 5mlsO2/100mls blood Physical conditioning is associated with a greater extraction of O2 from the blood. With long term training, capillary density and the numbers of mitochondria imcrease, as does the activity of the oxidative enzymes in the mitochondria. Table 29-4 in Ganong shows A-V O2 difference as 4.3 ml/100mls blood at rest. does the increased O2 extraction occur at rest or only with exercise? 11/2/2011 -- I don't think A-V difference is increased at rest, so that answer E sounds most appropriate to me. A-V difference is certainly increased in trained athletes. SYL

Answer 270

``` Aortic cross-clamping: Increase afterload decrease organ perfusion increase MAP prox to clamp LVF, AV regurg increase baroreceptor firing -> decrease symp activity -> decr PVR, decr HR ``` Guyton p227 - constrictor placed on aorta above renal arteries causes renal blood pressure to drop, resulting in secretion of renin, and formation of angiotensin and aldosterone causing upper body hypertensin. The result of this is an increase in pressure below the clamp to nearly normal to alleviate the ischaemia, but the upper body continues to have a very high pressure. There's no timeframe given in Guyton for this compensation, so in a synthetic setting like cross-clamping, I'm not sure whether it would take.

Answer 271

Oxygen extraction is the amount in mls of oxygen taken up by the tissues per litre of blood. O2 extraction at rest can be elicited from the AV oxygen Difference column from Table 32.1 Ganong 22nd Edition: Heart (114ml/L) > Brain (62ml/L) > Skeletal muscle (60ml/l) During exercise there is an increased oxygen demand from active tissues. The heart and skeletal muscles are the most active tissues during exercise. As O2 extraction by the heart is already so high at 55-65% compared to average 25% for the body as a whole,the increased O2 demand can only be met by increases in blood flow.(Brandis) Blood flow to the myocardium and active muscles increase with exercise whilst blood flow to the brain is unchanged. O2 EXTRACTION from active tisues is increased by a right shift of the oxygen dissociation curve with increased CO2, decreased PH,increased 2,3 DPG and increased temperature (Bohr Effect). This is marked in the skeletal muscles which also increase their O2 uptake by opening closed capillaries by dilation of arterioles and precapillary spincters through local mechanisms. More open capillaries decrease the diffusion distance of O2 to mitochrondia. The myocardium cannot develop an oxygen debt BUT skeletal muscle can (with severe exercise). This means that the O2 extraction in the skeletal muscles is increased but still is not enough to supply O2 demand. So the O2 extraction in skeletal muscle with severe exercise is very likely going to be greater than that in the myocardium. So, skeletal muscle is the likely correct answer.. From Ganong p632 22nd ed: Skeletal muscle: AV O2 difference increases 3 fold whilst O2 consumption increases 100 fold. So given that at rest AV O2 difference for the heart is 55-65% at 114ml/L then at rest skeletal muscle has about 25-30% extraction percentage. If skeletal muscle increases its O2 extraction 3x then its extraction percentage is about 90%. "The oxygen dissociation curve is shifted to the right by the Bohr effect, resulting in 90% oxygen extraction from blood perfusing muscles at maximal exercise." Power and Kam. 2nd Ed. pg 371. Hence option C is best.

Answer 272

Comment I think Jul 03 - answer is anaemia. need greater oxygen extraction per unit volume blood due to decreased oxygen carrying capacity, therefore decreasing pvO2 I think the first question would be "none of the above", as most of them need the position of the ODC to change or you would not be able to have a change in oxygen extraction if you could not change position of ODC for a given CO, oxygen uptake, etc For the first question, the only thing that could change the mixed venous O2 content given the constants - would be arterial oxygen content (if oxygen uptake, and CO are constant, then by the fick principle, mixed venous content would only change for arterial content) if the question uses the term "tension", then B would increase the pvO2 for a given venous content - as it R shifts the curve (as with all the constants - venous content does not change) Comment re Jul 03 Anaemia would not be correct because O2 tension is not affected by Hb. A person can have normal O2 tension with Hb of 5. However Anaemia would lead to a decrease O2 CONTENT (not partial pressure). yes you are correct, anaemia does not alter partial pressure. But this only applies to arterial side. As oxygen content is decreased that is supplying the tissue, with no change in oxygen consumption, there will be much less content left in venous blood, which will result in decreased partial pressure in venous blood. The PO2 is related to the CONTENT of O2 through the ODC which is unchanged (as given in the question). Lower CONTENT equals lower PO2 because the other unchanged variables means the Hb oxygen complex will have a lower %O2 and if you follow the ODC at a lower saturation: a lower PO2. Which makes sense really. Comment re Jul 03 I am thinking maybe hypercarbia could cause it perhaps if due to increased metabolism (eg Malignant hyperthermia) then it stands to reason that there would be increased in Co2 production / o2 consumption and then this would indeed reduce mixed venous o2 sats. but i guess on the D day with all options available and correctly worded it would be hopefully easier to chose the one best answer! As it stands its probably B or D Hypercarbia without change in VO2 or Hb would lead to an increase in the PO2 from the right shift of the ODC. Without changing the ODC (!) yeah right I'd like to try controlling for that in that lab. IMHO these are not the same question and should be dissociated. From Peter Kam's NSW primary course the answer to the first question with oxygen content is E but it is annotated that the question was abandoned due to an unclear best answer.

Answer 273

Ganong 22nd Ed P605: Baroreceptors in carotid sinus and aortic arch - afferent fibers pass via glosopharyngeal and vagus nerves to medulla. (A false) Most end in Nucleus Tractus Solitarius (?C1 area) Probably secrete glutamate as excitatory transmitter (B false) Excitatory (glutaminergic) projections to ventrolateral medulla (E false) then stimulate GABA-secreting inhibitory neurons (D true) and so inhibits vasomotor centre outflow D true Unsure about C, any neuroanatomy guru's out there? Ganong 22nd Ed pg 605: With regard option C: Pain causes increased arterial pressure and afferents from exercising muscle exert pressor effect via the C1 neurons in the rostral ventrolateral medulla. The somatosympathetic reflex. I think C1 area corresponds to part of the rostral ventrolateral medulla responsible for sympathetic outflow (vasomotor centre). This means: C1 area is not the location of the primary synapse. Pathway goes something like this: Arterial baroreceptors -> IX (from carotid) or X (from aortic) -> NTS --- Glu ---> caudal ventrolateral medulla --- GABA ---> C1 area. The end result is that there is inhibition of SNS outflow from C1 area (RVLM) if arterial baroreceptors are stimulated. The role of pain and exercise is probably something like: mechanoreceptors and nociceptors from periphery exert an excitatory effect on C1. This effectively resets the arterial baroreflex set point. Thus, the baroreflex is still in-tact, but the set point is increased in situations such as exercise and pain.

Answer 274

Ganong 22nd Ed P605: Baroreceptors in carotid sinus and aortic arch - afferent fibers pass via glosopharyngeal and vagus nerves to medulla. (A false) Most end in Nucleus Tractus Solitarius (?C1 area) Probably secrete glutamate as excitatory transmitter (B false) Excitatory (glutaminergic) projections to ventrolateral medulla (E false) then stimulate GABA-secreting inhibitory neurons (D true) and so inhibits vasomotor centre outflow D true Unsure about C, any neuroanatomy guru's out there? Ganong 22nd Ed pg 605: With regard option C: Pain causes increased arterial pressure and afferents from exercising muscle exert pressor effect via the C1 neurons in the rostral ventrolateral medulla. The somatosympathetic reflex. I think C1 area corresponds to part of the rostral ventrolateral medulla responsible for sympathetic outflow (vasomotor centre). This means: C1 area is not the location of the primary synapse. Pathway goes something like this: Arterial baroreceptors -> IX (from carotid) or X (from aortic) -> NTS --- Glu ---> caudal ventrolateral medulla --- GABA ---> C1 area. The end result is that there is inhibition of SNS outflow from C1 area (RVLM) if arterial baroreceptors are stimulated. The role of pain and exercise is probably something like: mechanoreceptors and nociceptors from periphery exert an excitatory effect on C1. This effectively resets the arterial baroreflex set point. Thus, the baroreflex is still in-tact, but the set point is increased in situations such as exercise and pain.

Answer 275

D correct according to Ganong chapter 29 Specifics (Ganong 22 ed, p566) Right atrial systole PRECEEDS left - part A wrong Right ventricular contraction starts AFTER L, BUT as pulmonary pressure is lower than aortic pressure, right ventricular ejection starts first - D is correct During expiration aortic and pulmonary valves close at the same time, but during inspiration, the pulmonary closes later. This makes sense when you think about the negative pressure that inspiration generates in the thorax and hence the lungs and the capillaries that supply them. The dagram on p 567 of Ganong hasn't got a 'c' wave listed on the atrial pressure curve, but in Brandis p41, the atrial c wave is caused by isovolumetric contraction of the left ventricle, which causes a bulging of the mitral valve into the left atrium, and that causes a small pressure rise. When the aortic valve opens, the pressure drops off again. It comes well before peak aortic pressure, so C is wrong (seriously - a picture is worth a thousand words; go look at the Brandis diagram.) I think C would be closer to correct if it was talking about the JVP waveform rather than the atrial pressures Not if you look at your wiggers diagram which shows the c wave at the beginning of systole and the x descent at the peak ejection flow and pressure closer to the middle of the systolic period.

Answer 276

Seems lots of answers possible - B,C,D all true? Well, this part depends on how long your neck is. These people in the photo (http://www.suite101.com/view_image.cfm/190025) must have lower cerebral perfusion pressure!! If the head is say 30cm above the heart, the gravity induced decrease in pressure (Ganong says 0.77mmHg for each cmH2O above heart, but I read somewhere else it should be around 0.73 (? depends on temperature), but really around 0.77 x 30 = 23mmHg below aortic MAP; hence, Ganong 22th ed p. 630 says head MAP is about 60~70mmHg BUT that is for normal MAP say 93 (120/80), so someone with MAP 100 should really have head arterial pressure of 80mmHg (100- (0.77 x 30) = 76.9). So I wouldnt go for option B; Option C would be correct if it is ~95~100mmHg, but not 70mmHg, and option D looks about right: ie. 100mmHg - 23mmHg - normal ICP (10mmHg) = 67mmHg which roughly equals to D.

Answer 277

11/2/2011 -- it would be great if the examiner could indicate the specific height above and below the heart level for a more accurate answer... SYL Comments Answer to CV50b is A: Venous pressure in foot 70mmHg B. False: Normal CVP = 4.6mmHg [WG21:p597] C. False. Need to account for hydrostatic pressure References There is a good picture in Ganong (21st Ed) page 591 Fig: 30-17. --- that picture forgets valves.

Answer 278

Question CV51: Answer is D Comment on above answer: Q: But there is a change in Pressure, so isn't work done? Answer: No, there is energy used (hydrolysis of ATP) to cause the contraction of the ventricular muscle, but there is NO external work performed, as no volume is ejected from the ventricle. Work is only one form of energy. Part of the energy used during isovolumetric contraction is converted into heat energy (& lost) and some is converted into potential energy (ie the energy in the increased pressure) and subsequently some of this is converted into kinetic energy of the moving blood after the aortic valve opens. Moreover the myocytes change length and shape, so work is done to change the shape of the heart though the valves are closed and volume is the same. However, for work to take place a force must change its point of application. Whilst force is applied by increasing ventricular pressure, the point of motion does not alter until ejection occurs. - I wonder if the second questions actually says isometric (in term of skeletal muscle) - then the different answers would make much more sense...... comment Question CV51: Answer is D ? I dont think thats quite correct. aortic pressure should be increasing as well . fig 29-3 ganong 21st ed. chapter 29 , aortic pressure clearly increasing during isovolumetric ventricular contraction. Aortic pressure falls because blood is still running off to the peripheries. Pressure falls until the ventricular pressure rises above aortic & opens the valve. The 23rd edition of ganong fig 31-3 (same diagram?) shows the ventricular volume stable & then decreasing (ejection) after the aortic valve opens. Aortic pressure drops until this time (although there is a little squiggle when the valve is opening)

Answer 279

Question CV51b - Answer is D as Work = PxV = 0 and there is no volume change. Comment on above answer: Q: But there is a change in Pressure, so isn't work done? Answer: No, there is energy used (hydrolysis of ATP) to cause the contraction of the ventricular muscle, but there is NO external work performed, as no volume is ejected from the ventricle. Work is only one form of energy. Part of the energy used during isovolumetric contraction is converted into heat energy (& lost) and some is converted into potential energy (ie the energy in the increased pressure) and subsequently some of this is converted into kinetic energy of the moving blood after the aortic valve opens. Moreover the myocytes change length and shape, so work is done to change the shape of the heart though the valves are closed and volume is the same. However, for work to take place a force must change its point of application. Whilst force is applied by increasing ventricular pressure, the point of motion does not alter until ejection occurs. - I wonder if the second questions actually says isometric (in term of skeletal muscle) - then the different answers would make much more sense......

Answer 280

C is most correct. An increase in pCO2 of 1mmHg increases CBF by 4%. Therefore a 5mmHg increase would increase CBF by 20%. This relationship is linear between the pCO2 range of 20-80mmHg. The most important regulatory mechanism for CBF is CMRO2, which, like CO2, exhibit a linear relationship to CBF. The question is how much - and as the last option is quite vague, I agree with C, but if the wording is different in the exam, I'd go with the metabolic answer.

Answer 281

The carotid body is a chemoreceptor and senses pO2, pCO2 and pH. Many people get the roles of the carotid body & carotid sinus mixed up. Having both options here should alert you to the correct answer. Mnemonics: Your mileage may vary! * You put chemicals into your body * You put chemicals into your body, to stop the pressure in your sinuses! * The body doesn't belong in the barrow (baro) - sick I know but memorable * (Rock stars put chemicals in their sinus, but) anaesthetists put chemicals in the body! * To monitor pressure in a system, need to have a cavity which senses the stretch. Cavity = Sinus. Easy! * Carotid body, 2nd letter is O = oxygen (ie chemoreceptor). Carotid sinus, 2nd letter is I = intra-arterial pressure (ie baroreceptor) (Alistair)

Answer 282

``` BLOOD VOLUME 4% Aorta + arteries + arterioles 5% capilaries 67% venous system 12% pulmonary 5% heart From Brandis p.92: Veins - 65% Arteries - 13% Arterioles - 2% Capillaries - 5% "Central" - 15% Seems Brandis' arteries (13%) >> (aorta + arteries + arterioles) in first source. Maybe 4% aorta only? Power and Kam 2nd ed. p120 When supine: 75% systemic circ 16% pulmonary circ 8% heart ``` ``` 15% Systemic arteries + arterioles 6% Systemic capillaries 60% Systemic veins 8% Pulmonary arteries 3% Pulmonary capillaries 5% Pulmonary veins i.e. 23% arterial 9% capillaries 65% venous When erect: Heart and pulmonary circulations' volumes fall 6% & 9%, and vein volume increases. Doesn't quite add up to 100%, unfortunately! Answer is B Comment: not sure how you can state that so confidently, when there is nothing in any of the references about "large veins"... is the question misremembered? ```

Answer 283

Hydrostatic pressure relates to fluid pressure. Therefore hydrostatic pressure in your legs is higher than in your head. When considering hydrostatic pressures in the capillaries, it is affected by arterial and venous pressures and precapillary and postcapillary resistances. Hydrostatic capillary pressures at the arterial end are 32mmHg and venous end 15mmHg. Given Poiseulles Law for laminar flow Q= TT(P1-P2)r4/8nl where Q =volume flow rate, (P1-P2)= driving pressure, r = radius, n = viscosity, l = tube length and 8/TT is the constant of proportionality then resistance R = driving pressure (P1-P2)/Flow (Q)=8nl/TTr4 then resistance is inversely proportional to radius4 A. Arteriolar Constriction : Will lead to decreased radius which will increase the precapillary resistance. The volume of fluid flowing into the capillaries will be reduced. As pressures is inversely proportional to volume then capillary hydrostatic pressures will be reduced. A is incorrect. B. Venous Constriciton: Will lead to decreased radius which will increase postcapillary resistance. Less volume of fluid flowing out of capillaries means more volume remains therefore capillary hydrostatic pressure will be increased. B is correct. C. Capillary dilatation: Will lead to increased radius of the capillaries which will decrease capillary resistance. This may decrease driving pressure across the capillaries from artery to veins and if venous pressure stays the same P2 then capillary hydrostatic pressuresP1 must decrease.(Note: capillary diameter changes are passive and are caused by alterations in pre and post capillary resistance as capillaries have no smooth muscle and therefore cannot actively vasoconstrict. Their endothelial cells do have actin and myosin so can change shape in response to chemical stimuli.) C likely incorrect. Also note that Increases in arterial or venous pressures increase capillary hydrostatic pressures as it can be thought of as being the same column of fluid.Venous pressure increases and venous resistance increases have greater effect than arterial changes on capillary hydrostatic pressures. Berne and Levy p122-123,157,166-169 Hydrostatic pressure is different at different parts of a vessel/different vessels. I do hope the actual question has more details than this.

Answer 284

Standard ECG recording is at 25mm/s, and 1cm =1mV. 1 big box on x-axis = 0.2sec, and consists of 5 small boxes each 0.04secs 1 big box on y-axis = 0.5mV, and consists of 5 small boxes each 0.1mV Hint: 1 big box = 5mm = 0.5cm

Answer 285

Standard ECG recording is at 25mm/s, and 1cm =1mV. 1 big box on x-axis = 0.2sec, and consists of 5 small boxes each 0.04secs 1 big box on y-axis = 0.5mV, and consists of 5 small boxes each 0.1mV Hint: 1 big box = 5mm = 0.5cm

Answer 286

C) venous return via factors like the muscle pump The other factors facilitate the increased CO. Alternative View: Ganong 22nd Ed pg 634: "A great increase in venous return takes place, although the increase in venous return is not the primary cause of the increase in cardiac output" Stroke volume does not increase much in an untrained individual Therefore increase in heart rate is the best answer = A For an untrained person who has a resting heart rate of 72 beats/min and a stroke volume of 70 ml, the resting cardiac output is 5.04 L/min. During the initial stages of exercise, increased cardiac output is due to an increase in both heart rate and stroke volume. When the level of exercise exceeds 40% to 60% of the individual's capacity, stroke volume has either plateaued or begun to increase at a much slower rate. Thus further increases in cardiac output are largely the result of increases in heart rate. Another view Power & Kam "the abrupt changes [in cardiac output] at the onset ... of exercise are caused by the effect of the muscle pump resulting in an increased venous return, as well as motor cortical activity and sensory nerve activity associated with movement. The slower changes in cardiac output reflect the time course of vasodilatation in the muscles and stimulation of the cardiovascular system." My interpretation is that he mechanisms for increased Cardiac Output are, in order: Central stimulation by motor cortex Muscle pump Locally mediated (neural initially then chemically) vasodilation of skeletal muscle vessels increasing demand Increase heart rate & contractility by central stimulation Increased LVEDV Decreased LVESV So an untrained person still has a similar capacity to increase venous return, but has less capacity to increase demand in muscles (less V02 max), and less capacity to sustain increased heart rate, contractility or stroke volume changes (personal opinion). Answer C ANOTHER view As per Levy and Pappano (9th Ed. Pg. 244), "The large volume of blood returning to the heart is pumped so rapidly through the lungs that CVP remains essentially constant. Thus the Frank-Starling mechanism of greater initial fibre length does not account for greater SV in moderate exercise. Increases in SV due to the Frank-Starling mechanism only occur in very vigorous exercise, where RAP and LVEDV increase. Guyton & Hall (10th Ed. Pg. 976) states that "...heart rate increase accounts by far for a greater proportion of increase in cardiac output than does an increase in stroke volume during strenuous exercise..." by virtue of a 270% increase in HR relative to a 50% increase in SV. Consequently, knowing that untrained individuals undertaking exercise do not have the same increase in LV cavity size and mass seen in althletes, it seems likely that the correct answer is A - increase in heart rate. Comment: This is one of those stupid questions that is either actually stupid or was just remembered incorrectly - or hasn't been repeated since 2001 for a good reason. On the one hand, if you kept VR constant during exercise, there would be no increase in cardiac output despite increases in HR or stroke volume. CO has to equal VR, or else there will be dire consequences. On the other hand, we all know that CO = HR x SV and SV doesn't have as large a magnitude of increase during exercise as HR in an untrained individual. Thus A appears right, although it's right because C is right. Summary Cardiac output and venous return must have the same value in any reasonable timeframe, because the circuit is a closed loop. The heart acts as a "demand pump", and this term means that (within physiological limits of cardiac output) the heart pumps out each minute ( the "cardiac output") exactly the amount of blood that is returned to it ("venous return") each minute. So to make the statement that "Cardiac output increases becauses venous return increases" is hollow - it has no real meaning. Its like saying you get older because you get older, or because time passes. Guyton makes this point in various articles. Logically its the identity relationship: A = A. Cardiac output and venous return are just different terms for the same thing. The bottom line is that the tissues determine cardiac output, not the heart. If you stimulate the heart to increase heart rate (eg with a pacemaker) and/or increase stroke volume (eg by using inotrophs) then what happens is the cardiac output stays the same, that is equal to the total blood returned to it each minute from the tissues. Initially this can be a difficult idea to grasp or believe. Nevertheless the experiments have been done (see Guyton's mongraph & book) and thats exactly what happens. What happens is that each "tissue" (= localised collection of cells) has a requirement for a certain blood flow to carry out its functions. For most tissues the most critical flow-related metabolite is oxygen, so the oxygen requirement sets the degree of vasoactivity in the tissue's arterioles. If the tissue needs more oxygen (eg exercising muscle), then by local mechanisms, the result is vasodilation in the muscle arterioles, resulting in increased blood flow. The volume of venous blood draining from the tissues increases, so the amount of blood returning to the heart now increases. This is sensed by the heart and by an increase in either heart rate or stroke volume or both, the blood pumped out of the heart increases. Of course, volume of volume returned to heart equals volume of blood pumped out of the heart in each reasonable time period are equal. (This cannot be assessed instantaneously because the pump mechanism does not result in uniform flow, but (obviously) irregular flow. At any very small time interval, the heart is either ejecting blood or it is not and flow out of the heart is discontinuous. One thing not mentioned above about blood flow in tissues is the effect of vasodilation or vasoconstriction on the driving pressure. In simple terms, the blood flow through a tissue can be modelled using a vascular version of Ohm's Law, namely: flow = (inlet pressure) - (outlet pressure) divided by the resistance of the tissue vascular bed. Now if vasodilation occurs, the tissue vascular resistance decreases, so (by the formula) the flow increases PROVIDED that the driving pressure is not decreased. For this simple analysis it is usual to assume input pressure (arterial pressure) in high, and outlet pressure (venous pressure) is low so that that pressure gradient across the tissue can be approximated as equal to arterial pressure. So what we need for this local tissue control (working via varying vasoactivity causing changes in vascular resistance) to work is that the arterial pressure to stay constant. And this is basically what happens. So the equation reduces to: flow = k / tissue vascular resistance (where k = constant representing constant mean arterial pressure). There is now only one variable on the right hand side and this is under local control. So the tissue can control its own blood flow (by varying vasoactivity) independently of what is happening elsewhere in the body. The system that keeps the constant pressure head is the autonomic nervous system (ANS), and indeed is the primary function of the ANS. Now back to the heart. The answer to this MCQ is of course dependent on the wording of the question: If its about what "causes" an increase in cardiac output with exercise, then the answer is the increased muscle blood flow consequent upon tissue autoregulation of its blood flow. If its about what mechanism the heart uses to be able to increase its pumping performance to handle the increased amount of blood returned to it, then the options are increased heart rate, increased stroke volume, or one more than the other, or both equally. Of course remember that an increase in LVEDV will result in an increased stroke volume BUT this is energetically disadvantageous for the heart because of the Law of la Place. So Starling's law is quick to be activated and of course is good for balancing the output from the left and right heart but not a good primary mechanism. An increased heart rate with quicker ejection does not have the same energy disadvantage. So on this basis you would predict that an increase in HR would be 'better' than an increase in SV, except initially where the really quick onset of the Starling's law mechanism is advantageous. So with these background comments, carefully read the wording in the texts to better understand what is being said. In all these "remembered" versions of question it not at all important to get worried about the actual wording, as afterall you are unlikely to have it. The important thing is to understand what the question is about and then to understand the point of the question by reading up on the topic, and/or discussing it with colleagues (or online in the wiki or Bulletin Board), so that whatever the actual wording you will know the answer. Thats a much better situation to be in than having the real question and knowing that answer "B" (say) is correct.

Answer 287

C | The rest are purely short term regulators

Answer 288

Answer is C c wave corresponds to isovolumetric ventricular contraction, a phase that follows completion of ventricular filling but precedes ventricular ejection (Ganong 22nd pg569, & figure 29-3 pg 567) Ref: Brandis, p.41

Answer 289

B Cardiac muscle - initial depolarisation due to opening of voltage gated sodium channels similar to that occuring in skeletal muscle, initial rapid repolarisation is due to closure of sodium channels, subsequent prolonged plateau phase is due to slower but prolonged opening of voltage gated calcium channels.

Answer 290

I think many of the answers provided here are wrong because they forget that the questions are phrased in the negative. I would change them but now I'm too confused so will try to come back later unless someone gets there first. I think the main point is a poorly-compliant aorta increases pulse pressure by reduced dampening. CV61 A - most correct but not convinced B - incorrect; a stiff non-compliant aorta will widen the pulse pressure C - incorrect - need to decrease diastolic

Answer 291

Alt version D correct E correct for reasons above Basic definitions: The pulse pressure, the difference between the systolic and diastolic pressures, is normally about 50 mm Hg. The mean pressure is the average pressure throughout the cardiac cycle. Because systole is shorter than diastole, the mean pressure is slightly less than the value halfway between systolic and diastolic pressure. It can actually be determined only by integrating the area of the pressure curve however, as an approximation, mean pressure equals the diastolic pressure plus one third of the pulse pressure. Therefore factors that increase the differnece between the systolic and diastolic pressure (ie lowering the diastolic pressure or increasing the systolic pressure ) will increase the pulse pressure. Further Comment: CV 61 question asks which factors will narrow. ie decrease the pulse pressure A. incorrect- more rapid ventricular ejection will INCREASE the pulse pressure B. correct- increased aortic compliance willl DECREASE pulse pressure C. correct- increase in diastolic P will DECREASE pulse pressure increased Aortic compliance will decrease the pulse pressure. http://www.cvphysiology.com/Blood%20Pressure/BP003.htm (Good summary and intereasting diagram) Pulse pressure widened by increased SV increased contractility (which for a given preload/afterload, will increase SV) decreased aortic compliance Systolic pressure is related to SV and aortic compliance, while diastolic pressure is related to the TPR, the SV, the compliance... I think the answers are:- CV61 - A 1st alt - C, D, E (there is not just one, but E would be the best) 2nd alt - E Comment: I dont think you can increase the aortic compliance to reduce the pulse pressure unless you are releasing a cross clamp (or bursting an aneurysm) when the pulse pressure is going to go to $^&£ anyway. A classic cause for increased pulse pressure is however the poorly compliant aorta that comes with age/atherosclerosis. Perhaps that is the point?

Answer 292

Version 2 E correct Basic definitions: The pulse pressure, the difference between the systolic and diastolic pressures, is normally about 50 mm Hg. The mean pressure is the average pressure throughout the cardiac cycle. Because systole is shorter than diastole, the mean pressure is slightly less than the value halfway between systolic and diastolic pressure. It can actually be determined only by integrating the area of the pressure curve however, as an approximation, mean pressure equals the diastolic pressure plus one third of the pulse pressure. Therefore factors that increase the differnece between the systolic and diastolic pressure (ie lowering the diastolic pressure or increasing the systolic pressure ) will increase the pulse pressure. Further Comment: CV 61 question asks which factors will narrow. ie decrease the pulse pressure A. incorrect- more rapid ventricular ejection will INCREASE the pulse pressure B. correct- increased aortic compliance willl DECREASE pulse pressure C. correct- increase in diastolic P will DECREASE pulse pressure increased Aortic compliance will decrease the pulse pressure. http://www.cvphysiology.com/Blood%20Pressure/BP003.htm (Good summary and intereasting diagram) Pulse pressure widened by increased SV increased contractility (which for a given preload/afterload, will increase SV) decreased aortic compliance Systolic pressure is related to SV and aortic compliance, while diastolic pressure is related to the TPR, the SV, the compliance... I think the answers are:- CV61 - A 1st alt - C, D, E (there is not just one, but E would be the best) 2nd alt - E Comment: I dont think you can increase the aortic compliance to reduce the pulse pressure unless you are releasing a cross clamp (or bursting an aneurysm) when the pulse pressure is going to go to $^&£ anyway. A classic cause for increased pulse pressure is however the poorly compliant aorta that comes with age/atherosclerosis. Perhaps that is the point?

Answer 293

Adrenaline is given IV in ventricular fibrillation but only after CPR has commenced and cardioversion has been attempted twice. The role of adrenaline is to contract the peripheral circulation and increase the central aortic pressure to improve coronary perfusion. "Successful defibrillation largely depends on the following 2 key factors: duration between onset of VF and defibrillation, and metabolic condition of the myocardium." "VF waveform usually begins with a relatively high amplitude and frequency, and, then, it degenerates to smaller and smaller amplitude until asystole after approximately 15 minutes, possibly from depletion of the heart's energy reserves. Consequently, early defibrillation is vital; emergency response teams can perform defibrillation before arrival at the ED." "Defibrillation success rates decrease 5-10% for each minute after onset of VF. In strictly monitored settings where defibrillation was most rapid, 85% success rates have been reported." - from [1]

Answer 294

A. Decreased diastolic BP Systolic, diastolic and pulse pressures all diminish in severe haemorrhage. However in moderate haemorrhage, pulse pressure is reduced but mean arterial pressure may be normal. B. Increased serum ADH Correct. Vasopressin (ADH) is actively secreted by the posterior pituitary gland in response to haemorrhage. The plasma concentration increases as the arterial blood pressure diminishes. C. Increased pulmonary vascular resistance Baroreceptor stimulation (from decreased cardiac output) can dilate pulmonary vessels reflexly. Comment - Something's wrong here. Baroreceptor firing increases when BP falls, true. But surely this would dilate most vessels. Is this confusion caused by the negative nature of this reflex? D. Decreased cerebral blood flow Cerebral blood flow is preserved during mild to moderate haemorrhage. Cerebral vascular resistance may actually diminish in contrast to general peripheral vasoconstriction. E. Increased urinary sodium concentration Decreased renal blood flow raises the blood levels of angiotensin II. This accelerates the release of aldosterone which stimulates sodium reabsorption by the renal tubules. Therefore, urinary sodium concentration is decreased. Comment - but I thought the absolute concentration of sodium would increase as levels of ADH would be maximal thus concentrating urine maximally? Comment - I agree with the comment above, but I think that B is the most correct answer Comment - I thought diastolic BP increased with this level of hypovolaemia - but only supporting ref I have is Faunce 8e p.88. Normal urine sodium is 75-300mmol/24hr (so around 50-200mmol/L). In hypovolaemic shock without ATN, urine sodium should be less than 20mmol/L. (RCPA Manual). However, several references also say SIADH increases urinary sodium. I am thinking maybe it's a matter of degree - with 20% blood loss maybe the aldosterone effect overrides the ADH effect, whereas in SIADH maybe you don't get the increased aldosterone, plus there is a much more increased ADH?? Anyone have any thoughts about the pulmonary vascular resistance or is it a red herring?? If PA pressure increases resistance decreases because of recruitment and distension but is the reverse true? So I'm gunning for B as well.

Answer 295

Compensatory mechanisms include increase in cardiac output increase in oxygen extraction (increase in 2,3 DPG and ODC shifts to right) redistribution of blood flow Opinion: I personally cannot see how A (and therefore any of them) can be correct. Rearranging the Fick equation VO2 = CO(Ca-Cv) => Ca - Cv = VO2/CO. Therefore as your cardiac output increases (as it will to compensate) and your consumption stays the same your arterial-venous content difference must reduce. To take a more extreme example, imagine a situation where you have almost no haemoglobin. Ca will be very small and therefore Ca-Cv will also be very small. Agreed. There is a percentage increase in oxygen extraction and hence a percentage increase in difference between CaO2 and CvO2. However, because CaO2 is already reduced, there is not a nett increase in oxygen extraction by the tissues. As mentioned above, increased CO must mean a drop in CaO2-CvO2. I agree also that option A) is incorrect. You have both a lower arterial and venous oxygen carrying capacity, but there shouldn't be an increased a-v oxygen content difference. The amount of oxygen removed should be the same if oxygen consumption is the same. Increased O2 extraction as in the abillity to offload O2 yes, because to offload the same amount of O2 as a non-anaemic you'll end up with a lower ppO2. Thoughts anyone? my 2c - I would go with C as being correct. AV O2 content difference should be same (systemic O2 consumption largely unchanged) or decreased, because there is an increased CO to offset the increased extraction ratio. Option A is incorrect because it talks about "content" rather than "tension". B and E are also wrong, as I think that the CO increase is unlikely to overcompensate (venous O2 tension should be somewhat decreased, if anything). D is wrong, as we know that HR increases. The increased cardiac output means that there is more myocardial work required to be done - thus, oxygen consumption goes up overall. I think that C is a reasonable answer. Comment: I wonder if i am about to contradict something I have written for similar questions... but late in the evening here I think the answer will be increased PO2 = B. In anaemia there is an increase in CO, there is a decrease in viscosity, there is a decrease in CONTENT carrying capacity but no change in consumption (excepting that the heart may be consuming more oxygen but this does not feature in the text books). To facilitate oxygen delivery with the reduced CaO2 then there is an increase in 2,3DPG which right shifts the ODC. Referring to MCQ's that discuss ODC and PO2 will clarify that a R shift in the ODC will increase the PO2. Actually those MCQ's can be tricky to read. None-the-less: Decreased affinity for O2 on the Hb (consider a closed system - a sealed test tube of blood or a momentary isolated drop of blood in the vascular system - the O2 if it isn't on the Hb because of the 2,3DPG then its dissolved in plasma at an increased concentration = PO2). I should also note that the maximum oxygen carrying capacity of blood is thought to be at a haematocrit of 0.3 (stoelting or miller??) Increase in cardiac output and HENCE, increased myocardial O2 consumption, which makes C a correct answer. And because of increased O2 consumption, O2 content difference would decrease. SYL 12/2/2011 Alternative view: A. Possibly correct - Difference A-V O2 content will only change if the absolute amount of O2 extracted by tissues increases. In other words, change in O2 consumption, as may occur with increased HR. But doesn't this overlap with option C? B. Wrong - Increased extraction ratio will decrease venous pO2 C. Possibly correct - Oxygen consumption may increase due to elevated HR D. Wrong - Heart rate will increase due to auto-regulatory fall in TPR E. Wrong - aren't B and E identical options? Most correct answer: C (more correct than A because it offers a direction for the change) Comment: Pretty tricky - best way to rearrange Fick equation for this is - CvO2 = CaO2 - Vo2/CO Since CaO2 goes down (lower Hb), and maybe VO2 goes up as SYL said above due to increased CO, AND CO definitely goes up, we have all 3 factors on the RHS of the equation changing, so who knows what CvO2 does? If anyone can find a reference I would be grateful. The OHDC is certainly right shifted so that would argue for a greater PO2 as someone said above, all else being equal .. which of course is not the case.. Possibly just badly remembered and there is a good answer we will see on the day :)

Answer 296

Compensatory mechanisms include increase in cardiac output increase in oxygen extraction (increase in 2,3 DPG and ODC shifts to right) redistribution of blood flow Opinion: I personally cannot see how A (and therefore any of them) can be correct. Rearranging the Fick equation VO2 = CO(Ca-Cv) => Ca - Cv = VO2/CO. Therefore as your cardiac output increases (as it will to compensate) and your consumption stays the same your arterial-venous content difference must reduce. To take a more extreme example, imagine a situation where you have almost no haemoglobin. Ca will be very small and therefore Ca-Cv will also be very small. Agreed. There is a percentage increase in oxygen extraction and hence a percentage increase in difference between CaO2 and CvO2. However, because CaO2 is already reduced, there is not a nett increase in oxygen extraction by the tissues. As mentioned above, increased CO must mean a drop in CaO2-CvO2. I agree also that option A) is incorrect. You have both a lower arterial and venous oxygen carrying capacity, but there shouldn't be an increased a-v oxygen content difference. The amount of oxygen removed should be the same if oxygen consumption is the same. Increased O2 extraction as in the abillity to offload O2 yes, because to offload the same amount of O2 as a non-anaemic you'll end up with a lower ppO2. Thoughts anyone? my 2c - I would go with C as being correct. AV O2 content difference should be same (systemic O2 consumption largely unchanged) or decreased, because there is an increased CO to offset the increased extraction ratio. Option A is incorrect because it talks about "content" rather than "tension". B and E are also wrong, as I think that the CO increase is unlikely to overcompensate (venous O2 tension should be somewhat decreased, if anything). D is wrong, as we know that HR increases. The increased cardiac output means that there is more myocardial work required to be done - thus, oxygen consumption goes up overall. I think that C is a reasonable answer. Comment: I wonder if i am about to contradict something I have written for similar questions... but late in the evening here I think the answer will be increased PO2 = B. In anaemia there is an increase in CO, there is a decrease in viscosity, there is a decrease in CONTENT carrying capacity but no change in consumption (excepting that the heart may be consuming more oxygen but this does not feature in the text books). To facilitate oxygen delivery with the reduced CaO2 then there is an increase in 2,3DPG which right shifts the ODC. Referring to MCQ's that discuss ODC and PO2 will clarify that a R shift in the ODC will increase the PO2. Actually those MCQ's can be tricky to read. None-the-less: Decreased affinity for O2 on the Hb (consider a closed system - a sealed test tube of blood or a momentary isolated drop of blood in the vascular system - the O2 if it isn't on the Hb because of the 2,3DPG then its dissolved in plasma at an increased concentration = PO2). I should also note that the maximum oxygen carrying capacity of blood is thought to be at a haematocrit of 0.3 (stoelting or miller??) Increase in cardiac output and HENCE, increased myocardial O2 consumption, which makes C a correct answer. And because of increased O2 consumption, O2 content difference would decrease. SYL 12/2/2011 Alternative view: A. Possibly correct - Difference A-V O2 content will only change if the absolute amount of O2 extracted by tissues increases. In other words, change in O2 consumption, as may occur with increased HR. But doesn't this overlap with option C? B. Wrong - Increased extraction ratio will decrease venous pO2 C. Possibly correct - Oxygen consumption may increase due to elevated HR D. Wrong - Heart rate will increase due to auto-regulatory fall in TPR E. Wrong - aren't B and E identical options? Most correct answer: C (more correct than A because it offers a direction for the change) Comment: Pretty tricky - best way to rearrange Fick equation for this is - CvO2 = CaO2 - Vo2/CO Since CaO2 goes down (lower Hb), and maybe VO2 goes up as SYL said above due to increased CO, AND CO definitely goes up, we have all 3 factors on the RHS of the equation changing, so who knows what CvO2 does? If anyone can find a reference I would be grateful. The OHDC is certainly right shifted so that would argue for a greater PO2 as someone said above, all else being equal .. which of course is not the case.. Possibly just badly remembered and there is a good answer we will see on the day :)

Answer 297

QT interval is measured form the start of the q wave to the end of the t wave. Therefore neither of the answers above is correct. The QT interval is inversely related to the heart rate. Normal QTcorrected

Answer 298

The wedge pressure is essentially the same as the pulmonary artery occlusion pressure, and the tracing obtained is similar to that of a left atrial pressure tracing ie. a, c and v waves present (answer D) From St Georges' Hospital in NSW A normal PAOP tracing is shown in Figure 6. Note that, like a central venous pressure trace, 'A' and 'V' waves are present, but that a 'C' wave is usually not present. (Answer C) I disagree with the above response, and agree with the first (correct answer = D) - Ganong 22nd edn p.567 diagram has tracing of LA pressure with a, c, and v waves. It doesn't say this a a 'wedge pressure trace', but essentially the question is asking about the LA isn't it? I agree with the first response. See Miller's, 7e, figure 40-27. Pulmonary artery wedge pressure has a similar morphology to right atrial pressure, although the a-c and v waves appear later in the cardiac cycle relative to the ECG. Pulmonary Artery Wedge Pressure (PAWP) Normal 5-12 mmHg (mean pressure) When the balloon on the PA catheter is inflated, the catheter floats further into the pulmonary catheter until it wedges against the vessel, giving us a view of what's beyond. Because there are no valves between the pulmonary artery and the mitral valve, the PAWP allows us to look at the pressure in the left atrium (LAP). PAWP is the most accurate reflection of left atrial pressure, therefore of left ventricular end-diastolic pressure (LVEDP), or preload. Anything that affects preload, from total blood volume, to venous return, to compliance of the left ventricle and its ability to receive that volume, will affect PAWP. Like the RA waveform, this reflection of LAP consists of a waves and v waves (c waves are not usually present). see a picture of a PAWP trace at the following link from which the above was taken: http://faculty.weber.edu/jkelly/3040%20unit%207.htm

Answer 299

Left ventricular work per beat (stroke work) is generally considered to be equal to the product of stroke volume and the mean aortic pressure (afterload) against which the ventricle ejects blood. Increased preload (C), venous return (E) or ejection fraction (B) will all increase stroke volume and therefore stroke work. Increased contractility (A) also leads to increased myocardial oxygen consumption. Conversely increased aortic compliance (D) reduces mean aortic pressure or afterload and thereby stroke work is decreased. Answer = D Comment Increased compliance leads to more efficient conversion of intermittent LV output into continous flow. This reduces the stroke work- Levy & Pappano Cardiovascular Physiology 9th edition p127-130

Answer 300

The total pulmonary vascular resistance (and the mean pressure of the pulmonary circulation) is approximately 1/7th that of the systemic circulation. The arterioles are the major resistance vessels of the arterial circuit NOT the capillaries. The unit used to express resistance is dyne.seconds/centimeters^5 or answer D. Dimensional analysis can also give this result: Resistance = Pressure / Flow = Force / Flow.Area = dyne / (cm^3/sec).(cm^2) = dyne.sec / cm^5 Query Hey what's wrong with option E? CO = MAP / TPR, so if we know MAP and CO why can't we calculate TPR? I mean, I know the factors given is used to calculate systemic vascular resistance, but there IS enough information to calculate TPR, right? Comment: I always thought SVR and TPR were the same??? The question refers to Total Peripheral Resistance, which is (MAoP-RAP)/CO ie. E is wrong. Pulmonary vascular resistance would be (MPAP-PCWP)/CO. Answer must be D. Total peripheral resistance = Systemic vascular resistance + Pulmonary vascular resistance No it's not! TPR isn't pvr + svr, its just another name for svr

Answer 301

If you look at the Wigger's diagram in Ganong, the only really obvious things are that the P wave starts at the end of diastasis and the onset of isovolumetric contraction is at the peak of the R wave. so A is most right (or most non-committal as well) Additional comments Strictly speaking, A is incorrect, as isovolumetric contraction starts before the QRS complex is completed. See Guyton. However, it is hard to find a better answer. B is incorrect, as T waves start before closure of the aortic valve (during ejection). D is incorrect as the ST segment begins at the completion of isovolumetric contraction. E is incorrect as P waves occur when the mitral valve is already open in diastasis. F is incorrect as the left atrial V wave peak (LA filling while MV closed) corresponds to just prior to AV opening at the END of isovolumetric relaxation. Additional comments Number 2 Power and Cam's Wiggers on page 130: T wave starts AT closure of aortic valve - Actually, on further review, it does start before aortic closure.... wrong too...... OK all options are wrong. Damn Additional comments Number 3 I might be pushed to go for option B if this comes up in the exam. I can think of good reasons to reject all the others, whereas I can't think of a good reason to reject B (based on principles) - and it makes sense that T wave starts as the muscle begins to repolarise, and at this stage, contractile activity would begin to weaken (hence, relaxation). The T wave probably does start just before AV closure, but that wouldn't stop me from picking B as being superior to the other options. Alternative view Examination of Figure 4.6 in Power and Kam (2nd Ed) reveals that whilst contraction begins at the same time as the action potential (P-wave or QRS-complex), the actual peak force of contraction is not reached until AFTER the plateau phase. Hence, electrical events must precede mechanical events. Hence, we can surmise much from this, as well as examination of Figure 9-5 in Guyton (10th Ed). This suggests the following: Atrial contraction occurs AFTER the P-wave; Isovolumetric ventricular contraction coincides with the R-wave; The ST-segment corresponds with semilunar valve opening and rapid ventricular ejection; The T-wave begins when the ventricle begins to relax (i.e. start of reduced ventricular contraction); The v-wave peaks immediately before AV valve opening - the END of isovolumetric relaxation. Therefore, the correct answer is not listed

Answer 302

Please also refer to CV84 Radial artery peak and pulse pressures greater than aortic trace so A is wrong Radial artery systolic and diastolic pressures greater than aortic Radial artery trace delayed onset of upstroke, steeper upstroke and shorter duration versus aortic trace Radial artery trace does NOT have a dicrotic notch so D is wrong

Answer 303

Guyton P227: "one of the most common causes of renal hypertension, especially in older persons, is patchy ischemic kidney disease" --> the parchy bits secrete renin --> stimulates angiotensionogen to creat angiotensin I --> angiotensin II --> remaining kidney mass stimulated to retain all salt and water.

Answer 304

C-True-"Due to a slower but prolonged opening of L-type voltage gated ca+ channels" which is balanced by the efflux of K+ from inwardly rectifying K channels D-"Final repolarisation (phase 3) to the resting membrane potential (phase 4)is due to closure of the ca+ channels and k+ efflux through various types of k+channels, (especially delayed rectifiers - these open at the same time as everything else but takes a long time). Phase 0 -opening of voltage gated Na+ channels Phase 1-initial repolarisation due to closure of Na+ channels comment my understanding is that the plateau is due to balance of Ca influx and K efflux Ca influx - through L-type channels K efflux - through several K currents (inward rectifying, the rapid and slow delayed rectifiers, and possibly even the transient outward current)

Answer 305

``` CV73 A True (systolic pressure increases,diastolic pressure stays the same or decreases B False (Diastolic pressure remains the same or decreases (initially rises then falls in middle age due to increased stiffness of the arteries) C False (compliance decreases) D False (LV wall thickens at the expense of the LV Cavity ) E False (Heart rate decreases due to increased vagal tone and decreased sensitivity to andrenergic receptors ,heart rate decreases by one beat/min/year over age 50) ```

Answer 306

CV73b A. Correct - Wide pulse pressure B. Wrong - Decreased aortic elasticity C. Wrong - Decreased ventricular compliance D. Wrong - Decreased diastolic pressure (but increased pulse pressure) E. ?

Answer 307

Hypoxia in most tissues will increase blood flow via vasodilation except in the lung where it causes vasoconstriction. The mechanism is likely to be a direct effect shunting blood away from a hypoxic area. If, as in the case of an underventilated alveoli, CO2 accumulates and there is a local drop in the pH this will also produce pulmonary vasoconstriction as opposed to vasodilation produced in other tissues Related facts of interest halothane inhibits hypoxic pulmonary vasoconstriction, leading to an increased alveolar:arterial oxygen gradient. Nitroprusside can worsen arterial hypoxemia in patients with chronic obstructive pulmonary disease as it interferes with hypoxic pulmonary vasoconstriction Cardiovascular System. Hypoxia causes reflex activation of the sympathetic nervous system via both autonomic and humoral mechanisms, resulting in tachycardia and increased cardiac output. Peripheral vascular resistance, however, decreases primarily via local autoregulatory mechanisms, with the net result that blood pressure generally is maintained unless hypoxia is prolonged or severe Comment: But if all the other organs vasodilate because of hypoxia, doesn't the lung have to accept the entire cardiac output anyway? So although the regional distribution is changed, the total flow should increase. The kidney however, receives Oxygen delivery 90% in excess of its metabolic requirements so I would have thought thats the one that wouldn't increase? Comment 2: If the capillary bed with hypoxia was only in the organ of interest then I would go for Lung...but as suggested if global hypoxia then the lung gets all the blood. A study in rabbits with carbonmonoxide hypoxia showed " the greatest dilator effects being observed in the portal bed, followed by skin, kidney, and muscle." J Physiol. 1967 September; 192(2): 549–559. 1967...cutting edge stuff. I guess we have to figure are the examiners trying to catch those who don't remember about Hypoxic Pulm Vasoconst or that the lung recieves all the blood....hope it was a poor descriminator and dropped....I keep thinking that about every second question though... REVIEW OF MEDICAL PHYSIOLOGY Ganong GOODMAN & GILMAN'S THE PHARMACOLOGICAL BASIS OF THERAPEUTICS - 11th Ed. (2006) Another comment: Is this a slightly tricky question in that HPV is primarily in response to decreased ALVEOLAR PO2 and not to capillary PO2? In which case I would probably go for Kidney for the reasons given above - capillary O2 could fall but kidney would still receive enough because of huge flow. I agree that Alveolar PO2 is the main determinant of HPV, but Nunn clearly states that it is also mediated by mixed venous (pulmonary arterial) PO2 and has a nice graph to demonstrate the relative contributions (6th ed p 101) - I would definitely go with Lung as the best answer, although would agree that the kidney (and probably also the liver) wouldn't change much. Lungs get whole CO and HPV is in response to alveolar hypoxia, I see what you mean about kidney but maybe Power and Kam 2008 p 215 "given it's high blood flow it's not surprising that any increase in oxygen demand is met by increased extraction rather than increased blood flow" also "graded hypoxia reduces hepatic arterial flow initially b4 returning to baseline and has minimal effect on portal vein flow" I am pretty sure the kidneys don't decrease their blood flow in response to hypoxaemia. I'm going for A. (also hate this question for wasting my study time with multiple checking of resources." I think everyone is overanalysing this question - the question asks about *blood flow* to *whole organs*. The lungs receive the whole venous return. The fraction of blood flow to the lungs cannot be altered. Of course there may be regional changes in pulmonary blood flow, but this is not what the question asks. Answer E

Answer 308

``` A - True (produced by platelets) B - False C - False D - False E - We'll take your word for it. :-) ```

Answer 309

CORRECT ANSWER IS C Hagen-Poiseuille Equation Q = P∏r^4/8nl Q = flow P = pressure difference r = radius n = viscosity l = length If diameter (therefore radius) decreases, so will the flow. Axial streaming will result in the viscosity dropping, so the flow will not decrease as much as expected, but it WILL decrease. You just can't get MORE fluid down SMALLER pipe. ANSWER A is WRONG Blood is a non-Newtonian fluid. ANSWER B is WRONG (A newtonian fluid is one of uniform composition, and I think it's actually a fluid where it's shear rate is proportional to it's shear stress) If viscosity decreases, flow will increase. ANSWER C is CORRECT Altenate view 1 From Levy and Pappano CVS Physiology (9th ed) pp117 - 122: Blood flow is nonnewtonian in very small vessels ie Poiseuille's law is not applicable. The apparent viscosity of blood diminishes as shear rate (flow) increases and as the tube dimension decreases. This makes the correct answer A Alternate view 2 I agree, the answer would be "A" as the capillary flow is non-Newtonian. Kam says: The apparent viscosity of blood falls progressively when the vessel diameter is reduced below 300 mcm. In capillary-sized tubes (6 mcm diameter) the viscosity of blood is as low as that of plasma. The red cells tend to occupy the axial, central, fast-moving stream, whilst the plasma flows slowly through the small vessel in the marginal streams. BACK TO FIRST VIEW ======in case of anaemia ,viscosity will decrease ,thats physiological advantage ,so resistance decrease and flow increase,i agree on C ,examiners aiming to test weather u know effect of viscosity on resistance and flow. In Answer to the above please see Ganong 22nd ed Fig 30-13 page 587 which shows velocity is lowest at capillary level. Therefore answer is C which holds true regardless.

Answer 310

Answer is A Afterload is the slope of the straight line connecting the LV End Diastolic Volume on x-axis and the End Systolic Point on the Pressure- Volume loop. Contractility is the slope of the straight line connecting the End Systolic Point to the x-axis. Again this is another gradient. Aortic valve opens at the diastolic BP, not closes. = aortic valve opens during ejection when pressure inside ventricle reaches just above diastolic ,though coming backward once the ejection slows and pressure inside ventricle is falling once reach diastolic ,aortic valve closes. so for the current warding that option can be true

Answer 311

Answer is A Afterload is the slope of the straight line connecting the LV End Diastolic Volume on x-axis and the End Systolic Point on the Pressure- Volume loop. Contractility is the slope of the straight line connecting the End Systolic Point to the x-axis. Again this is another gradient. Aortic valve opens at the diastolic BP, not closes. = aortic valve opens during ejection when pressure inside ventricle reaches just above diastolic ,though coming backward once the ejection slows and pressure inside ventricle is falling once reach diastolic ,aortic valve closes. so for the current warding that option can be true

Answer 312

omment 1 Not a good question or reproduction of a question as the performance of different types of exercise will have different effects. In general impulses to the vasomotor centre ↓ parasymp outflow and ↑ sympathetic discharge thus ↑ myocardial contractility, ↑SV, ↑HR and CO. The Baroreceptor reflex helps to buffer changes in BP resulting from ↑ CO and ↓ SVR in dynamic exercise. In exercise where perfusion of muscles is limited or absent (s.a sustained static contraction or work involving holding the arms above your head, vasodilating factors accumulate, stimulating sesory nerves and trigger the vasomotor centre to produce ↑ rise in BP. In isometric exercise (tonically contracting muscles) HR and SVR ↑; In isotonic (perf of ext work) HR ↑ SVR ↓ Net Effects can be summarised as; ↑ SV ↑ HR ↑CO ↓ SVR (↑in isometric) modest rise in BP Diastolic BP remains unchanged or falls Comment 2 Looking more broadly at CVS Changes in Exercise. There is an initial phase related to the Muscle Pump (increased VENOUS RETURN) and Motor Cortical Activity (SNS) and then a slower phase related to locally and centrally mediated vasodilation (increased DEMAND) and alterations in Regional Blood Flow. I attempt to summarise as follows (see Ref. Kam): Initial increase in Cardiac Output Increased venous return due to muscle pump Increased venous return due to venoconstriction Increased demand due to vasodilation Increased myocardial contractility and heart rate due to increased SNS outflow. Sustained by the associated decrease TPR. (as below) Changes in regional circulation Increased muscle blood flow due to neurally mediated relaxation of precapillary sphincters, and further arteriolar vasodilation due to local metabolites and neural factors (β1). Associated decrease in TPR. Vasoconstriction of splanchnic and renal circulations. Increased skin blood flow to dissipate heat. Increased Coronary Blood flow mediated by metabolic autoregulation, and β2 stimulation. No change in global Cerebral Blood Flow, but may be regional variations. Changes in Cardiac indices Heart rate increases linearly. Systolic Blood Pressure increases nonlinearly with increasing intensity of exercise. Diastolic Blood Pressure may show a small increase initially or at low levels of exercise but with increasing duration or intensity it will show a decrease. There is a nonlinear increase in Stroke Volume due to increased venous return and as such most of the increase occurs during light or moderate exercise (presumably most of this is due to effect of muscle and thoracic pumps).

Answer 313

Blood flow increases only to those muscles exercising. Splanchnic flow decreases thus A best. (Ganong). See also CV78 Comment 1: And SVR does increase in isometric exercise. 11/2/2011 -- Based on that Table 33-2 from Ganung [sic - ?Edition], doesn't it also show that skin blood flow is the same? SYL In Reply: I think this table has been removed from 23Ed - can't find any such table. In any case, "When the body temperature rises during exercise, the cutaneous blood vessels dilate in spite of continuing noradrenergic discharge in other parts of the body" Ganong pp581 (23Ed).

Answer 314

Hypokalaemia is correct Does anyone else think that the description of "ST segment flattening" is rather odd? Did they mean ST segment depression? A. Wrong - nil U wave B. Correct - U waves, PR prolongation C. Correct if the option was hypomagnesaemia - U waves, sometimes PR prolongation D. Hypocalcaemia - main feature is prolonged QT interval E. None of the above "Most correct" answer 'B'

Answer 315

A: wrong - is earlier than this (Ref: Berne and Levy Ed 9 pp 67 "Wiggers diagram") B: maybe.. but seems to coincide on the diagram with MV closure, not come afterward C: wrong - R wave is after this D: most likely

Answer 316

There are no transient inward currents (but there is transient outward K current) Berne and levy

Answer 317

Wonder if there was an option about slow repolarisation of papillary muscles/Purkinje fibres? That is the most common one I have seen.. but the paper I link to below lists 4 potential aetiologies Ganong 23ed Ch 30: The U wave is an inconstant finding, believed to be due to slow repolarization of the papillary muscles.

Answer 318

A - correct, see below B - false C - false, radial has higher peak D - correct, see below E - this is false, delayed due to distance from heart this quote from Miller e-version - chapter on CV monitoring "Thus, when compared with central aortic pressure, peripheral arterial waveforms have higher systolic pressure, lower diastolic pressure, and wider pulse pressure. Furthermore, there is a delay in arrival of the pressure pulse at peripheral sites, so the systolic pressure upstroke begins approximately 60 msec later in the radial artery than in the aorta. Despite morphologic and temporal differences between peripheral and central arterial waveforms, MAP in the aorta is just slightly greater than MAP in the radial artery." So I guess one of these options badly remembered

Answer 319

This is answered from one paragraph in Ganong. Ed 21, pp637 "training" A: Wrong - sounds plausible in that although HR = 220 -age, the ability to actually get there (to max HR) would require some training, but not best answer. B: Yes - "both at rest and at any given level of exercise, trained athletes have a larger SV and lower heart rate than untrained individuals" C: No - "increased number of capillaries with better distribution of blood to the muscle fibres" D: Unsure.. thre is a statement "less increase in blood flow to muscles as well, and because of this, less increase in HR and CO than untrained individuals" B still simplest and best answer though, given uncertain wording from remembered Q E: "..less increase in lactate production:

Answer 320

A: no, less than 200ms B: true - there can be a q-wave in any lead that is not pathological C: true - lead II needs a right arm, left leg, and an earth electrode, right leg D: false - lead II uses right arm and left leg leads I lean towards C... thoughts? ALternate: A: Wrong - less than 200ms B: Correct - non-pathological Q-waves can exist in a lead, provided contiguous leads do not have associated Q-waves C: Wrong - lead II needs a right arm, and left leg electrodes D: Wrong - lead II uses right arm and left leg leads I would answer B.

Answer 321

Conductance is the reciprocal of resistance NOT A Factors affecting resistance can be described by the Hagen Poiseuille equation with resistance proportion to 1/radius to the power of 4 Not B Reciprocal of the pressure difference - NOT C Resistance in parallel are added as 1/Rtotal = 1/R1 + 1/R2 etc therefore total conductance = the addition of individual conductance in parallel - E is correct

Answer 322

ANSWER B 'This circulatory shock syndrome caused by inadequate cardiac pumping is called cardiogenic shock or simply cardiac shock.' (Guyton, 11th ed. p263) Also it says in Guyton. Definition of cardiogenic shock: the circulatory shock that results from diminished cardiac pumping ability. Causes: MI, toxic states of the heart, severe heart valve dysfunction, heart arrhythmias. Could B be true instead?

Answer 323

"Indirectly proportional" means CBF decreases with increased [glucose]. Thats wrong. It can't be the other way around either as the brain does not set the plasma [glucose]. Cerebral metabolism uses carbohydrate (as an energy substrate) almost exclusively, so the brain's RQ = 0.99 Metabolic regulation => increased CBF with increased metabolic activity. All neuronal activity is aerobic. This flow-metabolism coupling is a form of "autoregulation" (as the control occurs locally), though when referring to the brain the term 'autoregulation' is predominantly used to refer to pressure autoregulation. Cerebral perfusion pressure (CPP) = MABP - ICP (if ICP > CVP) or Cerebral perfusion pressure = MABP - CVP (if ICP

Answer 324

This MCQ is MUST have been remembered incompletely BECAUSE the answer depends on the level of activity, as indicated in the abstract below. Noddeland H et al The pressure in the saphenous vein of the lower leg was monitored by ordinary ECG telemetry equipment supplied with a pressure transducer, a chopper and an extra filter. In eight healthy volunteers venous pressure averaged 80 +/- 5 (SD) mmHg in the standing position and 21 +/- 10 (SD) mmHg during slow walking. When the subjects were occupied with laboratory work in upright position the mean venous pressure was 40-50 mmHg, similar to that obtained sitting at a desk: 48 +/- 5 (SD) mmHg.

Answer 325

Answer to KD01a is A As the macula densa brings about the other changes which result in alteration to renal blood flow. Arterial pressure although important is autoregulated within a moderate range of systemic blood pressure (~80mmHg - 200mmHg). see Ganong fig 38-4. I find KD01a problematic: RBF is dependent on MAP and afferent and efferent resistance. It is autoregulated, although not perfectly. The JGA is a feedback loop which involves the macula densa (load detector: Na and Cl) communicating to the granular cells of the afferent arteriole to cause vasoconstriction (mediator: adenosine, purinergic receptors), and also release or reduce release of renin from these same cells. Thus, MAP, afferent vasodilation, efferent vasoconstriction all affect RBF. I guess that the JGA is the only one that coordinates these changes so is the "best" answer.

Answer 326

The Answer to KD01b is B A is WRONG: the kidney recieves 1.2-1.3Lof blood per minute. However numbers in A are correct for the effective renal plasma flow which is measured by PAH. Comment: The question says per kidney - there are two. B is RIGHT: p-aminohippuric acid (PAH)is used to measure renal blood flow as it is filtered by the glomeruli and secreted by the tubular cells, it has a high extraction ratio. Effective renal plasma flow is the amount of PAH inthe urine divided by the plasma PAH level. The renal blood flow is then calculated from this by dividing the ERPF by 1 - Hct. (page 704 in ganong) C is WRONG: sympathetic stimulation causes a decrease in renal blood flow, mediated by alpha1 and alpha 2 adrenergic receptors. I disagree and think A is correct - the answer is 600-650ml/min PER kidney for a total of 1.2-1.3L/min as is correctly stated above from Ganong. B and measurement by PAH is an INDIRECT but clinically useful measure. I agree with last statement. Vander states that "at low plasma concentrations, about 90% of the PAH entering the kidney is removed from the plasma and excreted in the urine...used as a measure of effective renal plasma flow". RBF = 1.1 L/min RPF = 605 mL/min i.e. 1100 x (1-Hct) ``` Juxtaglomerular cell Juxtaglomerular cells (JG cells, also known as granular cells) are the site of renin secretion. The JG cells are found in the afferent arterioles of the glomerulus and act as an intra-renal pressure sensor. Lowered pressure leads to decreased pressure on the JG cells, allowing them to swell. This swelling increases intracellular levels of cAMP which stimulates PKA which causes the secretion of renin. Renin then acts to increase systemic blood pressure (while maintaining GFR) via the renin-angiotensin system. ``` Macula densa The macula densa senses sodium chloride concentration in the distal tubule of the kidney and secretes a locally active (paracrine) vasopressor which acts on the adjacent afferent arteriole to decrease Glomerular Filtration Rate (GFR). Specifically, excessive filtration at the glomerulus or inadequate sodium uptake in the proximal tubule / thick ascending loop of Henle brings fluid to the distal convoluted tubule that has an abnormally high concentration of sodium. Na/K/2Cl cotransporters move sodium into the cells of the macula densa. The macula densa cells have an inadequate number of Na/K ATPases to excrete this added sodium, so the cell's osmolarity increases. Water flows into the cell to bring the osmolarity back down, causing the cell to swell. When the cell swells, a stretch-activated non-selective anion channel is opened on the basolateral surface. ATP escapes through this channel and is subsequently converted to adenosine. Adenosine causes constriction of the vascular smooth muscle cells of the afferent arteriole, reducing the amount of blood that reaches the nephron.

Answer 327

Answer to KD01c is A Renal cortical blood flow is 5mL/g/min Renal outer medulla blood flow is 0.6ml/g/min Renal inner medulla blood flow is 0.6ml/g/min Cerebral blood flow is 0.5mL/g/min ``` Juxtaglomerular cell Juxtaglomerular cells (JG cells, also known as granular cells) are the site of renin secretion. The JG cells are found in the afferent arterioles of the glomerulus and act as an intra-renal pressure sensor. Lowered pressure leads to decreased pressure on the JG cells, allowing them to swell. This swelling increases intracellular levels of cAMP which stimulates PKA which causes the secretion of renin. Renin then acts to increase systemic blood pressure (while maintaining GFR) via the renin-angiotensin system. ``` Macula densa The macula densa senses sodium chloride concentration in the distal tubule of the kidney and secretes a locally active (paracrine) vasopressor which acts on the adjacent afferent arteriole to decrease Glomerular Filtration Rate (GFR). Specifically, excessive filtration at the glomerulus or inadequate sodium uptake in the proximal tubule / thick ascending loop of Henle brings fluid to the distal convoluted tubule that has an abnormally high concentration of sodium. Na/K/2Cl cotransporters move sodium into the cells of the macula densa. The macula densa cells have an inadequate number of Na/K ATPases to excrete this added sodium, so the cell's osmolarity increases. Water flows into the cell to bring the osmolarity back down, causing the cell to swell. When the cell swells, a stretch-activated non-selective anion channel is opened on the basolateral surface. ATP escapes through this channel and is subsequently converted to adenosine. Adenosine causes constriction of the vascular smooth muscle cells of the afferent arteriole, reducing the amount of blood that reaches the nephron.

Answer 328

``` Juxtaglomerular cell Juxtaglomerular cells (JG cells, also known as granular cells) are the site of renin secretion. The JG cells are found in the afferent arterioles of the glomerulus and act as an intra-renal pressure sensor. Lowered pressure leads to decreased pressure on the JG cells, allowing them to swell. This swelling increases intracellular levels of cAMP which stimulates PKA which causes the secretion of renin. Renin then acts to increase systemic blood pressure (while maintaining GFR) via the renin-angiotensin system. ``` Macula densa The macula densa senses sodium chloride concentration in the distal tubule of the kidney and secretes a locally active (paracrine) vasopressor which acts on the adjacent afferent arteriole to decrease Glomerular Filtration Rate (GFR). Specifically, excessive filtration at the glomerulus or inadequate sodium uptake in the proximal tubule / thick ascending loop of Henle brings fluid to the distal convoluted tubule that has an abnormally high concentration of sodium. Na/K/2Cl cotransporters move sodium into the cells of the macula densa. The macula densa cells have an inadequate number of Na/K ATPases to excrete this added sodium, so the cell's osmolarity increases. Water flows into the cell to bring the osmolarity back down, causing the cell to swell. When the cell swells, a stretch-activated non-selective anion channel is opened on the basolateral surface. ATP escapes through this channel and is subsequently converted to adenosine. Adenosine causes constriction of the vascular smooth muscle cells of the afferent arteriole, reducing the amount of blood that reaches the nephron.

Answer 329

Answer to KD02 is A as PAH is both filtered and secreted For those, like me, who had no idea what 'PAH' is: p-amino-hippuric acid (PAH) is used to measure renal plasma flow due to its high renal extraction ratio (90%).

Answer 330

Answers to KD03 are C and E The ascending limb of the LoH is impermeable to water. However it has a number of transporters on its apical (luminal) surface which allow movement of electrolytes. From Ganong fig 38-6: Na+/K+/2Cl- channel - Na+, K+, Cl- uptake Na+/H+ exchanger - Na uptake, H+ extrusion K+ channels (ROMK) - K+ extrusion (recycling) Comment - I don't think the Na/K/2Cl transporter is active, only the Na/K ATPase on the basal membrane is active - thus E is the only correct answer Additional comment: As the question is worded, C is also correct. Remember that there are two types of active transport, primary & secondary. The Na/K/2Cl cotransporter mediates secondary active tranport of Cl (and K). (See Ganong 22nd ed. p. 711-2, 714-5 & Fig. 1-33 p. 35.) Additional comment: Also although only the basolateral Na/K/ATPase utilises ATP the Na/K/2Cl transporter is still "active" as it utilises the sodium gradient created by the Na/K/ATPase - secondary active transport. I agree that both C and E are correct.

Answer 331

answer to KD 04 is C Glucose undergoes secondary active re-uptake predominantly from the early proximal tubule. Both the glucose and Na+ bind to SGLT 2 receptors and are transported from the tubule lumen into the cell. The transport maximum (TmG) for glucose varies between nephrons. This results in "splay" as seen in Ganong fig 38-10. SGLT2 transporter has increased avidity to D-glucose isomer, than L-glucose

Answer 332

GFR - Glomerular filtration rate 180 l/day 125 mls/min typically urine 1-2 litres/day so approx 99% of the filtered water is reabsorbed in the renal tubules. (If only 90% was reabsorbed, then the urine volume would be 18 litres/day. (This level of urine production is pretty close to the maximum value found in diabetes insipidus.) As regards the molecular size of drugs: "The molecular size of drugs varies from very small (lithium ion, MW 7) to very large. However, the vast majority of drugs have molecular weights between 100 and 1000." Lower limit of 100 set by need for molecule to be able to achieve selective binding to receptors The upper limit of 1000 is "determined primarily by the requirement that drugs be able to move within the body (eg from site of administration to site of action). Drugs much larger than MW 1000 will not diffuse readily between compartments of the body." (from Katzung 7th ed) As regards glomerular filtration of drugs: All substances free in the blood stream will be filtered at ther glomerulus unless restricted (eg by large MW or by plasma protein binding). Substances up to MW 68,000 can be freely filtered though the charge on the molecule can affect this somewhat. As most drugs have MW

Answer 333

Several comments received re July 2000 paper: KD06 Reworded: The excretion of a substance that is freely filtered and secreted up to its maximum threshold is represented by: Note: Please draw new curve (A-F) consisting of a line of high gradient which then sharply changes into a line of lesser gradient (but not zero). AND Which one is filtered and actively secreted at kidney 4 graphs a-b,a-c, a-d, a-e and none of above. Correct answer corresponds to Ganong graph of PAH clearance a-d I think AND For a substance filtered and secreted by a saturable transport mechanism the correct curve is: four curves including a straight line, logarithmic washin curve, straight line then flat, straight line then abrupt change to less steep straight line. See page 712 of Ganong 22nd edn

Answer 334

Renal Handling of Urea Urea is a small (MW 60d), water soluble molecule, freely filtered About half filtered urea is reabsorbed in the proximal tubule An amount equal to that reabsorbed is then recirculated back into the descending loop of henle (via Urea Transporter UT family) From thick ascending limb, to medullary collecting duct, luminal urea permeability is essentially zero Finally, ~half the urea is reabsorbed again in the medullary collecting duct (this accounting for part of the medullary osmotic gradient and being the source of interstitial urea which is secreted into the descending loop of henle)

Answer 335

Answer is B. Word for word in Ganong: "The permeability of the glomerular capillaries is about 50 times that of the capillaries in skeletal muscle".

Answer 336

Arterial BP can stimulate R-A system. Thus answer is D D: Na content in distal tubule TGF is determined by Macula densa. This is at juntion of Prox Convoluted tubule and Ascending loop. Not in DCT. vander (Different texts refer to MD as being at the end of the LOH or at the start of the DCT. The Examiner's have noted that previously, and take either as correct.) Macula densa is part of JGA, I thought it was at the end of the ascending loop at the junction with DCT. Power and Kam says ascending LoH, Ganong 20e also says this, but then has a diagram with it labelled as distal tubule ;( Don't think the prox. tubule is involved at all. I would be tempted to say B on the basis that arterial pressure only influences GFR outside the autoregulatory range, and that you could say that the Na content arriving at the DCT (ie. at end of LoH) is effectively the content arriving at the MD. Thoughts?? I would have to go with D as well, although it's pretty nearly true, but the macula densa is just BEFORE the DCT. "The sensor (for TGF) is the macula densa. The amount of fluid enetering the DCT at the end of the thick ascending limb of the loop of Henle depends on the amount of Na and Cl in it" see Ganong 22nd ed pg 713 I'd go B as within normal limits GFR is independant of arterial BP. Kerry Brandis had a question on a multichoice at this years short cse (2007) regards this. He said the Macula Densa in in the first part of the DCT. Some people tried to argue the point but he seems to know his stuff.. ;)) arterial BP is also involved via myogenic control (additional to RAAS as above) Therefore D is MORE wrong and therefore the answer I´d say Table 38-4. (Ganong)Factors affecting the GFR. Changes in renal blood flow Changes in glomerular capillary hydrostatic pressure Changes in systemic blood pressure Afferent or efferent arteriolar constriction Changes in hydrostatic pressure in Bowman's capsule Ureteral obstruction Edema of kidney inside tight renal capsule Changes in concentration of plasma proteins: dehydration, hypoproteinemia, etc (minor factors) Changes in Kf Changes in glomerular capillary permeability Changes in effective filtration surface area Does arterial pressure really "regulate" GFR?? I think all the answers are factors which affect the GFR in certain circumstances, but in terms of true "regulation", I don't think arterial pressure does If arterial pressure changes, all the other factors will be involved in autoregulation of GFR I say B Na content in the distal tubule is the right answer I think. The macula densa is in the ascending thick loop of the LOH (see Ganong and Power/Kam). Therefore by the time the sodium reaches the distal tubule the ship has sailed! Not sure what is all the fuss about where the macula densa is -- thick end of ascending limb of LoH or distal tubule... does it really matter??? Because the answer up there is "Na content in distal tubule" -- essentially the Na content will be similar to the thick end of ascending limb for the transition to distal tubule, but the ultimately once the macula densa senses the Na content, autoregulation results. So for arguement sake if the macula densa is truly at the thick end of ascending limb of LoH, and the Na content is high in distal tubule, TGF still occurs, so the answer is true. I don't really believe BP per se autoregulates GFR -- it is all the other mechanism that autoregulates BP, or RBF if you like. Hence the answer should be B. For all those arguing that BP doesn't "regulate" GFR per se, I would like to point out that this question is actually worded as "involved in regulation". What does "involve" mean? Changes in BP would trigger autoregulation, does this constitute as involvement? Are you involved in a murder if you pull the trigger? Or is the victim the only person actually involved? I think the answer is clear. This question represents a lack of understanding on the examiner's part regarding the differences in various textbooks on the location of MD, and is coupled with poor wording with general terms like "involved in".

Answer 337

The permeability of gomerular capillaries approaches 100% for 4nm neutral or cationic molecules. Also: "The permeability of the glomerular capillaries is about 50 times that of the capillaries in skeletal muscle" (from Ganong). Ganong 21st Ed page 710: Proteins in the glomerular capillary wall are negatively charged..... (so) repel negatively charged substances. Thus filtration of anionic substances is less than that of cationic substances. So this option is wrong in both stems. Also: substances with molecular diameters 8nm not filtered and in between variable.

Answer 338

The permeability of gomerular capillaries approaches 100% for 4nm neutral or cationic molecules. Also: "The permeability of the glomerular capillaries is about 50 times that of the capillaries in skeletal muscle" (from Ganong). Ganong 21st Ed page 710: Proteins in the glomerular capillary wall are negatively charged..... (so) repel negatively charged substances. Thus filtration of anionic substances is less than that of cationic substances. So this option is wrong in both stems. Also: substances with molecular diameters 8nm not filtered and in between variable.

Answer 339

Maximum urine osmolality in a young healthy person is quoted as 1,200-1,400 mosmoles/kg. Minimal urine osmolality (in absence of ADH) is 30-60mOsm/kg "The urine osmolality is a measure of the concentration of the urine and is primarily determined by the level of antidiuretic hormone (ADH). In patients with normal renal function, the urine osmolality can range from a minimum of 50 to 100 mosmol/kg in the absence of ADH to a maximum of 900 to 1200 mosmol/kg with peak ADH effect." (from Up-to-Date reference below) Is it possible that the examiners could be so excrutiatingly petty to have made this a trick question with regard to the units?? Comment: there is nothing wrong to the unit because mosm/kg is essentially equal to mosm/l, if water is what we are talking about. Osmolarity - mOsmol/L... Number of Osmole per liter of solution. Affected by temperature. Osmolality - mOsmol/Kg... Number of Osmole per kg of solventy. Not affected by temperature. Nah.. A is correct

Answer 340

Answer : most significant is answer A. 85-90% of filtered phosphate is reabsorbed. Ganong p. 383 Urea - 53% reabsorbed. Creatinine - nil reabsorbed. Sulphate is also significantly reabsorbed according to Guyton and Lecture Notes on Human Physiol. Agree: however Kam's first part course lists urea but again question abandoned due to unclear best answer.

Answer 341

Correct answer A: Increased COb (All others decrease Pgc thus GFR). Addit: "Faunce 8th ed" pg 104 states that CO is NOT a direct determinant of GFR unless MAP falls outside the autoregulatory range. I tend to agree that increased CO would not change GFR, as MAP should be regulated by the autonomic NS via the baroreceptors, GFR is autoregulated also Comment: WITHIN the autoregulatory range, CO does play a role: "Outside the autoregulatory range, e.g. in circulatory shock, renal blood flow and GFR change with blood pressure. Even within the autoregulatory range, blood pressure does have some influence on renal haemodynamics. Furthermore, a number of extrinsic factors can override the intrinsic influence of autoregulation. These extrinsic factors can affect renal blood flow and PGC by altering the resistance of afferent and/or efferent arterioles; or they can change Kf by inducing mesangial cell contraction or relaxation, thereby altering the glomerular surface area available for filtration." From Acute Renal Failure in Practice Chapter 1: RENAL HAEMODYNAMICS AND GLOMERULAR FILTRATION by David Shirley, Giovambattista Capasso and Robert Unwin Also, Ganong 22nd edition Table 38.4 lists factors affecting GFR of which 2 (changes in renal blood flow and changes in systemic blood pressure) may approximate CO This is probably the 4th or 5th time this point has been argued... can we have help from the master? References In regard to Feb 00 option A-CO would decrease GFR as a result of INCREASED plasma flow which causes Increased rise of glomerular capillary oncotic pressure along glomerular capillaries-Result DECREASED GFR. REF p.30, tab 2.2 Vander's Renal physiology, 7th Ed comment I think with regards to KD14 [Feb00] [Apr01], in the context of this question, CO seems to be the best answer and most people would agree with me when I say that the remaining options given would definitely not increase GFR.

Answer 342

Apr 2001 version: E Addit: "Faunce 8th ed" pg 104 states that CO is NOT a direct determinant of GFR unless MAP falls outside the autoregulatory range. I tend to agree that increased CO would not change GFR, as MAP should be regulated by the autonomic NS via the baroreceptors, GFR is autoregulated also Comment: WITHIN the autoregulatory range, CO does play a role: "Outside the autoregulatory range, e.g. in circulatory shock, renal blood flow and GFR change with blood pressure. Even within the autoregulatory range, blood pressure does have some influence on renal haemodynamics. Furthermore, a number of extrinsic factors can override the intrinsic influence of autoregulation. These extrinsic factors can affect renal blood flow and PGC by altering the resistance of afferent and/or efferent arterioles; or they can change Kf by inducing mesangial cell contraction or relaxation, thereby altering the glomerular surface area available for filtration." From Acute Renal Failure in Practice Chapter 1: RENAL HAEMODYNAMICS AND GLOMERULAR FILTRATION by David Shirley, Giovambattista Capasso and Robert Unwin Also, Ganong 22nd edition Table 38.4 lists factors affecting GFR of which 2 (changes in renal blood flow and changes in systemic blood pressure) may approximate CO This is probably the 4th or 5th time this point has been argued... can we have help from the master? References In regard to Feb 00 option A-CO would decrease GFR as a result of INCREASED plasma flow which causes Increased rise of glomerular capillary oncotic pressure along glomerular capillaries-Result DECREASED GFR. REF p.30, tab 2.2 Vander's Renal physiology, 7th Ed comment I think with regards to KD14 [Feb00] [Apr01], in the context of this question, CO seems to be the best answer and most people would agree with me when I say that the remaining options given would definitely not increase GFR.

Answer 343

he formula is Starling's equation for fluid filtration in capillaries, including of course the glomerular capillaries. This equation has 6 variables. In the kidney a simplified version is used because 2 of variables cancel out from the full version of the equation. These 2 terms are the reflection coefficient (= 1 in the glomerulus so not required) and the oncotic pressure in Bowman's capsule (= 0 as glomerular filtrate is an ultrafiltrate ie protein-free). This leaves us with the following: GFR = Kf x (HPG - HPB - OPG) The version in the question includes OPB, so option B is the answer. Of course if OPB = 0 then options B & E are equivalent but if OPB is not zero (as occurs in glomerular damage eg nephrotic syndrome) then option B will cover that situation because the sign is correct so B is the "most correct" answer. Comment - B is the only answer - not just the "most correct" References E is correct. It is the DIFFERENCE between the hydrostatic pressures minus the DIFFERENCE between oncotic pressures. B is correct GFR = Kf [(HPG-HPB)-(OPG-OPB)] removing the internal brackets GFR= Kf[HPG-HPB-OPG+OPB). The negative sign preceding (OPG-OPB)is like a having minus 1 there (or a reflection coefficient of 1... which is what we assume it is) outside the bracket... to remove the bracket you have to multiplication by -1. therefore OPG becomes -ve and OPB must become +ve... and as indicated earlier it is easily confirmed by substitution when OPB is not zero as in the disease state.

Answer 344

Answer C in "normal" PTH states. Answer A in some hyperparathyroid states Ganong and Vander talk about two states. "Normal" PTH states and Hyperparathyroidism. Phosphate: In both states phosphate excretion is increased due to a reduction in proximal tubular reabsorption by PTH. Calcium: In "normal" PTH states, PTH increases renal-tubular calcium reabsorption leading to decreased urinary calcium excretion. In Hyperparathyroidism PTH still increases renal-tubular calcium reabsorption, however, this may be overwhelmed by the higher plasma calcium concentration and resultant increased calcium filtration. This may lead to an overall increased urinary calcium excretion. Ganong - "...PTH increases phosphate excretion in the urine. PTH also increases reabsorption of of Ca2+ in the distal tubules, although Ca2+ excretion is often increased in hyperparathyroidism because the increase in the amount filtered overwhelms the the effect of reabsorption." p 391 Vander - "PTH increases renal-tubular calcium reabsorption, mainly by an action on the distal convoluted tubule. At this location, it increases apical membrane calcium entry through calcium channels. The increased uptake of calcium from the lumen stimulates basolateral extrusion (by a combination of Ca-ATPase activity and Na-Ca antiporter activity) and thus decreases urinary calcium excretion." p 188 Vander - " A seeming paradox (in primary hyperparathryoidism) is that urinary calcium excretion is increased despite the fact that tubular calcium is enhanced by PTH. The reason is that elevated plasma calcium concentration induced by the effects of PTH on bone (and via calcitriol on the load of calcium entering the body from the GI tract) causes the filtered load of calcium to increase more than the reabsorptive rate." p189 for this is a physiology exam. plus the question is the PTH effect on KIDNEY. in hyperPTH lots of the plasma Ca come from bone so it is a secondary effect to cause increase urinary Ca excretion. go for C

Answer 345

Say 0.5-1 litre of urine per day out of a GFR of 180 litres/day gives a % reabsorption of 99.4-99.7% Option D (99.4%) is the best available answer. 99.9% is wrong as this calculates out to a urine volume less than that required to excrete the daily solute load in maximally concentrated urine. This can never be correct. ANSWER is C. 99%, see Vander table1-2 I agree with answer D, from table 38-5 Ganong. Answers/Questions lifted directly from this text? Yet another stupid question - C and D are both correct as correctly stated by the references above, it is just that Vander calls urine output 1.8L and Ganong 1L. Nice one ANZCA. "Here is our reference list of correct textbooks... you guess which one we took the question from!"

Answer 346

Answer is B. Afferent & efferent arterioles

Answer 347

A is true comment: thanks, not sure we needed it narrowed down. I agree. A is correct --> wrong: B or C is correct; don't deny that the look on your face is usually "?" while you're doing MCQs Hands up if you laughed at the above conversation?? Never thought I would laugh doing renal MCQs!!

Answer 348

Inulin is freely filtered substance that is not reabsorbed nor actively secreted, making this useful for calculation of glomerular filration rate (GFR). i.e. The renal clearance of inulin equals GFR. Clearance = ([Inulin] in urine/ [Inulin] in blood) x Volume urine per hour To possess a higher renal clearance, the subtance in question must be both freely filtered, must not be reabsorbed, and must be actvely secreted. Para-aminohippurate (PAH) satisfies these criteria (provided its plasma concentration is low) and finds its utility in the calculation of renal plasma flow (RPF) - (from which renal blood flow (RBF) can be estimated if you know the haematocrit). Effective RPF in mL/min = ([PAH] urine/ [PAH] plasma) x Volume urine per hour Whilst RBF in mL/min = RPF/(1-Hct) Additional minor comment: The wording of 'Alt version 2' is clearly wrong. If a substance is found in the urine at a higher concentration that inulin, then it could simply be present in plasma at significantly higher concentrations than the inulin, and none of the options need apply. I suspect the stem must have contained wording to the effect that the substance in question is cleared at a higher rate than inulin.

Answer 349

Inulin is freely filtered substance that is not reabsorbed nor actively secreted, making this useful for calculation of glomerular filration rate (GFR). i.e. The renal clearance of inulin equals GFR. Clearance = ([Inulin] in urine/ [Inulin] in blood) x Volume urine per hour To possess a higher renal clearance, the subtance in question must be both freely filtered, must not be reabsorbed, and must be actvely secreted. Para-aminohippurate (PAH) satisfies these criteria (provided its plasma concentration is low) and finds its utility in the calculation of renal plasma flow (RPF) - (from which renal blood flow (RBF) can be estimated if you know the haematocrit). Effective RPF in mL/min = ([PAH] urine/ [PAH] plasma) x Volume urine per hour Whilst RBF in mL/min = RPF/(1-Hct) Additional minor comment: The wording of 'Alt version 2' is clearly wrong. If a substance is found in the urine at a higher concentration that inulin, then it could simply be present in plasma at significantly higher concentrations than the inulin, and none of the options need apply. I suspect the stem must have contained wording to the effect that the substance in question is cleared at a higher rate than inulin.

Answer 350

Inulin is freely filtered substance that is not reabsorbed nor actively secreted, making this useful for calculation of glomerular filration rate (GFR). i.e. The renal clearance of inulin equals GFR. Clearance = ([Inulin] in urine/ [Inulin] in blood) x Volume urine per hour To possess a higher renal clearance, the subtance in question must be both freely filtered, must not be reabsorbed, and must be actvely secreted. Para-aminohippurate (PAH) satisfies these criteria (provided its plasma concentration is low) and finds its utility in the calculation of renal plasma flow (RPF) - (from which renal blood flow (RBF) can be estimated if you know the haematocrit). Effective RPF in mL/min = ([PAH] urine/ [PAH] plasma) x Volume urine per hour Whilst RBF in mL/min = RPF/(1-Hct) Additional minor comment: The wording of 'Alt version 2' is clearly wrong. If a substance is found in the urine at a higher concentration that inulin, then it could simply be present in plasma at significantly higher concentrations than the inulin, and none of the options need apply. I suspect the stem must have contained wording to the effect that the substance in question is cleared at a higher rate than inulin.

Answer 351

Inulin is freely filtered substance that is not reabsorbed nor actively secreted, making this useful for calculation of glomerular filration rate (GFR). i.e. The renal clearance of inulin equals GFR. Clearance = ([Inulin] in urine/ [Inulin] in blood) x Volume urine per hour To possess a higher renal clearance, the subtance in question must be both freely filtered, must not be reabsorbed, and must be actvely secreted. Para-aminohippurate (PAH) satisfies these criteria (provided its plasma concentration is low) and finds its utility in the calculation of renal plasma flow (RPF) - (from which renal blood flow (RBF) can be estimated if you know the haematocrit). Effective RPF in mL/min = ([PAH] urine/ [PAH] plasma) x Volume urine per hour Whilst RBF in mL/min = RPF/(1-Hct) Additional minor comment: The wording of 'Alt version 2' is clearly wrong. If a substance is found in the urine at a higher concentration that inulin, then it could simply be present in plasma at significantly higher concentrations than the inulin, and none of the options need apply. I suspect the stem must have contained wording to the effect that the substance in question is cleared at a higher rate than inulin.

Answer 352

Water movement from glomerular capillaries to Bowman's capsule is due to filtration (ie this is bulk flow). Similarly water flow along the tubules is due to bulk flow in response to a pressure gradient (again this is bulk flow). ALL water REABSORPTION is due to water moving passively down its osmotic gradient (ie this water movement is diffusion). Solvent drag is a slight variation of this diffusion esp in the proximal tubule. The answer to the "actual" exam question will depend on the exact wording of the question. FROM WIKI...OKAY OKAY I know it's level 4674 evidence..but it may be the missing key. Paracellular transport refers to the transfer of substances between cells of an epithelium.[1][2] It is in contrast to "transcellular transport", where the substancees travel through the cell, passing through both the apical membrane and basolateral membrane.[3][4][5] The distinction is in particular significance in renal physiology. Transcellular transport is more likley to involve energy expenditure than paracellular transport.[6] Capillaries of Blood Brain Barrier have transcellular transport only in contrast with normal capillaries which have both - transcellular and paracellular transport. This is due to the presence of tight junctions in Blood Brain Barrier and Astrocytes that surround the capillaries. ^ Diagram at citracal.com ^ t_16/14158724 at Dorland's Medical Dictionary ^ http://www.trinity.edu/rblyston/MicroA/Lectures/L07pt2-html/sld022.htm ^ Physiology at MCG 1/1ch2/s1ch2_37 ^ http://www.mun.ca/biology/desmid/brian/BIOL2060/CellBiol11/CB11_19.html ^ http://www.acbrown.com/renal/OutTubl.htm Reabsorption can be classified as either Facillitated diffusion (eg ADH induced aquaporins in DCT & CCD) or Osmosis Neither of these processes account for any excreted water. Filtration at the glomerulus accounts for the vast majority of water excreted and this filtration/bulk flow is influenced by osmotic pressures in Bowman's capsule and glomerular capillary. Most water travels paracellularly (except in DCT/CCD) ====================== i read the stem as water excretion ,? obligatory water loss. I'd be tempted to go "solvent drag". The water excreted or ?obligatory water loss is that which remains in the tubular lumen despite all forms of reabsorption. What keeps some water there in the cortical collecting ducts to be excreted a s urine despite sometimes abundant aquaporins being present is the drag of the remaining solvent in the lumen I would think. Feel free to correct me there. I think 'solvent drag' refers to solvent (ie water) dragging solute (ie NaCl) with it as it moves - so it's an effect caused by water movement, not one that causes water movement.

Answer 353

Angiotensin II increases reabsorption of Na+ and HCO3- by action on the proximal tubule. Answer is A. more Angiotensin II Increases Sodium and Water Reabsorption. Angiotensin II is perhaps the body’s most powerful sodium-retaining hormone. As discussed in Chapter 19, angiotensin II formation increases in circumstances associated with low blood pressure and/or low extracellular fluid volume, such as during hemorrhage or loss of salt and water from the body fluids. The increased formation of angiotensin II helps to return blood pressure and extracellular volume toward normal by increasing sodium and water reabsorption from the renal tubules through three main effects: 1. Angiotensin II stimulates aldosterone secretion, which in turn increases sodium reabsorption. 2. Angiotensin II constricts the efferent arterioles, which has two effects on peritubular capillary dynamics that raise sodium and water reabsorption. First, efferent arteriolar constriction reduces peritubular capillary hydrostatic pressure, which increases net tubular reabsorption, especially from the proximal tubules. Second, efferent arteriolar constriction, by reducing renal blood flow, raises filtration fraction in the glomerulus and increases the concentration of proteins and the colloid osmotic pressure in the peritubular capillaries; this increases the reabsorptive force at the peritubular capillaries and raises tubular reabsorption of sodium and water. 3. Angiotensin II directly stimulates 'sodium reabsorption in the proximal tubules, the loops of Henle, the distal tubules, and the collecting tubules'. One of the direct effects of angiotensin II is to stimulate the sodium-potassium ATPase pump on the tubular epithelial cell basolateral membrane. A second effect is to stimulate sodium-hydrogen exchange in the luminal membrane, especially in the proximal tubule. Thus, angiotensin II stimulates sodium transport across both the luminal and the basolateral surfaces of the epithelial cell membrane in the tubules. These multiple actions of angiotensin II cause marked sodium retention by the kidneys when angiotensin II levels are increased. Guyton 342 ,11th edition

Answer 354

Answer is D - Glomerulotubular balance refers to the kidney's ability to increase the reabsorption of solutes (predominantly in the proximal tubule) in response to an increase in GFR. This ensures that the proportion of solute reabsorbed remains constant. Agreed D is the best answer, but if F appeared in the absence of D it is also correct in that a constant proportion means that it is matched NOTE: do not confuse this with tubuloglomerular feedback which refers to the effects of the macula densa on the afferent arteriole in response to increased solute load in the ascending LOH

Answer 355

Best answer: A A. Erythropoietin from proximal tubule cells or peritubular capillaries in kidney B. ADH produced in supraoptic & paraventricular nuclei in the hypothalamus then transported to and released from the posterior pituitary C. Angiotensin II is produced in the blood D. ANP - from atria & ?other tissues E. Cholecalciferol - produced in skin, converted to 25-hydroxycholecalciferol in the liver, then converted to 1,25 dihydroxycholecalciferol in the kidney C is correct too, "the kidneys convert 20% of the circulating angiotensin I reaching them to angiotensin II" Ganong 27 p723

Answer 356

Answer A: Renal SNS activity causes a graded response incr sensitivity of JG cells to non-neural stim for renin release (β1) incr renin secretion by direct affect on JG cells (β1) incr Na resorption by direct NA action on renal tubular cells (PCT, DCT, asc LOH) via α1 & β1 renal vasoconstriction efferent then afferent arterioles RBF > GFR (thus FF incr) reduces GFR by causing mesangial cell contraction

Answer 357

Best Answer A --- errr.... how about best answer is clearly not going to be A? Most texts quote 65% of H2O resorption in PCT, 10% reabsorbed in the Descending thin limb of Henle, and the remaining proportion in the collecting duct system (5% during water loading, >24% during dehydration). The DCT is impermeable to H2O COMMENT: have mixed input about this; Brandis says DCT is responsive to ADH and can be permeable to H20; Power and Kam says it is not permeable; Ganung says it is relatively impermeable to H2) but not saying anything regarding to action of ADH. No H2O is actively transported - True but the movement of water is due to the active transport of sodium, i'd be leaning toward this answer given the other two options are so wrong. Ref: Vander, p.75., Table 6-2. comment hopefully option D or E was a blindingly clear answer! -> if it were it would have been remembered and not been forgotten as an irrelevant distractor. Comment.. If 65% is reabsorbed in PCT, and a further 10% subsequently reabsorbed, it sort of works that PCT reabsorbs 90% of the water that is destined to be reabsorbed (65/75 = 87%)... Seems a bit random given the questions are otherwise quite specific

Answer 358

Answer is C. Comments Rate of filtration = hydraulic permeability x surface area x net filtration pressure This is the same for all capillaries, so the answer is C. GFR depends on permeability, size and hydrostatic and osmotic forces. GFR: decreases with decrease in the size of the capillary bed (decrease in filtration surface area)- so A is wrong depends on capillary permeability and surface area, not just the net pressure gradient - so B is wrong depends on factors other than the permeability of the capillary - see Starling's hypothesis - so D is wrong No active transport is involved in (ultra)filtration - so E is wrong.

Answer 359

Answer: A Isotonic incr circulating volume-> incr Glomerular capillary pressure ->incr GFR (minor effect),and incr hydrostatic pressure of peri-tubular capillary thus incr renal interstitial pressure and decr Na resorption The macula densa regulates tubuloglomerular feedback via sensing ?Na &/or Cl in tubule (mediated by PG's or Adenosine depending on which text you read to act on JG cells) Increased circulating volume causing pressure diuresis will actually decrease activation of RAS and reduce ADH secretion (via RAS & direct baroceptor reflexes) JGA controls RAS and is in fact controlled (negatively) by pressure diuresis "This phenomenon, also known as pressure natriuresis, is associated with increased mean arterial pressure and plays an important role in the long term regulation of EBV and blood pressure. It is the result of an interaction of several factors. The imperfect nature of autoregulation of GFR and of GT balance contribute to pressure diuresis. The associated decrease in sympathetic activity and increased macula densa flow inhibit renin secretion and decrease the effects of angiotensin II and aldosterone." "Another factor is the effect of increased perfusion pressure to raise the intra-renal ISF hydraulic pressure which will then decrease net fluid reabsorption by enhancing back-flux through the leaky tight junctions of the proximal tubule. The resulting diuresis will function to decrease the ECF volume (and the EBV) and return arterial pressure towards normal." - from [1] Sorry? Additional: I think Vander is a little more understandable. He states that in response to elevated renal artery pressure the number of luminal Na-H antiporters and basolateral Na-K-ATPase pumps are reduced, thus reducing the reabsorption of sodium (and therefore) water. This is pressure natriuresis and diuresis. The problem with this mechanism is that it does not separately control solute concentration (sodium) and water concentration, which is why we need systems such as ADH and Aldosterone. Also note that this system will be blunted if AII levels are high, thus it requires AII levels to be lowered (which is what you expect for in response to hypertension; so that peripheral vasculature loses some of its constriction) in order to lose the sodium and water. Thus, these systems work in harmony. Ref: p.108-109 Vander. one more effect of pressure diuresis is increased medullary bld flow which will reduce the medullary osmolality and hence reduce the CD ability to absorb water.

Answer 360

Maximum urinary osmaolority is 1200-1400mOsm/l. So to excrete 600mOsm need minimum approx 500ml. correct answer is B No the correct answer is 1.15 litres. it is because the maximum urine concentration of 1400mosm/kg H20 is obtained in 0.5 litre of urine. so to obtain a urine concetration of 600mosm it would be =1400/600 multiplied by 0.5 litres ,which gives around 1.15 litre.in this answer i would choose 1 litre as the answer because it is closer to 1.15. Hahaha, that makes no sense. Your maths is even more wrong than your random 0.5kg value. I think this comment was by a CICM candidate trying to mislead us out of career jealousy Extra comment: I keep reading everywhere that osmolarity and osmolality are essentually the same and can be interchanged with regards to most physiological solutions, including urine. So if ~1200mOsmol/kg is equivalent to 1200mOsmol/L then Volume of urine = 600/1200 = 500mL Where did you find your information that 1400mOsmol/kg is obtained in 0.5L of urine? 1Kg of water or 1L of water are about the same. So a 1400mOsm per L or Kg is roughly the same. It is 1400mOs m PER LITRE (OR KG. 1400mOsm cannot be in 500mL or else it would then be a 2800mOsm solution by definition. Anyway, agree that 500mL is the answer. Max Conc is 1400mOsm/L and minimum fluid loss is about 500mL so I think the other person added the 2 together "The maximal urine osmolality under extreme conditions is about 1400mOsm/kg. Under these conditions, to excrete the solute load, the obligatory or minimum water loss as urine is about 500mL/day (i.e 1400/700)" Brandis 2007. Pg 159. (Using 700mOsm/day as daily solute load.) Substituting 600 for 700 equals 430mL. So option B closest. Re 1.15L. Huh? Stem says your trying to excrete 600 mosm. Thats an amount, not a concentration

Answer 361

GFR = hydraulic permeability x surface area x net filtration pressure NFP = glomerular capillary hydraulic pressure - Bowman's capsule hydraulic pressure - glomerular oncotic pressure (oncotic pressure in Bowman's capsule is essentially zero as little protein is filtered). Glomerular oncotic pressure increases along the capillary as water is filtered but very little protein is filtered. If renal plasma flow is decreased the glomerular oncotic pressure increases more rapidly and therefore the GFR decreases. Increased sympathetic stimulation constricts both the afferent and efferent arterioles decreasing the renal plasma flow. Obstructing the ureters causes increased tubular pressure all the way back to Bowman's capsule therefore decreasing GFR. Hypoproteinaemia decreases the glomerular capillary oncotic pressure increasing the net filtration pressure thereby increasing GFR. The answer is therefore C

Answer 362

Inulin is a polysaccharide that is filterable but is neither reabsorbed nor secreted. Therefore, the volume of plasma cleared of inulin per unit time is the same as the glomerular filtration rate (GFR). Clearance of inulin is thus equal to the glomerular filtration rate, not the renal blood flow. Creatinine is an end product of Creatine metabolism and is exported into blood continuously by skeletal muscle (rate proportional to skeletal muscle mass). Creatinine is freely filtered and not reabsorbed. A small amount is secreted by the proximal tubule, hence creatinine clearance is slightly higher than the GFR (i.e represents both a filtered and secreted component). The secreted fraction is normally ~ 10-20%, so the measured creatinine clearance overestimates GFR by the same percentage (an acceptable degree of error for routine assessment of GFR). Note the secreted component is a relatively larger fraction of amount excreted for a patient with a low GFR – therefore creatinine clearance more severely overestimates GFR in patients with a very low GFR. ‘B’ cannot be right as creatine itself is not cleared but rather metabolised to creatinine for clearance. The filtration fraction refers to the fraction of the renal plasma entering the glomeruli via the afferent arterioles that is filtered into Bowman’s space. This amounts to the ratio of GFR to RPF. As the gold standard for measurement of GFR is Inulin clearance and that for Renal Plasma Flow is PAH clearance, Inulin Clearance/PAH Clearance equates to GFR/RPF… the filtration fraction. I agree - so C is correct

Answer 363

Inulin is a polysaccharide that is filterable but is neither reabsorbed nor secreted. Therefore, the volume of plasma cleared of inulin per unit time is the same as the glomerular filtration rate (GFR). Clearance of inulin is thus equal to the glomerular filtration rate, not the renal blood flow. Creatinine is an end product of Creatine metabolism and is exported into blood continuously by skeletal muscle (rate proportional to skeletal muscle mass). Creatinine is freely filtered and not reabsorbed. A small amount is secreted by the proximal tubule, hence creatinine clearance is slightly higher than the GFR (i.e represents both a filtered and secreted component). The secreted fraction is normally ~ 10-20%, so the measured creatinine clearance overestimates GFR by the same percentage (an acceptable degree of error for routine assessment of GFR). Note the secreted component is a relatively larger fraction of amount excreted for a patient with a low GFR – therefore creatinine clearance more severely overestimates GFR in patients with a very low GFR. ‘B’ cannot be right as creatine itself is not cleared but rather metabolised to creatinine for clearance. The filtration fraction refers to the fraction of the renal plasma entering the glomeruli via the afferent arterioles that is filtered into Bowman’s space. This amounts to the ratio of GFR to RPF. As the gold standard for measurement of GFR is Inulin clearance and that for Renal Plasma Flow is PAH clearance, Inulin Clearance/PAH Clearance equates to GFR/RPF… the filtration fraction. I agree - so C is correct

Answer 364

Answer is A. [Ref: p294, 206f Ganong 22nd edition] A: Ornithine participates in urea cycle, and urea is produced from arginine when arginine breaks down into ornithine and urea. (Ornithine is like oxaceloacetate and is regenerated during the cycle) B: 50% of urea is reabsorbed

Answer 365

First question - A. - Na+ absorption in thick ascending limb P&K pg 206 - the loop of henle creates the high medullary interstitial osmolality which is essential for the production of concentrated urine. Active transport of sodium (with potassium and chloride) from the thick ascending limb into the interstitial fluid is the prime cause of the high intersitial osmolality. In the presence of ADH water is reabsorbed thru the walls of the CDs because of the high medullary interstitial osmolality. July 03 questions - Passive reabsorption of urea from collecting duct P&K pg 204, 208 - prox tubule reabsorbs half of filtered urea by passive diffusion. LoH, DCT, cortical CD are impermeable to urea so its concentration in the luminal fluid rises. Inner medullary collecting duct is permeable to urea, ADH increases permeability to urea. Another 10% is reabsorbed. Consequently the inner medullary interstitial urea concentration is high and accounts for 650mosm/kg H20 of the total osmolality of 1400 mosml/kg H20 of the fluid present (Na and Cl accounting for most of remainder). Addition: LOH is *not* impermeable to Urea: Urea is *secreted* by the loop of henle back into the lumen, so that there about the same amount as originally filtered (although now at 5x the concentration). This is why we talk about "cycling" of urea. See Vander -Fig 5-3. I would have thought the same answer would be correct for each question - ie, urea. The high medullary interstitial osmolality is responsible for the kidney's concentrating ability (obviously in the presence of ADH) The medullary interstitial osmolality is about 50% accounted for by urea, and 50% by NaCl (Vander's) Therefore, it would be more accurate to say the contributions are urea 50%, Na 25%, Cl 25% My answer would be urea The above comments state that about 50% urea is reabsorbed in the PCT, and a further 10% in the CD - but Vander's says that 50% is reabsorbed in the PCT, this is then secreted back into the tubule through the LOH, and finally 50% reabsorbed in the CD (50% of the filtered load is excreted) For the first version A is the best answer. it all start with thick LOH Na+ absorption. if we didn't have the initial 600mosmol concentration created urea can not be absorbed in the CD! I think the answer is Na for both (why would it be Na for the first and Urea for the second when the question is asking the same thing?) - Urea responsible for 650/1400mosmol/L present in medullary interstitium with Na responsible for the majority of the rest (i.e. more than 50% and therefore more important) - No Urea without Na already there as stated above - A (Version 1) + C (Version 2) are answers - does anyone dispute this? PK (2nd Ed) - pg235 - states that urea helps form the concentration gradient down from cortex to medulla BUT active transport of Na and Cl from the ascending limb of the loop of Henle is essential. Re: Na+ being the right answer for both, I'd agree if they had the same options but version 2 is Na+ from the vasa recta NOT Na+ transport in the ascending LOH. Comment: I'm really confused by the above discussion. According to Guyton: - 50% of medullary interstitial osmotic pressure is from urea and 50% is from NaCl (and a little from other solutes) - In the presence of ADH, urea diffuses passively down a concentration gradient into the medulla - The remembered options are referring to Na moving in various directions vs. urea alone - In conclusion I would say that passive reabsorption of urea is more important than Na+ alone in concentrating urine Thoughts? Am I way off? Edit: no you aren't way off.. I think urea is the correct answer too.. Na by itself (and Cl) cannot contribute more than around 25% of the hypertonicity

Answer 366

First question - A. - Na+ absorption in thick ascending limb P&K pg 206 - the loop of henle creates the high medullary interstitial osmolality which is essential for the production of concentrated urine. Active transport of sodium (with potassium and chloride) from the thick ascending limb into the interstitial fluid is the prime cause of the high intersitial osmolality. In the presence of ADH water is reabsorbed thru the walls of the CDs because of the high medullary interstitial osmolality. July 03 questions - Passive reabsorption of urea from collecting duct P&K pg 204, 208 - prox tubule reabsorbs half of filtered urea by passive diffusion. LoH, DCT, cortical CD are impermeable to urea so its concentration in the luminal fluid rises. Inner medullary collecting duct is permeable to urea, ADH increases permeability to urea. Another 10% is reabsorbed. Consequently the inner medullary interstitial urea concentration is high and accounts for 650mosm/kg H20 of the total osmolality of 1400 mosml/kg H20 of the fluid present (Na and Cl accounting for most of remainder). Addition: LOH is *not* impermeable to Urea: Urea is *secreted* by the loop of henle back into the lumen, so that there about the same amount as originally filtered (although now at 5x the concentration). This is why we talk about "cycling" of urea. See Vander -Fig 5-3. I would have thought the same answer would be correct for each question - ie, urea. The high medullary interstitial osmolality is responsible for the kidney's concentrating ability (obviously in the presence of ADH) The medullary interstitial osmolality is about 50% accounted for by urea, and 50% by NaCl (Vander's) Therefore, it would be more accurate to say the contributions are urea 50%, Na 25%, Cl 25% My answer would be urea The above comments state that about 50% urea is reabsorbed in the PCT, and a further 10% in the CD - but Vander's says that 50% is reabsorbed in the PCT, this is then secreted back into the tubule through the LOH, and finally 50% reabsorbed in the CD (50% of the filtered load is excreted) For the first version A is the best answer. it all start with thick LOH Na+ absorption. if we didn't have the initial 600mosmol concentration created urea can not be absorbed in the CD! I think the answer is Na for both (why would it be Na for the first and Urea for the second when the question is asking the same thing?) - Urea responsible for 650/1400mosmol/L present in medullary interstitium with Na responsible for the majority of the rest (i.e. more than 50% and therefore more important) - No Urea without Na already there as stated above - A (Version 1) + C (Version 2) are answers - does anyone dispute this? PK (2nd Ed) - pg235 - states that urea helps form the concentration gradient down from cortex to medulla BUT active transport of Na and Cl from the ascending limb of the loop of Henle is essential. Re: Na+ being the right answer for both, I'd agree if they had the same options but version 2 is Na+ from the vasa recta NOT Na+ transport in the ascending LOH. Comment: I'm really confused by the above discussion. According to Guyton: - 50% of medullary interstitial osmotic pressure is from urea and 50% is from NaCl (and a little from other solutes) - In the presence of ADH, urea diffuses passively down a concentration gradient into the medulla - The remembered options are referring to Na moving in various directions vs. urea alone - In conclusion I would say that passive reabsorption of urea is more important than Na+ alone in concentrating urine Thoughts? Am I way off? Edit: no you aren't way off.. I think urea is the correct answer too.. Na by itself (and Cl) cannot contribute more than around 25% of the hypertonicity

Answer 367

Power & Kam (p. 217) calculates a value of "0.684 mmol/day" for filtered H+. However, their arithmetic is less than convincing... The calculation below seems much closer to the mark [H+] in plasma = 40 x 10-9 moles/litre GFR = 180 l/day SO: Amount of [H+] filtered per day = 180 x 40 x 10-9 moles/day = 7,200 x 10-9 moles/day Now if you use 170 l/day for the GFR, this gives 6,800 nmoles/day. The answer then is 6,800 nmoles/day, or 6.8 micromoles/day or 0.0068 mmoles/day, which doesn't fit with any of the options. My guess is that the option 6.8 micromoles/day was on the paper and this has been remembered incorrectly. Most of the acid that requires excretion is eliminated either as titratable acids (about 30mmol) or ammonia. In addition a large amount of H+ is secreted to resorb filtered bicarbonate. If you're vegetarian you'd actually have a daily net H+ deficit. Brandis (viva book p164) states - the amount of H+ secretion associated with excretion of the anions of the fixed acids is very much smaller (70mmol/day) but this represents net excretion of this amount of H+ from the body. There is only a very tiny amount of H+ filtered per day (as concentration is so low) BUT there is a very large secretion of H+ into the proximal tubule (4,000-5,000 mmoles/day). Most of this is reacts with filtered bicarbonate in the brush border (carbonic anhydrase located here) and so very little of this very high secretion accounts for excretion from the body. In reply to the above paragraph: The question (as remembered here) is about H+ filtered per day, not secreted. So there are 2 possibilities for this question, either of which can resolve the confusion: Firstly, as I mentioned in my calculations above, if one of the actual options is 6.8 micromoles/day then that would be the correct answer Alternatively, if the actual question stem did say net hydrogen ion excretion (rather than filtered) then 68 mmol/day would be the correct option. For acid-base balance, the net amount excreted per day must equal the net fixed acid production per day. As the net production is 1 to 1.5mmol/kg/day (ie 70-100mmol/day) then 68 mmol/day is close enough. comment on above the excreted amount should be less than the filtered amount,how do u expect to excrete 68 mmol if u filter 0.0068 mmol. I agree with 68 mmol ,more physiological value . a comment on the comment above the excreted amount doesn't need to be less than the filtered amount as most of the fixed acid excretion is via secretion the question is referring to free hydrogen ions (being a tiny concentration in the plasma of 36-44 nmoles/L) and therefore the amount freely filtered is extremely small, and virtually negligible the predominant fixed acid excretion occurs by secretion of hydrogen ions into the tubular lumen via the Type A intercalated cells of the collecting duct (which is buffered by urinary buffers - mainly phosphates - and accounts for titratable acidity) and via ammonium secretion in the proximal tubule titratable acidity and ammonium excretion therefore make up the net daily excretion of approx. 70mmol/day, whereas freely filtered Hydrogen ions make up less than 0.01% of this - as stated in above calculations Power and Kam did come up with the 6800nmol/day... and when they convert it to mmol... it became 0.68mmol... not quite sure how that happened... (I wonder if they have corrected all the mistakes in the new addition? including the statement about the existence of an "antiAB antibody"!?)

Answer 368

A. no. Aldosterone acts at cortical collecting duct principal cells to increase Na+ reabsorption and K+ secretion and on type A intercalated cells to increase H+ secretion. It also induces protein production in cortical CD cells, incl epithelium Na channels in luminal membrane and basolateral membrane Na, K - ATPase pump. B. no. 15% of the filtered water is presented to the collecting ducts (PCT 65%, LoH 15%, DCT 5%) In the presence of ADH 14.7% of this is reabsorbed (urine conc of 1400mosmol/kg H20, volume ~500ml/day) In absence of ADH 2% is reabsorbed. 13% excreted. Urine conc may be as low as 30mosmol/kgH20, volume up to 23L/day) ganong 714-716 Umm YES: "Collecting Tubules" = PCT + DCT = 65 + 5 = 70% Umm NO actually. Collecting tubules connect the DCT to the collecting duct. C. yes? renin cleaves angiotensinogen to angiotensin I which ACE (predom in pulmonary capillary endothelium) converts to angiotensin II. Angiotensin II stimulates ADH secretion from posterior pituitary which increases permeability of collecting ducts to water. (can't depend on renin. since there are other mechanisms cause ADH release) D. no. cortical nephrons have short loops of Henle which start in the cortex and descend into the medulla. juxtamedullary nephrons have long LoH which extend into the medullary pyramids. the thick ascending limbs come back to the cortex, reaching the glomerulus from which it arose to form the macula densa, between afferent and efferent arterioles. (Ganong p700).

Answer 369

The loop of Henle creates the high medullary interstitial osmolality which is essential for the production of concentrated urine from the glomerular filtrate. The descending limb is permeable to water but not to sodium nor chloride. In the ascending limb sodium, potassium and chloride are actively reabsorbed by a cotransporter: Na-K-2CL Cotransporter on the apical membrane. Na-H Antiporter on the apical membrane. Na-K-ATPase on the basolateral membrane. So B is correct

Answer 370

Is the question supposed to be why urea is not used for GFR measure? Creatinine is used to estimate GFR, although not as accurate as inulin. This small accuracy is due to small secretion in the tubules, leading to over-estimation of GFR. Urea is not used for GFR measurement. Urea is absorbed in PCT (50%), actively secreted in LoH, and then reabsorbed in medullary CD. Urea reabsorption is further effected by urine flow, and level urea is effected by protein diet. Due to complexity of transportation of urea and unstable level in plasma, it is not used to measure GFR in clinical setting.

Answer 371

Answer: A Elevated BGL can result in filtered glucose exceeding the kidney's transport maximum for glucose, resulting in glycosuria Filtered load = plasma concentration x glomerular filtration rate (for glucose which is freely filtered) thus an increase in GFR makes glycosuria more likely With large increases in GFR and/or plasma glucose concentration that increase the filtered load of glucose above 375 mg/ min, the excess glucose filtered is not reabsorbed and passes into the urine.

Answer 372

The answer is glomerulotubular balance Increasing GFR results in an increase in tubular resborption of the proximal tubules so the percentage of filtered of Na reabsorbed remains fairly constant 65% Vander chapter 7 TUBULOGLOMERULAR FEEDBACK A form of metabolic autoregulation of RBF. Increasing MAP- increasing GFR - Increased filtrate delivery to tubules and to macunla densa-increased Na and Cl sensed by macula densa-increased production of adenosine - afferent arteriolar vasoconstriction- Reduced GFR GLOMERULOTUBULAR BALANCE Intrinsic function of the whole kidney whereby the tubular reabsorption is adjusted to changes in GFR proportionally. eg Increased GFR- more Na filtered. Hence Proximal tubules increase reabsorption so that reabsorbed Na remains at around 65%. 2015B Version ANSWER B or C The correct answer is Glomerulotubular balance. But this is not an option. Some sources use the term ‘tubuloglomerular balance’ instead of glomerulotubular balance, others use ‘tubuloglomerular balance’ to refer to tubuloglomerular feedback. Technically it is a form of autoregulation; ‘the intrinsic ability of a biological system to adjust or mitigate that system’s response to stimuli’. However, Guyton and Hall tend to discuss them as two separate things; ‘working together, the autoregulatory and glomerulotubular balance mechanisms prevent large changes in fluid flow in the distal tubules when the arterial pressure changes…’ p339. This is because most textbooks reserve the term ‘autoregulation’ for the cardiovascular phenomenon; ‘intrinsic ability of an organ to maintain a constant blood flow despite changes in perfusion pressure’ (www.cvphysiology.com) Dr mitta 18:18, 15 November 2015 (CST)

Answer 373

The answer is glomerulotubular balance Increasing GFR results in an increase in tubular resborption of the proximal tubules so the percentage of filtered of Na reabsorbed remains fairly constant 65% Vander chapter 7 TUBULOGLOMERULAR FEEDBACK A form of metabolic autoregulation of RBF. Increasing MAP- increasing GFR - Increased filtrate delivery to tubules and to macunla densa-increased Na and Cl sensed by macula densa-increased production of adenosine - afferent arteriolar vasoconstriction- Reduced GFR GLOMERULOTUBULAR BALANCE Intrinsic function of the whole kidney whereby the tubular reabsorption is adjusted to changes in GFR proportionally. eg Increased GFR- more Na filtered. Hence Proximal tubules increase reabsorption so that reabsorbed Na remains at around 65%. 2015B Version ANSWER B or C The correct answer is Glomerulotubular balance. But this is not an option. Some sources use the term ‘tubuloglomerular balance’ instead of glomerulotubular balance, others use ‘tubuloglomerular balance’ to refer to tubuloglomerular feedback. Technically it is a form of autoregulation; ‘the intrinsic ability of a biological system to adjust or mitigate that system’s response to stimuli’. However, Guyton and Hall tend to discuss them as two separate things; ‘working together, the autoregulatory and glomerulotubular balance mechanisms prevent large changes in fluid flow in the distal tubules when the arterial pressure changes…’ p339. This is because most textbooks reserve the term ‘autoregulation’ for the cardiovascular phenomenon; ‘intrinsic ability of an organ to maintain a constant blood flow despite changes in perfusion pressure’ (www.cvphysiology.com) Dr mitta 18:18, 15 November 2015 (CST)

Answer 374

A: WRONG sodium retention secondary to increased of renin activity and decreased in ANP. B: ? C: Decrease in RBF) see: reference 1 Also there is a good summary on AnaesthesiaUK: [1] which sums it up: Decreased CO, MAP, RBF, increased venous pressure Overall Na/H2O retention Increased ADH; activation of R-A-A Decreased ANP References from prescribed texts would be even better.

Answer 375

Certainly not Na and K as these are excreted everyday. Ca++ reabsorption also varies. Glucose is completely reabsorbed at normal plasma glucose levels, but will spill into the urine if hyperglycaemia is present (i.e. once Tm for a tubule is exceeded). Albumin is large and not filtered into the urine in normal circumstances.

Answer 376

Re first version above: A. prolonged vomiting --> metabolic alkalosis --> increased excretion and secretion but dehydration may reduced excretion B. metabolic acidosis - no change C. normal saline infusion - increase D. Aldosterone - increase E. Renal failure - clearly decrease K excretion - the most correct How does normal saline increase K excretion? Kam said body Na content or volume doesn't affect K excretion. Shift of hydrogen ions intracellularly: Intracellular acidosis enhances bicarbonate reabsorption in the collecting duct. Stimulation of the apical H+/K+ ATPase in the collecting duct: Increased activity of this ATPase leads to teleologically appropriate potassium ion reabsorption but a corresponding hydrogen ion secretion. This leads to a net gain of bicarbonate, maintaining systemic alkalosis. Stimulation of renal ammonia genesis: Ammonium ions (NH4 +) are produced in the proximal tubule from the metabolism of glutamine. During this process, alpha-ketoglutarate is produced, the metabolism of which generates bicarbonate that is returned to the systemic circulation. Impaired chloride ion reabsorption in the distal nephron: This results in an increase in luminal electronegativity, with subsequent enhancement of hydrogen ion secretion. Reduction in glomerular filtration rate (GFR): This has been proven in animal studies. Hypokalemia by unknown mechanisms decreases GFR, which in turn decreases the filtered load of bicarbonate. In the presence of volume depletion, this impairs renal excretion of the excess bicarbonate.

Answer 377

Regarding B... as Kam says low [H+] in alkalosis stimulate Na/K ATPase in principal cells and cause increased excretion of K+... shouldn't high [H+] inhibit Na/K ATPase and metabolic acidosis cause decreased excretion then? or maybe it is balanced out by increasing plasma [K+] by shifting intracellular K out. hence more filtered K balanced the decreased secretion? Also your intercalated A cell secrete H+ in exchange with K+... so when you secrete more H+ in acidosis... you should be absorbing more K? but then... if that is the case shouldn't alkalosis' effect get balanced out too? Just think Acidosis and Alkalosis should have the opposite effects... Getting myself totally confused... Shift of hydrogen ions intracellularly: Intracellular acidosis enhances bicarbonate reabsorption in the collecting duct. Stimulation of the apical H+/K+ ATPase in the collecting duct: Increased activity of this ATPase leads to teleologically appropriate potassium ion reabsorption but a corresponding hydrogen ion secretion. This leads to a net gain of bicarbonate, maintaining systemic alkalosis. Stimulation of renal ammonia genesis: Ammonium ions (NH4 +) are produced in the proximal tubule from the metabolism of glutamine. During this process, alpha-ketoglutarate is produced, the metabolism of which generates bicarbonate that is returned to the systemic circulation. Impaired chloride ion reabsorption in the distal nephron: This results in an increase in luminal electronegativity, with subsequent enhancement of hydrogen ion secretion. Reduction in glomerular filtration rate (GFR): This has been proven in animal studies. Hypokalemia by unknown mechanisms decreases GFR, which in turn decreases the filtered load of bicarbonate. In the presence of volume depletion, this impairs renal excretion of the excess bicarbonate.

Answer 378

Correct answer should be D Hypovolaemia/hyperosmolar state -> osmoreceptors in hypothalamus -> ADH from post pituitary -> aquaporins in collecting duct -> increased water reabsorption

Answer 379

D - 4.5 ``` [1] - KB Acid-Base text says 4.4 [2] - says 4.0 to 4.5 [3] - - this Physiology text says 4.4 [4] -says 4.5 [5] -says 4.4 [6] -Ganong 23rd ed (2010) p680 says 4.5 ```

Answer 380

65-70% in PCT irrespective of presence/absence of ADH. No other renal segment can reabsorb a higher percent (or amount) of the filtered water. This percent is high and doesn't vary much. However the amount of water reabsorbed in the medullary collecting duct does vary a lot (depending on [ADH]). A .... Vander (8th ed) p101 re ADH: Under all conditions the majority of the filtered volume is re-absorbed in the proximal tubule

Answer 381

Answer is D Upper sphincter consists of cricopharyngeus muscle (striated) and circular smooth muscle. It opens on swallowing. Lower sphincter is a physiological sphincter formed by the lowest 2-4cm of the oesophagus (tonic contractions of the circular muscle fibres). It opens upon food entering the oesophagus. Above information and greater detail: Power and Kam pp168-9

Answer 382

A: correct. B: some absorbed while K+ or H+ are transported in the opposite direction in exchange for the Na+ ions. C: glucose is transported by a sodium co-transport mechanism, so no Na absorption means no glucose absorption. D: incorrect. E: Cl- ions are passively "dragged" along by the positive electrical charges of the Na+ Comment Strictly speaking, option A is incorrect. Absorption of Na+ occurs by secondary active transport across the brush border membrane. It is the Na-K-ATPase pump at the basolateral membrane that drives the Na+ gradient. Comment Kam (p 179) says lists the mechanisms of transport at the apical membrane to be via: Na+Cl- CoTransport, Sodium channels, Sodium-glucose CoTransporter, and Sodium-AminoAcid CoTranporter. The driving force is the basolateral Sodium-Potassium ATPase. This sounds like Active Transport to me (albeit secondary) Comment Comment two: Well strictly speaking Active transport is both primary and secondary....and secondary active transport is reliant on an energy consuming active process...but I can see the hesitation. Question So... why is D incorrect? ANSWER: Because Na+ extrusion at the basolateral membrane is ACTIVE (Na+/K+ ATPase pump)

Answer 383

After entering the epithelial cell with the help of bile micelles performed as "ferry" fatty acids and monoglycerides are taken up into ER and used to form new triglycerides that are subsequently transported mainly in "Lymph chylomicrons", flowing upward through the thoracic lymph duct to empty into the circulatory blood. Only small quantities of short- and medium-chain fatty acids are absorbed directly into the portal blood. Remember the these fatty acids are less than 12 C in length. The most important enzyme for the digestion of triglycerides is pancreatic lipase. The most important emulsification of fat = lecithin (not bile salt); gastric lipase is generally unimportant, less than 10% digested

Answer 384

After entering the epithelial cell with the help of bile micelles performed as "ferry" fatty acids and monoglycerides are taken up into ER and used to form new triglycerides that are subsequently transported mainly in "Lymph chylomicrons", flowing upward through the thoracic lymph duct to empty into the circulatory blood. Only small quantities of short- and medium-chain fatty acids are absorbed directly into the portal blood. Remember the these fatty acids are less than 12 C in length. The most important enzyme for the digestion of triglycerides is pancreatic lipase. The most important emulsification of fat = lecithin (not bile salt); gastric lipase is generally unimportant, less than 10% digested

Answer 385

The major functions of vitamin B12 are: promotion of growth promotion of red blood cell formation and maturation The 'usual' cause of vitamin B12 deficiency is not lack of this vitamin in food but deficiency of formation of intrinsic factor, which is normally secreted by the parietal cells of the gastric glands and is essential for absorption of vitamin B12 by the mucosa of the terminal ileum. A- unlikely as this condition usually caused by deficiency of intrinsic factor. B- again this seems to be due to levels of b12-IF conplex, not a problem in the ileum per se. C- common cause of pernicious anaemia (anaemia=low Hb) therefore this seems to be the correct answer D- i think the red cells are produced, it is just that the Hb is low and the MCV is raised E- there can be some pathology in the dorsal columns due to B12, so this answer is possible, but I think less correct than C. Comment: - answer A is also correct - malnourished or alcoholic individuals and vegans can be B12 deficient. - answer E could also be correct - b12 deficiency also can cause demyelination of the dorsal and lateral corticospinal tracts. - perhaps the question was worded differently. ?? All of the above Causes of Cbl deficiency Inadequate dietary intake (ie, vegetarian diet) Atrophy or loss of gastric mucosa (eg, pernicious anemia, gastrectomy, ingestion of caustic material, hypochlorhydria, histamine 2 [H2] blockers) Functionally abnormal IF Inadequate proteolysis of dietary Cbl Insufficient pancreatic protease (eg, chronic pancreatitis, Zollinger-Ellison syndrome) Bacterial overgrowth in intestine (eg, blind loop, diverticula) Disorders of ileal mucosa (eg, resection, ileitis, sprue, lymphoma, amyloidosis, absent IF-Cbl receptor, Imerslünd-Grasbeck syndrome, Zollinger-Ellison syndrome, TCII deficiency, use of certain drugs) Disorders of plasma transport of cobalamin (eg, TCII deficiency, R binder deficiency) Dysfunctional uptake and use of cobalamin by cells (eg, defects in cellular deoxyadenosylcobalamin [AdoCbl] and methylcobalamin [MeCbl] synthesis) According to Guyton 11th Ed Pg 427 In Megaloblastic anemia, the erythroblasts cannot proliferate rapidly enough to form normal numbers of red blood cells. These red blood cells are mostly oversized, have bizarre shapes and have fragile membranes hence rupture easily.

Answer 386

15B Version Comments A. B12 is absorbed in terminal ileum: POSSIBLY CORRECT 'absorption in the ileum' although not specifically the terminal ileum (Power & Kam, p364) B. Due to a lack of ingested intrinsic factor: INCORRECT 'intrinsic factor secreted by stomach' (Power & Kam, p364) C. B12 is destroyed by gastric acid: INCORRECT, in fact the opposite is true - dietary B12 is liberated from dietary proteins by hydrogen ions (Stoelting, p724) D. Due to increased erythropoesis: INCORRECT, B12 acts as a hydrogen acceptor coenzyme (Guyton & Hall, 11th ed, p877) so should not be consumed or exhausted with increased erythropoesis E. Due to a lack of vitamin B12 intake for a few weeks: INCORRECT 'enough vitamin B12 can be stored (in the liver) to last for at least 1 year and maybe several years' (Guyton & Hall, 11th ed, p862) BEST ANSWER A Dr mitta 19:45, 12 November 2015 (CST)

Answer 387

A Likely true - from transporter figure (25-8 p478 Ganong) appears passive but via specific transport mechanism B False - apoferritin is in mucosal cells NOT lumen c True - absorption reduced by alkali D Likely True - absorption increased by gastric acid which keeps in ferrous state (but ferric reductase in brush border can convert 3+ to 2+ so could still be absorbed?) - Acid pH helps solubalize the iron, but I am not sure whether it is REQUIRED for absorbtion per se... E False - when incorporated into haem in mitochondria of RBC F Likely false - H.O. is enz involved in haem breakdown to biliverdin, Fe and CO. Neither Power and Kam nor Ganong state Vit c is cofactor G ? - There is haem transporter in brush border which allows haem absorption but as only approx 5% of dietry iron is absorbed perhaps "readily" is a bit too strong. Dietary heme iron is an important nutritional source of iron in carnivores and omnivores that is more readily absorbed than non-heme iron derived from vegetables and grain. Most heme is absorbed in the proximal intestine, with absorptive capacity decreasing distally. Comment A. Iron is absorbed mainly in the duodenum and jejenum, via a divalent metal transporter. It cross the apical membrane via facilitated diffusion (passive). It is absorbed in the ferrous (Fe2+) form, and the gastric pH and Vit C help convert ferric to ferrous form. The ferrous form is soluble at higher pH as well, meaning it is readily absorbed from small intestine. Iron is absorbed by pinocytosis which is in itself an active process, so A is not correct, I feel it is B or C depending on the exact wording on the day. references== Guyton 12th ed p419 Power and Kam p239-240 Ganong 22ed p115, 536, 477-78 Brandis (Iron, pg 198, 2007 ed)

Answer 388

Answer is B. A = ?FALSE. As iron levels INCREASE iron binds to the mRNA for apoferritin and increases translation, and therefore production, of apoferritin. I could not find a text that specifically states that in iron deficiency the synthesis of apoferritin decreases, but it makes sense that when iron levels decrease, so does the synthesis of apoferritin (if Fe is required to increase synthesis/production). B = TRUE. The transferrin saturation falls. C = FALSE. In iron deficiency, the plasma transferrin concentration increases, though the saturation is low. Therefore synthesis can not be decreased. D - FALSE. Ferritin is the storage form of iron, so in deficiency the levels will be DECREASED. E - FALSE. Hemosiderin is aggregated ferritin, which is deposited in tissues in iron overload states. Obviously in iron deficiency there will be LESS production of haemosiderin. IRON DEFICIENCY ANEMIA F.A. Rice, ART, CLS March 1, 1996 Please send comments to: F.A. Rice This lecture will concentrate on iron deficiency anemia. For discussion on anemia in general refer to the University of Washington or the Healthnet. Iron deficiency anemia is the most common anemia in the world. Iron is an essential component of the hemoglobin molecule, without iron the marrow is unable to produce hemoglobin. The red cell number falls and those which do reach the circulation are smaller than normal (microcytic) and lack hemoglobin, hence they are pale and under colored (hypochromic). The deficiency in iron may be absolute, that is, there is no iron available for the production of hemoglobin, this is true iron deficiency anemia. The deficiency may be relative, that is, the iron is present in storage in the marrow but is unavailable for hemoglobin production, this is anemia of chronic disease. Outline: Normal Iron Physiology total body iron iron sources and balance iron distribution iron absorption iron transport iron transfer into the RBC iron storage iron requirements causes of iron deficiency Iron deficiency anemia Development of anemia Laboratory diagnosis of iron deficiency routine procedures special procedures Normal Iron Physiology 1. Total Body Iron The amount of iron present varies with the body size, age and sex of the individual. -2 gm - 5 gm in average adult -4 gm average in the adult male -2.5 gm average in the adult female 2. Iron sources and balance The average normal North American diet contains approximately 15-20 mg of iron. Most is present in meat and green vegetables. Approximately 1.0 mg is absorbed each day and just about an equal amount is lost in the feces and sweat. As a result, the average adult is in a tenuous state of balance. This delicate balance is of little consequence as there is slightly more iron absorbed than lost and a store of iron is accumulated. If, however, the rate of iron loss increases, usually through blood loss such as chronic bleeding, these stores can be depleted and an absolute iron deficiency develops. 3. Iron distribution The majority of the iron present is present as hemoglobin iron. Approximately 25% of the iron is maintained as storage iron (ferritin and hemosiderin) primarily in the bone marrow. - hemoglobin 65 to 70% 1.5 to 3.0 gm - storage ferritin hemosiderin 20 to 30% 0.5 to 1.5 gm - other myoglobin heme enzyme remainder 4. Iron absorption Iron absorption occurs primarily in the duodenum. Most of this iron is in the ferric (+++) form and is complexed to other organic and inorganic molecules. The acid in the stomach and hydrolytic enzymes in the small intestine release the iron from these complexes. It is then reduced to the ferrous (++) form as it is more readily absorbed in this state. Absorption is increased by the presence of: glucose fructose some amino acids ascorbic acid (Vitamin C) These substances aid in the absorption process by either reducing ferric iron to the ferrous state or by helping bind the iron to the mucosal cell receptor sites. It is recognition of the positive effects of vitamin C which has resulted in many iron supplements being manufactured with this vitamin present. Heme iron, iron in meat myoglobin, is more easily absorbed than elemental iron. Iron absorption is decreased by the presence of: phosphate bicarbonate bile acids Once the ferrous iron (++) binds to receptors on the surface of mucosal cells it is moved into the cell. This is an energy dependent process. In the mucosal cell the iron is oxidized back to the ferric (+++) state and bound to apoferritin in the cell. This continues until all the apoferritin bound at which point newly absorbed iron is no longer oxidized but rather is passed through the cell and into the portal circulation still in the ferrous state. In the blood, iron is bound to transferrin in the ferric state. Bound to transferrin, the iron is transported to the marrow for use or storage. 5. Regulation of iron absorption The intestinal wall is covered with villi, finger-like projections, which are covered with absorptive mucosal cells. These cells are produced in the crypts of Lieberkhun, at the base of the villi, and move upwards to the villus tip to be disquamated (lost). Each cell is produced with a set amount of apoferritin. The more iron required by the body the less apoferritin manufactured in each cell. In other words, it is the amount of apoferritin within each mucosal cell which acts as the gatekeeper and regulates the amount of iron absorbed. 6. Iron transport Transferrin is the primary iron transport protein. It is a beta globulin and is produced in liver. It has a 1/2 life of 8-11 days. Each molecule of transferrin can bind and transport two molecules of iron in the ferric (+++) state. Transferrin prefers to carry iron to the marrow but will carry iron to other organs if the marrow is damaged or excess amounts of iron are already stored in the marrow. In rare instances when transferrin is absent (atransferrinemia) other proteins can bind iron but carry the iron to other organs such as liver, spleen and pancreas, little if any is carried to the marrow. As well as specific receptors for iron, transferrin has specific receptors for sites on the developing normoblast and RE cell. Once bound to the cell membrane, the transferrin changes shape and releases the iron. It then returns to the portal circulation to bind more iron. Under normal circumstances approximately 1/3 of the transferrin has iron bound to it. 7. Iron transfer across the Red Cell membrane Iron can be transfered to developing red cells either bound to transferrin or presented as ferritin to the developing cells as they cluster around "nurse cell" RE cells. The iron is moved into the developing red cell by a process similar to pinocytosis called ropheocytosis. Clusters of normoblasts around a nurse cell are called a "feed islands". 8. Iron storage Iron is stored as either ferritin or hemosiderin. -ferritin consists of an outer protein shell with iron complexed within the core. The outer shell consists of 22 apoferritin molecules while the core consists of an iron/phosphate complex consisting of 4,000 to 5,000 molecules of iron in each core. Ferritin is water soluble and a very small amount is dissolved in the plasma. The more ferritin stored the more dissolved in the plasma. The ferritin reference range for males is 40 to 300 ug/l and 20 to 150 ug/l for females. Ferritin is not visible by light microscopy, nor is it stained by the Prussian Blue reaction. It is preferentially used before hemosiderin, probably because it is soluble. -hemosiderin is aggregated ferritin molecules. The protein shell has been altered and as a result it is water insoluble. It can be seen by light microscopy as gold-brown granules and is demonstrated by the Prussian Blue stain. 9. Daily Iron Requirements The adult male requires approximately 1.0 mg/day. Just enough to cover normal iron loss. The adult female requires approximately 2.0 mg/day. Enough for daily loss and menstruation. Pregnant females require approximately 3.0 mg. Enough for normal, on going loss and fetal requirements. Children require approximately 2.0 mg/day. Enough for normal loss and extra to produce some residual iron stores and allow for increasing red cell mass. 10. Causes of Iron Deficiency diet - uncommon except in children failure to absorb increased utilization pregnancy adolescent growth atransferrinemia failure to utilize lead poisoning chronic diseases blood loss chronic blood loss is the most common cause of iron deficiency anemia. Iron deficiency anemia Development of anemia It must be remembered that anemia in iron deficiency develops slowly. The type and severity of the anemia varies with time. Development Stages: 1. depletion of iron stores, decreased ferritin levels, no anemia 2. increased transferrin levels, no anemia 3. fall in serum iron, no anemia 4. development of normocytic, normochromic anemia 5. development of microcytic, hypochromic anemia Laboratory Diagnosis of Iron Deficiency Routine procedures Hgb, Hct and RBC count are all decreased. The degree of decrease depends upon the length of time the marrow has been without sufficient supplies of iron. It must be remembered that at any stage the red cell number will not be proportionately as low as is the Hgb and Hct. This is due to the fact that the marrow can continue to produce empty cells. Indices - MCV - decreased, MCH - decreased, MCHC - decreased. The MCHC is the last to become lowered. This is due to the fact that as the marrow becomes more and more depleted of iron it produces smaller cells with a smaller amount of hemoglobin in each in an attempt to keep the concentration of hemoglobin in each normal. The RDW is increased which reflects the anisocytosis characteristic of iron deficiency. Morphology - changes from normal to simple microcytic to hypochromic microcytic as the iron deficiency progresses. When full blown there is marked anisocytosis and poikilocytosis with elliptocytes and target cells. NRBC may be seen on a rare occasion. White cell count and differential - normal Platelets - normal to increased. They are usually microcytic and stain very pale and can be missed on film. Retics - the relative count is decreased to normal while the production index (PI) is decreased. Special procedures A bone marrow examination is seldom, if ever, performed or needed for the diagnosis of an iron deficiency anemia. If however, a bone marrow is performed the following results would be present. Bone Marrow cellularity - normal to increased morphology - normoblastic with some dyserythropoiesis. Ragged reduced cytoplasm, vacuoles, multinuclearity, karryohexis, nuclear budding, abnormal mitosis. All these may be seen but are not the predominant features of the marrow. iron stain - absent. The absence of iron is considered to be the "gold standard" for the diagnosis of iron deficiency. siderocytes - absent Serum iron - decreased TIBC - increased %saturation - decreased Ferritin - decreased Free erythrocyte protoporphyrin (FEP) - increased NOTE! The above review provides strong evidence for using serum ferritin as an initial laboratory test for the evaluation of iron deficiency anemia. Bilirubin - normal to decreased Treatment Iron supplement. The response is monitored with the retic count, hemoglobin and hematocrit. A failure to respond may be due to: 1. continued bleeding 2. failure to take iron 3. wrong diagnosis 4. mixed deficiency 5. other causes - inflammation 6. malabsorption - unlikely 7. use of slow release iron

Answer 389

The answer is B - shedding of iron containing intestinal mucosal cells is the major route of iron excretion. The body does NOT control the excretion of iron. This is a bit of an unusual situation as for most other ions, control is exerted on excretion. For example, for Na+ and K+, the body maintains the body content by controlling renal excretion. What happens with iron is that there is a small loss each day which for the body to maintain balance means this much iron must be absorbed per day (on average). The loss each day is predominantly loss of blood (and other cells) which contain iron. This loss is small but larger in females (on average) because of menstrual blood losses, as erythrocytes have a significant iron content in haemoglobin. So the control of total body iron is exerted at the level of GIT absorption by "mucosal block" (also referred to as "mucosal intelligence"). Iron absorbed from the small bowel (predominantly the duodenum) is initially stored within the mucosal cell (as ferritin). An amount is absorbed into the blood & bound to transferrin. That which is absorbed into the mucosal cell but not taken up into the blood is lost as ferritin with the shedding of the intestinal mucosal cells. In some older texts (eg my old edition of Guyton) there is mention of excretion of transferrin (in the bile?) and binding of iron to it in the bowel but this is completely wrong!

Answer 390

Answer A Best Answer Vagal stimulation increase release ACH which acts on M3 on parietal cells – direct mechanism GRP – Gastric Release Peptide – indirect mechanism – this is released directly from nerve terminals. 1-Stimulates G cells to release Gastrin. 2-Acts directly on parietal cells Gastrin Receptors – minor mechanism. 3-Releases of histamine from Enterochromaffin cells which inturn acts on Parietal Cells – principle mechanism. Gastric Secretion is inhibited by Somatostatin Epidermal Growth Factor Beta adrenergic Agonist (so isn't C correct then?) Enteroglucagon Comment: i think Ach acts on M1 receptors on parietal cells, M3 is for increasing gut motility. Comment: "Acid secretion is stimulated by histamine via H2 receptors and by ACh via M3 muscarinic receptors." (Ganong 22nd ed, pg 493) Comment: "Pirenzepine inhibits gastric acid secretion at doses that have little effect on salivation or heart rate. Since the muscarinic receptors on the parietal cells do not appear to have a high affinity for pirenzepine, the M1 receptor responsible for alterations in gastric acid secretion is postulated to be localized in intramural ganglia (Eglen et al., 1996). Blockade of ganglionic muscarinic receptors (rather than those at the neuroeffector junction) also appears to underlie the ability of pirenzepine to inhibit the relaxation of the lower esophageal sphincter." (Goodman and Gilman's-chapter 7). So, D appears to be right too. M1 antagonists work as well as H2 receptor antag to reduce gastric acid secretions (but too many SE) just not via the M3 receptors in pariental cells.

Answer 391

A: Correct - Secretin is secreted from S cells in the upper intestinal mucosa, it increases secretion of bicarbonate by pancreatic duct cells and biliary tract. B: Gastrin is secreted by G cells located mainly in the gastric antrum. It stimulates the production of hydrochloric acid by parietal cells in the body of the stomach and so decreases the pH of gastric contents. It also stimulates the secretion of pepsinogen from chief (peptic) cells. C: Intrinsic factor is also secreted by parietal cells in the stomach. It binds with vitamin B12 in the stomach and prevents it from being digested before it can be absorbed in the terminal ileum. D: Cholecystokinin is released by cells in the duodenum and acts on both the pancreas and gallbladder; it simulates the release of trypsinogen, chymotrypsinogen, amylase and lipase from the pancreas; it stimulates the production of bile, the contraction of the gallbladder and the relaxation of the Sphincter of Oddi. It also mediates satiety in the CNS. CCK may potentiate the effect of secretin on increasing the duodenal pH. E: Gastrin releasing peptide also known as bombesin, is secreted from the enteric nerve of the gastric mucosa stimulating the release of gastrin from G cells. F: Pepsin is a proteolytic enzyme which is optimally active in a pH of 1.8 to 3.5; pepsinogen, the inactive precursor is secreted from gastric chief cells.

Answer 392

Carbohydrate leaves the stomach in a few hours, protein is slower than carbohydrates and fatty meals are the slowest. B is right

Answer 393

after Ganong... 'Stomach emptying depends on the type of food ingested. Food rich in carbohydrate leaves the stomach in a few hours. Protein-rich food leaves more slowly, and emptying is slowest after a meal containing fat. The rate of emptying also depends on the osmotic pressure of the material entering the duodenum. Hyperosmolality of the duodenal contents is sensed by "duodenal osmoreceptors" that initiate a decrease in gastric emptying which is probably neural in origin.' Ganong also notes that some people deliberately (or subconsciously!) ingest high quantities of fat before Coctail parties in order that the EOH is release more slowly into circulation as a result of v gastric emptying! ref Faunce pg 175 & Ganong So B is correct

Answer 394

from Ganong...'Secretion of pancreatic juice is primarily under hormonal control. Secretin acts on the pancreatic ducts to cause copious secretion of a very alkaline pancreatic juice that is rich in HCO3- and poor in enzymes. The effect on duct cells is due to an increase in intracellular cAMP. Secretin also stimulates bile secretion. CCK acts on the acinar cells to cause the release of zymogen granules and production of pancreatic juice rich in enzymes but low in volume. Its effect is mediated by phospholipase C '

Answer 395

Answer is C - by facilitated diffusion. Glucose is absorbed in the cells of the small intestine by two means - facilitated diffusion and by Na+ dependent facilitated transport. Na+ dependent transporters are present on the luminal side of the absorptive cells. The intracellular concentration of Na+ is kept low by the Na+/K+ ATPase at the basolateral membrane, allowing co-transport of glucose against a concentration gradient. Facilitative transporters are present on both the mucosal and serosal sides of the cell. Movement via these transporters is solely down the concentration gradient. Facilitative glucose transporters exist in several different isoforms known as GLUT 1 to GLUT 5. These are present in different tissues in the body and have different specific properties. Reference: Basic Medical Biochemistry - a clinical approach. Marks, Marks & Smith. I disagree Answer B Ganong (22 ed, pp 336 and 470) says glucose uptake by facilitated diffusion in muscle and fat under the influence of insulin. Glucose uptake by the small intestine is coupled with sodium via SGLT and secondary active transport Addit: I also agree Answer B is correct. The relevent piece in Ganong is quite clear that glucose is an example of secondary active transport linked to Na influx into epithelial cells, not facilitated diffusion. Yes. Answer B. It is secondary active transport by the SGLT. Facilitated diffusion can occur with fructose (GLUT-5) and passive diffusion with ribose/pentose sugars. This is the whole basis of glucose-containing rehydration solutions. As an aside, maximal glucose absorption is 120g/hr. The answer to the facilitated diffusion question is B; but a more thorough expose is included in Journal of Physiology (2001), 531.3, pp. 585-595.

Answer 396

salivary lipase and amylase begins lipid and carbohydrate digestion, acid & pepsins in the stomach begin protein digestion So... A is correct after http://gasboys.net

Answer 397

Calcium is actively transported from the intestine by calcium dependent ATPase regulated by 1,25 dihydroxycholecalciferol and some absorption occurs via passive diffusion. Calcium is 30-80% absorbed from the intestines. Phosphates and oxalates form insoluble salts with calcium preventing absorption. Calcium absorption is also facilitated by protein. About 30-50% ingested calcium is absorbed. Comment: None of the options appear to be the correct answer. What about option B? I agree, option B is correct. Ganong 17th Ed p440 states that "active transport is facilitated by 1,25-dihydroxycholecalciferol, the metabolite of Vit D that is produced in the kidneys. The metabolite induces the synthesis of a Ca-binding protein in the mucosal cells. The rate of production of 1,25-dihydroxycholecalciferol is increased when the plasma calcium level is decreased...Consequently, Ca absorption is adjusted to body needs". figure8.gif "Absorption of calcium from the luminal contents of the intestine involves both transcellular and paracellular pathways. The transcellular pathway dominates in the duodenum, and this is the pathway primarily regulated by 1,25 dihydroxyvitamin D (1,25(OH)2D). The figure (above) shows a model of our current understanding of how this process is regulated by 1,25(OH)2D. Calcium entry across the brush border membrane (BBM) occurs down a steep electrical-chemical gradient and requires no input of energy. Removal of calcium at the basolateral membrane must work against this gradient, and energy is required. This is achieved by the CaATPase, an enzyme induced by 1,25(OH)2D in the intestine. Calcium movement through the cell occurs with minimal elevation of the intracellular free calcium concentration by packaging the calcium in calbindin containing vesicles (149-151) that form in the terminal web following 1,25(OH)2D administration". - from [1] comment Guyton says that Calcium is absorbed from the small intestine via facilitated diffusion. Don't know if I believe him. Another one for you. Molina (Endocrine Physiology) says:- Ca absorption is via saturable transcellular pathways, and non-saturable paracellular pathways Ca entry through the apical membrane is a passive process This is facilitated by Ca-binding proteins (Vit D - dependent - I guess keeps the free intracellular [ca] low for further diffusion) Ca is actively extruded across the basolateral membrane Note Option B: Requires a carrier protein on the MUCOSAL side. Not true. The Carrier Protein (Ca-ATPase) is on the basolateral side, not the mucosal/apical side which is passive. And note the wording for the quote The metabolite induces the synthesis of a Ca-binding protein in the mucosal cells - it states IN MUCOSAL cells, not ON THE MUCOSAL SIDE. This is clearly incorrect if remembered correctly. I would choose Option A though it has its many problems, purely because C/D/E is clearly wrong, and passive diffusion DOES occur both apically and paracellularly, whereas there is definately NO carrier protein on the MUCOSAL SIDE of the enterocyte. Hopefully it was remembered wrong, or ANZCA makes the options more clear. Edit: if a Ca/H ATPase pump is required to get Ca levels low enough in the luminal cell for Ca to be absorbed on the luminal side, how is this not secondary active?

Answer 398

CORRECT ANSWER E A: INCORRECT: facilitated by protein (Ganong, 24e, ch 26) B: INCORRECT: 30-80% of ingested Ca2+ is absorbed (Ganong, 24e, ch 26) C: INCORRECT: forms insoluble salts with phosphate so absorption is inhibited (Ganong, 24e, ch 26) D: INCORRECT: Ca2+ enters unbound via a receptor (TRPV6) on the intestinal brush border, and the calcium binding protein (calbindin-D9k) is intracellular, not mucosal (Ganong, 24e, ch 21) E: CORRECT: - ‘passive entry across the apical membrane, cytosolic diffusion facilitated by vitamin D–dependent calcium-binding proteins, and active extrusion of Ca2+ across the opposing basolateral membrane mediated by a high-affinity Ca2+-ATPase and Na+/Ca2+ exchanger.’ (Molina, 4e) - ‘Calcium is absorbed in the mammalian small intestine by two general mechanisms: a transcellular active transport process, located largely in the duodenum and upper jejunum; and a paracellular, passive process that functions throughout the length of the intestine.' (Bonner) Dr mitta 06:47, 22 November 2015 (CST)

Answer 399

GI16 Ganong states that chyme in the jejunum normally contains few if any bacteria with more microorganisms being present in the ileum. The colon regularly contains large numbers of bacteria. It is thought the reason for the relative sterility of the jejunal is likely to be due to gastric acid and the comparatively rapid transit of the chyme may inhibit bacterial growth. Thus, E is the answer.

Answer 400

GI16 Ganong states that chyme in the jejunum normally contains few if any bacteria with more microorganisms being present in the ileum. The colon regularly contains large numbers of bacteria. It is thought the reason for the relative sterility of the jejunal is likely to be due to gastric acid and the comparatively rapid transit of the chyme may inhibit bacterial growth. Thus, E is the answer.

Answer 401

GI16b A. Wrong B. Correct C. Partly correct - The principle methods of defence against bacteria are mechanical (transit time, pH, epithelial sloughing) rather than IgA and cellular immunity D. Wrong E. Wrong - notable numbers in ileum Most correct answer: B (same answer as E above, but different distractors)

Answer 402

I bet the answer was immunoglobin synthesis - otherwise it'll be something obscure. Looks like you were right

Answer 403

Best answer is D Acid in the gastric antrum inhibits gastrin secretion Reference: Ganong 22nd ed p. 485 (including Table 26-3) Comment Gastrin secretion is stimulated by: Vagus, Amino acids/Peptides in stomach, Distention of stomach, Cathecolamines Gastrin secretion is inhibited by: Acid in stomach, Secretion, GIP, VIP

Answer 404

Answer is B A. Misoprostol is a gastroprotectant but causes diarrhoea (not constipation) (p. 973 & 995) B. Both ACh and gastrin directly and indirectly stimulate gastric acid secretion (p. 967-8) C. Proton pump inhibitors (e.g. omeprazole) irreversibly inactivate the H+,K+-ATPase molecule (p. 969) D. PPIs are the most potent suppressors of gastric acid secretion, while M1 antagonists (e.g. pirenzepine) have poor efficacy (p. 969 & 975) The pages numbers above indicate references in Goodman & Gilman 11th ed.

Answer 405

The answer is D | Once the liver's glycogen reserves are full, any additional carbohydrate will be converted to fat

Answer 406

See also: Vitamin D Bilirubin is conjugated in the liver Immunoglobulins are produced by plasma cells (Most plasma proteins are produced in the liver; some important exceptions are immunoglobulins & von Willebrand factor) Cholesterol is produced in the liver Cholecalciferol is produced in the skin So there are 2 correct options in the question version above: both immunoglobulins AND cholecalciferol are NOT produced in the liver. Perhaps the idea was that the term "cholecalciferol" is used to refer to the various forms of vitamin D3: cholecalciferol, 25-(OH)cholecalciferol & 1,25-(OH)2-cholecalciferol but this is a bit of a stretch. More likely the question has been remembered incorrectly. Comment: Power & Kam (pg 185) under the heading "Functions of the Liver" include "immunological functions associated with the synthesis of immunoglobulins, and the phagocytic action of Kupffer cells." Perhaps they refer to plasma cell production of immunoglobulins in the liver. Perhaps just go with answer D. Alternate view: Power and Kam says "immunological functions ASSOCIATED with the synthesis of immunoglobulins, etc..... " --> perhaps it refers to the functions related to the synthesis of immunoglobulins, not the actual synthesis of Ig..? Cholecalciferol is synthesized in the skin, converted to 25-OHcholecalciferol in liver. Hence, I think that both B and D are incorrect. References Power & Kam 1st Ed, p 185. Power & Kam 2nd Ed, p 209.

Answer 407

Bile salt micelles ([1]) are required for the absorption of Vitamin A,D,E and K (ie the fat soluble vitamins) Vitamin D absorption is dependent on micelles. However, most Vitamin D is from skin synthesis and via absorbed. Bile salts is an important constituent in forming micelles in the small intestine. So both bile acids and vitamin D seem correct. Glycerol is hydrophilic and is NOT absorbed via micelles (eg see [2] - esp abstract & introduction) ``` ADD Ganong (pg 502) states that "90-95% of the bile salts are absorbed from the small intestine...most from the terminal ileum." Therefore, stick with answer A (Vitamin D) ```

Answer 408

A wrong - it CAN be this, but more commonly the figures are as per below B is correct "The portal vein normally accounts for 3/4 of the blood supply, but the hepatic artery provides 3/4 of the oxygen consumed by the liver." Power and Kam pp 173. Side note - Brandis quotes oxygen DELIVERY which is a different thing, as majority portal C wrong - Ganong ed 21 pp 627 "Portal venous pressure is normally about 10 mmHg in humans" D wrong.. obviously, whatever systemic MAP is E wrong, Power & Kam "The liver blood flow is 1/4 of the cardiac output" pp 173

Answer 409

Answer is PGI2 (prostacyclin) - most potent inhibitor of platelet aggregation (by increasing platelet cAMP) and causes relaxation of vascular smooth muscle. "PGI2 plays an important role in vascular function because, like nitric oxide, it inhibits platelet adhesion to the vascular endothelium and is a strong vasodilator. Damaged endothelial cells do not produce PGI2, thereby making the vessel more susceptible to thrombosis and vasospasm. Thromboxanes (e.g., TXA2) and leukotrienes (e.g., LTC4) produce vasoconstriction and are important modulators of vascular function during tissue injury and inflammation. Prostaglandins have a vascular role during inflammation, and also play a more subtle role in normal flow regulation, most notably as modulators of other control mechanisms. Prostaglandins have both vasoconstrictor (e.g., PGF2a) and vasodilator activities (e.g., PGE2). Leukotrienes and prostaglandins can also make the vascular endothelium more "leaky" thereby promoting edema formation during inflammation." -from Arachidonic Acid Metabolites "Prostacyclin (PGI2) acts chiefly to prevent platelet adhesion. It is also an effective vasodilator. Prostacyclin's interactions in contrast to thromboxane, another eicosanoid, strongly suggest a mechanism of cardiovascular homeostasis between the two hormones in relation to vascular damage." - from Prostacyclin

Answer 410

Answer is PGI2 (prostacyclin) - most potent inhibitor of platelet aggregation (by increasing platelet cAMP) and causes relaxation of vascular smooth muscle. "PGI2 plays an important role in vascular function because, like nitric oxide, it inhibits platelet adhesion to the vascular endothelium and is a strong vasodilator. Damaged endothelial cells do not produce PGI2, thereby making the vessel more susceptible to thrombosis and vasospasm. Thromboxanes (e.g., TXA2) and leukotrienes (e.g., LTC4) produce vasoconstriction and are important modulators of vascular function during tissue injury and inflammation. Prostaglandins have a vascular role during inflammation, and also play a more subtle role in normal flow regulation, most notably as modulators of other control mechanisms. Prostaglandins have both vasoconstrictor (e.g., PGF2a) and vasodilator activities (e.g., PGE2). Leukotrienes and prostaglandins can also make the vascular endothelium more "leaky" thereby promoting edema formation during inflammation." -from Arachidonic Acid Metabolites "Prostacyclin (PGI2) acts chiefly to prevent platelet adhesion. It is also an effective vasodilator. Prostacyclin's interactions in contrast to thromboxane, another eicosanoid, strongly suggest a mechanism of cardiovascular homeostasis between the two hormones in relation to vascular damage." - from Prostacyclin

Answer 411

The answer appears to be A: PGE2. PGE2 has complex effects on mammalian airway smooth muscle tone and airflow which may be a result of its action on a variety of cells and also the variety of its receptor subtypes that have different signaling characteristics. It is active on sensory nerve terminals, smooth muscle, mucous glands and a variety of inflammatory cells. Its action on EP2 receptor is thought to mediate relaxation of airway smooth muscle but it also can also mediate the opposing effect of constriction via the EP1 receptor. PGE2 may be released from epithelial cells and smooth muscle cells in the airways to reduce the bronchoconstriction response seen in asthma.[1]. It also inhibits proliferation of airway smooth muscle cells which contributes to airway caliber.[2] Thromboxane is a potent constrictor of human airways in vitro and in vivo which acts via the TP receptor. PGD2 and PGF2 also act as bronchoconstrictors via TP receptors.[3] Leukotrienes are mediators of allergic responses and inflammation. Both LTB4 and LTD4 are involved in the immediate phase of asthma, mediating bronchospasm. LTB4 also acts as a chemotaxin which attracts leucocyes resulting the delayed phase seen in asthma.

Answer 412

From [1]: "Massive transfusion is associated with numerous deleterious effects and metabolic disturbances. The rate of mortality in patients who receive massive transfusion is 40%, which increases to 75% if coagulopathies develop." From the same reference: Other complications of massive transfusion are: hypocalcaemia and citrate toxicity caused by decreased citrate metabolism transient hyperkalaemia, particularly in neonates and patients with end-stage renal disease hypocalcaemia caused by hypovolemic activation of the renin-angiotension aldosterone pathway decreased oxygenation to the peripheral tissues due to decreased 2, 3-DPG levels in packed RBCs transfused close to their expiration date persistent acidosis caused by continued hemorrhaging, hypovolemia or ischemia hypothermia caused by infusion of RBCs stored at four degrees Celsius pulmonary edema air emboli microaggregates hindering blood delivery, caused by the formation of fibrin, white blood cells and platelets in packed RBCs undergoing prolonged storage transfusion related acute lung injury (TRALI), a noncardiogenic pulmonary edema occurring within six hours of transfusion. TRIM (transfusion related immunomodulation) - better outcomes with transplants BUT increased infection, cancer recurrence and SIRS/MODS infection (bacterial, viral, protozoal, prion disease, endotoxaemia) serological incompatability, rhesus conversion jaundice and impaired reticuloendothelial function coagulopathy/impaired haemostasis (dilution, depletion/DIC, decreased production) vasoactive reactions (kinin activation, damaged platelets and granulocytes) Addition: This list does not provide adequate information to answer the question - ie: which is MOST likely to happen, and in what TIMEFRAME? The five major problems encountered during massive transfusion are hypovolemia hypothermia, coagulopathy hyperkalaemia hypocalcemia. See: [2]. I would say the answers are C, B, and D respectively (although the last could be C as well..). Comment? Easiest way to remember - Complications Of Massive Transfusion = Coagulopathy Oxygenation Metabolic derangement Temperature Addit 2,3 DPG "binds with the β chains of haemoglobin, changing the protein conformation and reducing the oxygen affinity". Hence, transfusion of DPG deplete Packed Cells will increase the affinity of oxygen for Hb and thus impair oxygen unloading in the tissues. So for BL03c, the answer cannot be "D". It must be "C". [Ref Power & Kam p.67] commenting on the above line,agree,c is correct,decrease 23dpg,increase affinity of hb to o2,means decrease offloading.,ODC might not lt shift so much though due to rt shifting via acidosis.so least likely will be D answer. Another answer - probably most common is dilutional coagulopathy that we see. Are they trying to trick us with "coagulopathy due to hypocalcemia"? Would have to be pretty low for that to happen, much more common from simple dilution. - I agree, hypocalcaemia would produce cardiovascular collapse before it causes coagulopathy

Answer 413

From [1]: "Massive transfusion is associated with numerous deleterious effects and metabolic disturbances. The rate of mortality in patients who receive massive transfusion is 40%, which increases to 75% if coagulopathies develop." From the same reference: Other complications of massive transfusion are: hypocalcaemia and citrate toxicity caused by decreased citrate metabolism transient hyperkalaemia, particularly in neonates and patients with end-stage renal disease hypocalcaemia caused by hypovolemic activation of the renin-angiotension aldosterone pathway decreased oxygenation to the peripheral tissues due to decreased 2, 3-DPG levels in packed RBCs transfused close to their expiration date persistent acidosis caused by continued hemorrhaging, hypovolemia or ischemia hypothermia caused by infusion of RBCs stored at four degrees Celsius pulmonary edema air emboli microaggregates hindering blood delivery, caused by the formation of fibrin, white blood cells and platelets in packed RBCs undergoing prolonged storage transfusion related acute lung injury (TRALI), a noncardiogenic pulmonary edema occurring within six hours of transfusion. TRIM (transfusion related immunomodulation) - better outcomes with transplants BUT increased infection, cancer recurrence and SIRS/MODS infection (bacterial, viral, protozoal, prion disease, endotoxaemia) serological incompatability, rhesus conversion jaundice and impaired reticuloendothelial function coagulopathy/impaired haemostasis (dilution, depletion/DIC, decreased production) vasoactive reactions (kinin activation, damaged platelets and granulocytes) Addition: This list does not provide adequate information to answer the question - ie: which is MOST likely to happen, and in what TIMEFRAME? The five major problems encountered during massive transfusion are hypovolemia hypothermia, coagulopathy hyperkalaemia hypocalcemia. See: [2]. I would say the answers are C, B, and D respectively (although the last could be C as well..). Comment? Easiest way to remember - Complications Of Massive Transfusion = Coagulopathy Oxygenation Metabolic derangement Temperature Addit 2,3 DPG "binds with the β chains of haemoglobin, changing the protein conformation and reducing the oxygen affinity". Hence, transfusion of DPG deplete Packed Cells will increase the affinity of oxygen for Hb and thus impair oxygen unloading in the tissues. So for BL03c, the answer cannot be "D". It must be "C". [Ref Power & Kam p.67] commenting on the above line,agree,c is correct,decrease 23dpg,increase affinity of hb to o2,means decrease offloading.,ODC might not lt shift so much though due to rt shifting via acidosis.so least likely will be D answer. Another answer - probably most common is dilutional coagulopathy that we see. Are they trying to trick us with "coagulopathy due to hypocalcemia"? Would have to be pretty low for that to happen, much more common from simple dilution. - I agree, hypocalcaemia would produce cardiovascular collapse before it causes coagulopathy

Answer 414

From [1]: "Massive transfusion is associated with numerous deleterious effects and metabolic disturbances. The rate of mortality in patients who receive massive transfusion is 40%, which increases to 75% if coagulopathies develop." From the same reference: Other complications of massive transfusion are: hypocalcaemia and citrate toxicity caused by decreased citrate metabolism transient hyperkalaemia, particularly in neonates and patients with end-stage renal disease hypocalcaemia caused by hypovolemic activation of the renin-angiotension aldosterone pathway decreased oxygenation to the peripheral tissues due to decreased 2, 3-DPG levels in packed RBCs transfused close to their expiration date persistent acidosis caused by continued hemorrhaging, hypovolemia or ischemia hypothermia caused by infusion of RBCs stored at four degrees Celsius pulmonary edema air emboli microaggregates hindering blood delivery, caused by the formation of fibrin, white blood cells and platelets in packed RBCs undergoing prolonged storage transfusion related acute lung injury (TRALI), a noncardiogenic pulmonary edema occurring within six hours of transfusion. TRIM (transfusion related immunomodulation) - better outcomes with transplants BUT increased infection, cancer recurrence and SIRS/MODS infection (bacterial, viral, protozoal, prion disease, endotoxaemia) serological incompatability, rhesus conversion jaundice and impaired reticuloendothelial function coagulopathy/impaired haemostasis (dilution, depletion/DIC, decreased production) vasoactive reactions (kinin activation, damaged platelets and granulocytes) Addition: This list does not provide adequate information to answer the question - ie: which is MOST likely to happen, and in what TIMEFRAME? The five major problems encountered during massive transfusion are hypovolemia hypothermia, coagulopathy hyperkalaemia hypocalcemia. See: [2]. I would say the answers are C, B, and D respectively (although the last could be C as well..). Comment? Easiest way to remember - Complications Of Massive Transfusion = Coagulopathy Oxygenation Metabolic derangement Temperature Addit 2,3 DPG "binds with the β chains of haemoglobin, changing the protein conformation and reducing the oxygen affinity". Hence, transfusion of DPG deplete Packed Cells will increase the affinity of oxygen for Hb and thus impair oxygen unloading in the tissues. So for BL03c, the answer cannot be "D". It must be "C". [Ref Power & Kam p.67] commenting on the above line,agree,c is correct,decrease 23dpg,increase affinity of hb to o2,means decrease offloading.,ODC might not lt shift so much though due to rt shifting via acidosis.so least likely will be D answer. Another answer - probably most common is dilutional coagulopathy that we see. Are they trying to trick us with "coagulopathy due to hypocalcemia"? Would have to be pretty low for that to happen, much more common from simple dilution. - I agree, hypocalcaemia would produce cardiovascular collapse before it causes coagulopathy

Answer 415

Answer A However no immunoglobulins have a molecular weight of 70,000! There are two light chains present in all immunoglobulins (κ and λ). Immunoglobulins comprise two light and two heavy chains linked by disulfide bonds. There are five types of heavy chains : γ (IgG), α (IgA), μ (IgM), δ (IgD), ε (IgE). IgA: MW 170 000 Second most abundant Ig in the serum Predominate class of Ig found in secretions (tears, saliva, colostrum, mucous) Synthesised by mucosal epithelial cells Exists primarily (but not exclusively!) as a dimer in the secretory form Activation of complement by the Alternate Pathway. IgG: MW 160 000 Most abundant Ig in the serum (75% of serum Ig) Exists as a monomer Protect against microbial infections by binding, recruiting, activating PMNs, NK cells, Monocytes. IgM: MW 960 000 The first antibody produced after antigenic stimulus Exists as a pentamer Potent activator of complement and excellent cytotoxin. IgE: MW 180 000 Lowest serum Ig concentration Are able to activate and degranulate mast cells Exists as a monomer IgD: MW 180 000 Act as an antigen receptor on B lymphocytes Has no known function in the serum Monomer Comment: IgM is a pentamer (see Ganong or P&K) Frightening to think that this meaningless factoid could be responsible for a fail!

Answer 416

BL05 D. Levels are inversely proportional to haematocrit A. EPO increase the number of erythropoietin committed stem cells. Increase retics only B. The principal site of inactivation is the liver C. Has a half life of 5 hours (Power & Kam quote 6-9 hr on one page and 3-8 hr on another, but still not in the realm of days) E. Polypeptide B glycoprotein (DUNNO [it is a glycoprotein, but also don't know what sort]) Ganong 22nd Ed, p 459, Figure 24-5 : My interpretation of that graph is that plasma level of EPO is proportional to Hct The graph shows EPO levels decreasing as hct increases - this is inversely proportional Also : The increase in circulating red cells that it triggers takes 2-3 days to appear (ie 48 - 72 hrs) Completely irrelevant but in regards to glycoprotein B -> I cannot find a relevant reference - Glycoprotein B - found on herpes simplex virus - Gylcoprotein P involved in multidrug resistance in bacteria Nothing to do with EPO as far as I can see

Answer 417

BL05b A. Partly correct - Red cells take 2-3 days to appear B. Uncertain - Inactivated in the liver, but not metabolised. ?Inactivated by Kupffer cells C. Uncertain - Within two paragraphs Power & Kam states both that: Glycoprotein whose carbohydrate moeity prevent metabolism in liver; hepatic degradation is the main route of elimination. (Gangong agrees that it is mainly inactivated in the liver.) D. Depends on the option - Half-life is about 5 hours

Answer 418

BL05c A. Correct - Like renin, beta-adrenergic stimulation facilitates secretion B. Wrong - Stimulates production of erythrocytes. Megakaryocytes are stimulated by thrombopoeitin. C. Partly correct - Renal excretion may contribute, although MM is 30 kDa D. Wrong - produced in both kidney (75-90% in adult) and liver (remainder: 10-25% in adult). Liver is the chief source in foetus and neonates. Ganong says that it's also produced in the brain, uterus and oviducts. E. Wrong - Production facilitated by respiratory alkalosis at altitude, not metabolic and respiratory acidosis Most correct answer: A

Answer 419

BL05 D. Levels are inversely proportional to haematocrit A. EPO increase the number of erythropoietin committed stem cells. Increase retics only B. The principal site of inactivation is the liver C. Has a half life of 5 hours (Power & Kam quote 6-9 hr on one page and 3-8 hr on another, but still not in the realm of days) E. Polypeptide B glycoprotein (DUNNO [it is a glycoprotein, but also don't know what sort]) Ganong 22nd Ed, p 459, Figure 24-5 : My interpretation of that graph is that plasma level of EPO is proportional to Hct The graph shows EPO levels decreasing as hct increases - this is inversely proportional Also : The increase in circulating red cells that it triggers takes 2-3 days to appear (ie 48 - 72 hrs) Completely irrelevant but in regards to glycoprotein B -> I cannot find a relevant reference - Glycoprotein B - found on herpes simplex virus - Gylcoprotein P involved in multidrug resistance in bacteria Nothing to do with EPO as far as I can see BL05b A. Partly correct - Red cells take 2-3 days to appear B. Uncertain - Inactivated in the liver, but not metabolised. ?Inactivated by Kupffer cells C. Uncertain - Within two paragraphs Power & Kam states both that: Glycoprotein whose carbohydrate moeity prevent metabolism in liver; hepatic degradation is the main route of elimination. (Gangong agrees that it is mainly inactivated in the liver.) D. Depends on the option - Half-life is about 5 hours BL05c A. Correct - Like renin, beta-adrenergic stimulation facilitates secretion B. Wrong - Stimulates production of erythrocytes. Megakaryocytes are stimulated by thrombopoeitin. C. Partly correct - Renal excretion may contribute, although MM is 30 kDa D. Wrong - produced in both kidney (75-90% in adult) and liver (remainder: 10-25% in adult). Liver is the chief source in foetus and neonates. Ganong says that it's also produced in the brain, uterus and oviducts. E. Wrong - Production facilitated by respiratory alkalosis at altitude, not metabolic and respiratory acidosis Most correct answer: A

Answer 420

hmm i wonder if A is the right answer for this one No , I think you're all interpreting the the '?' wrong. What if it's meant to represent the bacterial carcass after a phagocyte is done with it? I think I'll go with B - HLF

Answer 421

"Antithrombin is a serpin (serine protease inhibitor) that inactivates a number of enzymes from the coagulation system, namely the activated forms of Factor X, Factor IX and Factor II (thrombin). Its affinity for these molecules (i.e. its effectivity) is enhanced by heparin." Ganong (22nd ed, p543) says antithrombin III inhibits "the active forms of IX, X, XI, and XII" and rather oddly not mentioning factor II (thrombin)!! The correct answer then is "All of the above". Note the Link below to Wikipedia states that it inhibits factor VIIa as well, but that is NOT mentioned elsewhere - NOT in Ganong anyway. Basically I belive ATIII inhibits purely intrinsic pathway only.

Answer 422

Vitamin K is a necessary cofactor for the enzyme that catalyzes the conversion of glutamic acid residues to gamma caboxyglutamic acid residues. The following proteins (II (prothrombin), VII,IX,X, protein C and Protein S) require conversion of glutamic acid residues to gamma carboxyglutamic acid residues. Ganong (22nd ed, p543) says antithrombin III inhibits "the active forms of IX, X, XI, and XII" and rather oddly not mentioning factor II (thrombin)!! Protein C and Protein S are Vitamin K dependent, and their actions include the inactivation of factor 5a.

Answer 423

Answer is A Desmopressin is a V2-selective analogue of vasopressin (ADH), with a longer lasting antidiuretic effect but minimal vasoconstrictor actions Desmopressin has a number of clinical uses, one of which is to increase factor VIII and von Willebrand factor levels in mild haemophilia or von Willebrand’s disease This action is due to V2 receptors (see product info in MIMS), as it is inhibited by the V2-selective antagonist SK&F 105494.

Answer 424

``` Answer is D Vitamin K is responsible for the post-translational modification of protein C (γ-carboxylation). Vitamin K is a necessary cofactor for the enzyme that converts glutamic acid residues to gamma-caboxyglutamic acid. The six Vit K dependent clotting factors are; II (prothrombin) VII IX X Protein C Protein S ```

Answer 425

Answer B Translation is the formation of a protein after decoding the mRNA, and occurs outside the nucleus. Post-translational modification is the alteration of existing amino acid residues in a protein, and includes phosphorylation, glycosylation, acetylation.

Answer 426

Haemoglobin is initially broken down (by the cells of the reticuloendothelial system - macrophages) to release haem and the globin. The globin is broken down to release the amino acids for recycling into other proteins or entry into intermediary metabolism. Haem cannot be reused so is broken down: The first reaction (catalysed by the microsomal enzme haem oxygenase in the macrophages) converts haem to biliverdin, (opening the porphyrin ring), Fe++ and carbon monoxide. The next step (catalysed by biliverdin reductase in macrophages) converts biliverdin to bilirubin. The lipid soluble bilirubin (unconjugated) is carried in plasma bound to albumin. Bilirubin is taken up by hepatocytes by facilitated diffusion Bilirubin is conjugated with UDP-glucuronic acid to produce bilirubin monoglucuronide. This is actively transported into the bile canaliculi and excreted. The iron is stored in the body (ferritin) and reutilised within the body eg in synthesis of haemoproteins. INTERESTING FACT: A point to note is that the reaction catalysed by haem oxygenase is the ONLY reaction in the body that produces carbon monoxide. So some of the carbon monoxide in the body is produced there. Of course in an urban environment exogenous sources of CO account for more than the endogenous production. Comment: Does this make B and C correct? Probably not... I think they would expect us to know the intermediate step between haem and bilirubin so C is incorrect (or too simplistic)

Answer 427

Platelet activation is initially commenced by collagen exposure resulting from cell wall injury. vWF binds next and helps to stabilise the platelet to the wall and helps protect it from the shearing forces of blood flow. Fibrinogen is a latter product in the coagulation pathway and serotonin is released as part of the platelet adhesion and release reaction. Ca2+ is necessary for activation of clotting factors and for intrinsic system. I think the answer is B. I disagree. I think the answer is 'A' because calcium is required for platelet activation. I believe high levels of interferons can also activate platelets and cause expression of tissue factor without vessel injury. See article below from pubmed: Movement of calcium ions and their role in the activation of platelets. (Massini P et al): The increase of the cytoplasmic Ca-concentration plays a central role in the initiation of platelet activation. Four kinds of movements of Ca-ions are presumed to occur during this process: Ca-ions liberated from membranes induce the rapid shape change. Vesicular organelles release Ca-ions into the cytoplasm which initiate the release reaction. The storage organelles called dense bodies, secrete their contents including Ca-ions to the outside during the release reaction. At the same time a rearrangement of the plasma membrane occurs, resulting in an increase in its permeability for Ca-ions as well as in an increase in the number of Ca-binding sites. Since most processes occurring during platelet activation are reversible, the platelet must be equipped with a mechanism which removes Ca-ions from the cytoplasm. A vesicular fraction obtained from homogenized platelets indeed accumulates Ca actively. This Ca-pump is stimulated by cyclic AMP and protein kinase; it may be involved in the recovery of platelets after activation. "Platelet activation can be induced by adhesion to proteins such as collagen, soluble agonists (epinephrine, ADP, serotonin and thrombin) and cell contact during platelet aggregation". Power + Kam. 2nd Ed. pg 282. Hence answer could be B, C and E. If E is serotonin and vWF from C is considered due to its role in platelet aggregation. B might be the best answer. Cell Based theory is a completely different explanation for platelet activation... I think the whole point of the Power & Kam page is that there are many things that can start platelet activation via receptor binding but that calcium is a key intracellular signalling molecule so you can have all the receptor activation you want but unless you have calcium nothing much happens. Definately A. Google books has a book Platelets in thrombotic and non-thrombotic disorders has a chapter on Platelet signalling: Calcium which suggests its importance. - agree: the question is cannot occur without and not can occur with! (no Ca - no activation) from QLD Primary long course notes - platelet activation cannot occur without an agonist, fibrinogen, and a divalent cation (Ca OR Mg) therefore answer is fibrinogen

Answer 428

C. Helper T-cells

Answer 429

BL15 first version: Answer C However D is also correct: there is a variable region on the heavy AND light chains IgG has 2 heavy and 2 light chains

Answer 430

BL15 2nd version: Answer A The variable regions on the heavy and light chain form the antigen binding sites of the Ig

Answer 431

ANSWER - ? A Platelet activation is initiated by binding to collagen, but can also be produced by ADP & thrombin. The question is asking which of the options is REQUIRED for platelet activation. Vessel wall damage exposes collagen, etc and will lead to activation, but is it strictly required? Can platelets be activated independent of vessel wall damage? Ca++ is involved within the cell in activating phospholipase-A2 and subsequent release of arachidonic acid and thromboxane-A2. However the step prior to this (phospholipase C activation) is responsible for platelet degranulation. Cyclooxygenase is involved in metabolism of arachidonic acid, but again this is subsequent to the degranulation step. Von Willebrand factor is present on the surface of platelets and is important for adhesion and aggregation, but is it strictly REQUIRED for activation? Prostaglandins are metabolites of arachidonic acid, and as such are not required for degranulation (which is a subsequent step). Does anybody have any bright ideas, or does that seem right? A seems to be the BEST answer, as all of the others refer to events which follow degranulation. I think B is a better answer as you can get platelets to activate invitro without vessels. Agreed, I also think B is a better answer, see pub med article below: Movement of calcium ions and their role in the activation of platelets. The increase of the cytoplasmic Ca-concentration plays a central role in the initiation of platelet activation. Four kinds of movements of Ca-ions are presumed to occur during this process: a) Ca-ions liberated from membranes induce the rapid shape change. b) Vesicular organelles release Ca-ions into the cytoplasm which initiate the release reaction. c) The storage organelles called dense bodies, secrete their contents including Ca-ions to the outside during the release reaction. d) At the same time a rearrangement of the plasma membrane occurs, resulting in an increase in its permeability for Ca-ions as well as in an increase in the number of Ca-binding sites. Since most processes occurring during platelet activation are reversible, the platelet must be equipped with a mechanism which removes Ca-ions from the cytoplasm. A vesicular fraction obtained from homogenized platelets indeed accumulates Ca actively. This Ca-pump is stimulated by cyclic AMP and protein kinase; it may be involved in the recovery of platelets after activation.

Answer 432

Correct answer: A Cytokines are low molecular weight proteins. They act on other cells via specific cell membrane receptors (i.e. they are signalling molecules) They activate tyrosine kinase in the cell They have paracrine/autocrine actions (i.e. act locally only) Blood concentrations are small but they are extremely potent Released from : 1. mast cells 2. macrophages 3. monocytes 4. lymphocytes 5. endothelial cells Production occurs in response to specific stimuli and involved in growth, differentiation and regulate immune responses ie inflammation/pathophysiology role Major groups of cytokines: 1. interleukins 2. interferons 3. chemokines (chemotactic cytokines) 4.haematopoietic colony stimulating factors 5. tumour necrosis factor Autacoids: chemical messengers that act on target cells close to their site of release. Aka Paracrine secretions... ? Answer C Brandis Pg 169

Answer 433

Answers E, F and G are false. FFP (fresh frozen plasma) is produced by separating plasma from Red cells, white cells and platelets from whole blood. (spinning down or by plasmaphoresis). It is anticoagulated with citrate and rapidly frozen. FFP contains all plasma proteins (incl albumin), all clotting factors including the labile factors V and VIII albeit in low concentrations (on thawing) and anticoagulant proteins C,S and Thrombomodulin. Quote from Red Cross website "Fresh Frozen Plasma (FFP) is separated and frozen within eighteen hours after collection of whole blood. A unit of FFP contains all coagulation factors including approximately 200 units of Factor VIII plus the other labile plasma coagulation factor, Factor V." FFP should be ABO group specific for the patient but does NOT need crossmatching. "Compatibility testing is not required, however ABO compatible plasma should be used wherever possible." (Red Cross website) It is indicated for treatment of protein C deficiency either as a prevention of thrombosis acutely (ie for surgery) or in Protein C deficiency DIC. Isn't H incorrect also? It contains anti-thrombin III which is an anti-thrombotic protein, it is used in treating ATIII deficiency Comment I think D is actually correct. FFP contains all the clotting factors but is slightly deficient in V and VIII. If the stem said it did not contain these at all then it would be false. However FFP has decreased levels of factor V (about 65%) and VIII (about 40%) as these deplete the most rapidly.

Answer 434

Answer is A. Complement can be activated either via the Classical pathway or the Alternative pathway (properdin pathway). Classical activation involves binding of C1 to immunoglobulin-antigen bound complexes resulting in the complement cascade. Alternative activation involves a circulating protein Factor I recognising repetitive sugar complexes found only on bacteria and viral outer walls. This interaction results in activation of C3 and C5 which then promotes opsinisation of bacteria and virons. Properdin acts to stabilise the activating enzyme complex. I disagree - if complement activation can occur via alternate pathway, the antigen-antibody complex is NOT REQUIRED !!! E - no plasma proteins = NO COMPLEMENT !!! I agree with EEE. Also, activation of alternate pathwya can occur by slow spontaneous hydrolysis of C3 (according to K. Brandis) - therefore doesn't need ab-ag complex.

Answer 435

Answer is C Macrophages are an integral part of INNATE IMMUNTY Macrophages are derived from bone marrow promonocytes which subsequently differentiate into blood monocytes and then to mature macrophages which settle in tissues. They have different names depending on the tissue they settle in: Lung = alveolar macrophages Liver = Kuppfer cells Kidney glomerulus = mesangial cells Brain = microglia Bone = osteoclasts etc, etc. Therefore the term "macrophages" includes all of the above (and more), and they are present in many different tissues. Killing/digestion of bacteria by macrophages is accomplished by several methods: Reactive oxygen species Nitric oxide production by iNOS (inducible Nitric Oxide Synthase) Other proteins including lysozyme, lactoferrin, defensins Hydrolytic enzymes Macrophages are activated by lymphokines (from T cells) and migrate in response to chemotactic stimuli. Note: Cytokines have been variously named as lymphokines, interleukins and chemokines, based on their presumed function, cell of secretion or target of action. Because cytokines are characterized by considerable redundancy and pleiotropism, such distinctions, allowing for exceptions, are obsolete [1].

Answer 436

Answer is C Macrophages are an integral part of INNATE IMMUNTY Macrophages are derived from bone marrow promonocytes which subsequently differentiate into blood monocytes and then to mature macrophages which settle in tissues. They have different names depending on the tissue they settle in: Lung = alveolar macrophages Liver = Kuppfer cells Kidney glomerulus = mesangial cells Brain = microglia Bone = osteoclasts etc, etc. Therefore the term "macrophages" includes all of the above (and more), and they are present in many different tissues. Killing/digestion of bacteria by macrophages is accomplished by several methods: Reactive oxygen species Nitric oxide production by iNOS (inducible Nitric Oxide Synthase) Other proteins including lysozyme, lactoferrin, defensins Hydrolytic enzymes Macrophages are activated by lymphokines (from T cells) and migrate in response to chemotactic stimuli. Note: Cytokines have been variously named as lymphokines, interleukins and chemokines, based on their presumed function, cell of secretion or target of action. Because cytokines are characterized by considerable redundancy and pleiotropism, such distinctions, allowing for exceptions, are obsolete [1].

Answer 437

Answer: found on all nucleated cells "The proteins encoded by HLAs are the proteins on the outer part of body cells that are (effectively) unique to that person. The immune system uses the HLAs to differentiate self cells and non-self cells." wikipedia.org The HLA's are encoded by the Major histocompatibility genes Two classes exist and are found in different cells. Class I is found on all nucleated cells. Class II is found on antigen presenting cells including B cells and activated T cells.

Answer 438

Answer: found on all nucleated cells "The proteins encoded by HLAs are the proteins on the outer part of body cells that are (effectively) unique to that person. The immune system uses the HLAs to differentiate self cells and non-self cells." wikipedia.org The HLA's are encoded by the Major histocompatibility genes Two classes exist and are found in different cells. Class I is found on all nucleated cells. Class II is found on antigen presenting cells including B cells and activated T cells.

Answer 439

The answer is B in both versions. Antigen Recognition: In the case of T cells, the antigen is taken up by an antigen-presenting cell and partially digested. A peptide fragment of it is presented to the appropriate receptors on T cells. Disagree: Answer is none of the above: Cytotoxic T cells recognise MHC Class I on all nucleated cells and kill directly - they do not require antigen presenting cells. I bet the first version had a better option or a non-of-the-above too. JB2012

Answer 440

The answer is B in both versions. Antigen Recognition: In the case of T cells, the antigen is taken up by an antigen-presenting cell and partially digested. A peptide fragment of it is presented to the appropriate receptors on T cells. Disagree: Answer is none of the above: Cytotoxic T cells recognise MHC Class I on all nucleated cells and kill directly - they do not require antigen presenting cells. I bet the first version had a better option or a non-of-the-above too. JB2012

Answer 441

The Correct Answer is E - None of the Above Thrombin is a multifunctional serine protease. It is the main endpoint of the 'current model' of coagulation, being generated from prothrombin by the Xa-Va (prothrombinase) complex. In the 'amplification' component of the coagulation process (occuring on platelet surface) thrombin acts to provide positive feedback thereby amplifying the initial procoagulant signal by: Enhancing Platelet Adhesion Activating Platelets Activating Factors V, VIII, XI By activating XI another amplification loop is produced generating additional IXa which activates X. The activation of V and VIII lead to further production and action of the tenase (FVIIIa and IXa) and prothrombinase (Xa and Va) complexes leading to large amounts of thrombin generation in the 'propagation' component of the coagulation process. Thrombin also activates factor XIII which stabilises the fibrin polymers with the formation of covalent cross links. Thrombin's anticoagulant action occurs via its binding to thrombomodulin. The thrombin/thrombomodulin complex activates protein C, which functions as an anticoagulant by the down regulation of thrombin generation via inactivation of factors Va and VIIIa. Inaddition to Protein C the thromin/thrombomodulin complex activates TAFI (Thrombin activated Finbrinolysis inhibitor) which inhibits plasmin. tPA is involved in fibrinolysis, activating plasminogen to plasmin. By forming a complex with fibrin and plasminogen it's catalytic activity is increased 1000 times. It is inhibited by Plasminogen activator inhibitor 1 (PAI-1)

Answer 442

?Correct answer is B. Formed in bone marrow in adults,( and thymus.) B lymphocytes also present in germinal centres of lymph nodes, the spleen and mucosa associated lymphoid tissue. Therefore A is false. Yes, B. Ganong 22nd pg525 Lymphocyte precursors come from the bone marrow (in adult life to a lesser extent than in foetal development). The transformation to B lymphocytes occurs in fetal liver and after birth, the bone marrow. Contrary View: Don't we mean A: Don't remain in the lymph system? They are throughout the blood!! Lymphocyctes (as distinct from precursors) are not just formed in marrow... Suspect the wording of the original was: which of the following statements is not true? (answer B) As worded both A and B slightly correct. Ganong 22nd Ed page 520 - "After birth, some lymphoctes are formed in the bone marrow. However, most are formed in the LN, thymys and spleen.." - "At any given time, only about 2% of body lymphocytes are in the peripheral blood"

Answer 443

The most accurate answer is D, cytotoxic T Cells. Ganong: Cytotoxic T cells destroy transplanted and other foreign cells, with their development aided and directed by helper T cells. E is more correct I think. There are several mechanisms of allograft rejection including humoral and cellular immunity reactions, hence no one type of cell can be implicated. However HLA is part of the process of the graft being recognised as foreign in all rejections. And all rejections are ultimately cytotoxic

Answer 444

HbA (alpha-2, beta-2) has 4 chains (2 alpha & 2 beta chains) and each polypeptide chain contains one haem-Fe++. Hence the answer is four haem porphyrin rings and four ferrous ions per molecule of Hb. Option C. The haem group consists of both a protoporphyrin ring and one ferrous ion on each chain, and there are 4 chains. "Haem (Fe-protoporphyrin IX) consists of a porphyrin ring with an iron atom bound to the four nitrogens at the centre"

Answer 445

(NB: 4 rings and 4 Fe++ is the right answer so these options have not been remembered correctly here)

Answer 446

answer: B. 65-70% in Hb, 25-30% in liver as ferritin, total 3-4 g (Brandis table p 199) HbA (alpha-2, beta-2) has 4 chains (2 alpha & 2 beta chains) and each polypeptide chain contains one haem-Fe++. Hence the answer is four haem porphyrin rings and four ferrous ions per molecule of Hb. Option C. The haem group consists of both a protoporphyrin ring and one ferrous ion on each chain, and there are 4 chains. "Haem (Fe-protoporphyrin IX) consists of a porphyrin ring with an iron atom bound to the four nitrogens at the centre"

Answer 447

BL27 Correct answer - D It shouldnt be A, B or E: as presumably more of anything will increase viscosity; and I have some notes from the melbourne course that state that because blood is a non-newtonian fluid: its viscosity decreases as flow increases (something about streaming and changing flow of RBCs) C - is incorrect "Viscosity of blood, relative to that of water increases as a function of tube diameter up to a diameter of about 0.3mm" (Berne and Levy, 3rd Ed, p.449) D - is correct. "The apparent viscosity of blood diminishes as the shear rate (ed: read velocity) increases, a phenomenon called shear thinning" (Berne and Levy, 3rd ed, p.450) Factors affecting blood viscosity: Hematocrit (as HCt increases, there is a disproportionate increase in viscosity) Temperature: As temperature decreases, viscosity increases (increases ~ 2% for each °C decrease in temperature) Flow rate of blood: Low flow rates -> marked increased in viscosity -> Increased cell-to-cell and protein-to-cell adhesive interactions -> Erythrocytes adhere to one another (rouleau formation) Vessel diameter: Small vessel diameters (e.g., in arterioles less than 300 microns), there is a paradoxical decrease in blood viscosity (Fahraeus-Lindqvist effect). This occurs because the hemotocrit decreases in small vessels relative to the hemotocrit of large feed arteries. Supporting view (based on Power & Kam): A. Wrong - viscosity depends on WCC, hence the dangerous Hyperviscosity Syndrome in leukaemia B. Wrong - similarly to A, viscosity increases with haematocrit C. Wrong - the apparent viscosity of blood falls in vessel diameters below 0.3 mm. As arterioles and capillaries are below this diameter, the blood viscosity measured in vivo is lower than that in vitro. This is due to the more axial flow of RBCs in small vessels. D. Correct - As shear forces increase with high flows, RBCs also flow more axially, decreasing haematocrit and reducing the apparent viscosity of whole blood. E. independent of fibrinogen I agree that the correct answer is: D

Answer 448

BL27b A. Partly correct - viscosity increases with haematocrit, but not sure that this is a proportional (i.e. linear) increase B. Partly correct - if other variables of laminar flow in a cylinder are measured, the apparent viscosity can be calculated from the Hagen-Poisuelle law C. Correct - decreases with flow, but not sure that this is a proportional (i.e. linear) increase D. Partly correct - viscosity does not depend on the diameter of the tube over a large range, but under diameters of 0.3 mm (i.e. physiologically relevant diameters) this is no longer true E. Correct - same as C Given the non-linear relationships, I would say that the "most correct" answer is: E

Answer 449

The major influences that predispose to thrombosis (Virchow's triad) are: injury to the endothelium, alterations in the normal blood flow, alterations in blood coagulability. Endothelial injury is the major factor and the only one that will lead to thrombosis by itself. However when the other two influences co-exist it is probably not required. Thus B is not true. C is not true either as thrombosis requires platelet activation. This is a poorly worded question and suggest that it may have been remembered incorrectly (& certainly incompletely), - but in pathological prothrombotic states (such as neoplasm), are platelets necessarily activated? Other question is, if you knock out the majority of your platelets with an antiplatelet agent (such as clopidogrel and aspirin), is it possible to form a clot? I think the answer is yes, which suggests that answer C may be correct, but an authoritative answer would be appreciated. All the definitions of thrombosis that I can find (after a quick google search) refer to blood clotting within a vessel - sounds pathological to me... I think I'd go with A. Agree with A. From Guyton 10th ed p 427: "An abnormal clot that develops in a blood vessel is called a thrombus". Hang on a minute. A cannot be the answer as A is true. The question is which is NOT true? (if this MCQ has been remembered correctly) A - Thrombosis is always pathological - True, thrombosis is pathological (not the same thing as clot formation) B - Thrombosis requires venous stasis - Not necessarily, although it MAY involve venous stasis (one factor in Virchow's triad), it doesn't have to C - Thrombosis does not involve platelet activation - Depends on wording. Does usually involve platelet activation although doesn't HAVE to. I'd go for B if we have to pick one that ISN'T true. Comment: Thrombi are not always pathological. 1000s of microthrombi form in the lower limbs each day, e.g. overnight, however these are small and filtered by the lungs (that is one of the important functions of the lung). Thus A is the correct option i.e. it is NOT true that all thrombi are pathological. Reference: Guyton Further comments The answer to this question will depend on the actual wording. However, the comments above raise some interesting points, particularly the difference between "normal" coagulation and "thrombosis" and whether thrombosis is always abnormal. The key thing in the below definitions (quoted from Wikipedia) is that thrombosis is the term used when the clot OBSTRUCTS the flow of blood in the vessel. So thrombosis is the result of "normal coagulation" when the outcome is an obstructed vessel. This surely must be a normal situation in response to injury, such as a bruise as the clot will not just seal the breach in the vessel wall but must sometimes (or even commonly) cause an obstruction to blood flow in the vessel. Is this "pathological"? Well I think it can be "physiological" in some situations (eg normal response to minor trauma), BUT if excessive and/or the result of abnormal activation processes then it would be "pathological". Thrombosis is the formation of a blood clot (thrombus; Greek: θρόμβος) inside a blood vessel, obstructing the flow of blood through the circulatory system. When a blood vessel is injured, the body uses platelets (thrombocytes) and fibrin to form a blood clot to prevent blood loss. Alternatively, even when a blood vessel is not injured, blood clots may form in the body if the proper conditions present themselves. If the clotting is too severe and the clot breaks free, the traveling clot is now known as an embolus.[1][2] Coagulation is the process by which blood forms clots. It is an important part of hemostasis, the cessation of blood loss from a damaged vessel, wherein a damaged blood vessel wall is covered by a platelet and fibrin-containing clot to stop bleeding and begin repair of the damaged vessel. Disorders of coagulation can lead to an increased risk of bleeding (hemorrhage) or obstructive clotting (thrombosis).

Answer 450

ADP does promote platelet aggregation to form a primary haemostatic plug - B is so far correct.

Answer 451

B Although a minor Cross match involves the testing of the Donors serum with the recipients RBCs according to Wikipedia.

Answer 452

C is most correct of remembered options.. Ref pg 531 Ganong 23rd ed. A: wrong - the diffrence is made up of globulin and fibrinogen, globulin can be divided into a1, a2, b1, b2 and gamma. Gamma are the immunoglobulins B: wrong never trust an answer with always in it ;) but there are obviously other causes like nephrotic syndrome Addit: Albumin is also an acute phase protein, so should decrease with any systemic inflammation.

Answer 453

A: is correct so not the asnwer B: also true C: this is wrong so correct answer D: this is also wrong so another correct answer suspect question, poorly remembered Type I Immediate - IgE / Mast Cell mediated (e.g. asthma, anaphylaxis) Type II Antibody mediated (1) Opsonisation (transfusion reactions, erythoblastosis foetalis) (2) Fc-receptor mediated (types of glomulonephritis) (3) Antibody-mediated cellular dysfunction (myasthenia gravis, Graves disease) Type III Immune-Complex Mediated (deposition of IgG-antigen complexes, e.g. serum sickness) Type IV Cell mediated (1) Delayed type (2) T-cell mediated cytotoxicity A. Wrong (i.e. true) B. Wrong (i.e. true) C. Correct (i.e. false) - Type II is antibody mediated D. Arguably wrong (i.e. arguably true) - Somewhere in the natural history, T Cells will have been presented antigens and modulate the balance of cellular-humoral responses in all cell and antibody hypersensitivities. E. ? The "most correct" answer is 'C'.

Answer 454

A is correct

Answer 455

A is correct | Power and Kam pg 274 2nd Ed

Answer 456

Answer: C Guyton 12e p. 457 "Plasmin digests fibrin fibers and some other protein coagulants such as fibrinogen, Factor V, Factor VIII, prothrombin [II] and Factor XII." - thus not VII.

Answer 457

Question doesn't seem to be too clear. If it's asking for changes after 24 hours of starvation (and not further along the track), then the most accurate answer is A for both versions. Effects of fasting First 24 hours: Glycogen stored in liver and muscle rapidly depleted (glycogenolysis). Small amounts of acetoacetate and beta-hydroxybutyrate produced by liver from free fatty acids. Small quantity of glucose produced by gluconeogenesis from lactate and glycerol by liver (predominantly) and kidney. 2-4 days: Glucose produced almost entirely by gluconeogenesis via amino acids, glycerol (from adipose tissue) and lactate from RBCs. Lactate and pyruvate (formed by glycolysis in renal medulla and RBCs) used to synthesize glucose by the Cori cycle. Further fasting: Ketone bodies gradually replace glucose as the fuel for brain and nervous tissue. Gluconeogenesis rate reduced as a protein-sparing mechanism (due to decreasing plasma glucagon concentration, to prefasting levels at 10 days).

Answer 458

Question doesn't seem to be too clear. If it's asking for changes after 24 hours of starvation (and not further along the track), then the most accurate answer is A for both versions. Effects of fasting First 24 hours: Glycogen stored in liver and muscle rapidly depleted (glycogenolysis). Small amounts of acetoacetate and beta-hydroxybutyrate produced by liver from free fatty acids. Small quantity of glucose produced by gluconeogenesis from lactate and glycerol by liver (predominantly) and kidney. 2-4 days: Glucose produced almost entirely by gluconeogenesis via amino acids, glycerol (from adipose tissue) and lactate from RBCs. Lactate and pyruvate (formed by glycolysis in renal medulla and RBCs) used to synthesize glucose by the Cori cycle. Further fasting: Ketone bodies gradually replace glucose as the fuel for brain and nervous tissue. Gluconeogenesis rate reduced as a protein-sparing mechanism (due to decreasing plasma glucagon concentration, to prefasting levels at 10 days).

Answer 459

Cortisol is a catabolic hormone and increases FFA, BSL AND amino acid levels in the plasma Parathyroid Hormone has no effect on fat, protein or carbohydrate metabolism. It regulates calcium (and phosphate) homeostasis in the body via its actions on the bone, kidney and gut. Growth Hormone causes increased protein synthesis, increased FFA and increased BSL (anti-insulin effect). Fat is preferentially used as an energy source Insulin has anabolic effects with increased glycogenogenesis (and reduced gluconeogenesis), protein synthesis and fat synthesis. Therefore, the answer is C (Growth Hormone)

Answer 460

Cortisol is a catabolic hormone and increases FFA, BSL AND amino acid levels in the plasma Parathyroid Hormone has no effect on fat, protein or carbohydrate metabolism. It regulates calcium (and phosphate) homeostasis in the body via its actions on the bone, kidney and gut. Growth Hormone causes increased protein synthesis, increased FFA and increased BSL (anti-insulin effect). Fat is preferentially used as an energy source Insulin has anabolic effects with increased glycogenogenesis (and reduced gluconeogenesis), protein synthesis and fat synthesis. Therefore, the answer is A (Cortisol)

Answer 461

Correct answer: D Primary adrenocortical hypofunction (Addison's Disease) results from a disease process that destroys the adrenal cortex. It used to be a relatively common complication of Tb, but now is usually due to autoimmune inflammation of the adrenal cortex. Features of the disease process include weight loss (decreased muscle bulk), lethargy and hypotension. Option A: Osteonecrosis of bone and marrow results from ischaemia, and aside from fracture, most cases are either idiopathic or follow corticosteroid administation. Option B: Mediated both by a primary oestrogen deficiency, and also adrenocorticoid excess. Glucocorticoids decrease plasma clacium by inhibiting osteoclast formation and activity, and can cause osteoporosis by decreased bone formation and increased bone resorption. Option C: Truncal obesity (ie redistribution of body fat) is a feature of Cushing's disease (adrenocortical hyperfunction) Option E: Delayed closure of the epiphyses is due to decreased pituitary growth hormone and decreased IGF-1. Source: Ganong pp380-1, 385-7. Robins Pathologic Basis of Disease 6th Ed p 1231

Answer 462

Answer is D The hypophysiotropic hormones include "growth hormone inhibitory hormone" (GHIH) (somatostatin) (Reference: Guyton 11th Table 74-1 p. 907) However several other texts also state that Growth hormone inhibiting hormone (somatostatin) suppresses the secretion of GH, ACTH, TSH and Prolactin. It also inhibits the secretion of glucagon and insulin (reference Power & Kam p 288, and Rhoades and Pflanzer p.406; Ganong 22nd ed p 249) I wonder whether Prolactin was an answer here and the question is misremembered? The hypothalamus primarily inhibits Prolactin release by releasing PIH (prolactin inhibitory hormone). See Guyton & Hall 10th ed p 848. Since the hypothalamus releases GHRH as well as GHIH it seems unfair to have growth hormone as the answer unless the stem is different.

Answer 463

Answer c decreased RAP->decreased ANP->loss of inhibition of renin secretion Comment: Decreased RAP also causes increased sympathetic stimulation to the renal artery -> Increased Renin secretion

Answer 464

Answer B and C: Angiotensin II and AVP. Ganong has table 24-1 which this question is clearly lifted from. Ang II and Vasopressin are both 'inhibitory'. Comment: in the second version of the question, all options except A would ultimately end up with decreased reinin secretion. See table on P458 of Ganong 22nd edition. Wording may have been "which option doesn't decrease renin release?"

Answer 465

Answer : E. Glucose is a factor that inhibits glucagon secretion. Ganong p 349. Comment: But there is a glucagon excess in hyperglycaemia from diabetic ketoacidosis... Urine output is increases (diuresis), so you become dehydrated, potassium and sodium is lost with the urine, although exchange of H+ and K+ with the cells gives a hyperkalaemia. Thus, I would think increased H+, increased K+ (serum), increased urine output, and increased glucagon, but DECREASED blood volume and DECREASED Na+ (ie: answer is B or D). Any ideas? Ref: Chapter 19, Ganong, 22nd Ed. Oxford Handbook seems to agree with me too... I have some ideas for the comment above... The question is about Hyperglycaemia not Diabetic Ketoacidosis. I suspect you mean p. 341 of Ganong "Effects of Hyperglycaemia" The most correct answer is E. Regarding Glucagon, "Glucagon secretion,.., is stimulated by hypoglycaemia and inhibited by hyperglycaemia. Amino acids, ..., also stimulate glucagon secretion, whilst fatty acids inhibit glucagon release. CCK, gastrin and secretin all stimulate glucagon secretion. Somatostatin inhibits glucagon release." [Kam, p.294] The hormonal response to DKA is due to an "apparent" intracellular glucose deficiency, with a relative plasma glucose excess, and so the compensatory mechanisms act to increase plasma glucose (ie. increased glucagon, sympathetic stimulation, and ? increased cortisol); this a pathological response (not physiological). further comment,i agree with option E ,and regarding volumes ,i think ECF will increase ,and if the dehydration is more intracellular. Hyperglycaemia causes a pseudohypernatremia (physiologically or pathologically) due to osmotic effect (and movement of water out of cells dilutes the [Na]). Is there any other times that increased BSL cause increased Na? Because everything else occurs, at some time or another (either physiologically or pathologically). Whereas I'm not sure increased Na does for the reasons above (either pseudohyponatremia or loss of Na due to DKA). I know the exam is on physiology, but there's plenty of pathophysiology being tested if you ask me!

Answer 466

ADH mechanism of action: Insertion of water channels into luminal (apical) membrane Antidiuretic Hormone is vasopressin. Its main renal effect is antidiuresis. In the presence of ADH, the water permeability of all regions of collecting ducts is high and only a small volume of maximally hyperosmotic urine is excreted. ADH acts in the collecting ducts on the principal cells. The receptors for ADH are located in the basolateral membrane of the principal cells; these are the vasopressin type 2 receptors. Binding of ADH to this G-protein coupled receptor results in activation of adenylate cyclase, which catalyses the intracellular production of cAMP. This induces migration of intracellular vesicles to and fusion with, the luminal membrane. The vesicles contain an isoform of the water channel protein, aquaporin 2, through which water can move. So the luminal membrane now becomes highly permeable to water. In the absence of ADH, the aquaporins are withdrawn from the luminal membrane by endocytosis. NOTE: the water permeability of the basolateral membranes of renal epithelial cells is always high because of the presence of other aquaporin isoforms – the permeability of the luminal membrane is hence rate limiting

Answer 467

ADH mechanism of action: Insertion of water channels into luminal (apical) membrane Antidiuretic Hormone is vasopressin. Its main renal effect is antidiuresis. In the presence of ADH, the water permeability of all regions of collecting ducts is high and only a small volume of maximally hyperosmotic urine is excreted. ADH acts in the collecting ducts on the principal cells. The receptors for ADH are located in the basolateral membrane of the principal cells; these are the vasopressin type 2 receptors. Binding of ADH to this G-protein coupled receptor results in activation of adenylate cyclase, which catalyses the intracellular production of cAMP. This induces migration of intracellular vesicles to and fusion with, the luminal membrane. The vesicles contain an isoform of the water channel protein, aquaporin 2, through which water can move. So the luminal membrane now becomes highly permeable to water. In the absence of ADH, the aquaporins are withdrawn from the luminal membrane by endocytosis. NOTE: the water permeability of the basolateral membranes of renal epithelial cells is always high because of the presence of other aquaporin isoforms – the permeability of the luminal membrane is hence rate limiting

Answer 468

Answer is D BMR is defined as the energy output or heat production in a subject in a state of mental and physical rest in a comfortable environment, 12 hours after a meal. The reason it is measured after 12hrs after a meal is because of the specific dynamic action (SDA) of food which tends to incease metabolic rate for about 4-6hrs after ingestion. "The SDA of a food is the obligatory energy expenditure that occurs during its assimilation into the body"(Ganong pg 281).

Answer 469

The fat soluble vitamins are vitamins A, D, E & K. All the rest are water-soluble.

Answer 470

Insulin receptors Membrane bound with two α-subunits and two β-subunits α-subunits are extracellular and bind insulin (so A false) β-subunits are transmembrane intracellular portion of β-subunits has tyrosine kinase activity Insulin binding extracellularly activates the tyrosine kinase activity of the β-subunits (B false; ED11b B true) This results in autophosphorylation of the tyrosine residues of the β-subunits The now activated (phosphorylated) β-subunits have effects on cytoplasmic proteins Phosphorylation of some: tyrosine, serine, threonine kinases De-phosphorylation of others These effects mainly take place on serine and threonine residues Insulin-Receptor Substrate proteins (IRS 1, IRS 2, IRS 3 and IRS 4) may be important in insulin-mediated intracellular pathways Net effect is regulation of metabolic cascades leading to activation or inactivation of key enzymes in response to insulin (a third-messenger system) In adipose and muscle cells GLUT4 (the glucose transporter protein) is activated by this mechanism (C and D true - D probably most correct): pre-made GLUT4 transporters are stored in cytoplasmic vesicles insulin receptor activation phosphorylates phosphoinositol-3 kinase this causes translocation of GLUT4 to the plasma membrane by fusion of vesicles with the membrane when insulin action ceases, GLUT4-containing patches of membrane are endocytosed into vesicles, ready for the next exposure to insulin Insulin-receptor binding causes aggregation of receptors in patches: This causes receptor-mediated endocytosis of the receptor patches Insulin-receptor complexes enter lysosomes where receptors are broken down and recycled t1/2 of receptors is 7 hours Exposure to increased insulin decreases receptor concentration (down-regulation) Exposure to reduced insulin increases receptor number (up-regulation)

Answer 471

Insulin receptors Membrane bound with two α-subunits and two β-subunits α-subunits are extracellular and bind insulin (so A false) β-subunits are transmembrane intracellular portion of β-subunits has tyrosine kinase activity Insulin binding extracellularly activates the tyrosine kinase activity of the β-subunits (B false; ED11b B true) This results in autophosphorylation of the tyrosine residues of the β-subunits The now activated (phosphorylated) β-subunits have effects on cytoplasmic proteins Phosphorylation of some: tyrosine, serine, threonine kinases De-phosphorylation of others These effects mainly take place on serine and threonine residues Insulin-Receptor Substrate proteins (IRS 1, IRS 2, IRS 3 and IRS 4) may be important in insulin-mediated intracellular pathways Net effect is regulation of metabolic cascades leading to activation or inactivation of key enzymes in response to insulin (a third-messenger system) In adipose and muscle cells GLUT4 (the glucose transporter protein) is activated by this mechanism (C and D true - D probably most correct): pre-made GLUT4 transporters are stored in cytoplasmic vesicles insulin receptor activation phosphorylates phosphoinositol-3 kinase this causes translocation of GLUT4 to the plasma membrane by fusion of vesicles with the membrane when insulin action ceases, GLUT4-containing patches of membrane are endocytosed into vesicles, ready for the next exposure to insulin Insulin-receptor binding causes aggregation of receptors in patches: This causes receptor-mediated endocytosis of the receptor patches Insulin-receptor complexes enter lysosomes where receptors are broken down and recycled t1/2 of receptors is 7 hours Exposure to increased insulin decreases receptor concentration (down-regulation) Exposure to reduced insulin increases receptor number (up-regulation)

Answer 472

ED 12 - Option A - Ganong 19th ed, p240: Major production of heat is via skeletal muscle contraction. Alternate opinion: Agree with vasodilation but for heat production not completely convinced. According to Guyton & Hall 10th ed p 822: ... the different factors that determine the rate of heat production, called the basal metabolic rate. The question states at rest so not too much skeletal muscle contraction happening. According to Wiki: Energy expenditure breakdown liver 27% brain 19% heart 7% kidneys 10% skeletal muscle 18% other organs 19% According to lecture notes on human physiology on google books Na/K Atpase contributes about 20-25% towards BMR. http://books.google.com.au/books?id=MK3Aze7Ef-UC&pg=PA596&lpg=PA596&dq=basal+metabolic+rate+atpase&source=bl&ots=FvZX4Vf6Yv&sig=SmSsBlRpmuSwNY4fw3Ffjum55wA&hl=en&ei=ea-JS9DkHIyC7QPcgsGzDQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CDIQ6AEwCQ#v=onepage&q=basal%20metabolic%20rate%20atpase&f=false Not high level sources I know but I tend to think that at rest the Na/K ATPase (which also exists in mucle) might edge out muscle overall? I'd say its likely to be very close at any rate. Yes but do Na/K ATPase pumps generate heat?? The contracting muscle (albeit running on ATPase) generates the heat. Guyton (pg819): "Skeletal muscle, even under resting conditions, accounts for 20-30% of the BMR"... Thus (A) for the first option (no comment about all the Wiki references... but Guyton trumps Wiki... I doubt the Exam writers use Wiki) That's 2 comments on wiki references then.

Answer 473

ED 12b - Option A Alternate opinion: Agree with vasodilation but for heat production not completely convinced. According to Guyton & Hall 10th ed p 822: ... the different factors that determine the rate of heat production, called the basal metabolic rate. The question states at rest so not too much skeletal muscle contraction happening. According to Wiki: Energy expenditure breakdown liver 27% brain 19% heart 7% kidneys 10% skeletal muscle 18% other organs 19% According to lecture notes on human physiology on google books Na/K Atpase contributes about 20-25% towards BMR. http://books.google.com.au/books?id=MK3Aze7Ef-UC&pg=PA596&lpg=PA596&dq=basal+metabolic+rate+atpase&source=bl&ots=FvZX4Vf6Yv&sig=SmSsBlRpmuSwNY4fw3Ffjum55wA&hl=en&ei=ea-JS9DkHIyC7QPcgsGzDQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CDIQ6AEwCQ#v=onepage&q=basal%20metabolic%20rate%20atpase&f=false Not high level sources I know but I tend to think that at rest the Na/K ATPase (which also exists in mucle) might edge out muscle overall? I'd say its likely to be very close at any rate. Yes but do Na/K ATPase pumps generate heat?? The contracting muscle (albeit running on ATPase) generates the heat. Guyton (pg819): "Skeletal muscle, even under resting conditions, accounts for 20-30% of the BMR"... Thus (A) for the first option (no comment about all the Wiki references... but Guyton trumps Wiki... I doubt the Exam writers use Wiki) That's 2 comments on wiki references then.

Answer 474

2002 Stem - Option C (Heat LOSS by vasodilation) Alternate opinion: Agree with vasodilation but for heat production not completely convinced. According to Guyton & Hall 10th ed p 822: ... the different factors that determine the rate of heat production, called the basal metabolic rate. The question states at rest so not too much skeletal muscle contraction happening. According to Wiki: Energy expenditure breakdown liver 27% brain 19% heart 7% kidneys 10% skeletal muscle 18% other organs 19% According to lecture notes on human physiology on google books Na/K Atpase contributes about 20-25% towards BMR. http://books.google.com.au/books?id=MK3Aze7Ef-UC&pg=PA596&lpg=PA596&dq=basal+metabolic+rate+atpase&source=bl&ots=FvZX4Vf6Yv&sig=SmSsBlRpmuSwNY4fw3Ffjum55wA&hl=en&ei=ea-JS9DkHIyC7QPcgsGzDQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CDIQ6AEwCQ#v=onepage&q=basal%20metabolic%20rate%20atpase&f=false Not high level sources I know but I tend to think that at rest the Na/K ATPase (which also exists in mucle) might edge out muscle overall? I'd say its likely to be very close at any rate. Yes but do Na/K ATPase pumps generate heat?? The contracting muscle (albeit running on ATPase) generates the heat. Guyton (pg819): "Skeletal muscle, even under resting conditions, accounts for 20-30% of the BMR"... Thus (A) for the first option (no comment about all the Wiki references... but Guyton trumps Wiki... I doubt the Exam writers use Wiki) That's 2 comments on wiki references then.

Answer 475

(A) ACTH Plasma angiotensinogen levels are increased by plasma corticosteroid, estrogen, thyroid hormone, and angiotensin II levels

Answer 476

Answer is B: The caloric value of carbohydrate is 4 kcal/g In comparison, fat is 9 kcal/g. Gripe: Is it too much to ask that they write questions using SI units???? They tell us to learn them!

Answer 477

Answer is B: The caloric value of carbohydrate is 4 kcal/g In comparison, fat is 9 kcal/g. Gripe: Is it too much to ask that they write questions using SI units???? They tell us to learn them!

Answer 478

Answer: E All of the above From Stoelting 4th ed p474: High dose oxytocin produces a direct relaxant effect of vascular smooth muscle that manifests as a decrease in blood bressure and is accompanied by reflex tachycardia and increased cardiac output. Oxytocin has a slight ADH like activity which when administered in high doses with an excessive quantity of fluid can cause water intoxication. (Although: Oxytocin CAN cause water intoxication would be a better question) Katzung 8th ed p640 mentions water intoxication as a possible side effect. Also mentioned here is that oxytocin has a weak pressor activity and that 'maternal deaths due to hypertensive episodes' are among the reported adverse reactions. Hence it appears there is a complex effect on BP PROBLEM The problem is that how can it increase and decrease blood pressure at the same time? Is this a time scale thing or one of it is wrong. Which means which one would you pick as the right answer, provided this choices are remembered correctly. Oxytocin is a nonapeptide that is synthesized in the PVN of the hypothalamus, transported to the posterior pituitary, and released to cause milk ejection and uterine contraction. It's similarity to Vasopressin results in some of the Adverse Effects: Water intoxication, fluid overload, with or without hyponatraemia (ADH-like effect), worsened by co-administration of intravenous fluids in pregnancy. Rapid intravenous administration causes tachycardia, hypotension and ECG changes. "Hypertension may occur rarely". My understanding is that the single best answer for a physiology exam would be "water intoxication", as this is a physiological result of the similarities of the hormones. Hypotension is a pharmacological effect of rapid administration of Syntocinon, and the occurence of hypertension sounds like more of a rare association than any extension of physiological effect. (Although I am aware that ADH can cause hypertension via V1-R). So it does not cause both hypotension and hypertension.

Answer 479

ADH synthesized in supraoptic and paraventricular nuclei, and released from posterior pituitary neurones in response to increased osmolarity (change of 1-2%)(among other triggers). An osmoreceptor is a sensory receptor primarily found in the hypothalamus of most homeothermic organisms that detects changes in osmotic pressure. Osmoreceptors can be found in several structures, including the organum vasculosum of the lamina terminalis (OVLT) and the subfornical organ (SFO).[1] Answer B. Osmoreceptors are located in anteroventral third ventricle, in anterior (not post) hypothalamus. Dcreased volume and pressure will always win over osmolality - the response to Volume Change (low pressure volumereceptors of the great veins) is less sensitive (requires 7-10% change) but more powerful than the sensitive response to a change in serum osmolarity (responds to 1-2% change).

Answer 480

To quote Kam : "G Proteins consist of α, β, and γ chains and bind guanosine disphosphate. Upon receptor stimulation, the α unit releases GDP in favour of GTP and dissociates from the βγ. The active α-GTP complex then affects ion channels or second messengers. G proteins are inactivated when the α unit hydrolyses GTP to GDP and rejoins the the βγ complex."

Answer 481

G proteins Ganong22 has an excellent picture of the structure of the receptor on p41 A and B both true therefore incorrect - All GPCRs consist of a single polypeptide chain which has 7 transmembrane α-helical segments. Each of these segments consists of between 20 to 28 hydrophobic AA residues separated by loops which contain non-hydrophobic residues C false therefore correct - Ganong states (pg43) that residues on the cytoplasmic surface of the receptor are sites of phosphorylation. When signal reaches a G protein, the protein exchanges GDP for GTP. The GTP-protein complex brings about the effect. The inherent GTPase activity of the protein then converts GTP to GDP. (D true therefore incorrect). Resource: Brandis, The physiology viva pg181 E is untrue therefore correct, because the GPCR is a serpentine structure, not heterotrimeric. G protein is heterotrimeric. Must read the question! Seems like both C and E are the correct answers to the question. GPCRs become desensitized when exposed to their ligand for a prolonged period of time. The key reaction of this downregulation is the phosphorylation of the intracellular (or cytoplasmic) receptor domain by protein kinases. [1]

Answer 482

G proteins Ganong22 has an excellent picture of the structure of the receptor on p41 A and B both true therefore incorrect - All GPCRs consist of a single polypeptide chain which has 7 transmembrane α-helical segments. Each of these segments consists of between 20 to 28 hydrophobic AA residues separated by loops which contain non-hydrophobic residues C false therefore correct - Ganong states (pg43) that residues on the cytoplasmic surface of the receptor are sites of phosphorylation. When signal reaches a G protein, the protein exchanges GDP for GTP. The GTP-protein complex brings about the effect. The inherent GTPase activity of the protein then converts GTP to GDP. (D true therefore incorrect). Resource: Brandis, The physiology viva pg181 E is untrue therefore correct, because the GPCR is a serpentine structure, not heterotrimeric. G protein is heterotrimeric. Must read the question! Seems like both C and E are the correct answers to the question. GPCRs become desensitized when exposed to their ligand for a prolonged period of time. The key reaction of this downregulation is the phosphorylation of the intracellular (or cytoplasmic) receptor domain by protein kinases. [1]

Answer 483

ED19 The interthreshold range is the range of core temperature over which no autonomic thermoregulatory responses occur. It is approx 0.2C at normal temperature in a non anaesthetised state. The threshold temps are influenced by circadian rhythms, food intake, thyroid hormones, drugs and thermal adaptation to warm or cold ambient temperatures. According to Power and Kam p329, General Anaesthesia increases the interthreshold range (B is false). Hence I believe A to be correct. Reference: Power & Kam p 327-329 DISAGREEMENT: option B is not 'decreases' the interthreshold range. It is 'LOWERS' it. The range widens, from 0.2 to 4 degrees, but is LOWER than before the anaesthetic. Option A is a bit nonsensical. Miller's p1534 doesn't mention 'constant input' from thermoreceptors as having any effect on the ITR.

Answer 484

ED19b [NOTE: A lot of this discussion refers to a mis-reading of the question and discusses ACTUAL body temperature, but the question asks about "SET POINT" body temperature - read carefully!] A. False - Human core temperature has a circadian fluctuation being lower in the morning and higher in the evening. B. ? C. False - Rectal temperature is normally higher than oral temperature by 0.5 degrees and is affected less by environmental changes than oral temperatures, so may not parallel a temperature taken elsewhere. D. False - Temperature increases due to heat produced by contracting muscles. E. True - Temperature decreases due to the vasodilating effects of anaesthetic agents which redistributes heat from core to periphery where it is lost. There will be further heat loss due to exposure of patient to the cool air of an operating theatre. It is a poor choice of terminology to speak of "set-point" although Kam does say the "Posterior hypothalamus is responsible for the ... set point around which the body temperature is maintained", but further discussion regarding effects of anaesthesia refers to altering the thermoregulatory threshold for responses to cold and heat by 2.5c (reduced) and 1.3c (increased). So using the older terminology, set-point decreases due to effect of anaesthesia upon hypothalamus. (So 19.b Answer E is correct but not for the reason stated above) The effect of general anaesthesia on patient's core temperature is a different question (which I think some people are confusing with the effect of GA on Thermoregulation). The core temperature response is a triphasic r'ship with an initial rapid drop due to vasodilation and redistribution of heat from the core to the peripheries (first 30 minutes), followed by a phase of temperature loss at a slower rate (as the pt become poikilothermic and will assume the temperature of his environment), and a final plateau phase at which the thermoregulatory response to cold has been triggered and the associated vasoconstriction and non-shivering thermogenesis (the only responses available under GA) conserve/generate enough heat to equal the ongoing heat loss (via conduction, radiation). This equilibrium occurs after 3-4 hours. (See aforementioned section in Kam, and Google "Sessler temperature general anaesthesia" for multiple articles on the topic.) --- I presume the set-point is always the middle of the interthreshold range. - Re: ED19b option C. I believe is true. The rectal temperature is representative of the temperature at the core of the body and varies least with changes in environmental temperature - Ganong Ch 18.4 Re: ED19b option E. I believe is false. Whilst GA will lower core temperature, the set-point will remain the same. what changes is the inter-threshold range and the triggering temperatures to thermoregulatory responses so the core temperature will not be controlled within such a tight range and will fall, but the stem specifically relates to the set-point and won't change under GA. - Miller's Ch48 by Sessler --- One more opinion: This question is not asking about body temperature!! It is asking about the middle of the inter threshold range!! This is certainly reduced in anaesthesia, but may also be changed in exercise. Therefore best answer is E JB2012 -- I don't think the set point changes with anaesthesia. The thermoneutral zone widens from 0.4 to 4 degrees C. You could call this change to the set point but then why not call it increases too? Can you post a reference to "the set point decreases in anaesthesia", and distinguish it from the thermoneutral zone widens?. TW

Answer 485

``` Not B and not C because those are mechanisms by which body temperature is brought towards set point. Most texts talk about the balance between the sodium and calcium concentrations (more intracellular Na -> rise in set point) being the actual mechanism by which the set point is set, but I am unsure whether the Na-K ATPase pump is involved. Assuming it is, then a decrease in Na/K/ATPase activity would result in more intracellular sodium and hence a a rise in body temperature should be expected. So A does not seem to be correct either leaving D as the correct answer. Comment: Please reference above statements. As I understand that whilst the drop in core temperature is due to vasodilation, the decrease in Threshold is due to anaesthesia induced hypothalamic dysfunction. I cannot find a source that gives a mechanism to explain this. There is only speculation (ref 3). "Redistribution of heat from warm central compartments (eg, abdomen, thorax) to cooler peripheral tissues (eg, arms, legs) from anesthetic-induced vasodilation explains most of the initial decrease in temperature, with actual heat loss being a minor contributor. Continuous heat loss to the environment appears to be primarily responsible for the slower subsequent decline. During steady-state equilibrium, heat loss equals metabolic heat production. Normally the hypothalamus maintains core body temperature within a very narrow range (the interthreshold range). Raising body temperature a fraction of a degree induces sweating and vasodilation, whereas lowering temperature triggers vasoconstriction and shivering. During general anesthesia, however, the body cannot compensate for hypothermia because anesthetics inhibit central thermoregulation by interfering with hypothalamic function. For example, isoflurane produces a dose-dependent decrease in the vasoconstrictive threshold (3°C for each percent of inhaled isoflurane)." [Ref : Lange Anaesthesiology] And "General anesthetics modulate autonomic nervous system function including thermoregulatory control, which resides in the preoptic area of the anterior hypothalamus. However, the mechanism by which anesthetics modulate hypothalamic function remains unknown." [Kushikata] Whereas Kam (p327) says "The posterior hypothalamus is responsible for the establishment of the establishment of a reference or set point..."; and "The set point may be determined by the ratio of sodium and calcium ions in the posterior hypothalamus". ```

Answer 486

Endothelin is a vasoconstrictor peptide derived from vascular endothelium, important in regulation of basal vascular tone and BP (therefore A is incorrect, and B is correct). It is produced from an inactive precursor and acts at specific endothelin receptors (type A receptors) to produce vasoconstriction. Type B receptors result in nitric oxide and prostacyclin production. Involved in cardiac failure and pulmonary hypertension, hence use of bosentan (endothelin antagonist) in treatment of pulmonary HT. B is probably most correct answer. Comment: ET1 vasc endothelium, brain kidneys ET2 intestine ET3 adrenal glands, blood, brain. Endothelins are released by shear stress, hence I assume by damaged vascular endothelium. hence all correct?

Answer 487

Endothelin is a vasoconstrictor peptide derived from vascular endothelium, important in regulation of basal vascular tone and BP (therefore A is incorrect, and B is correct). It is produced from an inactive precursor and acts at specific endothelin receptors (type A receptors) to produce vasoconstriction. Type B receptors result in nitric oxide and prostacyclin production. Involved in cardiac failure and pulmonary hypertension, hence use of bosentan (endothelin antagonist) in treatment of pulmonary HT. B is probably most correct answer. Comment: ET1 vasc endothelium, brain kidneys ET2 intestine ET3 adrenal glands, blood, brain. Endothelins are released by shear stress, hence I assume by damaged vascular endothelium. hence all correct?

Answer 488

Endothelin is a vasoconstrictor peptide derived from vascular endothelium, important in regulation of basal vascular tone and BP (therefore A is incorrect, and B is correct). It is produced from an inactive precursor and acts at specific endothelin receptors (type A receptors) to produce vasoconstriction. Type B receptors result in nitric oxide and prostacyclin production. Involved in cardiac failure and pulmonary hypertension, hence use of bosentan (endothelin antagonist) in treatment of pulmonary HT. B is probably most correct answer. Comment: ET1 vasc endothelium, brain kidneys ET2 intestine ET3 adrenal glands, blood, brain. Endothelins are released by shear stress, hence I assume by damaged vascular endothelium. hence all correct?

Answer 489

Growth Hormone DOES all of the above with increased FFA production and release from adipose tissue, therefore, can lead to ketosis. Hepatic glycogenolysis is promoted with reduced glucose uptake by tissues. Therefore, the answer is F Comment Answer A is worded funny. GH causes increased breakdown of TG to FFA. However, production of new FFA is not actually increased Any references for GH acting as a energy source during starvation? --- yeah, seems rather strange to me. I think this is a poorly remembered question GH does: increase plasma glucose, increase breakdown of TG to FFA thus increasing circulating FFA, which in turn can cause a ketosis as fat would be turned into acetyl CoA and preferentially used in the Krebs cycle Nutritional factors such as starvation, hypoglycaemia will stimulate an increase in GH secretion, however, GH itself cannot be used as an energy source

Answer 490

B Respiratory quotient and patterns of substrate utilization in human sepsis and trauma. "The lower mean respiratory quotient of septics indicates that they depend generally more than nonseptic trauma patients on fat as an energy substrate and confirms a previously obtained evidence of limited hepatic lipogenesis in sepsis. At the same time, however, it is suggested that fat utilization becomes impaired (and hepatic lipogenesis becomes prominent) in sepsis at a stage in which signs of impaired oxidative metabolism and major metabolic abnormalities also develop." .The range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0 (representing the value expected for pure carbohydrate oxidation) to ~0.7 (the value expected for pure fat oxidation).[1] See Stoelting 4th - p791 RER increases in strenuous exercise, can go above 1. A level of 1.2 tends to correspond with exhaustion In 23a isn't the decrease in respiratory quotient due to decrease in CO2 production due to incomplete utilization of glucose (to lactate rather than into TCA) because of tissue hypoperfusion and increased anaerobic glycolysis? But, why would increased anaerobic metabolism alter the RER or RQ, as CO2 is not generated (rather is generated from TCAC) and O2 not consumed (rather is used in last step of e- transport chain) during metabolism of glucose -> pyruvate -> lactate. So RER changes must be due to an alteration in the choice of substrate or to changes in ventilation. Consider the RQ in the brain is 0.99 due to the utilisation of glucose. (Ref. Kam p64. Brandis) and that "During exercise the RQ approaches a value of 2 because of... hyperventilation". Similarly for metabolic acidosis. (Kam p.320) So I would give answers as B, A, B RER is measured at the mouth and is VCO2/VO2 RQ is the ratio of CO2 produced to O2 consumed at the cellular level Thus, RQ is dependent on the type or types of fuel substrates being used by the cell RQ representing fat and CH2O metabolism Usually RER = RQ. But, if the subject is hyperventilating, has an acid-base disturbance, or is performing intense exercise RER ¹ RQ because RER would include CO2 produced as a result of buffering, whereas RQ includes only CO2 produced as a result of metabolism Non-protein Respiratory exchange ratio and respiratory quotient i) RER is an estimate of RQ, usually RER=RQ ii) RQ is the ratio of CO2 produced to O2 consumed at the cellular level iii) RER is the ratio of VCO2 to VO2 measured at the level of the mouth and/or nose, i.e. at the whole body level. iv) Because extra CO2 can result from buffering (remember H+ + HCO3- ß® H2CO3 ß ® H2O + CO2), CO2 production, measured at the level of the mouth, may not accurately affect CO2 production at the cellular level. v) Thus, whenever there is a significant change is arterial pH (such as during very intense exercise, sometimes during starvation, metabolic acidosis, hyperventilation, etc.), RER is not equal to RQ vi) RER = (VCO2/VO2) or (FECO2/(.2093-FEO2) vii) We say this is a non-protein RER because proteins can not be completely oxidized into CO2 and H2O, an additional element is present, nitrogen – thus we could measure O2 consumption needed for oxidizing a protein and the resulting CO2 production, but it would not accurately represent protein use by the body. Furthermore, under normal circumstances in humans (y/2)H2O + x CO2. Therefore, RQ = x / (x + y/4 - z/2). all have same question how it end up with different answers(although still have same options) errrrr

Answer 491

A Respiratory quotient and patterns of substrate utilization in human sepsis and trauma. "The lower mean respiratory quotient of septics indicates that they depend generally more than nonseptic trauma patients on fat as an energy substrate and confirms a previously obtained evidence of limited hepatic lipogenesis in sepsis. At the same time, however, it is suggested that fat utilization becomes impaired (and hepatic lipogenesis becomes prominent) in sepsis at a stage in which signs of impaired oxidative metabolism and major metabolic abnormalities also develop." The range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0 (representing the value expected for pure carbohydrate oxidation) to ~0.7 (the value expected for pure fat oxidation).[1] See Stoelting 4th - p791 RER increases in strenuous exercise, can go above 1. A level of 1.2 tends to correspond with exhaustion In 23a isn't the decrease in respiratory quotient due to decrease in CO2 production due to incomplete utilization of glucose (to lactate rather than into TCA) because of tissue hypoperfusion and increased anaerobic glycolysis? But, why would increased anaerobic metabolism alter the RER or RQ, as CO2 is not generated (rather is generated from TCAC) and O2 not consumed (rather is used in last step of e- transport chain) during metabolism of glucose -> pyruvate -> lactate. So RER changes must be due to an alteration in the choice of substrate or to changes in ventilation. Consider the RQ in the brain is 0.99 due to the utilisation of glucose. (Ref. Kam p64. Brandis) and that "During exercise the RQ approaches a value of 2 because of... hyperventilation". Similarly for metabolic acidosis. (Kam p.320) So I would give answers as B, A, B RER is measured at the mouth and is VCO2/VO2 RQ is the ratio of CO2 produced to O2 consumed at the cellular level Thus, RQ is dependent on the type or types of fuel substrates being used by the cell RQ representing fat and CH2O metabolism Usually RER = RQ. But, if the subject is hyperventilating, has an acid-base disturbance, or is performing intense exercise RER ¹ RQ because RER would include CO2 produced as a result of buffering, whereas RQ includes only CO2 produced as a result of metabolism Non-protein Respiratory exchange ratio and respiratory quotient i) RER is an estimate of RQ, usually RER=RQ ii) RQ is the ratio of CO2 produced to O2 consumed at the cellular level iii) RER is the ratio of VCO2 to VO2 measured at the level of the mouth and/or nose, i.e. at the whole body level. iv) Because extra CO2 can result from buffering (remember H+ + HCO3- ß® H2CO3 ß ® H2O + CO2), CO2 production, measured at the level of the mouth, may not accurately affect CO2 production at the cellular level. v) Thus, whenever there is a significant change is arterial pH (such as during very intense exercise, sometimes during starvation, metabolic acidosis, hyperventilation, etc.), RER is not equal to RQ vi) RER = (VCO2/VO2) or (FECO2/(.2093-FEO2) vii) We say this is a non-protein RER because proteins can not be completely oxidized into CO2 and H2O, an additional element is present, nitrogen – thus we could measure O2 consumption needed for oxidizing a protein and the resulting CO2 production, but it would not accurately represent protein use by the body. Furthermore, under normal circumstances in humans (y/2)H2O + x CO2. Therefore, RQ = x / (x + y/4 - z/2). all have same question how it end up with different answers(although still have same options) errrrr

Answer 492

B Respiratory quotient and patterns of substrate utilization in human sepsis and trauma. "The lower mean respiratory quotient of septics indicates that they depend generally more than nonseptic trauma patients on fat as an energy substrate and confirms a previously obtained evidence of limited hepatic lipogenesis in sepsis. At the same time, however, it is suggested that fat utilization becomes impaired (and hepatic lipogenesis becomes prominent) in sepsis at a stage in which signs of impaired oxidative metabolism and major metabolic abnormalities also develop." The range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0 (representing the value expected for pure carbohydrate oxidation) to ~0.7 (the value expected for pure fat oxidation).[1] See Stoelting 4th - p791 RER increases in strenuous exercise, can go above 1. A level of 1.2 tends to correspond with exhaustion In 23a isn't the decrease in respiratory quotient due to decrease in CO2 production due to incomplete utilization of glucose (to lactate rather than into TCA) because of tissue hypoperfusion and increased anaerobic glycolysis? But, why would increased anaerobic metabolism alter the RER or RQ, as CO2 is not generated (rather is generated from TCAC) and O2 not consumed (rather is used in last step of e- transport chain) during metabolism of glucose -> pyruvate -> lactate. So RER changes must be due to an alteration in the choice of substrate or to changes in ventilation. Consider the RQ in the brain is 0.99 due to the utilisation of glucose. (Ref. Kam p64. Brandis) and that "During exercise the RQ approaches a value of 2 because of... hyperventilation". Similarly for metabolic acidosis. (Kam p.320) So I would give answers as B, A, B RER is measured at the mouth and is VCO2/VO2 RQ is the ratio of CO2 produced to O2 consumed at the cellular level Thus, RQ is dependent on the type or types of fuel substrates being used by the cell RQ representing fat and CH2O metabolism Usually RER = RQ. But, if the subject is hyperventilating, has an acid-base disturbance, or is performing intense exercise RER ¹ RQ because RER would include CO2 produced as a result of buffering, whereas RQ includes only CO2 produced as a result of metabolism Non-protein Respiratory exchange ratio and respiratory quotient i) RER is an estimate of RQ, usually RER=RQ ii) RQ is the ratio of CO2 produced to O2 consumed at the cellular level iii) RER is the ratio of VCO2 to VO2 measured at the level of the mouth and/or nose, i.e. at the whole body level. iv) Because extra CO2 can result from buffering (remember H+ + HCO3- ß® H2CO3 ß ® H2O + CO2), CO2 production, measured at the level of the mouth, may not accurately affect CO2 production at the cellular level. v) Thus, whenever there is a significant change is arterial pH (such as during very intense exercise, sometimes during starvation, metabolic acidosis, hyperventilation, etc.), RER is not equal to RQ vi) RER = (VCO2/VO2) or (FECO2/(.2093-FEO2) vii) We say this is a non-protein RER because proteins can not be completely oxidized into CO2 and H2O, an additional element is present, nitrogen – thus we could measure O2 consumption needed for oxidizing a protein and the resulting CO2 production, but it would not accurately represent protein use by the body. Furthermore, under normal circumstances in humans (y/2)H2O + x CO2. Therefore, RQ = x / (x + y/4 - z/2). all have same question how it end up with different answers(although still have same options) errrrr

Answer 493

ED24: A is false. Acetyl CoA, generated via pyruvate is the product of glucose that enters citric acid cycle. B. is also false. Formation of lactate from pyruvate regenerates NADH back into NAD+. (NADP is recycled by the hexose monophosphate shunt of the E-M pathway in order to maintain reduced glutathione)

Answer 494

Alt version: A is correct. There is anaerobic metabolism inside RBCs which use the Embden-Meyerhof pathway to generate ATP as they lack mitochondria. A small amount of lactate is normally produced from pyruvate by lactate dehydrogenase (even in AEROBIC conditions) as this recycles NAD+ from NADH. The lactate produced is transported back to liver via bloodstream where 70% is removed and turned back into glucose. This process consumes 6 ATP per glucose reformed and constitutes the Cori cycle. The other 30% is removed by mitochondria-rich tissues such as skeletal and cardiac myocytes and proximal tubule cells. (Comments : Gladdens review article rejects much of the typical "textbook" explanation for the role of lactate in metabolism.)

Answer 495

Increased activity by adrenaline on beta2 receptors in the liver (Ganong p292 21st ed) Additional: The breakdown of glycogen in 1:4α linkage is catalyzed by phosphorylase, whereas another enzyme catalyzes the breakdown of glycogen in 1:6α linkage. Phosphorylase is activated in part by the action of epinephrine on β2-adrenergic receptors in the liver. This in turn initiates a sequence of reactions that provides a classic example of hormonal action via cAMP (See Figure 17-13, Ganong). Answer is C, and B respectively. However, Glucagon acts via GPCR, increased cAMP, activation of protein kinases, then enhanced phosphorylase a and b activity, resulting in increased hepatic glycogenolysis and gluconeogenesis (Kam p.294). So A and B are both correct statements (in second stem). Guyton 8th edition p 746 says: "All cells of the body are capable of storing at least some glycogen..." which suggests that they all contain some phosphorylase (otherwise the glycogen would be of no use to them...?) Ganong says that adrenaline stimulates phosphorylase in liver and Sk Muscle, whereas Glucagon only stimulates liver phosphorylase (therefore B in 2nd stem) - and as for the above, as mentioned, phosphorylase only acts on a specific type of linkage, another enzyme acts on another. Maybe all tissues don't need phosphorylase. The other thing is - maybe the 1st stem was supposed to be "everything true except..." , in that case D.

Answer 496

Increased activity by adrenaline on beta2 receptors in the liver (Ganong p292 21st ed) Additional: The breakdown of glycogen in 1:4α linkage is catalyzed by phosphorylase, whereas another enzyme catalyzes the breakdown of glycogen in 1:6α linkage. Phosphorylase is activated in part by the action of epinephrine on β2-adrenergic receptors in the liver. This in turn initiates a sequence of reactions that provides a classic example of hormonal action via cAMP (See Figure 17-13, Ganong). Answer is C, and B respectively. However, Glucagon acts via GPCR, increased cAMP, activation of protein kinases, then enhanced phosphorylase a and b activity, resulting in increased hepatic glycogenolysis and gluconeogenesis (Kam p.294). So A and B are both correct statements (in second stem). Guyton 8th edition p 746 says: "All cells of the body are capable of storing at least some glycogen..." which suggests that they all contain some phosphorylase (otherwise the glycogen would be of no use to them...?) Ganong says that adrenaline stimulates phosphorylase in liver and Sk Muscle, whereas Glucagon only stimulates liver phosphorylase (therefore B in 2nd stem) - and as for the above, as mentioned, phosphorylase only acts on a specific type of linkage, another enzyme acts on another. Maybe all tissues don't need phosphorylase. The other thing is - maybe the 1st stem was supposed to be "everything true except..." , in that case D.

Answer 497

Ganong 22ed p74 phosphocreatine is an energy rich phosphate compound that can supply the energy for muscle contraction for a short time by replenishing the stores of ATP when it is hydrolysed to creatine + phosphate groups Stores built up at rest and used when muscle is exercising C is correct in all versions Addition: creatine phosphate = phosphorylcreatine Creatine is synthesized in the liver from methionine, glycine, and arginine, and transported to the muscle cells for storage. In skeletal muscle, it is phosphorylated to form phosphorylcreatine, which is an important energy store for ATP synthesis. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise, the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction. Some athletes ingest creatine as a dietary supplement and claim that it enhances their performance in sprints and other forms of vigorous short-term exertion.

Answer 498

Ganong 22ed p74 phosphocreatine is an energy rich phosphate compound that can supply the energy for muscle contraction for a short time by replenishing the stores of ATP when it is hydrolysed to creatine + phosphate groups Stores built up at rest and used when muscle is exercising C is correct in all versions Addition: creatine phosphate = phosphorylcreatine Creatine is synthesized in the liver from methionine, glycine, and arginine, and transported to the muscle cells for storage. In skeletal muscle, it is phosphorylated to form phosphorylcreatine, which is an important energy store for ATP synthesis. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise, the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction. Some athletes ingest creatine as a dietary supplement and claim that it enhances their performance in sprints and other forms of vigorous short-term exertion.

Answer 499

Ganong 22ed p74 phosphocreatine is an energy rich phosphate compound that can supply the energy for muscle contraction for a short time by replenishing the stores of ATP when it is hydrolysed to creatine + phosphate groups Stores built up at rest and used when muscle is exercising C is correct in all versions Addition: creatine phosphate = phosphorylcreatine Creatine is synthesized in the liver from methionine, glycine, and arginine, and transported to the muscle cells for storage. In skeletal muscle, it is phosphorylated to form phosphorylcreatine, which is an important energy store for ATP synthesis. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise, the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction. Some athletes ingest creatine as a dietary supplement and claim that it enhances their performance in sprints and other forms of vigorous short-term exertion.

Answer 500

E. Increased CNS activity Therefore thinking about weight loss doesn't help :) I'm thinking very hard! However, "individuals living in the tropics have BMR 10% less than those living in temperate climates". (Kam p.320) Why isn't C (hot climate) the answer? Esp since part of the reason for decreased MR under anaesthesia/sedation/sleep is decreased brain activity. The question says which one "increases the least" - therefore, if the BMR is actually decreased (as opposed to increasing), then technically C wouldn't be the answer. Table 17.1 22nd Ed Ganong has a list of factors which affect BMR but no quatification given. "Emotional state" which I assume relates to increased CNS activity rates a mention. Interestingly on the same page (281) it mentions "When the temperature is high enough to raise the body temperatue, metabolic processes generally accelerate, and the metabolic rate rises about 14% for each degree Celsius of elevation." So increased temp, actually increases BMR. I think E is still the most likely of the options though - Ganong gives no quantification though. Agree with E. Definitionly any temperature above or below your thermoneutral zone will increase BMR. I can't find the quote but somewhere it says (?Power and Kam) that cerebral metabolic rate is remarkably constant and sitting physiology exams doesn't seem to increase it. Additional note: I would venture that the effect of "emotional state" (which needs to be calm/relaxed when measuring BMR) on metabolic rate has more to do with SNS activity than CNS activity. Comment 30/1/2011: Just wondering -- E didn't strike me as a stand-out answer at the first glance because I actually thought cold climate would reduce BMR. Isn't that why we cool victims of cardiac arrest? Obviously that also depends on how cold the climate is, what you are wearing, and whether one is shivering! No, I don't think that's right. Climate is different from body temperature. Outside the thermoneutral zone, increased energy is expended to maintain body temperature. If you *become* hypothermic, then MR does decrease - but the body tries very hard to prevent this. I agree with option E.

Answer 501

``` Answer is A Haematological effects of cortisol include: Mild increase in number of RBCs Increased Platelets and neutrophils Decrease in lymphocytes and eosinophiles (Reference: Power & Kam p. 303) ```

Answer 502

Increased vasopressin: Increased effective plasma osmotic pressure Decreased extracellular fluid volume Pain, emotion, exercise Nausea and vomiting Standing Clofibrate, carbamazepine (very common to see hyponatraemia in mental illness - not all psychogenic polydipsia!!) Angiotensin II Morphine increases secretion of ADH as per Sasada and Smith. Decreased vasopressin: Decreased effective plasma osmotic pressure Increased extracellular fluid volume Alcohol [presumably one reason why etOH gives dehydration; and the urge to urinate!]

Answer 503

Answer is B A. Calcitriol increases Ca2+ absorption in the small intestine, but I can't find a reference to suggest that it similarly regulates PO4 absorption B. Calcitriol "acts directly on the parathyroid glands to decrease preproPTH mRNA" (Ganong 22nd p. 392). See also Ganong 22nd Fig. 21-8. C. Vitamin D3 is ingested in the diet (Ganong 22nd p. 388), but I can't find a reference that describes the regulation of this process D. See comment for option B I disagree. Calcitriol is 1,25-Vit D (hormone, not a vitamin) whose main role is the absorption of calcium and phosphate by the intestine (Vander, p.187). Both options A and B are true, but A is the correct answer. Regarding Option C: cholecalciferol, D3, is produced in the skin from 7-dehydrocholesterol, but presumably also available from dietary sources. It is hydroxylated to 25-(OH)-cholecalciferol, the predominant circulating but inactive form. In the kidney PCT it is converted to 1,25(OH)2 by renal 1-hydroxylase (stimulated by PTH). 24,25-(OH)2-cholecalciferol is the inactive breakdown product. (Kam p.300) B may be true, but the question is asking about "effects on calcium". Therefore, the main effect is to increase the calcium level by increasing the GIT absorption - answer appears to be A

Answer 504

basal insulin release 1u /hr at higher rate the response to stimuli (*5-10)fold total daily 40 unit

Answer 505

Answer: C (refers to top version above) The Nernst equation, intoduced by the German theoretical chemist W Nernst (1864-1941) states that either the maximum electomotive force that can be generated by a given ratio of concentrations of an ion, or that the maximum ratio of concentration that can be sustained by a potential difference imposed from an external force. Example: Chloride ions are present at a higher concerntation in the ECF than the cell interior, and thus diffuse down the concerntation gradient into the cell. The interior of the cell is negative in relation to the exterior, and chloride ions are pushed out of the cell along the electrical gradient. An equilibirum is reached when the Chloride influx and efflux are the same - thus there is no net movement of ions The membrane potential at which this is equilibirium exists is the equlilibrium potential The Nernst equation is used to calculate the magnitude of the equilibrium potential at a given concentration difference across the cell membrane.

Answer 506

answer c ..potential at which there no net movement of ions...the Goldman equation is a modification of the Nerst equation taking into account the concentration and permeability of Cl Na & K (see Faunce p5 or Peter Kam) Comment: Isn't option D also correct? ie Gibbs Donnan effect is the balance of electrical gradient with chemical gradient... (No, Gibbs Donnan refers to the effect of non-diffusible anions upon the distribution of the diffusible ions across a semi-permeable membrane. This is different to the Nernst Potential). "Nernst Equation: Equation for calculating the membrane potential at which individual ions are at equilibrium across the membrane." Hence C. Yentis A - Z. 3rd Ed

Answer 507

primary motor centre for shivering is located in the posterior hypothalamus, near the wall of the 3rd ventricle. normally inhibited by heat centre of hypothalamus excited by cold signals from skin/spinal cord. when excited, signals are sent bilaterally down brain stem, then lateral columns of spinal cord, and on to anterior horn motor neurons. signals are nonrhythmic. they cause increase in muscle tone, which is thought to trigger feedback oscillation of the muscle spindle stretch reflex mechanism, causing shivering. Addition: Mechanisms activated by cold Increase heat production Shivering Hunger Increased voluntary activity Increased secretion of norepinephrine and epinephrine Decrease heat loss Cutaneous vasoconstriction Curling up Horripilation (goose bumps - no apparent useful purpose) Mechanisms activated by heat Increase heat loss Cutaneous vasodilation Sweating Increased respiration Decrease heat production Anorexia Apathy and inertia Ref: Table 14-6, Ganong.

Answer 508

Answer D Denervation of muscle causes: Muscle Atrophy Denervation hypersensitivity to circulating Ach – foetal γ subunit is up regulated over muscle cells other than the motor end plate. Muscle fibrillations Note: fasciculations are related pathological discharge of motor neurons. Ganong Ed21 p76 Doesn't that make C (increased) a better answer? -jdf Aside: Note that the extrajunctional ACh receptors (which contain a foetal γ subunit in place of an adult e subunit) proliferate over the surface of a muscle in the context of burns patients or neuromuscular disorders, and when activated release K+ into the circulation, thus making these patients at risk of hyperkalaemia in the setting of depolarising muscle blockers (eg suxamethonium). I have mentioned this here so you can correlate the physiology with the pharmacology. Damage to lower motor neurons is indicated by abnormal EMG potentials, fasciculations, paralysis, weakening of muscles, and neurogenic atrophy of skeletal muscle. Extensive info at: [1]

Answer 509

Kam says "The posterior hypothalamus is also responsible for establishing the reference or set point around which the body temperature is maintained. The set point may be determined by the ratio or sodium and calcium ions in the posterior hypothalamus." (P&K p.327) The efferent limb for shivering comes from the posterior hypothalamus, while that for sweating/vasodilation comes from the anterior. Comment Sensory receptors (sensors)in the body temperature control system are located in the skin, deep tissues, spinal cord, extrahypothalamic portions of the brain and the the hypothalamus itself - each contributing 20% of information for integration by the hypothalamus (the "thermostat"). Effector mechanisms are activated when the temperature is too high or low relative to the setpoint of the thermostat. Stimulation of the anterior hypothalamus initiates mechanisms for heat loss while stimulation of the posterior hypothalamus initiates mechanisms for heat conservation and production. (Memory Aid: Think of Fanning Yourself in Front) Note that Kam says posterior hypothalamus regulates the setpoint but the wording of the question makes it ambiguious as to which answer it is References Ganong, W. (2005) Review of Medical Physiology 22nd Ed pg.254 The McGraw-Hill Companies, Inc. USA Brandis, K The Physiology Viva Questions & Answers (Revised Ed)pg.285

Answer 510

Kam says "The posterior hypothalamus is also responsible for establishing the reference or set point around which the body temperature is maintained. The set point may be determined by the ratio or sodium and calcium ions in the posterior hypothalamus." (P&K p.327) The efferent limb for shivering comes from the posterior hypothalamus, while that for sweating/vasodilation comes from the anterior. Comment Sensory receptors (sensors)in the body temperature control system are located in the skin, deep tissues, spinal cord, extrahypothalamic portions of the brain and the the hypothalamus itself - each contributing 20% of information for integration by the hypothalamus (the "thermostat"). Effector mechanisms are activated when the temperature is too high or low relative to the setpoint of the thermostat. Stimulation of the anterior hypothalamus initiates mechanisms for heat loss while stimulation of the posterior hypothalamus initiates mechanisms for heat conservation and production. (Memory Aid: Think of Fanning Yourself in Front) Note that Kam says posterior hypothalamus regulates the setpoint but the wording of the question makes it ambiguious as to which answer it is References Ganong, W. (2005) Review of Medical Physiology 22nd Ed pg.254 The McGraw-Hill Companies, Inc. USA Brandis, K The Physiology Viva Questions & Answers (Revised Ed)pg.285

Answer 511

CTZ has both 5-HT3 and D2 receptors (see Kam, Fig 5-5, p.175). Answer is A. Additional: "The chemoreceptor trigger zone is a bilateral set of centers in the brainstem lying under the floor of the fourth ventricle. Electrical stimulation of these centers does not induce vomiting, but application of emetic drugs does - if and only if the vomition centers are intact. The chemoreceptor trigger zones function as emetic chemoreceptors for the vomiting centers - chemical abnormalities in the body (e.g. emetic drugs, uremia, hypoxia and diabetic ketoacidosis) are sensed by these centers, which then send excitatory signs to the vomition centers. Many of the antiemetic drugs act at the level of the chemoreceptor trigger zone." - from [1] Physiology of Vomiting "The results suggest that both H1- and H2-histamine receptors in the emetic chemoreceptor trigger zone of the area postrema are concerned in histamine-induced emesis." - from [2] Abstract about H1 & H2 receptors in CTZ "Dolasetron, granisetron, ondansetron and tropisetron selectively and competitively bind to 5-HT3 receptors, blocking serotonin binding at vagal afferents in the gut and in the regions of the CNS involved in emesis, including the chemoreceptor trigger zone and the nucleus tractus solitarii." - from [3] "Domperidone, a peripheral dopamine2-receptor antagonist, regulates the motility of gastric and small intestinal smooth muscle and has been shown to have some effects on the motor function of the esophagus. It also has antiemetic activity as a result of blockade of dopamine receptors in the chemoreceptor trigger zone." - from [4]. I know A is definately correct, but isn't C as well. Circumventricular organs are those which have incompete BBB and thus allow direct contact with substances in blood without need for specialised transport carriers. However they are still bathed in CSF, and substances within the CSF also have an affect on the CTZ. Just try Googling "substances in CSF acting on CTZ" and look at the top 10 results, all which conclude that CTZ detects substances in blood AND CSF from multiple sources Anyone want to comment??

Answer 512

Answer A. is correct. (Not necessarily) Substance P is a transmitter released by primary afferents in the dorsal horn, and act on neurokinin receptors (see Kam, p.337). There are other neurotransmitter peptides released from primary pain afferents in addition to Sub P such as neurokinin A and CGRP. (Sub P and NK-A act upon NK-Rc). Glutamate acts upon AMPA-Rc. There are two main classes of second-order dorsal horn neurons : "nociceptive-specific" and "wide-dynamic range". "The nociceptive-specific neurons are located within the superficial laminae of the dorsal horn and respond selectively to noxious stimuli. The WDR neurons are generally located in deeped laminae and respond to both noxious and non-noxious input". (Kam p.336). So Option B could be correct (depending upon the actual stem). Dull and sharp pain travel via different fibres: sharp pain via A-delta fibres and dull pain via C fibres. Answer E may be about the gating mechanism?: A-beta fibres and C fibres both synapsing on enkephalinergic interneurons and nociceptors. Small myelinated fibres terminal in lamina I and V whereas unmyelinated afferents terminate in Lamina II (suggests Aδ and C fibres do not terminate at same receptor). Comments A-delta fibres release glutamate, C fibres release Substance P. So they can't both act on the same receptor. The former acts on Glutamate receptors, and the latter on Neurokinin 1 & Substance P receptors.

Answer 513

The total volume of CSF is 150 mls. The daily production is 550 mls/day so the CSF turns over about 3 to 4 times per day. The CSF is formed by the choroid plexus (50%) and directly from the walls of the ventricules (50%). Ependymal cells (type of glial cell), also named ependymocytes, line the cavities of the CNS and make up the walls of the ventricles. These cells create and secrete cerebrospinal fluid and beat their cilia to help circulate that CSF. [1]. CSF flows through the foramens of Magendie & Luschka into the subarachnoid space of the brain and spinal cord. It is absorbed by the arachnoid villi (90%) and directly into cerebral venules (10%)." from Fluid Physiology text on AnaesthesiaMCQ.com Defn: "The cerebrospinal fluid (CSF) is located within the ventricles, spinal canal, and subarachnoid spaces. The volume of CSF in humans is 140-150 ml, only 30-40 ml actually in the ventricular system, with a production rate of 21 ml/hr. Function: CSF cushions the brain, regulates brain extracellular fluid, allows for distribution of neuroactive substances, and is the "sink" that collects the waste products produced by the brain." Production: The principal source of CSF are the choroid plexi of the lateral, third and fourth ventricles, and the volume varies between 10-20% of brain weight (Bradbury, 1979). The turnover rate of total CSF is species dependent and varies between approximately 1 hr for rat and 5 hr for human (Davson and Segal, 1996). Circulation: The majority of the CSF is in the subarachnoid space, where the arachnoid membranes bridge the sulci of the brain, in the basal cisterns and around the spinal cord. CSF moves within the ventricles and subarachnoid spaces under the influence of hydrostatic pressure generated by it production. from [2] Suspect answer is C. note: choroidal blood vessels is in the eye. Nothing to do with CSF production pg 289 Kerry Brandis

Answer 514

Mammalian nerve fibres have been divided into A, B & C groups (Erlanger & Gasser). Type A fibres are further divided in to α, β, γ and δ. Aα fibres: proprioception/somatic motor function Aβ fibres: touch/pressure Aγ fibres:motor to muscle spindles Aδ fibres: pain/cold/touch Nociceptive systems include myelinated Aδ fibres (fast pain) and unmyelinated C fibres (slow pain).Type A and B fibres are myelinated whereas type C fibres are unmyelinated. Conduction velocities for A fibres (ranging from 12-120m/s) are much faster than C fibres (0.5-2.3 m/s). Substance P is a polypeptide that acts as a transmitter subserving slow severe pain (C fibres). It has been found in high concentrations in the endings of primary afferent neurons in the spinal cord and is probably the mediator at the first synapse in slow pain pathways. Glutamate is the synaptic transmitter released by primary afferent fibres subserving fast mild pain. Therefore the answer:C appears to be the best answer based on elimination of the other options. References Ganong, W.F. (2005) Review of Medical Physiology 22nd Ed pg. 60, 61, 111, 112, 142, 143. The McGraw-Hill Companies, Inc. USA

Answer 515

From Kam, Aδ fibres result in short-lasting, pricking-type pain. Myelinated Conduction velocity 10 - 30 m.sec-1 12-30m/s from Ganong p142 Axon diameter 2-5 µm Cell body in Dorsal Root Ganglion -> laminae I and V of dorsal horn (hence D incorrect) Therefore C. I thought A-delta and C fibres in lamina I and II of dorsal horn, A-beta in III and IV, wide dynamic range in V

Answer 516

``` A Glycine - inhibitory B. Glutamate - excitatory C. Gamma amino butyric acid - inhibitory D. Serotonin - inhibitory (has some excitatory effects) E. Dopamine - inhibitory ``` Glutamate appears the most correct

Answer 517

``` A - F - mv B - F - at 37 deg C - F D - F - calculates the membrane potential E - T ?? ```

Answer 518

Looks like this comes from a table in Ganong: 2.1 Ed 21, pp 61 "Nerve fibre types in mammalian nerve" Fibre type A (largest diameter) AP spike duration listed as 0.4-0.5 ms A is correct Nah... 4ms

Answer 519

F - 14 mmHg - hlf

Answer 520

``` For the Aug11 version: A. Correct - K+ 60% (Ganong says 62%) B. Wrong - Glucose 64% C. Wrong - HCO3- 101% (about the same, although pH is lower in CSF) D. Wrong - Na+ 98% (about the same) E. Wrong - Cl- 114% Ganong 23ed, p571, Table 34-2. ```

Answer 521

``` For the Aug11 version: A. Correct - K+ 60% (Ganong says 62%) B. Wrong - Glucose 64% C. Wrong - HCO3- 101% (about the same, although pH is lower in CSF) D. Wrong - Na+ 98% (about the same) E. Wrong - Cl- 114% ```

Answer 522

A. Correct - resting membrane potential -90MV (Ganong 22nd ed p68) B. Correct - Action Potential Duration (APD) 2-4ms (Ganong 22nd ed p68) C. Correct - should be ARP (not ERP) Absolute Refractory Period 1-3msec (Ganong 22nd ed p68) D. Incorrect - Conduction Velocity is 5m/s (Ganong 22nd ed p68) E. All of above correct - if D was 5m/s then the correct answer would be E, otherwise A, B and C appear correct

Answer 523

A. ?Correct - "The [Myosin] heads contain an actin-binding site and a catalytic site that hydrolyses ATP" (Ganong p53 20th ed) and Kam states that "actin potentiates the ATPase activity of myosin" BUT goes on to say that troponin I inhibits the "actomyosin ATPase" suggesting that it's the complex of actin and myosin that hydrolyses the ATP. (Kam p24) B. Most correct - width A bands is constant, Z lines move closer together when muscle contracts and farther apart when stretched (Ganong 22nd ed p69) C. Incorrect - In the powerstroke the myosin head attaches to the actin at 90 degrees and subsequently bend to 45 degrees producing relative movement between the filaments ( Pinnock 2nd edn p287; Guyton) D. Incorrect - Interaction is between actin and myosin. Calcium binds to Troponin C, Troponin displaced laterally exposing binding site for actin. Myosin head binds to actin. E. Incorrect - Contraction and relaxation both active processes, ATP consuming. Ca-Mg ATPase pumps move calcium back into sarcoplasmic reticulum (Ganong 22nd edn p70) Aside: Types of Troponin: Troponin molecules are small globular units located at intervals along the tropomyosin molecules. Troponin T binds the other troponin components to tropomyosin [T for Tether] Troponin I inhibits the interaction of myosin with actin. [I for Inhibitory] Troponin C contains the binding sites for the Ca2+ that initiates contraction. [C for Calcium] Ref: Ganong, Ch 3. I'd go with A - the myosin head acts as an ATPase. Do Z lines move closer together during isometric contraction? My guess is No... B could be a trick question...... COMMENT: Regarding the above suggestion about isometric contraction - The Z-lines must move closer together in the muscle fibres (causing muscle fibre shortening), or how can you generate increased tension in the muscle? The muscle fibres/sarcomeres must shorten, and the elastic elements within the muscle must increase in length, but overall the length of the muscle does not change. (See Fig 4.9 in Power & Kam) Further Comment: The Z lines move closer together but notice that the word closer is absent. The Z lines certainly do not move together as that is physically impossible. As worded I think A is the most correct answer. "Actin potentiates the ATPase activity of myosin." P+K 2nd Ed. pg 27. Suggest A then B most correct. while the head of myosin does hydrolyse ATP (as actomyosin ATPase), the question asks 'during muscle contraction'... hydrolysis of ATP occurs after muscle contraction to 're-cock' the myosin head. So in terms of the question asked - B is correct References Option B IS the correct answer: Pg 27, Fig 1.35 Kam (2nd Ed) and is consistent with the NSW primary course practise exam.

Answer 524

A gamma efferents supply muscle spindles (intrafusal fibres) A beta efferents supply both intrafusal and extrafusal fibres. A alpha efferents supply extrafusal fibres. The muscle spindle functions as feedback mechanism to maintain muscle length. Answer is A. --- How does this translate to A exactly? I am not sure what smoothing contraction actually means. References See muscle spindle for references Ganong p130-131

Answer 525

A. Alpha (or Ia/Ib) fibres are actually afferent fibres. Ia fibres have two branches, one to the nuclear bag fibre and one to the nuclear chain fibre of the muscle spindle. Experimentally these have been shown to link directly onto motor neurones supplying the extrafusal fibres (ie regular contractile fibres) of the same muscle and contribute to the stretch reflex. This stretch reflex may contribute to clonus. (pg 130, 135 Ganong) Ib fibres originate from Golgi tendon organs. These link to inhibitory interneurones to the motor neurones supplying the extrafusal fibres of the same muscle and to excitatory connections on the muscles antagonists. Stimulation of these fibres cause relaxation of the muscle, and as regulator of muscle force may contribute to clonus. (Ganong p133, 134, 135) INCORRECT B. A finding characteristic of states in which increased gamma efferent discharge is present is clonus. (Ganong p134) INCORRECT C. When the muscle is stretched in the case of hyperactive spindles the burst of discharges supplying the motor neurone causes the muscle to contract, stopping the spindle discharge. This causes a reduction in Ia stimulation and increase in Ib stimulation, causing the muscle to relax. However, the muscle is still stretched, and the cycle is repeated. The delay in the circuit is what gives clonus it's characteristic beat. (Ganong p134) INCORRECT D. Increased tone is generally a result of increased gamma efferent discharge, thus more likely to produce clonus. (Ganong p134) INCORRECT E. Therefore alt question stem must be correct, and E. is CORRECT COMMENT 2 types of Alplha fibers 1) Provide the somatic motor innervation to extrafusal fibers 2) Afferent fibers arising from muscle spindles

Answer 526

``` A. False: similar answer also required for MU02E; calcium uptake by sarcoplasmic reticulum is an active process - energy ATP consuming - which is performed by Ca2+Mg2+ATPase pump (Ganong 22nd ed, p70) B. False: The T-tube system is a grid-like network of invaginations of the sarcolemma (the cell membrane). The T-tubes are therefore in continuity with the extracellular space. Their function is to allow rapid propagation of the action potential from the cell membrane on the outside of the muscle cell to the contractile elements within. In skeletal muscle the action potential is solely due to Na+ influx (depolarisation) and K+ efflux (repolarisation) - not calcium, so this option is false. The T-tubules are arranged between paired terminal cisterns of the sarcoplasmic reticulum (the so-called 'triad' arrangement), which contain Ca2+. T-tubule depolarisation activates voltage-gated dihydropyridine receptors in the T-tubule membrane, which then directly activate non-voltage gated calcium channels known as ryanodine receptors in the sarcoplasmic reticulum. Opening of these channels causes an large outflow of Ca2+ from the sarcoplasmic reticulum into the cytoplasm, activating the contractile mechanism. (Ganong 21st ed p65-71) C. False: Calcium binds to Troponin C, not to tropomyosin. Thin muscle filaments are polymers made up of the following components (Ganong 21st ed p67): Actin chains which are paried in a long double helix Tropomyosin molecules - long filaments which sit in the double helix groove formed by the actin chains; at rest tropomyosin covers the myosin binding sites on actin Troponin - globular molecules located at intervals along the tropomyosin molecules Troponin T ("T" for tether) binds troponin I and troponin C to tropomyosin Troponin I ("I" for inhibitory) is tightly bound to actin at rest, and preventing contraction Troponin C ("C" for calcium) has Ca2+ binding sites; Ca2+ binding releases troponin I from actin, allowing tropomyosin to move off the actin binding sites - thus permitting contraction through the interaction of myosin heads with actin D. True: similar answer also required MU02B; width of the A bands is constant, whereas the Z lines move closer together when the muscle contracts and farther apart when stretched (Ganong 22nd ed p69) ```

Answer 527

A. Smooth muscle found in most hollow viscera is functionally syncytial (def: pertaining to a multinucleate mass of protoplasm produced by the merging of cells - Dorlands Medical Dictionary ) and contains pacemakers that discharge irregularly. Smooth muscle found in the eye does not. But the majority does so CORRECT (Ganong, pg 65) B. Visceral smooth muscle is unique (unlike other smooth muscle) in that it contracts when stretched in the absence of any extrinsic innervation. INCORRECT (Ganong pg 83) C. Ca 2+ is involved in the initiation of contraction of smooth muscle. Visceral smooth muscle has poorly developed sarcoplasmic reticulum, and the increase in the intracellular [Ca 2+] that initiates contraction is due primarily to Ca 2+ influx from the ECF via voltage-gated and ligand-gated ion channels. INCORRECT (Ganong pg 83) D. Increase in membrane potential results in decrease in number of action potentials, decrease in membrane potential results in increased number. INCORRECT (Ganong pg 83) I think this could be a bit tricky / badly worded. There are no pacemaker potentials as such, just a wandering baseline There might be cells that function as pacemakers, but I don't think that is what the question is referring to I think D is correct - If the membrane potential increases, isn't it referring to a less negative membrane potential, and therefore closer to threshold for spike potentials? As I said - badly worded perhaps, Ganong is the same.... "Smooth muscle lacks cross striations. The type found in most hollow viscera is functionally syncytial and contains pacemakers that discharge irregularly." Ganong 22nd Ed. pg 65. So option A? C is definitely correct. Ca2+ is released from the SR and also enters across the membrane. Can't say for sure about the others. Im going for A. C is wrong because smooth muscle hardly have any SR at all. most Ca2+ from extracellular. D is wrong as well as has been correctly stated above increase in membrane action potential decreases the number of spike potential and vice versa. ganong. Alternative: A. Wrong - smooth muscle has a wandering membrane potential, neither a true RMP nor a true pacemaker causing rhythmic APs B. Wrong - extracellular Ca++ entry is dependent of many process other than nerve APs C. Partly correct - as argued above, SR is poorly developed and Ca++ is principally extracellular D. Correct - Increasing membrane potential above the threshold for voltage-gated ion channels will result in multiple spiking action potentials I would choose "most correct" answer: 'D'.

Answer 528

A. Correct - entry of calcium from ECF via voltage-gated calcium channels, sarcoplasmic reticulum poorly formed ( Table 10.2 pg 239 Physioglogy Viva Kerry Brandis) B. I had thought smooth muscle contraction commenced more slowly, was of longer duration, and achieved a greater force than skeletal muscle (but can't remember the reference) C. D. Incorrect - Contraction must last for long periods, tonic contractions typical (Table 10.2 pg 239 Physioglogy Viva Kerry Brandis), and effect of "latch-bridge" mechanism. E. Correct - Guyton 9th ed. pg 102 - "the latent period (for smooth muscle) is some 50 times as great as that for skeletal muscle). According to Guyton there is a slower onset and more prolonged contraction of smooth muscle compared to skeletal muscle because of the slowness of attachment and detachment of cross-bridges and a slower excitation-contraction coupling mechanism] F. Correct - no sarcomere in smooth muscle, no regular arrangement between fibrils of myosin, actin, tropomyosin, not arrange in regular arrays ( Pinnock 2nd edn p295; p297) G. Correct - smooth muscle 1/3 amount myosin; 2x amt actin, appears that greater amount of actin relative to mysoin (Pinnock 2nd edn p296) H. Incorrect - only few mitochondria in smooth muscle (high numbers in cardiac muscle), cellular metabolism depends largely on glycolysis ( Pinnock 2nd edn p295&297) I. Incorrect - less well developed sarcoplasmic reticulum in smooth muscle (Table 10.2 p239 KB Physiology Viva)

Answer 529

A. Correct - Smooth muscle obtains its calcium from a different source compared with skeletal muscle. "Visceral smooth muscle generally has a poorly developed sarcoplasmic reticulum, and the increase in intracellular calcium concentration... is due primarily to Ca2+ influx from the ECF via voltage-gated Ca2+ channels" (Ganong ed 20, p.79) B. Incorrect - Smooth muscle DOES contain tropomyosin, but does not contain troponin (Ganong ed 20, p78) C. Incorrect - "Actin and myosin-II are present, and they slide on each other to produce contraction" (Ganong ed 20 p.78) D. Incorrect - contractions are relatively slow, develop high forces and are maintained for longer durations when compared to striated muscle fibres ( Pinnock 2nd edn p346) Skeletal muscle Ca++ from "sarcoplasmic reticulum" (= endoplasmic reticulum in skeletal muscle) Under voluntary control Has striations Smooth muscle Ca++ from ECF Not under voluntary control No striations

Answer 530

A is correct. Force developed within skeletal muscle is dependent on the load which affects the muscle fibre length. On a microscopic level, the muscle fibre length affects the amount of overlap between the actin and myosin. Addition: Muscular contraction involves shortening of the contractile elements, but because muscles have elastic and viscous elements in series with the contractile mechanism, it is possible for contraction to occur without an appreciable decrease in the length of the whole muscle. Such a contraction is called isometric ("same measure" or length). Contraction against a constant load, with approximation of the ends of the muscle, is isotonic ("same tension"). Since work is the product of force times distance, isotonic contractions do work whereas isometric contractions do not. In other situations, muscle can do negative work while lengthening against a constant weight. The energy supplied to a muscle must equal its energy output. The energy output appears in work done by the muscle, in energy-rich phosphate bonds formed for later use, and in heat. The overall mechanical efficiency of skeletal muscle (work done/total energy expenditure) ranges up to 50% while lifting a weight during isotonic contraction and is essentially 0% during isometric contraction. Ref: Ganong, Chapter 3. Would it be true to say that no external work is done during isometric contraction, but stretching of the elastic and viscous structures eg tendons, will mean that work really is performed by the muscle. Afterall energy is consummed...a great deal in a sustained isometric contraction. Disagree with Answer = A. The force must be independent of muscle fibre length D. Isotonic = same force regardless of whether the muscle is contracting or not. For what it's worth, Guyton says "The characteristics of isotonic contraction depend on the load against which the muscle contracts, as well as the inertia of the load" (pp80, 11Ed). I'm going with A. Re the comment above, the force is absolutely not independent of muscle fibre length!! There's a whole SAQ (Physiol-01A2) on that, so you'd better know something about that relationship!!

Answer 531

A. Incorrect - Thick filaments are made from myosin, NOT actin. Actin, tropomysin and troponin form the thin filaments in the I band. (Ganong p. 62) B. ?Correct - Myosin chains attach tail to tail to form the M line. Actin is anchored to the Z line via actinin (Ganong 20ed p64). "Titin, a large protein, connects the Z lines to the M lines and provides the scaffolding for the sarcomere" (Ganong p64). Therefore this answer is slightly dodgy, but in the absence of a better answer is the most correct. C. Incorrect. Sarcomere is the area between 2 adjacent Z bands

Answer 532

Isotonic contraction is "contraction against a constant load, with approximation of the ends of the muscle" [Ganong 20ed p68] (i.e. the same tension but the muscle gets shorter c.f. isometric where no appreciable change in length occurs) A. Incorrect. Sarcomere length can decrease as the actin and myosin overlap B. Correct. The A band ALWAYS stays constant as it is fixed by the length of the myosin chains and is therefore a change is NOT associated with isotonic contraction (or isometric for that matter) C. Incorrect. I band can shorten as it is the area containing only thin actin filaments D. Incorrect. Z lines move closer together in isotonic contraction E. Incorrect. M lines move closer together in isotonic contraction

Answer 533

A. inncorect as T tubules contain ECF. Sarcoplasmic reticulum release Ca ion directly. D. Correct - Z lines move closer as myosin and actin filaments increase in their overlap Addition: The thick filaments are lined up to form the A bands, whereas the array of thin filaments forms the less dense I bands. The lighter H bands in the center of the A bands are the regions where, when the muscle is relaxed, the thin filaments do not overlap the thick filaments. The Z lines transect the fibrils and connect to the thin filaments. Mus1ani.gif

Answer 534

Answer is A. Long absolute refractory period "Because it has a prolonged AP, cardiac muscle is in its refractory period and will not contract in response to a second stimulus until near the end of the initial contraction. Therefore, cardiac muscle cannot be tetanized like skeletal muscle." (Ganong 22nd 568) AlsoA. The area between two adjacent Z lines is called a sarcomere. INCORRECT (Ganong pg 65) B. Z lines transect the fibrils and connect to thin filaments, which are made up of actin, tropomyosin and troponin (subunits I, C and T). INCORRECT (Ganong pg 65) D. Titin (large protien) connects Z lines to the M lines and provides scaffolding for the sarcomere. Desmin binds the Z lines to the plasma membrane. INCORRECT (Ganong pg 67) E. Smooth muscle fibres are about 2 - 5 micrometers in diameter and 20 - 500 micrometers in length, compared to skeletal muscle which are 20 times as thick and thousands of times as long. INCORRECT (Guyton, pg 95) Kam p27 - "As contraction lasts hardly longer than the action potential (Fig 1.39), it is impossible to summate contractions or tetanize cardiac muscle." This is achieved through the plateau phase of the action potential.

Answer 535

A. The area between two adjacent Z lines is called a sarcomere. INCORRECT (Ganong pg 65) B. Z lines transect the fibrils and connect to thin filaments, which are made up of actin, tropomyosin and troponin (subunits I, C and T). INCORRECT (Ganong pg 65) D. Titin (large protien) connects Z lines to the M lines and provides scaffolding for the sarcomere. Desmin binds the Z lines to the plasma membrane. INCORRECT (Ganong pg 67) E. Smooth muscle fibres are about 2 - 5 micrometers in diameter and 20 - 500 micrometers in length, compared to skeletal muscle which are 20 times as thick and thousands of times as long. INCORRECT (Guyton, pg 95)

Answer 536

The soleus muscle has fatigue resistant (red) muscle fibres - and, in contrast, the gastrocnemius tends to have white muscle fibres. Answer D is correct. Fatigue resistant muscles are red because of high myoglobin content. These type of muscle fibres have high mitochondrial content (therefore aerobic metabolism), slow myosin ATPase (therefore strong sustained contractions up to 100ms), and low levels of glycolytic enzymes (unable to function anaerobically by glycolysis). Fast twitch muscle fibres are at least twice the diameter of slow twitch fibres. References See the chapter in Kerry Brandis - " The physiology viva: Questions and answers" Guyton 8th edition pg 945

Answer 537

Re: first version above: A: Sodium conductance increases first then potassium conductance (same as in nerve AP's ) B: Units of conductance are mmho/cm2 (not cm3) (Correction: the SI unit for Electrical Conductance (g) - the reciprocal of Electrical Impednance (Z) - is "siemens" (S). This replaces the inverse ohms (mho). I note that Ganong and other Physiology texts use "mmho per cm2 of membrane" as units for membrane conductance. (Ref. Bureau International des Poids et Mesures [1]) References Ganong 22nd ed, p59

Answer 538

Re: first version above: A: Sodium conductance increases first then potassium conductance (same as in nerve AP's ) B: Units of conductance are mmho/cm2 (not cm3) (Correction: the SI unit for Electrical Conductance (g) - the reciprocal of Electrical Impednance (Z) - is "siemens" (S). This replaces the inverse ohms (mho). I note that Ganong and other Physiology texts use "mmho per cm2 of membrane" as units for membrane conductance. (Ref. Bureau International des Poids et Mesures [1]) References Ganong 22nd ed, p59

Answer 539

C - Correct. ". . . with increasing voluntary effort, more and more (motor units) are brought into play. This process is sometimes called recruitment of motor units." (from Ganong 22 ed, Chapter 3, p76) A motor unit consists of a single nerve fibre, and all the muscle fibres that nerve fibre innervates. Therefore, can't have recruitment of muscle fibres without recruitment of more motor units (same for nerves). So C is best answer. ``` Factors responsible for Grading of muscle contraction recruitment of motor units frequency of discharge of motor nerves length of muscle fibre the asynchronous firing of motor units - from Ganong ```

Answer 540

C - Correct. ". . . with increasing voluntary effort, more and more (motor units) are brought into play. This process is sometimes called recruitment of motor units." (from Ganong 22 ed, Chapter 3, p76) A motor unit consists of a single nerve fibre, and all the muscle fibres that nerve fibre innervates. Therefore, can't have recruitment of muscle fibres without recruitment of more motor units (same for nerves). So C is best answer. ``` Factors responsible for Grading of muscle contraction recruitment of motor units frequency of discharge of motor nerves length of muscle fibre the asynchronous firing of motor units - from Ganong ```

Answer 541

C. 3 millisecs DISCUSSION: Answer = B? The question is asking about a large (i.e. type A) nerve fibre Ganong (22nd Edn - p.61, Table 2-1) says that type A nerve fibres have spike duration of 0.4 - 0.5 msec, with absolute refractory period of 0.4 - 1 msec. Do we consider the duration of the action potential as the duration of the spike, or spike PLUS absolute refractory period? Obviously if it is just the spike duration, then answer B would be closer. However, even adding the 2 times does not give 3 msec...it only gives half that time....so isn't B a better answer? ?D 30msec. Fig 2.7 22nd Ed Ganong demonstrates complete action potential of a large mammalian myelinated fibre, drawn to show the proportions of the components without time or voltage distortion. The after hyperpolarisation lasts about 40ms and prevents the AP returning to baseline - the returning to RMP being the definition of an Action Potential given by Yentis 3rd Ed. I may be slow but I fail to see the practicality of these questions. Just keep it simple. Ganong 22Ed pg 61, table 2.1 -> Large fibres (the largest is A-alpha) but note that these numbers refer to the spike potential which = ~ 70% of the AP duration => AP duration = 0.6-0.7ms (all from Ganong so it must be right)

Answer 542

D. The overall mechanical efficiency of skeletal muscle (work done/ total energy expenditure) ranges up to 50% while lifting a weight during isotonic contraction. (Ganong p75)

Answer 543

Correct answer: B The muscle spindle senses muscle length and changes in length. (Thus correct answer B). The annulospiral ending is the sensory nerve terminal whose discharge rate increases as the sensory ending is stretched. It is named so as it is composed of a set of rings in a spiral configuration. This terminal is wrapped around the muscle spindle (intrafusal fibres), but NOT the fibres that make up the bulk of the muscle (extrafusal fibres). Source: Muscle spindle stretch reflexes (CAL) - link above. Both the annulospiral and flowerspray nerve endings respond to stretch, but in different ways: the flowerspray endings respond to changing length as it happens, whereas once motion has stopped the annulospiral endings continue to register the change in length (ie they respond to dynamic and static changes respectively). Ganong 22nd ed p.131

Answer 544

A. Incorrect - extrajunctional and junctional nictoinic receptors both have two alpha subunits, gamma unit substituted for epsilon unit in extrajunctional nicotinic receptor (Miller 6th edn, p871) B. Incorrect - extrajunctional receptors have a 2-10 fold longer opening time than mature junctional receptors (BJA CEPD reviews Vol 2, No.5 2002) C. Incorrect - extrajunctional or immature receptors may be expressed anywhere in the muscle membrane (Miller 6th edn p871) D. ? E. If A, B, C and D all incorrect, then E would be correct answer. Query: Aren't "extrajunctional" receptors by definition NOT in the junction? They have variously been called extrajunctional, foetal, immature... I think C is most correct - sure they can grow anywhere but they are not specifically under the NMJ (probably by sheer probability they may well not end up under the endplate) Stoelting 4th edn 212-4 "The NMJ contains three types of nicotinic cholinergic receptors (nAChRs); two are postsynaptic on the skeletal muscle surface - one junctional and one extrajunctional - and one is presynaptic on the nerve ending." "nAChRs are present in large numbers on postjunctional membranes. Extrajunctional nAChRs appear throughout skeletal muscles whenever there is deficient stimulation of the skeletal muscle membrane by the nerve." My take on that is that the NMJ has all three receptors, but mostly the adult type...mostly. Regarding Option C: Extrajunction nAChR are found around the endplate mixed with the usual junction receptors. They can be found anywhere on the muscle cell membrane. Not sure what option C means cause all nAChR are synthetised in the muscle cytoplasm and transported to the endplate References Miller 6th edn. Chapter 22 - Neuromuscular Physiology and Pharmacology British Journal of Anaesthesia, CEPD Reviews, Vol 2, No.5, 2002 - Physiology of the neuromuscular junction (Subscription to www.oxfordjournals.org [1] required to access or try your local friendly medical librarian!!)

Answer 545

Correct answer 'A', ATP in all muscle types. ATP is the energy currency of the cell, fuelling myosin ATPase activity in all muscle cell types. Creatine is an intramuscular phosphate store, to allow rapid re-phosphorylation of ADP to ATP. Ganong p74: Muscle contraction requires energy, and muscle has been called "a machine for converting chemical energy into mechanical work." The immediate source of this energy is ATP, and this is formed by the metabolism of carbohydrates and lipids. I am concerned that perhaps phosphorylcreatine is important initially. Recall that oxygen usage in exercise initially exceeds O2 uptake, and after exercise this "oxygen debt" is taken up by O2 uptake that is more than the immediate needs. The reason for this is that phosphorylcreatine has been used in exercising muscle to provide an alternative form of energy compared to ATP. Specifically, phosphorylcreatine is hydrolyzed to creatine and phosphate groups with the release of considerable energy. At rest, some ATP in the mitochondria transfers its phosphate to creatine, so that a phosphorylcreatine store is built up. During exercise, the phosphorylcreatine is hydrolyzed at the junction between the myosin heads and the actin, forming ATP from ADP and thus permitting contraction to continue. Comment I think the correct answer is that skeletal and cardiac muscle use phosphorylcreatine initially Source: [1] Part of abstract: "In order to explain the insulin-like effect of exercise, it was proposed in 1951 that contracting muscle fibers liberate creatine, which acts to produce an acceptor effect--later called respiratory control--on the muscle mitochondria. The development of this notion paralleled the controversy between biochemists and physiologists over the delivery of energy for muscle contraction. With the demonstration of functional compartmentation of creatine kinase on the mitochondrion, it became clear that the actual form of energy transport in the muscle fiber is phosphorylcreatine. The finding of an isoenzyme of creatine phosphokinase attached to the M-line region of the myofibril revealed the peripheral receptor for the mitochondrially generated phosphorylcreatine. This established a molecular basis for a phosphorylcreatine-creatine shuttle for energy transport in heart and skeletal muscle and provided an explanation for the inability to demonstrate experimentally a direct relation between muscle activity and the concentrations of adenosine triphosphate and adenosine diphosphate." Ganong (22nd ed), p295 "Creatine is synthesised in the liver from methionine, glycine and arginine. In skeletal muscle, it is phosphorylated to form phosphorylcreatine. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction." Power & Kam, p25 "ATP is the only energy source that contractile proteins can use, but muscles only store enough for eight twitches. During muscle contraction ATP is rapidly regenerated from ADP by utilization of the more plentiful creatine phosphate stores. The net result is the hydrolysis of creatine phosphate, not ATP." I agree - I think ATP is the energy source in all 3 muscles... just depends on where the ATP is replenished from phosphorylcreatine anaerobic glycolysis aerobic glycolysis Me too... ATP is initial source... In muscle extracts, ATP and PCr are held in equilibrium by creatine kinase (ADP + PCr -> ATP + Cr: equilibrium constant 20-100 which favours formation of ATP and therefore PCr might not be the immediate energy source for muscle contraction). In 1962 Cain and Davies were able to inhibit creatine kinase in frog muscle with 1-fluoro-2.4-dinitrobenzene (FDNB) and thereby measured the decrease in ATP concentration during contraction. This provided the proof that ATP is the direct energy source for contraction. Without inhibiting creatine kinase, decrease in ATP concentration could not be measured because the formed ADP was immediately rephosphorylated to ATP via the creatine kinase (Lohmann) reaction. As a matter of fact, during contraction of normal muscle, the ATP concentration remains fairly constant while the PCr concentration decreases. It is important to realize that the ATP concentration in muscle is quite low (5-8 mmoles/g muscle), enough only for a few contractions. The used ATP is immediately resynthesized from PCr. However, the PCr concentration is also low (20-25 mmoles/g), enough for some additional contractions. Glycogen offers a rapid but still limited energy store (endogenous glycogen concentration about 75 mmoles glucose units in glycogen/g muscle), whereas oxidative phosphorylation is the slowest and most efficient process for ATP production.

Answer 546

Correct answer 'A', ATP in all muscle types. ATP is the energy currency of the cell, fuelling myosin ATPase activity in all muscle cell types. Creatine is an intramuscular phosphate store, to allow rapid re-phosphorylation of ADP to ATP. Ganong p74: Muscle contraction requires energy, and muscle has been called "a machine for converting chemical energy into mechanical work." The immediate source of this energy is ATP, and this is formed by the metabolism of carbohydrates and lipids. I am concerned that perhaps phosphorylcreatine is important initially. Recall that oxygen usage in exercise initially exceeds O2 uptake, and after exercise this "oxygen debt" is taken up by O2 uptake that is more than the immediate needs. The reason for this is that phosphorylcreatine has been used in exercising muscle to provide an alternative form of energy compared to ATP. Specifically, phosphorylcreatine is hydrolyzed to creatine and phosphate groups with the release of considerable energy. At rest, some ATP in the mitochondria transfers its phosphate to creatine, so that a phosphorylcreatine store is built up. During exercise, the phosphorylcreatine is hydrolyzed at the junction between the myosin heads and the actin, forming ATP from ADP and thus permitting contraction to continue. Comment I think the correct answer is that skeletal and cardiac muscle use phosphorylcreatine initially Source: [1] Part of abstract: "In order to explain the insulin-like effect of exercise, it was proposed in 1951 that contracting muscle fibers liberate creatine, which acts to produce an acceptor effect--later called respiratory control--on the muscle mitochondria. The development of this notion paralleled the controversy between biochemists and physiologists over the delivery of energy for muscle contraction. With the demonstration of functional compartmentation of creatine kinase on the mitochondrion, it became clear that the actual form of energy transport in the muscle fiber is phosphorylcreatine. The finding of an isoenzyme of creatine phosphokinase attached to the M-line region of the myofibril revealed the peripheral receptor for the mitochondrially generated phosphorylcreatine. This established a molecular basis for a phosphorylcreatine-creatine shuttle for energy transport in heart and skeletal muscle and provided an explanation for the inability to demonstrate experimentally a direct relation between muscle activity and the concentrations of adenosine triphosphate and adenosine diphosphate." Ganong (22nd ed), p295 "Creatine is synthesised in the liver from methionine, glycine and arginine. In skeletal muscle, it is phosphorylated to form phosphorylcreatine. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction." Power & Kam, p25 "ATP is the only energy source that contractile proteins can use, but muscles only store enough for eight twitches. During muscle contraction ATP is rapidly regenerated from ADP by utilization of the more plentiful creatine phosphate stores. The net result is the hydrolysis of creatine phosphate, not ATP." I agree - I think ATP is the energy source in all 3 muscles... just depends on where the ATP is replenished from phosphorylcreatine anaerobic glycolysis aerobic glycolysis Me too... ATP is initial source... In muscle extracts, ATP and PCr are held in equilibrium by creatine kinase (ADP + PCr -> ATP + Cr: equilibrium constant 20-100 which favours formation of ATP and therefore PCr might not be the immediate energy source for muscle contraction). In 1962 Cain and Davies were able to inhibit creatine kinase in frog muscle with 1-fluoro-2.4-dinitrobenzene (FDNB) and thereby measured the decrease in ATP concentration during contraction. This provided the proof that ATP is the direct energy source for contraction. Without inhibiting creatine kinase, decrease in ATP concentration could not be measured because the formed ADP was immediately rephosphorylated to ATP via the creatine kinase (Lohmann) reaction. As a matter of fact, during contraction of normal muscle, the ATP concentration remains fairly constant while the PCr concentration decreases. It is important to realize that the ATP concentration in muscle is quite low (5-8 mmoles/g muscle), enough only for a few contractions. The used ATP is immediately resynthesized from PCr. However, the PCr concentration is also low (20-25 mmoles/g), enough for some additional contractions. Glycogen offers a rapid but still limited energy store (endogenous glycogen concentration about 75 mmoles glucose units in glycogen/g muscle), whereas oxidative phosphorylation is the slowest and most efficient process for ATP production.

Answer 547

Correct answer 'A', ATP in all muscle types. ATP is the energy currency of the cell, fuelling myosin ATPase activity in all muscle cell types. Creatine is an intramuscular phosphate store, to allow rapid re-phosphorylation of ADP to ATP. Ganong p74: Muscle contraction requires energy, and muscle has been called "a machine for converting chemical energy into mechanical work." The immediate source of this energy is ATP, and this is formed by the metabolism of carbohydrates and lipids. I am concerned that perhaps phosphorylcreatine is important initially. Recall that oxygen usage in exercise initially exceeds O2 uptake, and after exercise this "oxygen debt" is taken up by O2 uptake that is more than the immediate needs. The reason for this is that phosphorylcreatine has been used in exercising muscle to provide an alternative form of energy compared to ATP. Specifically, phosphorylcreatine is hydrolyzed to creatine and phosphate groups with the release of considerable energy. At rest, some ATP in the mitochondria transfers its phosphate to creatine, so that a phosphorylcreatine store is built up. During exercise, the phosphorylcreatine is hydrolyzed at the junction between the myosin heads and the actin, forming ATP from ADP and thus permitting contraction to continue. Comment I think the correct answer is that skeletal and cardiac muscle use phosphorylcreatine initially Source: [1] Part of abstract: "In order to explain the insulin-like effect of exercise, it was proposed in 1951 that contracting muscle fibers liberate creatine, which acts to produce an acceptor effect--later called respiratory control--on the muscle mitochondria. The development of this notion paralleled the controversy between biochemists and physiologists over the delivery of energy for muscle contraction. With the demonstration of functional compartmentation of creatine kinase on the mitochondrion, it became clear that the actual form of energy transport in the muscle fiber is phosphorylcreatine. The finding of an isoenzyme of creatine phosphokinase attached to the M-line region of the myofibril revealed the peripheral receptor for the mitochondrially generated phosphorylcreatine. This established a molecular basis for a phosphorylcreatine-creatine shuttle for energy transport in heart and skeletal muscle and provided an explanation for the inability to demonstrate experimentally a direct relation between muscle activity and the concentrations of adenosine triphosphate and adenosine diphosphate." Ganong (22nd ed), p295 "Creatine is synthesised in the liver from methionine, glycine and arginine. In skeletal muscle, it is phosphorylated to form phosphorylcreatine. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction." Power & Kam, p25 "ATP is the only energy source that contractile proteins can use, but muscles only store enough for eight twitches. During muscle contraction ATP is rapidly regenerated from ADP by utilization of the more plentiful creatine phosphate stores. The net result is the hydrolysis of creatine phosphate, not ATP." I agree - I think ATP is the energy source in all 3 muscles... just depends on where the ATP is replenished from phosphorylcreatine anaerobic glycolysis aerobic glycolysis Me too... ATP is initial source... In muscle extracts, ATP and PCr are held in equilibrium by creatine kinase (ADP + PCr -> ATP + Cr: equilibrium constant 20-100 which favours formation of ATP and therefore PCr might not be the immediate energy source for muscle contraction). In 1962 Cain and Davies were able to inhibit creatine kinase in frog muscle with 1-fluoro-2.4-dinitrobenzene (FDNB) and thereby measured the decrease in ATP concentration during contraction. This provided the proof that ATP is the direct energy source for contraction. Without inhibiting creatine kinase, decrease in ATP concentration could not be measured because the formed ADP was immediately rephosphorylated to ATP via the creatine kinase (Lohmann) reaction. As a matter of fact, during contraction of normal muscle, the ATP concentration remains fairly constant while the PCr concentration decreases. It is important to realize that the ATP concentration in muscle is quite low (5-8 mmoles/g muscle), enough only for a few contractions. The used ATP is immediately resynthesized from PCr. However, the PCr concentration is also low (20-25 mmoles/g), enough for some additional contractions. Glycogen offers a rapid but still limited energy store (endogenous glycogen concentration about 75 mmoles glucose units in glycogen/g muscle), whereas oxidative phosphorylation is the slowest and most efficient process for ATP production.

Answer 548

Correct answer 'A', ATP in all muscle types. ATP is the energy currency of the cell, fuelling myosin ATPase activity in all muscle cell types. Creatine is an intramuscular phosphate store, to allow rapid re-phosphorylation of ADP to ATP. Ganong p74: Muscle contraction requires energy, and muscle has been called "a machine for converting chemical energy into mechanical work." The immediate source of this energy is ATP, and this is formed by the metabolism of carbohydrates and lipids. I am concerned that perhaps phosphorylcreatine is important initially. Recall that oxygen usage in exercise initially exceeds O2 uptake, and after exercise this "oxygen debt" is taken up by O2 uptake that is more than the immediate needs. The reason for this is that phosphorylcreatine has been used in exercising muscle to provide an alternative form of energy compared to ATP. Specifically, phosphorylcreatine is hydrolyzed to creatine and phosphate groups with the release of considerable energy. At rest, some ATP in the mitochondria transfers its phosphate to creatine, so that a phosphorylcreatine store is built up. During exercise, the phosphorylcreatine is hydrolyzed at the junction between the myosin heads and the actin, forming ATP from ADP and thus permitting contraction to continue. Comment I think the correct answer is that skeletal and cardiac muscle use phosphorylcreatine initially Source: [1] Part of abstract: "In order to explain the insulin-like effect of exercise, it was proposed in 1951 that contracting muscle fibers liberate creatine, which acts to produce an acceptor effect--later called respiratory control--on the muscle mitochondria. The development of this notion paralleled the controversy between biochemists and physiologists over the delivery of energy for muscle contraction. With the demonstration of functional compartmentation of creatine kinase on the mitochondrion, it became clear that the actual form of energy transport in the muscle fiber is phosphorylcreatine. The finding of an isoenzyme of creatine phosphokinase attached to the M-line region of the myofibril revealed the peripheral receptor for the mitochondrially generated phosphorylcreatine. This established a molecular basis for a phosphorylcreatine-creatine shuttle for energy transport in heart and skeletal muscle and provided an explanation for the inability to demonstrate experimentally a direct relation between muscle activity and the concentrations of adenosine triphosphate and adenosine diphosphate." Ganong (22nd ed), p295 "Creatine is synthesised in the liver from methionine, glycine and arginine. In skeletal muscle, it is phosphorylated to form phosphorylcreatine. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction." Power & Kam, p25 "ATP is the only energy source that contractile proteins can use, but muscles only store enough for eight twitches. During muscle contraction ATP is rapidly regenerated from ADP by utilization of the more plentiful creatine phosphate stores. The net result is the hydrolysis of creatine phosphate, not ATP." I agree - I think ATP is the energy source in all 3 muscles... just depends on where the ATP is replenished from phosphorylcreatine anaerobic glycolysis aerobic glycolysis Me too... ATP is initial source... In muscle extracts, ATP and PCr are held in equilibrium by creatine kinase (ADP + PCr -> ATP + Cr: equilibrium constant 20-100 which favours formation of ATP and therefore PCr might not be the immediate energy source for muscle contraction). In 1962 Cain and Davies were able to inhibit creatine kinase in frog muscle with 1-fluoro-2.4-dinitrobenzene (FDNB) and thereby measured the decrease in ATP concentration during contraction. This provided the proof that ATP is the direct energy source for contraction. Without inhibiting creatine kinase, decrease in ATP concentration could not be measured because the formed ADP was immediately rephosphorylated to ATP via the creatine kinase (Lohmann) reaction. As a matter of fact, during contraction of normal muscle, the ATP concentration remains fairly constant while the PCr concentration decreases. It is important to realize that the ATP concentration in muscle is quite low (5-8 mmoles/g muscle), enough only for a few contractions. The used ATP is immediately resynthesized from PCr. However, the PCr concentration is also low (20-25 mmoles/g), enough for some additional contractions. Glycogen offers a rapid but still limited energy store (endogenous glycogen concentration about 75 mmoles glucose units in glycogen/g muscle), whereas oxidative phosphorylation is the slowest and most efficient process for ATP production.

Answer 549

Correct answer 'A', ATP in all muscle types. ATP is the energy currency of the cell, fuelling myosin ATPase activity in all muscle cell types. Creatine is an intramuscular phosphate store, to allow rapid re-phosphorylation of ADP to ATP. Ganong p74: Muscle contraction requires energy, and muscle has been called "a machine for converting chemical energy into mechanical work." The immediate source of this energy is ATP, and this is formed by the metabolism of carbohydrates and lipids. I am concerned that perhaps phosphorylcreatine is important initially. Recall that oxygen usage in exercise initially exceeds O2 uptake, and after exercise this "oxygen debt" is taken up by O2 uptake that is more than the immediate needs. The reason for this is that phosphorylcreatine has been used in exercising muscle to provide an alternative form of energy compared to ATP. Specifically, phosphorylcreatine is hydrolyzed to creatine and phosphate groups with the release of considerable energy. At rest, some ATP in the mitochondria transfers its phosphate to creatine, so that a phosphorylcreatine store is built up. During exercise, the phosphorylcreatine is hydrolyzed at the junction between the myosin heads and the actin, forming ATP from ADP and thus permitting contraction to continue. Comment I think the correct answer is that skeletal and cardiac muscle use phosphorylcreatine initially Source: [1] Part of abstract: "In order to explain the insulin-like effect of exercise, it was proposed in 1951 that contracting muscle fibers liberate creatine, which acts to produce an acceptor effect--later called respiratory control--on the muscle mitochondria. The development of this notion paralleled the controversy between biochemists and physiologists over the delivery of energy for muscle contraction. With the demonstration of functional compartmentation of creatine kinase on the mitochondrion, it became clear that the actual form of energy transport in the muscle fiber is phosphorylcreatine. The finding of an isoenzyme of creatine phosphokinase attached to the M-line region of the myofibril revealed the peripheral receptor for the mitochondrially generated phosphorylcreatine. This established a molecular basis for a phosphorylcreatine-creatine shuttle for energy transport in heart and skeletal muscle and provided an explanation for the inability to demonstrate experimentally a direct relation between muscle activity and the concentrations of adenosine triphosphate and adenosine diphosphate." Ganong (22nd ed), p295 "Creatine is synthesised in the liver from methionine, glycine and arginine. In skeletal muscle, it is phosphorylated to form phosphorylcreatine. The ATP formed by glycolysis and oxidative phosphorylation reacts with creatine to form ADP and large amounts of phosphorylcreatine. During exercise the reaction is reversed, maintaining the supply of ATP, which is the immediate source of the energy for muscle contraction." Power & Kam, p25 "ATP is the only energy source that contractile proteins can use, but muscles only store enough for eight twitches. During muscle contraction ATP is rapidly regenerated from ADP by utilization of the more plentiful creatine phosphate stores. The net result is the hydrolysis of creatine phosphate, not ATP." I agree - I think ATP is the energy source in all 3 muscles... just depends on where the ATP is replenished from phosphorylcreatine anaerobic glycolysis aerobic glycolysis Me too... ATP is initial source... In muscle extracts, ATP and PCr are held in equilibrium by creatine kinase (ADP + PCr -> ATP + Cr: equilibrium constant 20-100 which favours formation of ATP and therefore PCr might not be the immediate energy source for muscle contraction). In 1962 Cain and Davies were able to inhibit creatine kinase in frog muscle with 1-fluoro-2.4-dinitrobenzene (FDNB) and thereby measured the decrease in ATP concentration during contraction. This provided the proof that ATP is the direct energy source for contraction. Without inhibiting creatine kinase, decrease in ATP concentration could not be measured because the formed ADP was immediately rephosphorylated to ATP via the creatine kinase (Lohmann) reaction. As a matter of fact, during contraction of normal muscle, the ATP concentration remains fairly constant while the PCr concentration decreases. It is important to realize that the ATP concentration in muscle is quite low (5-8 mmoles/g muscle), enough only for a few contractions. The used ATP is immediately resynthesized from PCr. However, the PCr concentration is also low (20-25 mmoles/g), enough for some additional contractions. Glycogen offers a rapid but still limited energy store (endogenous glycogen concentration about 75 mmoles glucose units in glycogen/g muscle), whereas oxidative phosphorylation is the slowest and most efficient process for ATP production.

Answer 550

B. Slow Ca channels Cardiac muscle: has a very long action potential compared with skeletal muscle has a plateau (phase 2) during the action potential creating a prolonged period of depolarisation. (Skeletal muscle AP does not have a plateau) Both of the above phenomenon are caused by slow calcium channels which are slower to open than the fast calcium channels in skeletal muscle and are open for longer. This prevents cardiac muscles being able to go into tetany. Cardiac muscle Contains large numbers of mitochondria Is not able to go into oxygen debt requires a constant supply of oxygen --- What about Dihydropyridine receptors in skeletal muscle? They are a L-type calcium channel found in T-tubules, part of the excitation-contraction coupling process. - Dihydropyridine receptor interacts with ryanodine receptor on SR, resulting in Ca release from SR. (No L-type channel with it). Ganong p74

Answer 551

Alt: B seems wrong - spindles measure length and rate of change of length, golgi tendon organs measure tension. (Kam p27) I agree with the above. Consider a muscle undergoing isometric contraction. Although tension increases, muscle spindle length remains the same. Hence Ia discharge remains constant. Muscle spindles are found within the belly of muscles and run in parallel with the main muscle fibres.The spindle senses muscle length and changes in length. It has sensory nerve terminals whose discharge rate increases as the sensory ending is stretched. This nerve terminal is known as the ANNULOSPIRAL ending, so named because it is composed of a set of rings in a spiral configuration. These terminals are wrapped around specialised muscle fibres that belongs to the muscle spindle (INTRAFUSAL FIBRES) and are quite separate from the fibres that make up the bulk of the muscle (EXTRAFUSAL FIBRES). Muscle spindles: (from Power & Kam) "... detect muscle length and movement, whilst tendon organs sense tension." "... are innervated by primary and secondary afferents and γ motor nerves." "The γ motor fibres supply the contractile ends of the spindle ..." "The sensory nerves are excited by any stretching of the non-contractile centre ..." The primary endings (Ia) are more sensitive to movement and are called 'Dynamic'. The secondary endings (II) are more sensitive to absolute length, and are called 'Static'. Spindle change in length reflexly excite the motor neuron to the extrafusal fibres of the muscle in which they lie. γ efferents are involved in "The γ Loop", a reflex contraction of the skeletal muscle due to activity in the Extra-Pyramidal System. This is referred to as "α-γ linkage" in Ganong (see fig 6-3 and text). The control of γ efferent discharge adjusts the sensitivity of the spindles, which aids postural control. Summary A - Wrong: "the ends of the spindle capsule attached to the tendons at either end of the muscle" Ganong Ch923Ed B - Wrong: measure length, not tension C - Best answer. HAve contractile elements, although they are too weak to contract the muscle and instead play an important role in spindle regulation when they contract. See also: Muscle spindle

Answer 552

B Comments Ganong 23ed pg 100 DHPR physically triggers RYR receptor (sarcoplasmic Ca2+ channel) opening. DHPR is a voltage gated channel, acts on RYR but not as a Ca++ channel in the Skeletal muscle. Calcium entry (RYR) then functions as calcium induced calcium release (positive feedback) mechanism from skeletal SR. (I think) Comments welcome.

Answer 553

Hyperventilation is a result of the direct action of progesterone on the medullary respiratory control centre. Progesterone increases the sensitivity to carbon dioxide and as such the set point is reduced. Normal CO2 levels therefore trigger increased ventilation. Result: Hyperventilation of pregnancy. Partially compensated respiratory alkalosis.\ Soon after conception, there is a progressive increase in minute ventilation which peaks at 50% above normal by the 2nd trimester. This is associated with 40% increase in tidal volume and 15% increase in respiratory rate. As dead space remains the same, alveolar ventilation is 70% higher at end of gestation. Increased ventilation results in decreased arterial and alveolar PaCO2. Average values during gestation: PaCO2 of 32mmHg and PaO2 of 105mmHg. Development of alkalosis is averted by compensatory decreases in bicarbonate.

Answer 554

The typical ABGs in late pregnancy have a low pCO2: 30-32mmHg due to maternal hyperventilation. The body compensates for this low pCO2 by dropping arterial [HCO3] (renal mechanism) and this tends to return the pH back towards normal (7.4). Because the hyperventilation tends to persist the amount of compensation that occurs is often sufficient to return the maternal pH to within the normal range, though typically at the upper end, say 7.42-7.44. So option C above looks like the answer. See also "Acid-base effects of pregnancy" [1] The following article is old (1965) but looks very relevant (I haven't seen it personally but it is referenced in [2]): Prowse CM, Gaenster EA. Respiratory and acid-base changes during pregnancy. Anesthesiology 1965;26:381-384. I'd go for B Normal maternal ABG is pH roughly normal (7.40 - 7.42), CO2 about 35, O2 100 - 105, bicarb 20. New comment: The answer has to be C. There is a respiratory alkalosis which is not fully corrected, but that isn't relevant - the only correct PaCO2 is in option C. I'm not sure where the last comment is sourced from.

Answer 555

The action of prostaglandins on the the ductus arteriosus is to maintain patency. As such, NSAIDs should be avoided in the third trimester in fear of closing the ductus pre-delivery (obviously disasterous). The source of the PGs is the placenta so once cord clamping occurs, PG concentration reduces causing contraction of the spiralling smooth muscles in the ductus. The definite method of closure is poorly understood but is felt to relate to the direct action of oxygen (high oxygen content of pulmonary vein returning from now-ventilated lungs) on the endothelial and smooth muscle lining of the ductus. This in turn causes a reduction in paracrine prostaglandin release and closure of the ductus. Other hypotheised contributors... Development of turbulent flow through the ductus as flow is halted and reversed. This occurs as aortic pressure increases (due to increased SVR) and pulmonary artery pressure decreases (due to reduced PVR). So answer is somewhere between B and D, but would probably er on the D side as it seems more explanatory. Kam says, on page 357, that the ductus arteriosus "constricts in response to the increasing PaO2 after the first breath and closure of the foramen ovale, and to decreasing concentrations of circulating and locally produces prostaglandins E1 and E2. The physiological closure occurs within 10-15 hours, and the permanent closure takes place in 2-3 weeks by thrombosis and fibrosis." --- I agree that the answer is D, but the comment about high oxygen content of pulmonary vein seems a bit confusing. Sure - the lungs oxygenate the blood ex-utero, and obviously the oxygenated blood has to travel from the lungs to left heart via pulmonary veins, but the important point is that the aorta now contains highly oxygenated blood and the flow in the ductus is reversed (compared to in-utero). That is - blood with high PO2 now flows from aorta to pulmonary artery (L -> R shunt), and the high PO2 causes the vascular smooth muscle to contract. References Power and Kam 2nd Ed, p 412.

Answer 556

The correct answer is B. Blood from the placenta flows via the umbilical vein to the left branch of the hepatic portal vein. 60% of this blood from the umbilical vein then bypasses the liver via the ductus venosus to enter the IVC directly. This allows the oxygenated blood from the placenta to be conducted directly to the right atrium, from whence it is shunted across the foramen ovale (with the help of the crista terminalis, a ridge of muscle in the right atrium) to directly enter the left atrium and then supply the brain and heart with oxygentated blood. Therefore, option D is wrong (blood from the SVC flows from right atrium to right ventricle and therefore does not cross the foramen ovale. Option E is also wrong. The ductus arteriosus does not direct oxygentated blood to the head (this is achieved by the crista terminalis directing blood through the foramen ovale). The ductus arteriosus receives 90% of right ventricular output, allowing this blood to bypass the lungs with their high pulmonary vascular resistance. Blood from the ductus arteriosus then enters the descending aorta, and thus the less well oxygenated blood is directed to the lower limbs. Option C is incorrect - oxygen saturation is 80% in the umbilical vein (oxygen tension is 30 mmHg), which transports oxygentated blood from the placenta to the fetus. Apr 2001: With regard to the foetal circulation: A. Blood from SVC goes into the left atrium via the ductus arteriosis B. Ductus venosus drains into the IVC directly C. Oxygen saturation is 40% in umbilical vein D. ? The correct answer is B. Blood from the SVC enters the right atrium directly, and oxygen saturation is 80% in the umbilical vein; therefore options A and C are incorrect. Mar 02: In the foetal circulation: A. Umbilical vein straight into IVC B. SVC blood to LA via foramen ovale C. Only has foetal haemoglobin D. ? E. ? All of these options are incorrect. Blood from the umbilical vein must first enter the left branch of the hepatic portal vein before 60% is directed to the IVC via the ductus venosus,and 40% enters the liver. Therefore A is incorrect. B is wrong because it is actually 60% of IVC (NOT SVC) blood which goes from right atrium to left atrium via the foramen ovale. Blood from the SVC mixes with the remaining 40% of IVC blood in the right atrium, and this blood then flows to the right ventricle. C is incorrect because at no stage (neither in the fetus nor after birth) is haemoglobin entirely composed of the fetal form. Before birth, HbF constitutes 90% of all Hb (this percentage declines after 35 weeks gestation such that at birth HbF is only 75-80% of the total Hb). By 6 months of age, HbF is replaced by adult Hb. See also MF09 Study Hint Extremely difficult to find an adequate picture of the fetal circulation. Skip Ganong and Kam (which is actually wrong), look at West for a simplified version (without ductus venosum) which emphasises the essential features, and then look at the BEST picture ever in Berne and Levy which properly describes all the parts. Important aspects are: ductus venosus, ductus arteriosus, foramen ovale and the crista dividens (also called the crista terminalis). Also that the pulmonary and systemic circulations are in parallel during fetal life rather than in series as after birth. Numbers to recall are the 60:40 split of IVC blood to L:R ventricles, and 10:90 split of Ao blood to pulmonary:ductus arteriosus. A few typical PO2's wouldn't go astray. Please see this beautiful animation to clear the concept http://www.indiana.edu/~anat550/cvanim/fetcirc/fetcirc.html [1]

Answer 557

Both B and C appear to be correct. Metabolism of brown fat in the neonate is one form of non-shivering thermogenesis. It is important as the neonate can sustain large heat losses as a consequence of a high surface area:volume ratio (2-2.5 times that of adults). Brown fat is found in the axillae, near major blood vessels in the neck (not the "great vessels"; they are in the chest), in the mediastinum, between the scapulae and in the perinephric tissue. It comprises about 11% of total body fat. Cold ambient temperatures increase sympathetic activity and release of noradrenaline in brown fat. Lipase activity is increased due to increase in adenyl cyclase and protein kinase activity via β3 receptor stimulation by noradrenaline. Triglycerides are hydrolysed to free fatty acids and glycerol which provides the substrate for mitochondrial oxidative phosphorylation. Some uncoupling of OxPhos occurs so that more heat is produced compared to other body cells, and presumably ATP is not produced (but I can't find a specific ref for this). Oxygen is required. Thus option A is incorrect (ATP is not produced). (P&K 363 Guyton 861) D is also incorrect as the reaction requires mitochondia (brown fat has high mitochondria content). B is (partially) correct in describing the distribution of brown fat but is not the best answer C correctly attributes the autonomic nervous system (in this case, the sympathetic nervous system, β3-NorAdR) as being essential to the process of heat production by brown fat metabolism. Regarding Option A: Hydrolysis of triglycerides to FFA and glycerol doesn't involve ATP (either generation or consumption). The generation of heat in brown fat is by the usual Krebs cycle and generation of a proton gradient across the mitochondrial membrane, but the big difference is the presence of an "uncoupling protein" which uncouples the Kreb's cycle and other intermediary metabolism from ATP generation by providing an alternative pathway for the protons to re-enter the mitochondrial matrix, other than via ATP synthase. Normally ATP has to be consumed (eg shivering) in order for the Krebs cycle etc to progress, but this is not the case in brown fat. However brown fat does still produce ATP (mentioned in Ganong), but just less than you would expect given its consumption of nutrients. So i guess option A is technically correct, but i wouldn't consider it the single best answer Regarding option B: Brown fat may be around the great vessels of the neck, but i don't necessarily think there role is to insulate them as such, but rather to transfer heat to them (as compared with white fat; given their different morphology, they may actually not be very good insulators?). I guess it's like comparing a heater to a blanket. So again, probably not the single best answer Regarding Option C: CORRECT. Brown fat has a dense sympathetic innervation, so i think this IS the single best answer Regarding Option D: INCORRECT. The uncoupling of oxidative phosphorylation occurs in the wall of the mitochondria Power and Kam (2nd ed) p375: increased by beta3 sympathetic activity

Answer 558

Both B and C appear to be correct. Metabolism of brown fat in the neonate is one form of non-shivering thermogenesis. It is important as the neonate can sustain large heat losses as a consequence of a high surface area:volume ratio (2-2.5 times that of adults). Brown fat is found in the axillae, near major blood vessels in the neck (not the "great vessels"; they are in the chest), in the mediastinum, between the scapulae and in the perinephric tissue. It comprises about 11% of total body fat. Cold ambient temperatures increase sympathetic activity and release of noradrenaline in brown fat. Lipase activity is increased due to increase in adenyl cyclase and protein kinase activity via β3 receptor stimulation by noradrenaline. Triglycerides are hydrolysed to free fatty acids and glycerol which provides the substrate for mitochondrial oxidative phosphorylation. Some uncoupling of OxPhos occurs so that more heat is produced compared to other body cells, and presumably ATP is not produced (but I can't find a specific ref for this). Oxygen is required. Thus option A is incorrect (ATP is not produced). (P&K 363 Guyton 861) D is also incorrect as the reaction requires mitochondia (brown fat has high mitochondria content). B is (partially) correct in describing the distribution of brown fat but is not the best answer C correctly attributes the autonomic nervous system (in this case, the sympathetic nervous system, β3-NorAdR) as being essential to the process of heat production by brown fat metabolism. Regarding Option A: Hydrolysis of triglycerides to FFA and glycerol doesn't involve ATP (either generation or consumption). The generation of heat in brown fat is by the usual Krebs cycle and generation of a proton gradient across the mitochondrial membrane, but the big difference is the presence of an "uncoupling protein" which uncouples the Kreb's cycle and other intermediary metabolism from ATP generation by providing an alternative pathway for the protons to re-enter the mitochondrial matrix, other than via ATP synthase. Normally ATP has to be consumed (eg shivering) in order for the Krebs cycle etc to progress, but this is not the case in brown fat. However brown fat does still produce ATP (mentioned in Ganong), but just less than you would expect given its consumption of nutrients. So i guess option A is technically correct, but i wouldn't consider it the single best answer Regarding option B: Brown fat may be around the great vessels of the neck, but i don't necessarily think there role is to insulate them as such, but rather to transfer heat to them (as compared with white fat; given their different morphology, they may actually not be very good insulators?). I guess it's like comparing a heater to a blanket. So again, probably not the single best answer Regarding Option C: CORRECT. Brown fat has a dense sympathetic innervation, so i think this IS the single best answer Regarding Option D: INCORRECT. The uncoupling of oxidative phosphorylation occurs in the wall of the mitochondria Power and Kam (2nd ed) p375: increased by beta3 sympathetic activity

Answer 559

Both B and C appear to be correct. Metabolism of brown fat in the neonate is one form of non-shivering thermogenesis. It is important as the neonate can sustain large heat losses as a consequence of a high surface area:volume ratio (2-2.5 times that of adults). Brown fat is found in the axillae, near major blood vessels in the neck (not the "great vessels"; they are in the chest), in the mediastinum, between the scapulae and in the perinephric tissue. It comprises about 11% of total body fat. Cold ambient temperatures increase sympathetic activity and release of noradrenaline in brown fat. Lipase activity is increased due to increase in adenyl cyclase and protein kinase activity via β3 receptor stimulation by noradrenaline. Triglycerides are hydrolysed to free fatty acids and glycerol which provides the substrate for mitochondrial oxidative phosphorylation. Some uncoupling of OxPhos occurs so that more heat is produced compared to other body cells, and presumably ATP is not produced (but I can't find a specific ref for this). Oxygen is required. Thus option A is incorrect (ATP is not produced). (P&K 363 Guyton 861) D is also incorrect as the reaction requires mitochondia (brown fat has high mitochondria content). B is (partially) correct in describing the distribution of brown fat but is not the best answer C correctly attributes the autonomic nervous system (in this case, the sympathetic nervous system, β3-NorAdR) as being essential to the process of heat production by brown fat metabolism. Regarding Option A: Hydrolysis of triglycerides to FFA and glycerol doesn't involve ATP (either generation or consumption). The generation of heat in brown fat is by the usual Krebs cycle and generation of a proton gradient across the mitochondrial membrane, but the big difference is the presence of an "uncoupling protein" which uncouples the Kreb's cycle and other intermediary metabolism from ATP generation by providing an alternative pathway for the protons to re-enter the mitochondrial matrix, other than via ATP synthase. Normally ATP has to be consumed (eg shivering) in order for the Krebs cycle etc to progress, but this is not the case in brown fat. However brown fat does still produce ATP (mentioned in Ganong), but just less than you would expect given its consumption of nutrients. So i guess option A is technically correct, but i wouldn't consider it the single best answer Regarding option B: Brown fat may be around the great vessels of the neck, but i don't necessarily think there role is to insulate them as such, but rather to transfer heat to them (as compared with white fat; given their different morphology, they may actually not be very good insulators?). I guess it's like comparing a heater to a blanket. So again, probably not the single best answer Regarding Option C: CORRECT. Brown fat has a dense sympathetic innervation, so i think this IS the single best answer Regarding Option D: INCORRECT. The uncoupling of oxidative phosphorylation occurs in the wall of the mitochondria Power and Kam (2nd ed) p375: increased by beta3 sympathetic activity

Answer 560

Question MF06 According to diagram in Cardiovascular Physiology - Berne and Levy the oxygen saturation in the fetal circulation is as follows: Thoracic IVC 67% Right Atrium - Mixture of SVC blood (Sats 25%)and IVC blood (Sats 67%) Ascending Aorta 62% Pulmonary Vv 42% Ductus Arteriosus 52% Venous return to the Right Atrium is approx 2/3's IVC and 1/3 SVC. Therefore sats approx 52%. Correct Answer: A - Thoracic IVC.

Answer 561

Question MF06b A. Incorrect. Uterine vein Po2 40mmHg > Umbilical Vein Po2 30mmHg B. Correct. Umbilical Vein PO2 30mmHg > Umbilical Artery PO2 20mmHg

Answer 562

Alt version Jul03 A. Correct - The pO2 of the umbilical artery 20mmHg, pO2 maternal uterine vein 40mmHg B. Incorrect - Foetal Haemoglobin has an increased affinity for O2 (compared with HbA) which assists with the loading of O2 at a reduced pO2 - L shift of ODC.

Answer 563

Re MF07a: Both A and C appear to be correct. I will be interested to view other responses to determine whether one of these options is "more correct" than the other! Onset of ventilation in the neonate leads to increased blood flow through the lungs due to A direct effect of outwards traction on the pulmonary vessels due to lung expansion, (decreased PVR related to lung vol, remember diagram in West) and A(slightly later) effect of increased oxygenation causing pulmonary vascular dilation (loss of Hypoxic Pulmonary Vasoconstriction), Both these factors will cause a decrease in pulmonary vascular resistance and increased blood flow through the lungs. This will have the effect of Decreasing right ventricular pressure due to decreased right ventricular afterload (option C), and Increasing left atrial pressure due to increased blood flow from the pulmonary vasculature to the left atrium (option A) Option B is incorrect. The mechanism of closure of the ductus venosus is unknown. Comment I think C is more correct as the clamping of the umbilical artery, thus increased SVR would probably have a role in increasing LAP as well.

Answer 564

Re MF07b: This question is similar to the above and again I have selected answers which relate to increased left atrial pressure (ie, option C, closure of the foramen ovale, which occurs when LAP exceeds RAP), and decreased right ventricular pressure (ie, option A, decreasing right ventricular outflow pressure) - as being correctly ascribed to the effect of spontaneous ventilation/first breath of a neonate. RV outflow pressure will decrease with the onset of respiration due to a decrease in afterload (decreased pulmonary vascular resistance). Left atrial pressure will rise due to the increased flow of blood to the LA from the pulmonary vasculature and increased LVEDP due to clamping of umbilical vessels (P&K 357), and this will cause functional closure of the foramen ovale when left atrial pressure exceeds right atrial pressure. I suppose that this pressure differential will take longer to be established than the comparatively instant fall in RV outflow pressure and therefore option A is probably the more correct answer in this instance. Or rather the change in LAP is not solely due to onset of ventilation and so is not the single best answer. D and E are both incorrect. The increase in systemic vascular resistance is caused by the exclusion of the low resistance placenta from the circulation by clamping the umbilical vessels. The increase in left ventricular pressure is caused by increased afterload due to clamping of the umbilical vessels. Comment - I agree that the ventilation directly causes the fall in R-sided pressures. The increased blood flow through the pulmonary circulation subsequently increases LAP (but I would say the increased SVR also has a role to play in elevating L-sided pressures). The increased LAP is due to the higher blood flow, not necessarily directly related to the ventilation of the lung COMMENT: I agree with the above comment that BEST ANSWER (MF07b) = A. The other things do happen, but C/D/E are more correctly attributed to the increased SVR. Ventilation directly causes decreased PVR and therefore lower RV pressure.

Answer 565

Answer: 30mL/kg This is the same as the adult value for FRC, and is established soon after birth. At term, foetal lung contains ~20ml/kg of fluid FRC rises rapidly at birth - 17ml/kg at 10 min FRC increases to almost 30 ml/kg by 30-60 min

Answer 566

Answer: 30mL/kg This is the same as the adult value for FRC, and is established soon after birth. At term, foetal lung contains ~20ml/kg of fluid FRC rises rapidly at birth - 17ml/kg at 10 min FRC increases to almost 30 ml/kg by 30-60 min

Answer 567

A. correct (ductus venosus PO2 is at foetal maximum of 30mmHg) B. Incorrect (inferior vena cava is in the thorax) Ductus arteriosus connects pulmonary artery to aorta, therefore bypass lungs. It is irrelevant to brain circulation.

Answer 568

Correct answer : A At birth, circulation changes from parallel system to a system in series. Consequence of resistance changes throughout neonate circulation Placenta provides low-resistance and is eliminated when cord is clamped. Results in increased systemic vascular resistance (->increased BP) increased LVEDP decreased RAP due reduced vena cavae flow Lung expansion decreases PVR (removal of hypoxic vasoconstriction) along with increasing PaO2 , pH and decreasing PaCO2. The closure of the ductus arteriosus and the decreased PVR mean that the output of the right ventricle all passes through the lung. LAP increases due to increased pulmonary blood flow and increased LVEDP Foramen ovale closes (functional) when LAP exceeds RAP. Permanent closure at 4-6/52 post birth Ductus arteriosus constricts with increasing PaO2, closure of foramen ovale and decreasing concentrations of PGE1 and PGE2. Physiological closure: 10-15hours & Permanent closure: 2-3 weeks Persisting foetal circulation may occur with hypoxia or acidosis. Ductus venosus closes several hours post birth.

Answer 569

Tidal volume of neonates is 7mls/kg. So, in summary: TV 7mL/kg, FRC 30mL/kg, FVC 40mL/kg how on earth do you get a neonate to perform a forced vital capacity breath? Comment Please reference VC for neonates being 40ml/kg. I could not find it in references quoted. If VC for neonates is 40ml/kg, it would have to change the other figures used for neonatal lung capacities which sounds incorrect. (this figure comes from Frances Ware's notes from Brisbane course) FVC in Neonates: [3] (unfortunately no figure supplied for 'normal' VC) I think the correct answer for FVC is B. 150ml (67ml/kg * 2.3kg) Comment Vital capacity increases from birth to 3 years of age 45mL/kg as neonate 55mL/kg for infants [4] Comment Regarding the calculations above, where did the weight of 2.3kg for a neonate come from...i had thought 3.5 was more the "normal" weight? See P&K 2e: "FRC is 40% of TLC in the neonate." So TLC = 75ml/kg. So therefore VC = (75 - 30)ml/kg. And in a 2.3kg infant VC = 100mL Comment Doesn't the VC include TV, IRV, ERV? That means you can't calculate it by just taking the FRC away from the TLC, as the ERV is not counted in your answer Is it possible that the VC is smaller than an adult per weight due to the effect of having a closing capacity above FRC? Comment: Lung volumes for neonate (taken from melb course notes) TV 7ml/kg Dead space 2.2ml/kg VC 35ml/kg (compared with 60ml/kg for adult) FRC 30ml/kg Specific compliance same as adult As VC is the sum of ERV, TV and IRV, and we know TV is the same as the adult, clearly ERV or IRV is less in the neonate. Maybe ERV is less but RV more making FRC the same but VC less and also keeping TLC the same??

Answer 570

So TV in 2.3kg neonate = 7 x 2.3 = 16.1ml Answer C for MF11b. So, in summary: TV 7mL/kg, FRC 30mL/kg, FVC 40mL/kg how on earth do you get a neonate to perform a forced vital capacity breath? Comment Please reference VC for neonates being 40ml/kg. I could not find it in references quoted. If VC for neonates is 40ml/kg, it would have to change the other figures used for neonatal lung capacities which sounds incorrect. (this figure comes from Frances Ware's notes from Brisbane course) FVC in Neonates: [3] (unfortunately no figure supplied for 'normal' VC) I think the correct answer for FVC is B. 150ml (67ml/kg * 2.3kg) Comment Vital capacity increases from birth to 3 years of age 45mL/kg as neonate 55mL/kg for infants [4] Comment Regarding the calculations above, where did the weight of 2.3kg for a neonate come from...i had thought 3.5 was more the "normal" weight? See P&K 2e: "FRC is 40% of TLC in the neonate." So TLC = 75ml/kg. So therefore VC = (75 - 30)ml/kg. And in a 2.3kg infant VC = 100mL Comment Doesn't the VC include TV, IRV, ERV? That means you can't calculate it by just taking the FRC away from the TLC, as the ERV is not counted in your answer Is it possible that the VC is smaller than an adult per weight due to the effect of having a closing capacity above FRC? Comment: Lung volumes for neonate (taken from melb course notes) TV 7ml/kg Dead space 2.2ml/kg VC 35ml/kg (compared with 60ml/kg for adult) FRC 30ml/kg Specific compliance same as adult As VC is the sum of ERV, TV and IRV, and we know TV is the same as the adult, clearly ERV or IRV is less in the neonate. Maybe ERV is less but RV more making FRC the same but VC less and also keeping TLC the same??

Answer 571

VC in a neonate is 40ml/kg. FVC in a 2.3kg neonate = 40 x 2.3 = 92 Answer A for MF11c Comment I thought neonatal lung volumes were same as the adults, thus FVC should be 67mL/kg thus 67 x 2.3 = 154.1mL (B) Comment I'm going with original answer (40mL/kg); thus A: see: [1][2] So, in summary: TV 7mL/kg, FRC 30mL/kg, FVC 40mL/kg how on earth do you get a neonate to perform a forced vital capacity breath? Comment Please reference VC for neonates being 40ml/kg. I could not find it in references quoted. If VC for neonates is 40ml/kg, it would have to change the other figures used for neonatal lung capacities which sounds incorrect. (this figure comes from Frances Ware's notes from Brisbane course) FVC in Neonates: [3] (unfortunately no figure supplied for 'normal' VC) I think the correct answer for FVC is B. 150ml (67ml/kg * 2.3kg) Comment Vital capacity increases from birth to 3 years of age 45mL/kg as neonate 55mL/kg for infants [4] Comment Regarding the calculations above, where did the weight of 2.3kg for a neonate come from...i had thought 3.5 was more the "normal" weight? See P&K 2e: "FRC is 40% of TLC in the neonate." So TLC = 75ml/kg. So therefore VC = (75 - 30)ml/kg. And in a 2.3kg infant VC = 100mL Comment Doesn't the VC include TV, IRV, ERV? That means you can't calculate it by just taking the FRC away from the TLC, as the ERV is not counted in your answer Is it possible that the VC is smaller than an adult per weight due to the effect of having a closing capacity above FRC? Comment: Lung volumes for neonate (taken from melb course notes) TV 7ml/kg Dead space 2.2ml/kg VC 35ml/kg (compared with 60ml/kg for adult) FRC 30ml/kg Specific compliance same as adult As VC is the sum of ERV, TV and IRV, and we know TV is the same as the adult, clearly ERV or IRV is less in the neonate. Maybe ERV is less but RV more making FRC the same but VC less and also keeping TLC the same??

Answer 572

A. Correct. The infant has half the cholinesterase activity of the older child or adult (however there is still adequate enzyme to metabolise the drug) (from Brandis course notes 2006, p 26) B. Correct. Muscle relaxants are highly ionised, hence penetrate cell membranes poorly. They are virtually trapped in the ECF, so Vd correllates with ECF. ECF decreases with age. (from Brandis course notes 2006, p 25) C. Incorrect. Less alpha 1 acid glycoprotein in the noenate (from Brandis course notes 2006, p 3) D. Incorrect. Less CP450 in the newborn, rapid increase in first few weeks of life (from Brandis course notes 2006, p 6) comment,i thought this should be a pharm. Q. References Brandis course notes- Foetal and noenatal pharmacology, 2006, Dr Frances Ware

Answer 573

ABO antibodies are usually of the IgM type and are unable to cross the placenta. Rh antibodies are often of the IgG type, which is able to cross the placenta. ABO/Rh- incompatibility causes haemolytic disease of the newborn due to maternal transfer of IgG antibodies. So, B for the first and B for the second. The 'natural' antibodies to A and B blood group antigens are mostly IgM, which do not cross the placenta. Comment: While this is true, it doesn't really answer the question. There are no 'natural' antibodies to Rh antigens anyway. According to Obsterics and the Newborn (Beischer, Mackay and Colditz) 3rd ed. "Although ABO blood group antigens cause the greatest number of incompatibility reactions (65-70%), rhesus antigens are much more potent in terms of eliciting maternal antibody response". I'd suggest a poorly worded question, however, perhaps the best answer is D. Edit: I am with the IgM answer. Rh alloimmunisation occurs with prior sensitisation of an Rh-ve mother to an Rh+ve fetus (or transfusion). These will cross the placenta (IgG) and cause trouble in a lot of cases. The fact the mothers and babies can have different ABO types without problem indicates to me that this is the simplest and best answer. And on further reading, I think the entire premise of the question is wrong, as it appears according to the reference below that ABO incompatibilities are in fact more common, albeit less serious. "The natural antibodies (isoagglutinins) to A and B blood group substances, however, are mostly of the IgM class (typical of anti-carbohydrate responses) and therefore do not cross the placenta. IgG antibodies to the A and B blood group antigens may develop in some individuals, and the resulting ABO incompatibility actually accounts for about two thirds of all discernable cases of HDN; such cases, however, are generally very mild and require little or no treatment. Thus, while ABO incompatibility is actually much more common than Rh incompatibility, it is much less likely to cause significant disease." from http://jeeves.mmg.uci.edu/immunology/CoreNotes/Chap10.pdf page 82 (right at the end).. I know it is not Guyton, Ganong or P&K feel free to add a reference from them if you can find one!

Answer 574

ABO antibodies are usually of the IgM type and are unable to cross the placenta. Rh antibodies are often of the IgG type, which is able to cross the placenta. ABO/Rh- incompatibility causes haemolytic disease of the newborn due to maternal transfer of IgG antibodies. So, B for the first and B for the second. The 'natural' antibodies to A and B blood group antigens are mostly IgM, which do not cross the placenta. Comment: While this is true, it doesn't really answer the question. There are no 'natural' antibodies to Rh antigens anyway. According to Obsterics and the Newborn (Beischer, Mackay and Colditz) 3rd ed. "Although ABO blood group antigens cause the greatest number of incompatibility reactions (65-70%), rhesus antigens are much more potent in terms of eliciting maternal antibody response". I'd suggest a poorly worded question, however, perhaps the best answer is D. Edit: I am with the IgM answer. Rh alloimmunisation occurs with prior sensitisation of an Rh-ve mother to an Rh+ve fetus (or transfusion). These will cross the placenta (IgG) and cause trouble in a lot of cases. The fact the mothers and babies can have different ABO types without problem indicates to me that this is the simplest and best answer. And on further reading, I think the entire premise of the question is wrong, as it appears according to the reference below that ABO incompatibilities are in fact more common, albeit less serious. "The natural antibodies (isoagglutinins) to A and B blood group substances, however, are mostly of the IgM class (typical of anti-carbohydrate responses) and therefore do not cross the placenta. IgG antibodies to the A and B blood group antigens may develop in some individuals, and the resulting ABO incompatibility actually accounts for about two thirds of all discernable cases of HDN; such cases, however, are generally very mild and require little or no treatment. Thus, while ABO incompatibility is actually much more common than Rh incompatibility, it is much less likely to cause significant disease." from http://jeeves.mmg.uci.edu/immunology/CoreNotes/Chap10.pdf page 82 (right at the end).. I know it is not Guyton, Ganong or P&K feel free to add a reference from them if you can find one!

Answer 575

Answer is C Static compliance is less than adult Dynamic compliance is less than adult Specific compliance =Compliance/FRC = 0.05 cmH2O-1 in both adult and neonatal lung. Watch the units: cmH2O to power of -1 comment; if both static and dynamic is less than the adult,and FRC is the same, doesnt that make the specific less anyway. Comment on the prev comment: FRC in a neonate is not the same as an adult (it's only the same when indexed by weight). In summary: Compliance & FRC are low in a neonate, but the ratio of the 2 (i.e. specific compliance) is the same as in an adult. Comment on the comment above: Compliance = Volume/Pressure FRC can be either Volume/Weight or just Volume If the unit of specific compliance is cmH2O^-1, then the FRC value used here must be just Volume (of FRC, which increases the heavier you are). This means this question has nothing to do with "neonates" but simply with a "lighter person". In addition, I would've thought there would be huge variations in the weight of neonates. So how can Specific Compliance just be the same? Comment: It's definitely specific compliance being the same. The last comment makes some assumptions that aren't immediately clear, but are somewhat misleading. Specific compliance is 0.05cm.H2O litres (breath) per litre (FRC), simplified to 0.05 cm.H2O. ie if FRC is 75mL in a neonate of 2.5kg, then compliance is 0.05x0.75 = 0.00375L/cmH2O. In an adult of FRC 2.1L, compliance is 0.05*2.1=0.105L/CmH2O, which we know to be correct. References & related material The Physiology Viva (revised edition) - K Brandis - Ch 4 Respiratory Physiology - p117

Answer 576

Answer is C Static compliance is less than adult Dynamic compliance is less than adult Specific compliance =Compliance/FRC = 0.05 cmH2O-1 in both adult and neonatal lung. Watch the units: cmH2O to power of -1 comment; if both static and dynamic is less than the adult,and FRC is the same, doesnt that make the specific less anyway. Comment on the prev comment: FRC in a neonate is not the same as an adult (it's only the same when indexed by weight). In summary: Compliance & FRC are low in a neonate, but the ratio of the 2 (i.e. specific compliance) is the same as in an adult. Comment on the comment above: Compliance = Volume/Pressure FRC can be either Volume/Weight or just Volume If the unit of specific compliance is cmH2O^-1, then the FRC value used here must be just Volume (of FRC, which increases the heavier you are). This means this question has nothing to do with "neonates" but simply with a "lighter person". In addition, I would've thought there would be huge variations in the weight of neonates. So how can Specific Compliance just be the same? Comment: It's definitely specific compliance being the same. The last comment makes some assumptions that aren't immediately clear, but are somewhat misleading. Specific compliance is 0.05cm.H2O litres (breath) per litre (FRC), simplified to 0.05 cm.H2O. ie if FRC is 75mL in a neonate of 2.5kg, then compliance is 0.05x0.75 = 0.00375L/cmH2O. In an adult of FRC 2.1L, compliance is 0.05*2.1=0.105L/CmH2O, which we know to be correct. References & related material The Physiology Viva (revised edition) - K Brandis - Ch 4 Respiratory Physiology - p117

Answer 577

Correct Answer: C - Increased affinity of HbF (left shift of ODC)for Oxygen allows maintenance of adequate pO2 despite lower maternal pO2 values

Answer 578

``` At term, maternal paCO2 is 32-33mmHg due to maternal hyperventilation. This respiratory alkalosis is almost fully compensated so the pH is typically returned to the normal range. Typically after 8 - 10 weeks pH = 7.45 PaCO2 = 26 - 32 mmHg HCO3 = 18 - 21 mmol/L Base deficit = -2 → -3 mmol/L [Feb11] Answer: D ```

Answer 579

``` At term, maternal paCO2 is 32-33mmHg due to maternal hyperventilation. This respiratory alkalosis is almost fully compensated so the pH is typically returned to the normal range. Typically after 8 - 10 weeks pH = 7.45 PaCO2 = 26 - 32 mmHg HCO3 = 18 - 21 mmol/L Base deficit = -2 → -3 mmol/L [Feb11] Answer: D ```

Answer 580

The Thermoneutral Zone is the range of ENVIRONMENTAL temperatures over which metabolic heat production is minimal (minimal O2 consumption) and thermoregulation is maintained by vasomotor activity. (Power & Kam, p376) Term neonates: 32-34 deg celsius. Adults 25-30 deg celsius. Preterm babies need higher range due to higher evaporative heat losses. See also related SAQ on temperature regulation

Answer 581

The Thermoneutral Zone is the range of ENVIRONMENTAL temperatures over which metabolic heat production is minimal (minimal O2 consumption) and thermoregulation is maintained by vasomotor activity. (Power & Kam, p376) Term neonates: 32-34 deg celsius. Adults 25-30 deg celsius. Preterm babies need higher range due to higher evaporative heat losses. See also related SAQ on temperature regulation

Answer 582

Heat loss by conduction, convection and radiation is only possible if there is a gradient between the body and the ambient temperature. If the ambient temperature is increased, the only means to lose heat is by evaporation-which is dependent on the humidity. ie evaporation is greater on a less humid day. The answer is D. Evaporation.

Answer 583

Heat loss by conduction, convection and radiation is only possible if there is a gradient between the body and the ambient temperature. If the ambient temperature is increased, the only means to lose heat is by evaporation-which is dependent on the humidity. ie evaporation is greater on a less humid day. The answer is D. Evaporation.

Answer 584

Only molecules with different charge distributions can absorb infrared radiation. Non-polar molecules argon, helium, oxygen and xenon DO NOT absorb infrared radiation. Essentially, infrared is good for every gas in a circuit except for oxygen. The Clarke electrode can measure O2 tension in gas as well as blood. See Davis, PD and Kenny, GNC 5th edition, p205 The Datex (now GE) monitors measure oxygen using a rapid-response paramagnetic analyser. This is capable of rapid breath by breath analysis so inspired and expired oxygen concentrations are measured and displayed on the monitor screen. In Australia GE/Datex monitors are widely used in Anaesthesia so this is the "commonly used" method. The more traditional paramagnetic analysers (Servomox brand) are used in reference labs; they are highly accurate but slow.

Answer 585

Only molecules with different charge distributions can absorb infrared radiation. Non-polar molecules argon, helium, oxygen and xenon DO NOT absorb infrared radiation. Essentially, infrared is good for every gas in a circuit except for oxygen. The Clarke electrode can measure O2 tension in gas as well as blood. See Davis, PD and Kenny, GNC 5th edition, p205 The Datex (now GE) monitors measure oxygen using a rapid-response paramagnetic analyser. This is capable of rapid breath by breath analysis so inspired and expired oxygen concentrations are measured and displayed on the monitor screen. In Australia GE/Datex monitors are widely used in Anaesthesia so this is the "commonly used" method. The more traditional paramagnetic analysers (Servomox brand) are used in reference labs; they are highly accurate but slow.

Answer 586

Only molecules with different charge distributions can absorb infrared radiation. Non-polar molecules argon, helium, oxygen and xenon DO NOT absorb infrared radiation. Essentially, infrared is good for every gas in a circuit except for oxygen. The Clarke electrode can measure O2 tension in gas as well as blood. See Davis, PD and Kenny, GNC 5th edition, p205 The Datex (now GE) monitors measure oxygen using a rapid-response paramagnetic analyser. This is capable of rapid breath by breath analysis so inspired and expired oxygen concentrations are measured and displayed on the monitor screen. In Australia GE/Datex monitors are widely used in Anaesthesia so this is the "commonly used" method. The more traditional paramagnetic analysers (Servomox brand) are used in reference labs; they are highly accurate but slow.

Answer 587

A - N2O also will reignite the splint. INCORRECT B - correct. eg bleomycin + 100% o2. Or neonates - FiO2 1.0 is extremely harmful. C needs 2 ATM O2 D is 99.9 or something like that F is correct This question was in the pharm july 2006 exam; answer is a ; see pharm section for info Note: I agree O2 may reignite a glowing splint...but is it the only one. I am always dubious about questions with "always" or "only one" or "never".....this never/always never happens in biological systems. I wonder what hydrogen would do next to a glowing splint....I would "never" hope to be in a blimp with this combination of gas and splint. BTW...what is a glowing splint...sounds orthopaedic to me? COMMENT: B is definitely true and it doesn't mention a time limit in the question, although I agree that F is also true (obviously, given the number of options from more than one MCQ). I would be wary about option A - with the wording ("the only gas..."), and it would seem that someone else found something about N2O reigniting a glowing splint, so A seems to be false....though I have not checked out the link. There are different grades of "Medical Grade Oxygen" A-> G from 99.0 to 99.995. Most anaesthetic machines have purity > 99.5% Reference: My consultant Oxygen is manufactured by the fractional distillation of air. COMMENT: Hydrogen gas would also reginite a glowing splint. Reference: http://en.wikipedia.org/wiki/Hindenburg_disaster

Answer 588

A naked 70kg man in a OT which has a temperature that is lower than core body temperature will lose heat by radiation, conduction and convection. Radiation does not require contact with the surrounding environment and this form would contribute the most. See also comments in AC79 Most heat loss in OT is via radiation (40-50%). Note: This may be a trick question - radiative heat loss goes by the fourth power of absolute temperature, whereas convective heat loss goes by the temperature difference between body and outside air temperature, and thus there would be a breakeven point at some low temperature... References See table 9.1, p 103 in Davis & Kenny 5th edition. See Figure 5: [1][Excellent!!] See Perioperative thermoregulation

Answer 589

MetHb interferes with reducedHb at 660nm and HbO2 at 940nm -> R tends to 1, therefore reads Sp02 as 85% Answer A The comment is correct. I find it easier to think of it as; metHb has high absorption at both the red and infrared wavelength. So the ratio of pulse added absorption A660/A940 approaches 1 the higher the conc of MetHb. The processor determines this ratio as equivilant to a HbO % of 85%. As the reading tends towards 85% as metHb levels rise, whether this displayed value (eg 85% value at high metHb levels) is an under- or an over- estimate depends on what the actual saturation level is. Foetal haemoglobin has a very similar absorption at 660nm and 940nm as has HbA, so the displayed value is not altered by the presence of varying amounts of HbF. Carboxyhemoglobin has an absorption spectra similar to that of oxyhaemogloin at 660 nm which results in an overestimation of the oxyegen saturation. Patients with sickle cell (HbS). HbS has no reported effects on puls oximetry. It should benoted however that there is a shift in the ODC to the right so pateints with a sickle cell may have a lower than expected saturation for a given P02.

Answer 590

MetHb interferes with reducedHb at 660nm and HbO2 at 940nm -> R tends to 1, therefore reads Sp02 as 85% Answer A The comment is correct. I find it easier to think of it as; metHb has high absorption at both the red and infrared wavelength. So the ratio of pulse added absorption A660/A940 approaches 1 the higher the conc of MetHb. The processor determines this ratio as equivilant to a HbO % of 85%. As the reading tends towards 85% as metHb levels rise, whether this displayed value (eg 85% value at high metHb levels) is an under- or an over- estimate depends on what the actual saturation level is. Foetal haemoglobin has a very similar absorption at 660nm and 940nm as has HbA, so the displayed value is not altered by the presence of varying amounts of HbF. Carboxyhemoglobin has an absorption spectra similar to that of oxyhaemogloin at 660 nm which results in an overestimation of the oxyegen saturation. Patients with sickle cell (HbS). HbS has no reported effects on puls oximetry. It should benoted however that there is a shift in the ODC to the right so pateints with a sickle cell may have a lower than expected saturation for a given P02.

Answer 591

A. Both have a molecular weight approaching 44 (CO2: 44.0095 and N20:44.0128). One mole of each will have the same weight. B. Density is mass per volume. Equal volumes of gas, at the same temperature and pressure, contain the same number of molecules (Avagadro's Law). One mole of each gas will have the same mass, and the same volume(22.4L STPD if ideal) and so therefore the same density. C. Viscosity is the measure of the resistance of a fluid to deform under shear stress. It is individual to different fluids. It is measured in the Poise (1 P = 1 g·cm−1·s−1), or the Pascal Second in SI(1 Pa·s = 1 kg·m−1·s−1 = 10 P) Note that the Pascal Second is the more commonly used unit. CO2=0.07 cP at boiling point (-78°C)= 0.0001372 Poise @ STP (reference AirLiquide Gas encyclopedia) N20=0.000136 Poise @ STP (reference AirLiquide Gas encyclopedia) D: See (B) above. Both have the same volume at STP. The answer is C. Comment: This question sucks. CO2 and N2O have similar weights. They DO NOT have the same MW nor density. A, B, and C are all correct answers to this question. C is the most correct, but even then, the difference is 0.88% compared to the mass difference of 0.007%.

Answer 592

If gas is allowed to expand by adiabatic expansion, the gas does not draw heat energy from its surroundings but rather decreases the kinetic energy of the gas molecules itself -> this causes cooling of the gas and is the principle of the cryoprobe. E is correct I disagree... believe the answer is D - Temperature Yentis - "Adiabatic change involves a fast change in vol/pressure such that there is not time for the temp to equilibrate with the surroundings" Wikipaedia (I know its dodgy... but its a 2nd reference) - "an adiabatic process or an isocaloric process is a process in which no heat is transferred to or from the working fluid. The term "adiabatic" literally means an absence of heat transfer." depends a bit on the wording I guess (surprised?) Adiabetic no temperature change in surroundings... but there would be a change in the medium (eg gas expanding/cooling) Adiabatic - No HEAT transfer. If expansion (i.e. increase volume), both pressure and temperature decrease. Therefore, E is correct. Adiabatic Work - Ideal Gas Definition: "Occurring without gain or loss of heat. When a gas is compressed under adiabatic conditions, its pressure increases and its temperature rises without the gain or loss of any heat. Conversely, when a gas expands under adiabatic conditions, its pressure and temperature both decrease without the gain or loss of heat. The adiabatic cooling of air as it rises in the atmosphere is the main cause of cloud formation. "[[1]] Or in other words, "The definition of an adiabatic expansion, for now, is dq = 0. That is, no heat goes in or out of the system. However, dw ≠ 0. As the gas expands it does work on the surroundings. Since the gas is cut off from any heat bath it can not draw heat from any source to convert into work. The work must come from the internal energy of the gas so that the internal energy decreases. Since the internal energy of an ideal gas in only dependent on T that means that the temperature of the gas must decrease." [[2]] --- In other words, although the gas does not transfer heat to or from it's environment, the temperature of the gas itself changes...often quite a lot! Ref Daves Measurement Ch4 5Ed. -- Davis pg 41 – for the perfect gas laws to apply, heat must be transferable to the surroundings. Adiabatic means “without heat transfer”. Thus temp must rise or fall. Answer is E. References Adiabatic Expansion Cooling of Gases (for the clever folk) Davis pg 41

Answer 593

Simple application of the Alveolar Gas Equation: Alveolar pO2 = [(380-47)*0.21] - [40/0.8] = References The answer is A = 20 mmHg Alveolar gas equation : PAo2=PIo2 - (PAco2/R) + F PAo2=partial pressure of o2 in the alveoli(A) PIo2=Po2 of inspired air =(20.93/100)* (atmospheric pressure in mmHg - water vapour pressure in mmHg) Atmospheric pressure is 760 mmHg at sea level at 37c,(or 380 as given in the question in high altituded or any other given value to be substituted).water vapor pressure of moist inspired gas(which is fully saturated with water vapor at 37c is 47 mmHg) R = respiratory exchange ratio = 0.8 F =small correction factor ,typically 2 mmHg ,which can be ignored The equation is very important to estimate the (A-a gradient) as well,even on different levels of inspired oxygen

Answer 594

Poiseuille's equation relates to laminar flow. Laminar Flow = presure difference(P) divided by resistance(R) ie flow varies inversely with resistance. Poiseuille's law corresponds to Ohm's law for electrical circuits (V = IR), where the pressure drop ΔP is analogous to the voltage V and volume flow rate is analogous to the current I. The resistance R is determined as follows: R = (8 x viscosity x length) / (pi x r4) This concept is useful because the effective resistance in a tube is inversely proportional to the fourth power of the radius. This means that halfing the size of the tube increases the resistance to fluid movement by 16 times. Flow is inversely proportional to Resistance Viscosity and length are independent parameters and change in one cannot affect the other Alternate opinion: Ohm's law describes the relationship between flow and resistance not the Poiseuille equation. If you rearrange the Poiseuille equation viscosity = pressure diff * pie * r 4 / 8 * flow * length. ie that viscosity is inversely proportional to length. Thus B. Counter-alternate opinion: The option says that viscosity "varies" inversely with length. This is mathematically true, but viscosity is a property of the gas, and not a variable that may change just by alteration of the length of the tube. So, the word "varies" invalidates this response, IMHO. I would pick option A. Another opinion: I agree with everyone! A and B are both incorrect. The Hagen-Poiseuille Law does NOT state that resistance varies inversely with flow; that is Ohm's Law. Viscosity is a constant (for a particular fluid), and does not vary inversely with length! That would be like saying pi varies with radius. I suggest C, D or E is correct... References Wikipedia - Pouseuille's Law The Physics of Fluid Flow Hmmm!

Answer 595

Turbulence is irregular fluid flow with eddie currents and vortices. Turbulent flow is more likely with a Reynold's Number of >2000 and is defined by the below equation: Reynold's Number = Velocity x Density x Diameter/Viscosity Turbulent flow will be more likely in larger vessels, greater fluid velocities and more dense substances, but less viscous substances. Therefore, the answer is B (and decreased viscosity) Addition: Remember that Re number is a measure of inertial to viscous forces, where turbulence is more likely when inertial > viscous, whereas laminar flow is more likely when viscous > inertial...

Answer 596

A pneumotachograph is a device used to measure gas flow (usually expired gasses). It is based on the Hagen-Poiseuille Equation (in laminar flow, flow is proportional to pressure drop) The gasses pass through a gauze screen / capillary network, with pressure transducers on either side of the screen. The screen: maintains laminar flow Provides a fixed resistance that results in a pressure drop proportional to flow Can be heated to provide a constant temperature (constant viscosity and density) and to prevent moisture accumulation. The pressure difference is transduced into an electrical signal by means of a pressure transducer. Once a certain flow velocity is exceeded, turbulences appear (Re increases) - this makes the pneumotachograph inaccurate. Anaesthetic gases will alter viscosity, and provide a potential source of error. A. TRUE. Can be used to measure rapid changes in airflow B. FALSE. Measures pressure difference to determine flow. C. FALSE. Is inaccurate above a certain flow velocity. D. FALSE. Is a fixed orifice flowmeter. E. TRUE. Can be used to measure volume, if flow is integrated with respect to time. F. FALSE. Gas viscosity (and therefore flow) vary with temperature. Heated gauze offsets this problem.

Answer 597

Direct fick? "In theory, the Fick technique is the gold standard for cardiac output measurement but it is invasive and methodological error is not uncommon---for example, when high inspired oxygen concentrations are being administered." [1]. However, I note other papers note that the dye dilution method is the gold standard, while others suggest that there is no gold standard...

Answer 598

Direct fick? "In theory, the Fick technique is the gold standard for cardiac output measurement but it is invasive and methodological error is not uncommon---for example, when high inspired oxygen concentrations are being administered." [1]. However, I note other papers note that the dye dilution method is the gold standard, while others suggest that there is no gold standard...

Answer 599

Electrical impedance, or simply impedance, is a measure of opposition to a sinusoidal alternating electric current. The concept of electrical impedance generalises Ohm's law to AC circuit analysis. Unlike electrical resistance, the impedance of an electric circuit can be a complex number, but the same unit, the ohm, is used for both quantities. - from Wikipedia The term impedance is used in preference to resistance when there is a dependency on the frequency of the electric current flowing. It is obtained mathematically by combining its reactance,X (the ability of a device to resist AC current) and its resistance, R (the ability of a device to resist DC current). The unit of impedence is the same as resistance (ie. Ohm) but it is denoted by the symbol z. Z = the squre root of ( R squared + X squared) The impedence of a resistor to the flow of alternating current does not vary with the frequency of the current. With a capacitor the impedence decreases as the frequency of AC increases. Its resistance to DC current is very high however as it is simply an open circuit. With an inductor the impedence increases as the frequency of AC increases. This is because the build up and collapse of magnetic field around it tends to slow down changes in current flow. Inductors therefore tend to block AC current but transmit DC current. Question13 Only C is correct (Impedence is constant across a resistor) Question 13b Only C is correct (Impedence increases in an inductor) Question 13c A is correct - resistor has no change In Summary Impedance generalises the concept of Resistance to AC circuits.(ref Wiki) In a Resistor, the Impedance equals the resistance. Across a Capacitor, Impedance is inversely proportional to the frequency of the AC current. Across an Inductor, the Impedance is directly proportional to the frequency of the AC current. Hence, answers are C,C,A.

Answer 600

Electrical impedance, or simply impedance, is a measure of opposition to a sinusoidal alternating electric current. The concept of electrical impedance generalises Ohm's law to AC circuit analysis. Unlike electrical resistance, the impedance of an electric circuit can be a complex number, but the same unit, the ohm, is used for both quantities. - from Wikipedia The term impedance is used in preference to resistance when there is a dependency on the frequency of the electric current flowing. It is obtained mathematically by combining its reactance,X (the ability of a device to resist AC current) and its resistance, R (the ability of a device to resist DC current). The unit of impedence is the same as resistance (ie. Ohm) but it is denoted by the symbol z. Z = the squre root of ( R squared + X squared) The impedence of a resistor to the flow of alternating current does not vary with the frequency of the current. With a capacitor the impedence decreases as the frequency of AC increases. Its resistance to DC current is very high however as it is simply an open circuit. With an inductor the impedence increases as the frequency of AC increases. This is because the build up and collapse of magnetic field around it tends to slow down changes in current flow. Inductors therefore tend to block AC current but transmit DC current. Question13 Only C is correct (Impedence is constant across a resistor) Question 13b Only C is correct (Impedence increases in an inductor) Question 13c A is correct - resistor has no change In Summary Impedance generalises the concept of Resistance to AC circuits.(ref Wiki) In a Resistor, the Impedance equals the resistance. Across a Capacitor, Impedance is inversely proportional to the frequency of the AC current. Across an Inductor, the Impedance is directly proportional to the frequency of the AC current. Hence, answers are C,C,A.

Answer 601

Electrical impedance, or simply impedance, is a measure of opposition to a sinusoidal alternating electric current. The concept of electrical impedance generalises Ohm's law to AC circuit analysis. Unlike electrical resistance, the impedance of an electric circuit can be a complex number, but the same unit, the ohm, is used for both quantities. - from Wikipedia The term impedance is used in preference to resistance when there is a dependency on the frequency of the electric current flowing. It is obtained mathematically by combining its reactance,X (the ability of a device to resist AC current) and its resistance, R (the ability of a device to resist DC current). The unit of impedence is the same as resistance (ie. Ohm) but it is denoted by the symbol z. Z = the squre root of ( R squared + X squared) The impedence of a resistor to the flow of alternating current does not vary with the frequency of the current. With a capacitor the impedence decreases as the frequency of AC increases. Its resistance to DC current is very high however as it is simply an open circuit. With an inductor the impedence increases as the frequency of AC increases. This is because the build up and collapse of magnetic field around it tends to slow down changes in current flow. Inductors therefore tend to block AC current but transmit DC current. Question13 Only C is correct (Impedence is constant across a resistor) Question 13b Only C is correct (Impedence increases in an inductor) Question 13c A is correct - resistor has no change In Summary Impedance generalises the concept of Resistance to AC circuits.(ref Wiki) In a Resistor, the Impedance equals the resistance. Across a Capacitor, Impedance is inversely proportional to the frequency of the AC current. Across an Inductor, the Impedance is directly proportional to the frequency of the AC current. Hence, answers are C,C,A.

Answer 602

Fick's equation for diffusion relates variables that govern the diffusion rate of molecules such as oxygen and CO2. A. False - inversely proportional to thickness B. False - directly proportional to concentration gradient C. False - directly proportional to surface area D. True - inversely proportional to thickness As the concentration difference in molecules across the membrane increases, diffusion is increased. With an increase in length of diffusion pathway, rate of diffusion decreases. Also: diffusion is proportional to the gas solubility in the tissue diffusion is inversely proportional to the square root of the MW of the gas Reference West 7th ed page 26

Answer 603

The most accurate measure is by transoesophageal Doppler echocardiographic quantification of CO. TOE–CO. Department of Anaesthesiology and Surgical Intensive Care Medicine and 2 Department of Thoracic and Cardiovascular Surgery, University of Münster Hospital, Albert-Schweitzer-Straße 33, D-48149, Münster, Germany Electrical velocimetry can also be interchanged with TOE - CO according to the same paper. I guess the answer is E.

Answer 604

Note: Indocyanine green is used to measure flow in the liver because it is metabolised there. There is little uptake of this dye in other tissues. A - incorrect; the hepatic vein measures the output, but the portal vein only supplies about 75% of the input to the liver with the hepatic artery supplying the other 25%. Also note that it is exceeding difficult and invasive to measure the concentration in the portal vein. B - incorrect; the hepatic artery and portal vein flow INTO the liver and do not measure the output C - correct; the radial arterial concentration approximates the concentration entering the liver and the hepatic vein measures the concentration after the liver uptake. D - incorrect; see answer A E - incorrect; the radial artery is used but the concentration in the right atrium does not approximate the concentration in the hepatic vein as it is diluted with blood from the rest of the body which does not take up indocyanine green. This question is simply an application of the law of conservation of mass interpreted with the Fick principle. The law of conservation of mass in simple terms says that the amount of substance leaving a system (i.e. what you get out of a vein) equals the amount of substance entering a system (i.e. what you get out of the artery) minus any uptake in the system itself (the difference between the artery and the vein). The Fick principle applies this in the form that the amount of substance taken up by a system is equal to the product of the flow through the system and the arterio-venous difference in concentration. The following information can be found under the subtitle "Use for estimating hepatic blood flow" on Up to Date: Indocyanine Green (ICG) provides a good estimate of hepatic blood flow. Infusion of ICG at a rate below the liver's capacity to clear it leads to a steady state within one hour at which time the clearance rate is equal to the infusion rate. Hepatic blood flow can be estimated by use of the Fick equation: \mathrm{Estimated\:Hepatic\:Blood\:Flow}=\frac{R}{[\mathrm{Indocyanine\:Green_{arterial}]-[Indocyanine\:Green_{Hepatic\:Vein}]} where: R equals the hepatic removal rate of ICG, which equals the infusion rate at steady state ICGa equals the concentration of ICG in the hepatic artery and portal vein blood (One of the assumptions underpinning this method of measuring hepatic blood flow is that the concentration of ICG in the hepatic artery equals that of the portal vein) ICGhv equals the concentration of ICG in the hepatic vein blood ICGa is the same as the ICG concentration in peripheral venous blood since other organs do not clear ICG (hence it's suitability for this use). Regarding options A and D, since the arterial concentration taken at the radial artery should be the same as hepatic artery and portal vein, they are potential answers are they not (except for the relative difficulties in accessing them)?

Answer 605

D Water 4.18kJ/kg/K Blood 3.6kj/kg/K Gases 1.01kj/kg/K Normal Saline 4139kJ/kg/K Gelofusine 4082kJ/kg/K Hartmann's 4153kJ/kg/K Specific heat capacity, also known simply as specific heat (Symbol: C or c) is the measure of the heat energy required to raise the temperature of a given amount of a substance by one degree. Commonly, the amount is specified by mass; for example, water has a mass-specific heat capacity of about 4184 joules per kelvin per kilogram July 00 Version Saline has elevated boiling and depressed freezing point. Hence, I think that the answer for July 00 version is Saline. The SHC of other fluids used during CPB was also measured and found to be 4139 J kg-1 degrees C-1 and 4082 J kg-1 degrees C-1 for normal saline and Gelofusine, respectively http://bja.oxfordjournals.org/cgi/reprint/84/1/28 this states water is greater 4.179 compared to saline 4.139

Answer 606

C Water 4.18kJ/kg/K Blood 3.6kj/kg/K Gases 1.01kj/kg/K Normal Saline 4139kJ/kg/K Gelofusine 4082kJ/kg/K Hartmann's 4153kJ/kg/K Specific heat capacity, also known simply as specific heat (Symbol: C or c) is the measure of the heat energy required to raise the temperature of a given amount of a substance by one degree. Commonly, the amount is specified by mass; for example, water has a mass-specific heat capacity of about 4184 joules per kelvin per kilogram July 00 Version Saline has elevated boiling and depressed freezing point. Hence, I think that the answer for July 00 version is Saline. The SHC of other fluids used during CPB was also measured and found to be 4139 J kg-1 degrees C-1 and 4082 J kg-1 degrees C-1 for normal saline and Gelofusine, respectively http://bja.oxfordjournals.org/cgi/reprint/84/1/28 this states water is greater 4.179 compared to saline 4.139

Answer 607

These questions appear to be very devious as the solubilities, and the order of ranking, are different in BLOOD and PLASMA. (Added note: This difference may just be because the question has been remembered differently by different people). These figures are taken from various sources, as I can't find any single source amongst the standard texts I have. However they do seems consistant across these sources; but please feel free to add or change here if you find a good listing... For BLOOD Blood:gas solubility coefficients - N20 > CO2 > 02 > N2 --- 0.45 > 0.06 > 0.003 > 0.0017 For PLASMA Water:gas solubility coefficients - CO2 > N2O > 02 > N2 --- 0.57 > 0.38 > 0.024 > 0.012 Helium, should they decide to include it, is even more insoluble with water:gas 0.008 As a side point, for those (like me!) who have found the CO2 solubility units confusing as different in course notes, West's physiology etc, they are: 0.06 mls CO2 / 100ml / mmHg OR 0.03 mmol / L / mmHg The O2 figure that we see in the respiratory questions of 0.003 is in the unit mls O2 / 100ml / mmHg

Answer 608

These questions appear to be very devious as the solubilities, and the order of ranking, are different in BLOOD and PLASMA. (Added note: This difference may just be because the question has been remembered differently by different people). These figures are taken from various sources, as I can't find any single source amongst the standard texts I have. However they do seems consistant across these sources; but please feel free to add or change here if you find a good listing... For BLOOD Blood:gas solubility coefficients - N20 > CO2 > 02 > N2 --- 0.45 > 0.06 > 0.003 > 0.0017 For PLASMA Water:gas solubility coefficients - CO2 > N2O > 02 > N2 --- 0.57 > 0.38 > 0.024 > 0.012 Helium, should they decide to include it, is even more insoluble with water:gas 0.008 As a side point, for those (like me!) who have found the CO2 solubility units confusing as different in course notes, West's physiology etc, they are: 0.06 mls CO2 / 100ml / mmHg OR 0.03 mmol / L / mmHg The O2 figure that we see in the respiratory questions of 0.003 is in the unit mls O2 / 100ml / mmHg

Answer 609

A- density is less B- helium is non-flammable C- air Viscosity (1 bar and 0 °C (32 °F)) : 0.0001695 Poise; helium Viscosity (1.013 bar and 0 °C (32 °F)) : 0.0001863; nitrogen Viscosity (1.013 bar and 0 °C (32 °F)) : 0.0001657 Poise looks like viscosity would be increased, but minimally so D- Due to its reliance on the ability of the fluid or gas to displace the float, the graduations on a given rotameter will only be accurate for a given substance. The main property of importance is the density of the fluid. Either separate rotameters for different substances must be used, or the read out adjusted. E- unless in the addition process oxygen was diluted, then there would be theoretically better gas flow as less dense gas would favour laminar and therefore better flow, therefore more oxygen transfer F- can't see why, anyone? therefore ? C, best of a bad bunch Alternative explanation a -false- density is less b -false- helium is non-flammable c -True- viscosity is increases /?minimally d-false-rotameters are calibrated for a specific gas based on its properties of density and viscosity. At the bottom of the rotameter the flow is laminar and dependant on viscosity. At the top flow is turbulent and more dependant on density. e-?true- whether oxygen transfer increases or decreases will depend on the level of the obstruction. In upper airway obstruction where flow is turbulent heliox may restore laminar flow due to its reduced density and increase O2 uptake. The wording of the question however says pulmonary obstruction which to me means lower airway obstruction. In this case flow is already laminar so adding helium will not help and will serve to reduce oxyegn concentration and therfore uptake. f-false HELIOX ( 79% He and 21% oxygen) will reduce work of beathing and improve oxygenation in patients with an UPPER AIRWAY OBSTRUCTION such as tumor ( ref: Peck & williams).

Answer 610

Negative exponential processess continue indefinitely (in theory), so the values of half-life and time constant have been substituted for total length. Half-life is the time taken for a negative exponential process to reach half of its initial value. Time constant is the time taken for a negative exponential process to reach zero if it continued at its initial rate of change. One time constant equals: 0.37 of original amount half-life/ln2 half-life/0.693 volume/flow compliance*resistance Figures for a negative exponential process, in terms of time constants: After 1 time constant, 37% is left and 63% been removed After 2 time constants, 13% is left and 87% been removed After 3 time constants, 5% is left and 95% has been removed After 4 time constants, 2% is left and 98% has been removed The only correct answer is A.

Answer 611

Answer : A - Laminar flow Hagen-Poiseuille's law : The principle that the volume of a homogeneous fluid passing per unit time through a capillary tube is directly proportional to the pressure difference between its ends and to the fourth power of its internal radius, and inversely proportional to its length and to the viscosity of the fluid. It is the physical law concerning the voluminal laminar stationary flow . Laminar flow may change to turbulent if a constriction is reached which results in the velocity increasing. Although turbulent flow can occur if there is a sharp increase in flow through a tube, there a several other factors influencing the type of flow such as the viscoscity and debsity of the fluid and the diameter of the tube. These factors can be combined to give an index known as the Reynold's number: Reynolds number = velocity x density x diameter / viscosity A reynold's number greater than 2000 is likely to produce turbulent flow. There is no equation/law governing turbulent flow, however turbulent flow is proportional to: the square root of pressure gradient 1/length radius squared 1/ density of fluid

Answer 612

A Pulmonary Artery Cathether is used for monitoring/measuring: Pulmonary Artery Pressures. Mixed Venous Oxygen Saturations and Blood Gas Analysis. Temperature. Pulmonary Artery Occlusion Pressure ∝ LVEDP ∝ LVEDV. Cardiac Output by Thermodilution.

Answer 613

Answer is B: oxygen partial pressure. This is the Clarke electrode. The two electrodes are placed in KCl solution and enclosed in a gas permeable membrane. When exposed to oxygen, the reaction at the platinum cathode is: O2 + 4e- +2H2O --> 4OH- The current flow created in the circuit depends on the concentration of dissolved oxygen only, not total oxygen content. Current flow = k1 x [Dissolved O2] The concentration of dissolved oxygen is different from oxygen content' if oxygen is present in other forms. With blood, this 'other form' is oxygen present in chemical combination with haemoglobin. It is the dissolved oxygen only that is 'seen' by the elextrode and is responsible for current flow. By Henry's law, the concentration of dissolved oxygen is directly proportional to pO2, so the Clark electrode is calibrated to give the pO2 value. [Dissolved O2] = k2 x pO2 NOTE: The constant of proportionality (k2 in eqn above) of the Henry's law equation is temperature dependent, and so colder water at the same pO2 value contains more dissolved oxygen. Hence for the calibration from current flow to pO2 value to be correct, the temperature has to be constant (typically 37C) and the equation use the proportionality constant for that temperature.

Answer 614

B - Hagen-Poiseuille equation: \mathrm{Q}=\frac{\Delta P \pi r^4}{8 \eta l} (For turbulent flow, the Hagen-Poiseuille equation does not apply and the important fluid property becomes answer C, density, not viscosity.) Comment: issues with the wording perhaps? For laminar flow, flow IS affected by viscosity; reducing viscosity of fluid/air can produce turbulent flow

Answer 615

The Answer is C. Henry's law CO2 exists in blood in 3 forms: dissolved, as carbamino compounds, and as bicarbonate. Henry's Law deals with gases dissolved in solution. When the system is at equilibrium, the law states: Amount of gas dissolved (mmol/l) = k x partial pressure of gas above solution (mmHg) where k is a solubility constant of the gas in the liquid phase, and its value for CO2 is 0.03 mmol/l.mmHg Therefore Henry's law states that the amount of CO2 dissolved in blood will be directly proportional to the partial pressure of C02 above the liquid interface. Other laws The other laws apply to gases in the gas phase, as follows: Boyle's Law states that at a constant temperature, the volume of a given mass of gas varies inversely with the absolute pressure. This is often expressed as: P1V1 = P2V2 Dalton's Law states that for gases that are in a mixture, each gas exerts a pressure equal to the pressure it would exert if it alone occupied the volume. This pressure is referred to as "the partial pressure of that gas". The sum of the partial pressures of all the gases present is equal to the total gas pressure. Avogadro's Law states that equal volumes of gases at the same temperatures and pressure contain equal numbers of molecules. Charles' law states that at constant pressure, the volume of a given mass of gas varies directly with the absolute temperature.

Answer 616

Answer: D - Cortex Electroencephalography is the neurophysiologic measurement of the electrical activity of the brain by measuring voltage differences between different parts of the cerebral cortex. EEG has limited use for Depth of Sedation/Anesthesia monitoring until the raw signal is processed by eg BIS, Entropy wikipedia Brisbane Long Course Notes From Power & Kam. EEG is the recording of the spontaneous electrical activity of the brain, generated by the superficial layer of pyramidal cells by changes in Post-synaptic Potentials (EPSE + IPSE's). The electrical potentials are the summation of the post-synaptic potentials of the pyramidal cells in response to rhythmic discharges from the thalamic nuclei (alpha waves). Input from the Reticular formation causes desynchronisation

Answer 617

A. (not enough information) B. Wrong - strain gauge uses change in electrical resistance of its four elements as they deform with pressure waves C. Wrong - Katharometer assess differences between two gas mixtures by assessing change in resistance of heating coils that heat the two gas samples D. Correct - Bourdon gauge measures pressure through mechanical unfurling of a gas-filled coil E. Correct - Thermocouple uses the potential difference (voltage) across a junction between two dissimilar metals at the same temperature However this question is worded (do / do not use), there will be two correct answers. Perhaps the question was mis-recalled, and 'thermocouple' was actually an option for 'thermistor'. If so, Boudon gauge would be the correct answer. Strain guage [1] A strain gauge (alternatively: strain gage) is a device used to measure deformation (strain) of an object. Invented by Edward E. Simmons in 1938, the most common type of strain gauge consists of an insulating flexible backing which supports a metallic foil pattern. The gauge is attached to the object by a suitable adhesive. As the object is deformed, the foil is deformed, causing its electrical resistance to change. This resistance change, usually measured using a Wheatstone bridge, is related to the strain by the quantity known as the gauge factor. A katharometer is an instrument for determining the composition of a gas mixture. It functions by having two parallel tubes both containing gas and heating coils. The gases are examined by comparing the rate of loss of heat from the heating coils into the gas. The coils are arranged in a bridge circuit so that resistance changes due to unequal cooling can be measured. One channel normally holds a reference gas and the mixture to be tested is passed through the other channel. The thermal conductivity of a gas is inversely related to its molecular weight. Hydrogen has approximately six times the conductivity of nitrogen for example. Katharometers are used medically in lung function testing equipment and in gas chromatography. The results are slower to obtain compared to a mass spectrometer, but the device is inexpensive, and has good accuracy when the gases in question are known, and it is only the proportion that must be determined. In electronics, thermocouples are a widely used type of temperature sensor and can also be used as a means to convert thermal potential difference into electric potential difference. In 1821, the German-Estonian physicist Thomas Johann Seebeck discovered that when any conductor (such as a metal) is subjected to a thermal gradient, it will generate a voltage. Any attempt to measure this voltage necessarily involves connecting another conductor to the "hot" end. This additional conductor will then also experience the temperature gradient, and develop a voltage of its own which will oppose the original. A Bourdon gauge uses a coiled tube which as it expands due to pressure increase causes a rotation of an arm connected to the tube.

Answer 618

Cardiogenic shock leads to a low output state. Consequently; the CO or CI will drop- due to low output PCWP will likely increase due to low output EF is likely to fall MAP may initially be maintained due to increased SVR SVR will increase- to try and compensate for the low output So I think the answere is A in both cases References: Brandis K, The Physiology Viva Ganong WF, Review of Medical Physiology, Lange medical books/McGraw-Hill Guyton and Hall, Textbook of Medical Physiology, 10th Ed, Saunders, 2000 COMMENT: Question states Cardiogenic shock which implies low MAP Therefore first answer is B. Comment: (Not at all.... the MAP is not a value measured by the PAC, neither is the EF) (SVR mathematically determined from MAP and RAP. SVR = (MAP - RAP * 79.9)/CO ) The typical picture of cardiogenic shock is characterised by: an increased HR, an increased SVR, a reduced SV, CO, CI, systemic hypotension (defined for cardiogenic shock as SBP

Answer 619

Cardiogenic shock leads to a low output state. Consequently; the CO or CI will drop- due to low output PCWP will likely increase due to low output EF is likely to fall MAP may initially be maintained due to increased SVR SVR will increase- to try and compensate for the low output So I think the answere is A in both cases References: Brandis K, The Physiology Viva Ganong WF, Review of Medical Physiology, Lange medical books/McGraw-Hill Guyton and Hall, Textbook of Medical Physiology, 10th Ed, Saunders, 2000 COMMENT: Question states Cardiogenic shock which implies low MAP Therefore first answer is B. Comment: (Not at all.... the MAP is not a value measured by the PAC, neither is the EF) (SVR mathematically determined from MAP and RAP. SVR = (MAP - RAP * 79.9)/CO ) The typical picture of cardiogenic shock is characterised by: an increased HR, an increased SVR, a reduced SV, CO, CI, systemic hypotension (defined for cardiogenic shock as SBP

Answer 620

Measured PCWP overestimates LVEDV if there is abnormal mitral valve (MR or MS) Measured PCWP underestimates LVEDV if the LV is noncompliant Measured PCWP overestimates LVEDV if there is increased PEEP (or airway pressure) LV systolic dysfunction does not interfere with the end diastolic point in the pressure-volume loop. Clinically, LV systolic dysfunction can be an indication for the use of PA catheters for the purposes of optimising LV preload (eg for cardiac surgery). Thus the answer is D References Finnis ICU notes Muralidhar Indian Journal of Anaesthesia article Ganong p573 Fig 29-7

Answer 621

B velocity is an intrinsic property of the tissue A Increasing frequency increases attenuation (ie decreases penetration) but increases resolution C Some energy is reflected at each interface D Wavelength is result of frequency (ie velocity = frequency x wavelength) E Tissue can absorb or scatter sound energy variably

Answer 622

nswer E. Pressure is the force applied or distributed over a surface in a direction perpendicular to that surface, and is expressed as force per unit area. The SI unit is the pascal (Pa) and 1 pascal is a pressure of 1 newton acting over an area of 1 square metre. A standard atmosphere atm represents the mean atmospheric pressure at mean sea level at the latitude of Paris ≈ 760 mmHg at OoC. The torr, named after the inventor of the barometer Evangelista Torricelli is a standardised mmHg and often used interchangeably. 1 Pa = 1 N/M2 1 bar = 100,000 Pa 1 atm = 760 mmHg 1 mmHg = 133.322 Pa 1 mmHg ≈ 0.735 cmH20 ``` SI = Le Système international d'unités, the modern metric system of measurement formulated by the 11th General Conference on Weights and Measures in France 1960. The base SI units are a choice of seven well-defined units which by convention were regarded as dimensionally independent; more recently the metre and candela have been redefined in terms of other SI units. SI base unit Base quantity Name Symbol length metre m mass kilogram kg time second s electric current ampere A thermodynamic temperature kelvin K amount of substance mole mol luminous intensity candela cd ``` All other SI units are derived from these base units via a system of quantity equations. For convenience, 22 of these derived units are given special names/symbols; which can be expressed in terms of the base units. Examples of SI derived units Derived quantity Name Base units pressure pascal m-1kg s-2 force newton m kg s-2 power watt m2kg s-3 frequency hertz s-1 Although mmHg and bar are non-SI units of pressure, they are accepted for use with the SI. For high pressure gas supplies, the bar is retained for general use. In clinical practice, one pascal is an inconveniently small unit, so kilopascals (kPa) or megapascals (MPa) are usually used. Another comment: my mnemonic for remembering the base SI units is METTALL (ie. mass, electric current, time, temperature, amount, luminosity, length). References Basic Physics and Measurement in Anaesthesia 5th ed. http: //www.physics.nist.gov/cuu/Units/units.html http: //www.bipm.org/en/si/ http: //en.wikipedia.org/wiki/SI

Answer 623

For 1st version above: ANSWER E. Newton Newton = m·kg·s-2 and is not a base SI unit (but an SI derived unit). Force = Mass x Acceleration

Answer 624

``` For the second version 15A15B A. Joule: derived unit (m2·kg·s-2) B. Litre: neither base nor derived according to NIST (web address in references), but it is derived according to Davis & Kenny (p296) C. Pascal: derived (m-1·kg·s-2) D. Ampere: base unit E. Newton: derived (m·kg·s-2) Therefore I would say the correct answer is D Dr mitta 22:37, 11 November 2015 (CST) ```

Answer 625

Answer B. A: Ag/AgCl anode, Platinum Cathode B: True according to Davis et al, but Protein deposition on cathode occurs therefore plastic membrane separates blood from an electrolyte surrounding Cathode. C: 0.6 Volt Potential, current is what is measured. D: Electrode must be maintained at 37deg C E: Standardised Gas misxtures are used for calibration Reaction at Cathode = O2 + 4e- + 2H2O = 4OH-

Answer 626

Answer is E I think. The wavelength definitely changes, it seems completely different to mass spec to me, more than just monoatomic molecules can be measured and "...the relative concentrations of the agents are obtained from the detectors measuring the radiation from the individual agents." Davis and Kenny 5th Ed Pg 226 When light encounters molecules in the air, the predominant mode of scattering is elastic scattering, called Rayleigh scattering. This scattering is responsible for the blue color of the sky; it increases with the fourth power of the frequency and is more effective at short wavelengths. It is also possible for the incident photons to interact with the molecules in such a way that energy is either gained or lost so that the scattered photons are shifted in frequency. Such inelastic scattering is called Raman scattering. (for more see references below) "Gaseous analysis: carbon dioxide, nitrous oxide, and volatile agents. In their gaseous states, these can be measured by a number of techniques including: infrared absorption spectroscopy; photoacoustic spectroscopy; silicone rubber and piezoelectric absorption; refractometry; Raman scattering; and mass spectrometry. Most in-theatre side-sampling benches presently utilize infrared absorption." http://ceaccp.oxfordjournals.org/content/9/1/19.full

Answer 627

Answer: B as in "poor R wave progression"

Answer 628

Air is air so always will be 21% In the absence of other options, not sure what is wanted. However surgical "tool air" has a higher pipeline pressure in hospitals (1,200kPa from memory) than "medical air" pipelines (nominally 400 kPa). Medical air supplies have to be very reliable with back-up and monitoring systems. Not sure what the rules for tool air are but there may be some differences in the compressors used. The differences & the answer will be in the relevant A/NZ standard.

Answer 629

The more traditional paramagnetic analysers (Servomex brand) use the nitrogen-filled spheres in a dumbbell arrangement are used in reference labs; they are highly accurate but slow. Not suitable for breath-by-breath analysis on an Anaesthetic machine.

Answer 630

B inc temp will dec viscosity and dec density but in turbulent flow, density matters more than viscosity According to Reynold's number equation, incresing radius reduces the number, therefore less likely turbulent flow, should be C.

Answer 631

``` A fire needs 3 things: oxidising agent (eg oxygen) fuel heat ETT is the fuel, diathermy gives the heat, and both O2 and N2O are oxidising agents. References ``` [Fire/trachy/full text] Tykocinski M, Thomson P, Hooper R (2006) Airway fire during tracheotomy. ANZ J Surg. 2006 Mar;76(3):195-7. [PMID 16626366] Varcoe et al. Airway fire during tracheostomy. Case Report. ANZ Journal of Surgery 2004; 74(6): 507–508 [PMID 15191500] [1] Fires in the Operating Theatre. This recommends disconnecting the circuit and using a self-inflating bag to ventilate with room air. Muchatuta N & Sale S - Fires & explosions, Anaesthesia & Intensive Care Medicine, 8(11) pp457-460 (via ANZCA Journals) http://www.anzca.edu.au/resources/books-and-

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