Flashcards in Renal and Acid-Base Physiology Deck (39)
Secretion of K+ by the distal tubule will be decreased by
(A) metabolic alkalosis
(B) a high-K+ diet
(D) spironolactone administration
(E) thiazide diuretic administration
The answer is D [V B 4 b].
Distal K+ secretion is decreased by factors that decrease the driving force for passive diffusion of K+ across the luminal membrane. Because spironolactone is an aldosterone antagonist, it reduces K+ secretion. Alkalosis, a diet high in K+, and hyperaldosteronism all increase [K+] in the distal cells and thereby increase K+ secretion. Thiazide diuretics increase flow through the distal tubule and dilute the luminal [K+] so that the driving force for K+ secretion is increased.
Jared and Adam both weigh 70 kg. Jared drinks 2 L of distilled water, and Adam drinks 2 L of isotonic NaCl. As a result of these ingestions, Adam will have a
(A) greater change in intracellular fluid (ICF) volume
(B) higher positive free-water clearance (CHO2)
(C) greater change in plasma osmolarity
(D) higher urine osmolarity
(E) higher urine flow rate
The answer is D [I C 2 a; VII C; Figure 5.15; Table 5.6].
After drinking distilled water, Jared will have an increase in intracellular fluid (ICF) and extracellular fluid (ECF) volumes, a decrease in plasma osmolarity, a suppression of antidiuretic hormone (ADH) secretion, and a positive free-water clearance (CHO2), and will produce dilute urine with a high flow rate. Adam, after drinking the same volume of isotonic NaCl, will have an increase in ECF volume only and no change in plasma osmolarity. Because Adam's ADH will not be suppressed, he will have a higher urine osmolarity, a lower urine flow rate, and a lower CH O 2 than Jared.
A 45-year-old woman develops severe diar-rhea while on vacation. She has the following arterial blood values:
pH = 7.25
PCO2 = 24 mm Hg
[HCO3 -] = 10 mEq/L
Venous blood samples show decreased blood [K+] and a normal anion gap.
The correct diagnosis for this patient is
(A) metabolic acidosis
(B) metabolic alkalosis
(C) respiratory acidosis
(D) respiratory alkalosis
(E) normal acid–base status
The answer is A [IX D 1 a–c; Tables 5.8 and 5.9].
An acid pH, together with decreased HCO3- and decreased Pco2, is consistent with metabolic acidosis with respiratory compensation (hyperventilation). Diarrhea causes gastrointestinal (GI) loss of HCO3 -, creating a metabolic acidosis.
A 45-year-old woman develops severe diarrhea while on vacation. She has the following arterial blood values:
pH = 7.25
Pco2 = 24 mm Hg
[HCO3-] = 10 mEq/L
Venous blood samples show decreased blood [K+] and a normal anion gap.
Which of the following statements about this patient is correct?
(A) She is hypoventilating
(B) The decreased arterial [HCO3-] is a result of buffering of excess H+ by HCO3-
(C) The decreased blood [K+] is a result of exchange of intracellular H+ for extracellular K+
(D) The decreased blood [K+] is a result of increased circulating levels of aldosterone
(E) The decreased blood [K+] is a result of decreased circulating levels of antidiuretic hormone (ADH)
The answer is D [IX D 1 a–c; Tables 5.8 and 5.9].
The decreased arterial [HCO3-] is caused by gastrointestinal (GI) loss of HCO3- from diarrhea, not by buffering of excess H+ by HCO3 -. The woman is hyperventilating as respiratory compensation for metabolic acidosis. Her hypokalemia cannot be the result of the exchange of intracellular H+ for extracellular K+, because she has an increase in extracellular H+, which would drive the exchange in the other direction. Her circulating levels of aldosterone would be increased as a result of extracellular fluid (ECF) volume contraction, which leads to increased K+ secretion by the distal tubule and hypokalemia.
Use the values below to answer the following question.
Glomerular capillary hydrostatic pressure = 47 mmHg
Bowman space hydrostatic pressure = 10 mmHg
Bowman space oncotic pressure = 0 mmHg
At what value of glomerular capillary oncotic pressure would glomerular filtration stop?
(A) 57 mmHg
(B) 47 mmHg
(C) 37 mm Hg
(D) 10 mm Hg
(E) 0 mm Hg
The answer is C [II C 4, 5].
Glomerular filtration will stop when the net ultrafiltration pressure across the glomerular capillary is zero; that is, when the force that favors filtration (47 mmHg) exactly equals the forces that oppose filtration (10 mmHg + 37 mmHg).
The reabsorption of filtered HCO3-
(A) results in reabsorption of less than 50% of the filtered load when the plasma concentration of HCO3- is 24 mEq/L
(B) acidifies tubular fluid to a pH of 4.4
(C) is directly linked to excretion of H+ as NH4+
(D) is inhibited by decreases in arterial Pco2
(E) can proceed normally in the presence of a renal carbonic anhydrase inhibitor
The answer is D [IX C 1 a, b].
Decreases in arterial Pco2 cause a decrease in the reabsorption of filtered HCO3 - by diminishing the supply of H+ in the cell for secretion into the lumen. Reabsorption of filtered HCO3- is nearly 100% of the filtered load and requires carbonic anhydrase in the brush border to convert filtered HCO3- to CO2 to proceed normally. This process causes little acidification of the urine and is not linked to net excretion of H+ as titratable acid or NH4+.
The following information was obtained in a 20-year-old college student who was participating in a research study in the Clinical Research Unit:
[Inulin] = 1 mg/mL
[X] = 2 mg/mL
[Inulin] = 150 mg/mL
[X] = 100 mg/mL
Urine flow rate = 1 mL/min
Assuming that X is freely filtered, which of the following statements is most correct?
(A) There is net secretion of X
(B) There is net reabsorption of X
(C) There is both reabsorption and secretion of X
(d) The clearance of X could be used to measure the glomerular filtration rate (GFR)
(E) The clearance of X is greater than the clearance of inulin
The answer is B [II C 1].
To answer this question, calculate the glomerular filtration rate (GFR) and CX. GFR = 150 mg/mL × 1 mL/min ÷ 1 mg/mL = 150 mL/min. CX = 100 mg/mL × 1 mL/min ÷ 2 mg/mL = 50 mL/min. Because the clearance of X is less than the clearance of inulin (or GFR), net reabsorption of X must have occurred. Clearance data alone cannot determine whether there has also been secretion of X. Because GFR cannot be measured with a substance that is reabsorbed, X would not be suitable.
To maintain normal H+ balance, total daily excretion of H+ should equal the daily
(A) fixed acid production plus fixed acid ingestion
(B) HCO3 - excretion
(C) HCO3 - filtered load
(D) titratable acid excretion
(E) filtered load of H+
The answer is A [IX C 2].
Total daily production of fixed H+ from catabolism of proteins and phospholipids (plus any additional fixed H+ that is ingested) must be matched by the sum of excretion of H+ as titratable acid plus NH4 + to maintain acid–base balance.
One gram of mannitol was injected into a woman. After equilibration, a plasma sample had a mannitol concentration of 0.08 g/L. During the equilibration period, 20% of the injected mannitol was excreted in the urine. The woman's
(A) extracellular fluid (ECF) volume is 1 L
(B) intracellular fluid (ICF) volume is 1 L
(C) ECF volume is 10 L
(D) ICF volume is 10 L
(E) interstitial volume is 12.5 L
The answer is C [I B 1 a].
Mannitol is a marker substance for the extracellular fluid (ECF) volume. ECF volume = amount of mannitol/concentration of mannitol = 1 g – 0.2 g/(0.08 g/L) = 10 L.
A 58-year-old man is given a glucose tolerance test. In the test, the plasma glucose concentration is increased and glucose reabsorption and excretion are measured. When the plasma glucose concentration is higher than occurs at transport maximum (Tm), the
(A) clearance of glucose is zero
(B) excretion rate of glucose equals the filtration rate of glucose
(C) reabsorption rate of glucose equals the filtration rate of glucose
(D) excretion rate of glucose increases with increasing plasma glucose concentrations
(E) renal vein glucose concentration equals the renal artery glucose concentration
The answer is D [III B; Figure 5.5].
At concentrations greater than at the transport maximum (Tm) for glucose, the carriers are saturated so that the reabsorption rate no longer matches the filtration rate. The difference is excreted in the urine. As the plasma glucose concentration increases, the excretion of glucose increases. When it is greater than the Tm, the renal vein glucose concentration will be less than the renal artery concentration because some glucose is being excreted in urine and therefore is not returned to the blood. The clearance of glucose is zero at concentrations lower than at Tm (or lower than threshold) when all of the filtered glucose is reabsorbed but is greater than zero at concentrations greater than Tm.
A negative free-water clearance (-CH2O) will occur in a person who
(A) drinks 2 L of distilled water in 30 minutes
(B) begins excreting large volumes of urine with an osmolarity of 100 mOsm/L after a severe head injury
(C) is receiving lithium treatment for depression and has polyuria that is unresponsive to the administration of antidiuretic hormone (ADH)
(D) has an oat cell carcinoma of the lung, and excretes urine with an osmolarity of 1,000 mOsm/L
The answer is D [VII D; Table 5.6].
A person who produces hyperosmotic urine (1,000 mOsm/L) will have a negative free-water clearance (–CH2O) [CH2O = V – Cosm]. All of the others will have a positive CH2O because they are producing hyposmotic urine as a result of the suppression of antidiuretic hormone (ADH) by water drinking, central diabetes insipidus, or nephrogenic diabetes insipidus.
A buffer pair (HA/A-) has a pK of 5.4. At a blood pH of 7.4, the concentration of HA is
(A) 1/l00 that of A-
(B) 1/10 that of A-
(C) equal to that of A-
(D) 10 times that of A-
(E) 100 times that of A-
The answer is A [IX B 3].
The Henderson-Hasselbalch equation can be used to calculate the ratio of HA/A-:
pH = pK + log A-/HA
7.4 = 5.4 + log A-/HA
7.4 - 5.4 = log A-/HA
2 = log A-/HA
100 = A-/HA or HA/A- is 1/100
Which of the following would produce an increase in the reabsorption of isosmotic fluid in the proximal tubule?
(A) Increased filtration fraction
(B) Extracellular fluid (ECF) volume expansion
(C) Decreased peritubular capillary protein concentration
(D) Increased peritubular capillary hydrostatic pressure
(E) Oxygen deprivation
The answer is A [II C 3; IV C 1 d (2)].
Increasing filtration fraction means that a larger portion of the renal plasma flow (RPF) is filtered across the glomerular capillaries. This increased flow causes an increase in the protein concentration and oncotic pressure of the blood leaving the glomerular capillaries. This blood becomes the peritubular capillary blood supply. The increased oncotic pressure in the peritubular capillary blood is a driving force favoring reabsorption in the proximal tubule. Extracellular fluid (ECF) volume expansion, decreased peritubular capillary protein concentration, and increased peritubular capillary hydrostatic pressure all inhibit proximal reabsorption. Oxygen deprivation would also inhibit reabsorption by stopping the Na+–K+ pump in the basolateral membranes.
Which of the following substances or combinations of substances could be used to measure interstitial fluid volume?
(B) D2O alone
(C) Evans blue
(D) Inulin and D2O
(E) Inulin and radioactive albumin
The answer is E [I B 2 b–d].
Interstitial fluid volume is measured indirectly by determining the difference between extracellular fluid (ECF) volume and plasma volume. Inulin, a large fructose polymer that is restricted to the extracellular space, is a marker for ECF volume. Radioactive albumin is a marker for plasma volume.
At plasma para-aminohippuric acid (PAH) concentrations below the transport maximum (Tm), PAH
(A) reabsorption is not saturated
(B) clearance equals inulin clearance
(C) secretion rate equals PAH excretion rate
(D) concentration in the renal vein is close to zero
(E) concentration in the renal vein equals PAH concentration in the renal artery
The answer is D [III C; Figure 5.6].
At plasma concentrations that are lower than at the transport maximum (Tm) for para-aminohippuric acid (PAH) secretion, PAH concentration in the renal vein is nearly zero because the sum of filtration plus secretion removes virtually all PAH from the renal plasma. Thus, the PAH concentration in the renal vein is less than that in the renal artery because most of the PAH entering the kidney is excreted in urine. PAH clearance is greater than inulin clearance because PAH is filtered and secreted; inulin is only filtered.
Compared with a person who ingests 2 L of distilled water, a person with water deprivation will have a
(A) higher free-water clearance (CHO2)
(B) lower plasma osmolarity
(C) lower circulating level of antidiuretic hormone (ADH)
(D) higher tubular fluid/plasma (TF/P) osmolarity in the proximal tubule
(E) higher rate of H2O reabsorption in the collecting ducts
The answer is E [VII D; Figures 5.14 and 5.15].
The person with water deprivation will have a higher plasma osmolarity and higher circulating levels of antidiuretic hormone (ADH). These effects will increase the rate of H2O reabsorption in the collecting ducts and create a negative free-water clearance (–CH2O). Tubular fluid/plasma (TF/P) osmolarity in the proximal tubule is not affected by ADH.
Which of the following would cause an increase in both glomerular filtration rate (GFR) and renal plasma flow (RPF)?
(B) A ureteral stone
(C) Dilation of the afferent arteriole
(D) Dilation of the efferent arteriole
(E) Constriction of the efferent arteriole
The answer is C [II C 4; Table 5.3].
Dilation of the afferent arteriole will increase both renal plasma flow (RPF) (because renal vascular resistance is decreased) and glomerular filtration rate (GFR) (because glomerular capillary hydrostatic pressure is increased). Dilation of the efferent arteriole will increase RPF but decrease GFR. Constriction of the efferent arteriole will decrease RPF (due to increased renal vascular resistance) and increase GFR. Both hyperproteinemia (↑ π in the glomerular capillaries) and a ureteral stone (↑ hydrostatic pressure in Bowman space) will oppose filtration and decrease GFR.
A patient has the following arterial blood values: pH = 7.52 Pco2 = 20 mm Hg [HCO3 −] = 16 mEq/L Which of the following statements about this patient is most likely to be correct?
(A) He is hypoventilating
(B) He has decreased ionized [Ca2+] in blood
(C) He has almost complete respiratory compensation
(D) He has an acid–base disorder caused by overproduction of fixed acid
(E) Appropriate renal compensation would cause his arterial [HCO3 -] to increase
The answer is B [IX D 4; Table 5.8].
First, the acid–base disorder must be diagnosed. Alkaline pH, low Pco2, and low HCO3 - are consistent with respiratory alkalosis. In respiratory alkalosis, the [H+] is decreased and less H+ is bound to negatively charged sites on plasma proteins. As a result, more Ca2+ is bound to proteins and, therefore, the ionized [Ca2+] decreases. There is no respiratory compensation for primary respiratory disorders. The patient is hyperventilating, which is the cause of the respiratory alkalosis. Appropriate renal compensation would be decreased reabsorption of HCO3 -, which would cause his arterial [HCO3 -] to decrease and his blood pH to decrease (become more normal).
Which of the following would best distinguish an otherwise healthy person with severe water deprivation from a person with the syndrome of inappropriate antidiuretic hormone (SIADH)?
(A) Free-water clearance (CH O 2)
(B) Urine osmolarity
(C) Plasma osmolarity
(D) Circulating levels of antidiuretic hormone (ADH)
(E) Corticopapillary osmotic gradient
The answer is C [VII B, D 4; Table 5.6].
Both individuals will have hyperosmotic urine, a negative free-water clearance (−CH O 2 ), a normal corticopapillary gradient, and high circulating levels of antidiuretic hormone (ADH). The person with water deprivation will have a high plasma osmolarity, and the person with syndrome of inappropriate antidiuretic hormone (SIADH) will have a low plasma osmolarity (because of dilution by the inappropriate water reabsorption).
Which of the following causes a decrease in renal Ca2+ clearance?
(B) Treatment with chlorothiazide
(C) Treatment with furosemide
(D) Extracellular fluid (ECF) volume expansion
The answer is B [Table 5.11].
Thiazide diuretics have a unique effect on the distal tubule; they increase Ca2+ reabsorption, thereby decreasing Ca2+ excretion and clearance. Because parathyroid hormone (PTH) increases Ca2+ reabsorption, the lack of PTH will cause an increase in Ca2+ clearance. Furosemide inhibits Na+ reabsorption in the thick ascending limb, and extracellular fluid (ECF) volume expansion inhibits Na+ reabsorption in the proximal tubule. At these sites, Ca2+ reabsorption is linked to Na+ reabsorption, and Ca2+ clearance would be increased. Because Mg2+ competes with Ca2+ for reabsorption in the thick ascending limb, hypermagnesemia will cause increased Ca2+ clearance.
A patient arrives at the emergency room with low arterial pressure, reduced tissue turgor, and the following arterial blood values:
pH = 7.69
[HCO3 -] = 57 mEq/L
Pco2 = 48 mm Hg
Which of the following responses would also be expected to occur in this patient?
(B) Decreased K+ secretion by the distal tubules
(C) Increased ratio of H2PO4 - to HPO4 -2 in urine
(D) Exchange of intracellular H+ for extracellular K+
The answer is D [IX D 2; Table 5.8].
First, the acid–base disorder must be diagnosed. Alkaline pH, with increased HCO3 - and increased Pco2, is consistent with metabolic alkalosis with respiratory compensation. The low blood pressure and decreased turgor suggest extracellular fluid (ECF) volume contraction. The reduced [H+] in blood will cause intracellular H+ to leave cells in exchange for extracellular K+. The appropriate respiratory compensation is hypoventilation, which is responsible for the elevated Pco2. H+ excretion in urine will be decreased, so less titratable acid will be excreted. K+ secretion by the distal tubules will be increased because aldosterone levels will be increased secondary to ECF volume contraction.
A woman has a plasma osmolarity of 300 mOsm/L and a urine osmolarity of 1200 mOsm/L. The correct diagnosis is
(A) syndrome of inappropriate antidiuretic hormone (SIADH)
(B) water deprivation
(C) central diabetes insipidus
(D) nephrogenic diabetes insipidus
(E) drinking large volumes of distilled water
The answer is B [VII B; Figure 5.14].
This patient's plasma and urine osmolarity, taken together, are consistent with water deprivation. The plasma osmolarity is on the high side of normal, stimulating the posterior pituitary to secrete antidiuretic hormone (ADH). Secretion of ADH, in turn, acts on the collecting ducts to increase water reabsorption and produce hyperosmotic urine. Syndrome of inappropriate antidiuretic hormone (SIADH) would also produce hyperosmotic urine, but the plasma osmolarity would be lower than normal because of the excessive water retention. Central and nephrogenic diabetes insipidus and excessive water intake would all result in hyposmotic urine.
A patient is infused with para-aminohippuric acid (PAH) to measure renal blood flow (RBF). She has a urine flow rate of 1 mL/min, a plasma [PAH] of 1 mg/mL, a urine [PAH] of 600 mg/mL, and a hematocrit of 45%. What is her “effective” RBF?
(A) 600 mL/min
(B) 660 mL/min
(C) 1,091 mL/min
(D) 1,333 mL/min
The answer is C [II B 2, 3].
Effective renal plasma flow (RPF) is calculated from the clearance of para-aminohippuric acid (PAH) [CPAH = UPAH × V/PPAH = 600 mL/min]. Renal blood flow (RBF) = RPF/1 – hematocrit = 1091 mL/min.
Which of the following substances has the highest renal clearance?
(A) Para-aminohippuric acid (PAH)
The answer is A [III D].
Para-aminohippuric acid (PAH) has the greatest clearance of all of the substances because it is both filtered and secreted. Inulin is only filtered. The other substances are filtered and subsequently reabsorbed; therefore, they will have clearances that are lower than the inulin clearance.
A woman runs a marathon in 90°F weather and replaces all volume lost in sweat by drinking distilled water. After the marathon, she will have
(A) decreased total body water (TBW)
(B) decreased hematocrit
(C) decreased intracellular fluid (ICF) volume
(D) decreased plasma osmolarity
(E) increased intracellular osmolarity
The answer is D [I C 2 f; Table 5.2].
By sweating and then replacing all volume by drinking H2O, the woman has a net loss of NaCl without a net loss of H2O. Therefore, her extracellular and plasma osmolarity will be decreased, and as a result, water will flow from extracellular fluid (ECF) to intracellular fluid (ICF). The intracellular osmolarity will also be decreased after the shift of water. Total body water (TBW) will be unchanged because the woman replaced all volume lost in sweat by drinking water. Hematocrit will be increased because of the shift of water from ECF to ICF and the shift of water into the red blood cells (RBCs), which causes their volume to increase.
Which of the following causes hyperkalemia?
(C) Insulin injection
(D) Decreased serum osmolarity
(E) Treatment with β-agonists
The answer is A [Table 5.4].
Exercise causes a shift of K+ from cells into blood. The result is hyperkalemia. Hyposmolarity, insulin, β-agonists, and alkalosis cause a shift of K+ from blood into cells. The result is hypokalemia.
Which of the following is a cause of metabolic alkalosis?
(B) Chronic renal failure
(C) Ethylene glycol ingestion
(D) Treatment with acetazolamide
(F) Salicylate poisoning
The answer is E [Table 5.9].
A cause of metabolic alkalosis is hyperaldosteronism; increased aldosterone levels cause increased H+ secretion by the distal tubule and increased reabsorption of “new” HCO3 -. Diarrhea causes loss of HCO3 - from the gastrointestinal (GI) tract and acetazolamide causes loss of HCO3 - in the urine, both resulting in hyperchloremic metabolic acidosis with normal anion gap. Ingestion of ethylene glycol and salicylate poisoning leads to metabolic acidosis with increased anion gap.
Which of the following is an action of parathyroid hormone (PTH) on the renal tubule?
(A) Stimulation of adenylate cyclase
(B) Inhibition of distal tubule K+ secretion
(C) Inhibition of distal tubule Ca2+ reabsorption
(D) Stimulation of proximal tubule phosphate reabsorption
(E) Inhibition of production of 1,25-dihydroxycholecalciferol
The answer is A [VI B; Table 5.7].
Parathyroid hormone (PTH) acts on the renal tubule by stimulating adenyl cyclase and generating cyclic adenosine monophosphate (cAMP). The major actions of the hormone are inhibition of phosphate reabsorption in the proximal tubule, stimulation of Ca2+ reabsorption in the distal tubule, and stimulation of 1,25-dihydroxycholecalciferol production. PTH does not alter the renal handling of K+.
A man presents with hypertension and hypokalemia. Measurement of his arterial blood gases reveals a pH of 7.5 and a calculated HCO3− of 32 mEq/L. His serum cortisol and urinary vanillylmandelic acid (VMA) are normal, his serum aldosterone is increased, and his plasma renin activity is decreased. Which of the following is the most likely cause of his hypertension?
(A) Cushing syndrome
(B) Cushing disease
(C) Conn syndrome
(D) Renal artery stenosis
The answer is C [IV C 3 b; V B 4 b].
Hypertension, hypokalemia, metabolic alkalosis, elevated serum aldosterone, and decreased plasma renin activity are all consistent with a primary hyperaldosteronism (e.g., Conn syndrome). High levels of aldosterone cause increased Na+ reabsorption (leading to increased blood pressure), increased K+ secretion (leading to hypokalemia), and increased H+ secretion (leading to metabolic alkalosis). In Conn syndrome, the increased blood pressure causes an increase in renal perfusion pressure, which inhibits renin secretion. Neither Cushing syndrome nor Cushing disease is a possible cause of this patient's hypertension because serum cortisol and adrenocorticotropic hormone (ACTH) levels are normal. Renal artery stenosis causes hypertension that is characterized by increased plasma renin activity. Pheochromocytoma is ruled out by the normal urinary excretion of vanillylmandelic acid (VMA).