West Book questions

Question 1

Concerning the blood-gas barrier of the human lung,

A. The thinnest part of the blood-gas barrier has a thickness of about 3 mm.
B. The total area of the blood-gas barrier is about 1 square meter.
C. About 10% of the area of the alveolar wall is occupied by capillaries.
D. If the pressure in the capillaries rises to unphysiologically high levels, the
blood-gas barrier can be damaged.
E. Oxygen crosses the blood-gas barrier by active transport.

Answer 1

D is correct.

The capillary walls are so thin that if the pressure in them rises too much, they are damaged and leak plasma or blood, a condition known as stress failure. The other choices are incorrect because the thin- nest part of the blood-gas barrier is about 0.3 μm thick, its total area exceeds 50 square meters, almost all of the area of the alveolar wall is occupied by capillaries, and oxygen crosses the barrier by passive diffusion.

Question 2

When oxygen moves through the thin side of the blood-gas barrier from the alveolar gas to the hemoglobin of the red blood cell, it traverses the following layers in order:

A. Epithelial cell, surfactant, interstitium, endothelial cell, plasma, red cell membrane. B. Surfactant, epithelial cell, interstitium, endothelial cell, plasma, red cell membrane. C. Surfactant, endothelial cell, interstitium, epithelial cell, plasma, red cell membrane. D. Epithelium cell, interstitium, endothelial cell, plasma, red cell membrane.
E. Surfactant, epithelial cell, interstitium, endothelial cell, red cell membrane.

Answer 2

B is correct.

Question 3

What is the PO2 (in mm Hg) of moist inspired gas of a climber on the summit of
Mt. Everest (assume barometric pressure is 247 mm Hg)?

A. 32
B. 42
C. 52
D. 62
E. 72

Answer 3

B is correct.

The calculation is 0.2093 × (247 − 47).

Question 4

Concerning the airways of the human lung,

A. The volume of the conducting zone is about 50 ml.
B. The volume of the rest of the lung during resting conditions is about 5 liters.
C. A respiratory bronchiole can be distinguished from a terminal bronchiole
because the latter has alveoli in its walls.
D. On the average, there are about three branchings of the conducting airways
before the first alveoli appear in their walls.
E. In the alveolar ducts, the predominant mode of gas flow is diffusion rather
than convection.

Answer 4

E is correct.

The combined cross sectional area of the alveolar ducts is so great (Figure 1-5) that gas diffusion is the main mode of transport rather than convection. The other choices are incorrect. The volume of the con- ducting airways is about 150 ml, the volume of the lung at FRC is about 3 liters, a respiratory bronchiole but not a terminal bronchiole has alveoli in its walls, and there are about 16 branches of the conducting airways before the first alveoli appear.

Question 5

Concerning the blood vessels of the human lung,

A. The pulmonary veins form a branching pattern that matches that of the airways.
B. The average diameter of the capillaries is about 50 mm.
C. The bronchial circulation has about the same blood flow as the pulmonary
circulation.
D. On the average, blood spends about 0.75 second in the capillaries under
resting conditions.
E. The mean pressure in the pulmonary artery is about 100 mm Hg.

Answer 5

D is correct (see Figure 3-2).

The other choices are incorrect because the branching pattern of the arteries, not the veins, matches the airways, the average diameter of the capillaries is about 7 to 10 μm, the flow in the bronchial circulation is very small compared to the pulmonary circulation, and the mean pressure in the pulmonary artery is about 15 mm Hg.

Question 6

The only variable in the following list that cannot be measured with a simple
spirometer and stopwatch is?

A. Tidal volume.
B. Functional residual capacity. C. Vital capacity.
D. Total ventilation.
E. Respiratory frequency.

Answer 6

B is correct.

The FRC includes the residual volume and cannot be measured with a simple spirometer. All the other choices can be measured with a spirometer and stopwatch (see Figure 2-2).

Question 7

Concerning the pulmonary acinus,

A. Less than 90% oxygen uptake of the lung occurs in the acini.
B. Percentage change in volume of the acini during inspiration is less than that of
the whole lung.
C. Volume of the acini is less than 90% of the total volume of the lung at FRC.
D. Each acinus is supplied by a terminal bronchiole.
E. The ventilation of the acini at the base of the upright human lung at FRC is
less than those at the apex.

Answer 7

D is correct.

An acinus is that portion of the lung supplied by a terminal bronchiole. The other choices are incorrect because all the oxygen uptake occurs in the acini, the change in volume of the acini during breathing is greater than that of the whole lung because the volume of the conducting airways remains almost constant, the volume of the acini is about 95% of the total volume of the lung at FRC (FRC is about 3 liters, conducting airways are about 150 ml), and the ventilation of the
acini is greater at the base than the apex of the upright lung at FRC (see
Figure 7-8).

Question 8

In a measurement of FRC by helium dilution, the original and final helium concentrations were 10% and 6%, and the spirometer volume was kept at 5 liters. What was the volume of the FRC in liters?

A. 2.5
B. 3.0
C. 3.3
D. 3.8
E. 5.0

Answer 8

C is correct.

If the volume of the FRC is denoted as V, the amount of helium initially in the spirometer is 5 × 0.1, and the amount after dilution is (5 + V) × 0.06. Therefore, V = 0.5/0.06 − 5 or 3.3 liters.

Question 9

A patient sits in a body plethysmograph (body box) and makes an expiratory effort against his closed glottis. What happens to the following: pressure in the lung airways, lung volume, box pressure, box volume?

 Airway Pressure  /  Lung Volume  /  Box Pressure  /  Box Volume A)         Down             /          Up            /           Up           /       Down B)         Down             /          Up            /        Down         /         Up  C)            Up               /        Down         /           Up           /       Down D)            Up               /        Down         /        Down        /          Up E)             Up              /            Up           /        Down        /        Down

Answer 9

D is correct.

When the patient makes an expiratory effort, he compresses the gas in the lung so that airway pressure increases and lung volume decreases slightly. The reduction of volume in the lung means that the box gas volume increases and therefore, its pressure decreases according to Boyle’s law.

Question 10

If CO2 production remains constant and alveolar ventilation is increased threefold, the alveolar PCO2 after a steady state is reached will be what percentage of its former value?

A. 25
B. 33
C. 50
D. 100
E. 300

Answer 10

B is correct.

The alveolar ventilation equation states that if CO2 production is constant, the alveolarPCO2 is inversely related to the alveolar ventilation. Therefore, if the ventilation is increased 3 times, thePCO2 will be reduced to a third of its former value, that is, 33%.

Question 11

In a measurement of physiologic dead space using Bohr’s method, the alveolar and mixed expired PCO2 were 40 and 30 mm Hg, respectively. What was the ratio of dead space to tidal volume?

A. 0.20
B. 0.25
C. 0.30
D. 0.35
E. 0.40

Answer 11

B is correct.

The equation states that the ratio equals (PA − PE)/PA, or (40 − 30)/40, that is 0.25.

Question 12

Using Fick’s law of diffusion of gases through a tissue slice, if gas X is 4 times as soluble and 4 times as dense as gas Y, what is the ratio of the diffusion rates of X to Y?

A. 0.25
B. 0.5
C. 2
D. 4
E. 8

Answer 12

C is correct.

The law states that the diffusion rate is proportional to the solubility but inversely proportional to the square root of the density. Therefore, the ratio of X to Y is 4/(􏰀4) or 4/2, that is, 2.

Question 13

An exercising subject breathes a low concentration of CO in a steady state. If the alveolar PCO is 0.5 mm Hg and the CO uptake is 30 ml·min−1, what is the diffusing capacity of the lung for CO in ml·min−1·mm·Hg−1?

A. 20
B. 30
C. 40
D. 50
E. 60

Answer 13

E is correct.

The equation is CO2 uptake divided by alveolar Pco , or 30/0.5,
that is, 60 ml·min−1·mm Hg −1

Question 14

In a normal person, doubling the diffusing capacity of the lung would be expected to

A. Decrease arterial PCO2 during resting breathing.
B. Increase resting oxygen uptake when the subject breathes 10% oxygen. C. Increase the uptake of nitrous oxide during anesthesia.
D. Increase the arterial PO2 during resting breathing.
E. Increase maximal oxygen uptake at extreme altitude.

Answer 14

E is correct.

The question is really asking for the conditions under which oxygen uptake or CO2 output are diffusion limited. The only correct answer is maximal oxygen uptake at extreme altitude (see Figure 3-3B). None of the other choices refer to situations where gas transfer is diffusion limited. The only possible alternative choice is B, but resting oxygen uptake is unlikely to be diffusion limited when a subject breathes 10% oxygen. Furthermore, in all these questions, we are looking for the one best answer, and this is clearly E.

Question 15

If a subject inhales several breaths of a gas mixture containing low concentrations of carbon monoxide and nitrous oxide,

A. The partial pressures of carbon monoxide in alveolar gas and end-capillary blood will be virtually the same.
B. The partial pressures of nitrous oxide in alveolar gas and end-capillary blood will be very different.
C. Carbon monoxide is transferred into the blood along the whole length of the capillary.
D. Little of the nitrous oxide will be taken up in the early part of the capillary.
E. The uptake of nitrous oxide can be used to measure the diffusing capacity of
the lung.

Answer 15

C is correct.

This question is testing the concepts of diffusion and perfu- sion limitation. Carbon monoxide is a diffusion-limited gas, so it is trans- ferred into the blood along the whole length of the capillary, and there is a large difference in partial pressure between alveolar gas and end-capillary blood (Figure 3-2). The opposite is true for nitrous oxide.

Question 16

Concerning the diffusing capacity of the lung,

A. It is best measured with carbon monoxide because this gas diffuses very slowly across the blood-gas barrier.
B. Diffusion limitation of oxygen transfer during exercise is more likely to occur at sea level than at high altitude.
C. Breathing oxygen reduces the measured diffusing capacity for carbon monoxide compared with air breathing.
D. It is decreased by exercise.
E. It is increased in pulmonary fibrosis, which thickens the blood-gas barrier.

Answer 16

C is correct.

Breathing oxygen reduces the measured diffusing capacity for carbon monoxide because the oxygen competes with carbon monoxide for hemoglobin, and therefore, the rate of reaction of carbon monoxide with hemoglobin (θ) is reduced. The other choices are incorrect because the reason for using carbon monoxide to measure the diffusing capacity of the lung is because it is a diffusion-limited gas, not because it diffuses slowly across the blood-gas barrier (its diffusion rate is not very different from that of oxygen). Diffusion limitation of oxygen transfer during exercise is more likely to occur at high altitude than sea level, and the diffusing capacity is increased by exercise and decreased by pulmonary fibrosis.

Question 17

The diffusing capacity of the lung for carbon monoxide is increased by

A. Emphysema, which causes loss of pulmonary capillaries.
B. Asbestosis, which causes thickening of the blood-gas barrier.
C. Pulmonary embolism, which cuts off the blood supply to part of the lung. D. Exercise in a normal subject.
E. Severe anemia.

Answer 17

D is correct.

Exercise increases the diffusing capacity because of recruitment and distension of pulmonary capillaries. Emphysema, asbestosis, pulmonary embolism, and severe anemia reduce the diffusing capacity because of a reduction in surface area of the blood-gas barrier, an increase in its thickness, or a reduction of the volume of blood in the pulmonary capillaries.

Question 18

The ratio of total systemic vascular resistance to pulmonary vascular resistance is about?

A. 2: 1
B. 3: 1
C. 5: 1
D. 10: 1
E. 20: 1

Answer 18

D is correct.

The flows in the systemic and pulmonary circulations are the same, but the mean pressure difference across the pulmonary circulation is about (15 − 5) mm Hg whereas that for the systemic circulation is about (100 − 2) mm Hg (see Figure 4-1). Therefore, the ratio is about 10:1.

Question 19

Concerning the extra-alveolar vessels of the lung,

A. Tension in the surrounding alveolar walls tends to narrow them.
B. Their walls contain smooth muscle and elastic tissue.
C. They are exposed to alveolar pressure.
D. Their constriction in response to alveolar hypoxia mainly takes place in the veins. E. Their caliber is reduced by lung inflation.

Answer 19

B is correct (Figure 4-3).

The other choices are incorrect because the ten- sion in the surrounding alveolar walls tends to pull the extra-alveolar vessels open, these vessels are not exposed to alveolar pressure, hypoxic pulmonary vasoconstriction occurs mainly in the small arteries, and the caliber of the extra-alveolar vessels is increased by lung inflation (see Figures 4-2 and 4-6).

Question 20

A patient with pulmonary vascular disease has mean pulmonary arterial and venous pressures of 55 and 5 mm Hg, respectively, while the cardiac output is 3 liters·min−1. What is his pulmonary vascular resistance in mm Hg·liters−1·min?

A. 0.5
B. 1.7
C. 2.5
D. 5
E. 17

Answer 20

E is correct.
The pulmonary vascular resistance is given by the pressure difference divided by the flow, or (55 − 5) divided by 3, that is, approximately 17 mm Hg·liter−1·min.

Question 21

The fall in pulmonary vascular resistance on exercise is caused by

A. Decrease in pulmonary arterial pressure.
B. Decrease in pulmonary venous pressure.
C. Increase in alveolar pressure.
D. Distension of pulmonary capillaries.
E. Alveolar hypoxia.

Answer 21

D is correct.

Distension of pulmonary capillaries lowers their vascular resistance. However, a decrease in both pulmonary arterial and pulmonary venous pressure reduces capillary pressure (other things remaining equal), and resistance therefore rises. The same is true of an increase in alveolar pressure, which tends to compress the capillaries. Alveolar hypoxia increases vascular resistance because of hypoxic pulmonary vasoconstriction.

Question 22

In a measurement of cardiac output using the Fick principle, the O2 concentrations of mixed venous and arterial blood are 16 and 20 ml·
100 ml−1, respectively, and the O2 consumption is 300 ml·min−1. The cardiac output in liters·min−1 is?

A. 2.5
B. 5
C. 7.5
D. 10
E. 75

Answer 22

C is correct.

The Fick principle states that the cardiac output is equal to the oxygen consumption divided by the arterial-venous oxygen concentration difference. The latter is (20 − 16) ml·100 ml−1 or (200 − 160) ml·liter−1. Therefore, the cardiac output is equal to 300/(200 − 160) or 7.5 liters·min−1.

Question 23

In zone 2 of the lung,

A. Alveolar pressure exceeds arterial pressure.
B. Venous pressure exceeds alveolar pressure.
C. Venous pressure exceeds arterial pressure.
D. Blood flow is determined by arterial pressure minus alveolar pressure. E. Blood flow is unaffected by arterial pressure.

Answer 23

D is correct.

In zone 2, flow is determined by arterial minus alveolar pressure. The other choices are incorrect because arterial pressure exceeds alveolar pressure, alveolar pressure exceeds venous pressure, and of course arterial pressure exceeds venous pressure.

Question 24

Pulmonary vascular resistance is reduced by?

A. Removal of one lung.
B. Breathing a 10% oxygen mixture.
C. Exhaling from functional residual capacity to residual volume. D. Acutely increasing pulmonary venous pressure.
E. Mechanically ventilating the lung with positive pressure.

Answer 24

D is correct.

Acutely increasing pulmonary venous pressure will raise capillary pressure and result in recruitment and distension of the capillaries. The other choices are incorrect because removing one lung greatly reduces the vascular bed, 10% oxygen breathing results in hypoxic pulmonary vasoconstriction, reducing lung volume to residual volume increases the resistance of the extra-alveolar vessels, and mechanically ventilating the lung with positive pressure increases the alveolar pressure and therefore tends to compress the capillaries.

Question 25

Hypoxic pulmonary vasoconstriction

A. Depends more on the PO2 of mixed venous blood than alveolar gas.
B. Is released in the transition from placental to air respiration.
C. Involves CO2 uptake in vascular smooth muscle.
D. Partly diverts blood flow from well-ventilated regions of diseased lungs. E. Is increased by inhaling low concentrations of nitric oxide.

Answer 25

B is correct.

The great reduction in pulmonary vascular resistance during the transition from placental to air respiration is largely brought about by the release of hypoxic pulmonary vasoconstriction. The other choices are incorrect because the PO2 of alveolar gas is much more important than the PO2 of mixed venous blood, CO2 uptake is irrelevant, the constriction partly diverts blood flow from poorly ventilated, not well-ventilated regions of diseased lungs, and the inhalation of nitric oxide partly reverses hypoxic pulmonary vasoconstriction.

Question 26

If the pressures in the capillaries and interstitial space at the top of the lung are 3 and 0 mm Hg, respectively, and the colloid osmotic pressures of the blood and interstitial fluid are 25 and 5 mm Hg, respectively, what is the net pressure in mm Hg moving fluid into the capillaries?

A. 17
B. 20
C. 23
D. 27
E. 33

Answer 26

A is correct.
The movement of fluid between the capillary lumen and interstitium obeys Starling’s Law. In the example given, the hydrostatic pressure difference moving fluid out of the capillary is (3 − 0), and the colloid osmotic pressure tending to move fluid into the capillary is (25 − 5) mm Hg. Therefore, the net pressure in mm Hg moving fluid into the capillaries is 17 mm Hg.

Question 27

The metabolic functions of the lung include

A. Converting angiotensin II to angiotensin I. B. Producing bradykinin.
C. Secreting serotonin.
D. Removing leukotrienes.
E. Generating erythropoietin.

Answer 27

D is correct.

Leukotrienes are almost completely removed from the blood in the pulmonary circulation (see Table 4-1). The other choices are incorrect because angiotensin I is converted to angiotensin II, bradykinin is largely inactivated, serotonin is almost completely removed, and erythropoietin is unchanged.

Question 28

A climber reaches an altitude of 4,500 m (14,800 ft) where the barometric
pressure is 447 mm Hg. The PO2 of moist inspired gas (in mm Hg) is

A. 47
B. 63
C. 75
D. 84
E. 98

Answer 28

D is correct.

The PO2 of moist inspired gas is given by 0.2093 × (447 − 47), that is, about 84 mm Hg.

Question 29

A man with normal lungs and an arterial PCO2 of 40 mm Hg takes an overdose of barbiturate that halves his alveolar ventilation but does not change his CO2 output. If his respiratory exchange ratio is 0.8, what will be his arterial PO2
(in mm Hg), approximately?

A. 40
B. 50
C. 60
D. 70
E. 80

Answer 29

B is correct.

To answer this question we first use the alveolar ventilation equation, which states that if the CO2 output is unchanged, thePCO2 is inversely proportional to the alveolar ventilation. Therefore, since alve- olar ventilation was halved, the arterialPCO2 was increased from 40 to 80 mm Hg. Then we use the alveolar gas equation PA O2 PIO2 PA CO2 / R+F , and we ignore F because it is small. Therefore, PAO2 = 149 − 80/0.8, which is approximately equal to 50 mm Hg.

Question 30

In the situation described in Question 2, how much does the inspired O2 concentration (%) have to be raised to return the arterial PO2 to its original level?

A. 7
B. 11
C. 15
D. 19
E. 23

Answer 30

A is correct.

The last equation above shows that to return the arterial PO2 to its normal value of about 100, we need to raise the inspired PO2 from 149 to 199 mm Hg. Recall that the inspired PO2 equals the fractional concentration of oxygen × (760 − 47). Therefore, the fractional concentration = 199/713 or 0.28 approximately. Thus, the inspired oxygen concentration as a percentage has to be increased from 21 to 28, that is, by 7%. Note that this example emphasizes how powerful the effect of increasing the inspired oxygen concentration on the arterial PO2 is when hypoxemia is caused by hypoventilation.

Question 31

A patient with normal lungs but a right-to-left shunt is found at catheterization to have oxygen concentrations in his arterial and mixed venous blood of 18 and 14 ml ·100 ml−1, respectively. If the O2 concentration of the blood leaving the pulmonary capillaries is calculated to be 20 ml · 100 ml−1, what is his shunt as a percentage of his cardiac output?

A. 23
B. 33
C. 43
D. 53
E. 63

Answer 31

B is correct.

This question is about the shunt equation shown in Figure 5-3. The shunt as a fraction of cardiac output is given by (Cc′ − Ca)/(Cc′ − Cv). Inserting the values gives the shunt as (20 − 18)/(20 − 14) or 2/6, that is, 33%.

Question 32

If a climber on the summit of Mt. Everest (barometric pressure 247 mm Hg) main- tains an alveolar PO2 of 34 mm Hg and is in a steady state (R ≤ 1), his alveolar PCO2 (in mm Hg) cannot be any higher than

A. 5
B. 8
C. 10
D. 12
E. 15

Answer 32

B is correct.

The inspired PO2 = 0.21 × (247 − 47) or 42 mm Hg. Therefore, using the alveolar gas equation as stated above and neglecting the small factor F, the alveolar PO2 is given by 42 − PCO2 /R where R is equal to or less than 1. Therefore to maintain an alveolar PO2 of 34 mm Hg, the alveolar PCO2 cannot exceed 8 mmHg.

Question 33

A patient with severe chronic obstructive pulmonary disease, which causes marked ventilation-perfusion inequality, has an arterial PO2 of 50 mm Hg and an arterial PCO2 of 40 mm Hg. The PCO2 is normal despite the hypoxemia because

A. Ventilation-perfusion inequality does not interfere with CO2 elimination.
B. Much of the CO2 is carried as bicarbonate.
C. The formation of carbonic acid is accelerated by carbonic anhydrase.
D. CO2 diffuses much faster through tissue than O2.
E. The O2 and CO2 dissociation curves have different shapes.

Answer 33

E is correct.

This question is testing knowledge about the effects of ventilation-perfusion inequality on O2 and CO2 transfer by the lung. VA/Q inequality impairs the transfer of both O2 and CO2 so that, other things being equal, this patient would have both a low arterial PO2 and high PCO 2 . However, by increasing the ventilation to the alveoli, thePCO2 can be brought back to normal, but the PO2 cannot. The reason for this is the different shapes of the O2 and CO2 dissociation curves. The other choices are incorrect because, as already stated, VA/Q does interfere with CO2 elimination. The statements that much of the CO2 is carried as bicarbonate, the formation of carbonic acid is accelerated by carbonic anhydrase, and CO2 diffuses much faster through tissue than O2 are true but are not the explanation for the normal PO2 despite the hypoxemia.

Question 34

The apex of the upright human lung compared with the base has

A. A higher PO2.
B. A higher ventilation.
C. A lower pH in end-capillary blood.
D. A higher blood flow.
E. Smaller alveoli.

Answer 34

A is correct.

The apex of the upright human lung has a high ventilation- perfusion ratio (see Figures 5-8, 5-9 and 5-10). Therefore, the apex has a higher alveolar PO2 than the base. The other choices are incorrect because the ventilation of the apex is lower than that of the base, the pH in end- capillary blood is higher because of the reduced PCO2 at the apex, the blood flow is lower as already stated, and the alveoli are larger because of the regional differences of intrapleural pressure (Figure 7-8).

Question 35

If the ventilation-perfusion ratio of a lung unit is decreased by partial bronchial obstruction while the rest of the lung is unaltered, the affected lung unit will show

A. Increased alveolar PO2.
B. Decreased alveolar PCO2.
C. No change in alveolar PN2.
D. A rise in pH of end-capillary blood.
E. A fall in oxygen uptake.

Answer 35

E is correct.

A decreased ventilation-perfusion ratio reduces the alveolar
PO2 and therefore the oxygen uptake by the lung unit. The other choices are incorrect because the unit will show a decreased alveolar PO2 as already stated, an increased alveolar PCO2 , a change in alveolar Pn2 (in fact a small rise), and a reduction in the pH of end-capillary blood because of the
increased PCO2 .

Question 36

A patient with lung disease who is breathing air has an arterial PO2 and PCO2 of 49 and 48 mm Hg, respectively, and a respiratory exchange ratio of 0.8. The approximate alveolar-arterial difference for PO2 (in mm Hg) is

A. 10
B. 20
C. 30
D. 40
E. 50

Answer 36

D is correct.

First, we calculate the ideal alveolar PO2 using the alveolar gas equation. This is PAO2 PIO2 PACO2 / R+F, and we ignore the small factor F. Therefore, the ideal alveolar PO2 = 149 − 48/0.8, that is, 89 mm Hg. How- ever, the arterial PO2 is given as 49 so that the alveolar-arterial difference for
PO2 is 40 mm Hg.

Question 37

The presence of hemoglobin in normal arterial blood increases its oxygen
concentration approximately how many times?

A. 10
B. 30
C. 50
D. 70
E. 90

Answer 37

D is correct.

Normal arterial blood has a PO2 of about 100 mm Hg. The concentration of oxygen in the absence of hemoglobin is the dissolved oxygen, which is 100 × 0.003, or 0.3 ml O2·100 ml−1 blood. However, normal arterial blood contains about 15 g·100 ml−1 of hemoglobin, and each gram can combine with 1.39 ml O2. Since the oxygen saturation of normal arterial blood is about 97%, the total oxygen concentration is given by (1.39 × 15 × 97/100) + 0.3 ml O2·100 ml−1 blood. This is about 20.5 as opposed to the dissolved oxygen concentration of 0.3 ml O2·100 ml−1 blood. Therefore, the presence of hemoglobin increases the oxygen concentration about 70 times.

Question 38

An increase in which of the following increases the O2 affinity of hemoglobin?

A. Temperature
B. PCO2
C. H+ concentration
D. 2,3-DPG
E. Carbon monoxide added to the blood

Answer 38

E is correct.

A small amount of carbon monoxide added to blood increases its oxygen affinity, that is, it causes a leftward shift of the O2 dissociation curve (see Figure 6-2). All the other choices reduce the oxygen affinity of hemoglobin, that is, they shift the dissociation curve to the right (see Figure 6-3).

Question 39

A patient with carbon monoxide poisoning is treated with hyperbaric oxygen that increases the arterial PO2 to 2000 mm Hg. The amount of oxygen dissolved in the arterial blood (in ml·100 ml−1) is

A. 2
B. 3
C. 4
D. 5
E. 6

Answer 39

E is correct.

Since the solubility of oxygen is 0.003 ml O2 ·100 ml−1 blood,
an arterial PO2 of 2000 mm Hg will increase the concentration of the dis- solved oxygen to 6 ml O2 ·100 ml−1 blood. Note that this actually exceeds the normal arterial-venous difference for the oxygen concentration.

Question 40

A patient with severe anemia has normal lungs. You would expect

A. Low arterial PO2.
B. Low arterial O2 saturation.
C. Normal arterial O2 concentration.
D. Low oxygen concentration of mixed venous blood.
E. Normal tissue PO2.

Answer 40

D is correct.

In a patient with severe anemia but normal lungs, the oxygen concentration of arterial blood will be reduced, and therefore, if the cardiac output and oxygen uptake are normal, the oxygen concentration of mixed venous blood will also be reduced. The other choices are incorrect because the arterial PO2 and O2 saturation will be normal if the patient has normal lungs, but of course the arterial oxygen concentration will be reduced, and the tissue PO2 will therefore be abnormally low. Note that a patient with severe anemia usually has some increase in cardiac output, but nevertheless, the oxygen concentration of mixed venous blood will be low. As in all these questions, the one best answer is being sought, and this is clearly D. See Table 6-1 for a summary of these changes.

Question 41

In carbon monoxide poisoning, you would expect

A. Reduced arterial PO2.
B. Normal oxygen concentration of arterial blood.
C. Reduced oxygen concentration of mixed venous blood.
D. O2 dissociation curve is shifted to the right.
E. Carbon monoxide has a distinct odor.

Answer 41

C is correct.

Because the oxygen concentration of arterial blood is reduced, this must also be true of mixed venous blood, other things being equal. The other choices are incorrect. If the patient has normal lungs, the arterial PO2 will be normal, but of course the oxygen concentration of arterial blood will be reduced. Carbon monoxide shifts the O2 dissociation curve to the left, that is, it increases the oxygen affinity of the hemoglobin. Carbon monoxide has no odor, which is one reason why it is so dangerous. See Table 6-1 for the changes.

Question 42

The laboratory reports the following arterial blood gas values in a patient with severe lung disease who is breathing air: PO2 60 mm Hg, PCO2 110 mm Hg, pH 7.20. You conclude

A. The patient has a normal PO2.
B. The patient has a normal PCO2.
C. There is a respiratory alkalosis.
D. There is a partially compensated respiratory alkalosis.
E. The values for PO2 and PCO2 are internally inconsistent.

Answer 42

E is correct.

Since the patient is breathing air, the inspired PO2 is about 149 mm Hg. Using the alveolar gas equation, the alveolarPO2 will be
about 149 − 110, that is, 39 mm Hg for an R value of 1, and even less for an R value of less than 1. This is below the stated arterial PO2 , which can- not be correct. In addition, the other four choices are clearly wrong. The patient does not have a normal PO2 or PCO 2 , and there is an acidosis rather than an alkalosis.

Question 43

Most of the carbon dioxide transported in the arterial blood is in the form of

A. Dissolved.
B. Bicarbonate.
C. Attached to hemoglobin.
D. Carbamino compounds.
E. Carbonic acid.

Answer 43

B is correct.

As the first column of Figure 6-4 shows, about 90% of the CO2 transported in the arterial blood is in the form of bicarbonate. About 5% is dissolved and another 5% is transported as carbamino compounds. The most important of these is carbaminohemoglobin.

Question 44

A patient with chronic lung disease has arterial PO2 and PCO2 values of 50 and 60 mm Hg, respectively, and a pH of 7.35. How is his acid-base status best described?

A. Normal
B. Partially compensated respiratory alkalosis
C. Partially compensated respiratory acidosis
D. Metabolic acidosis
E. Metabolic alkalosis

Answer 44

C is correct.

The abnormally high PCO 2 of 60 mm Hg and the reduced pH of 7.35 are consistent with a partially compensated respiratory acidosis. Figure 6-8A shows that if the PCO 2 rises to 60 mm Hg and there is no renal compensation, the pH is less than 7.3. Therefore, the patient shows some compensation. The fact that the pH has not fully returned to the normal value of 7.4 means that the respiratory acidosis is only partially compensated. The other choices are incorrect because clearly the gas exchange with the high PCO 2 is not normal, there is an acidosis rather than an alkalosis because the pH is reduced, and this is not a metabolic acidosis because the PCO is elevated.

Question 45

The PO2 (in mm Hg) inside skeletal muscle cells during exercise is closest to

A. 3
B. 10
C. 20
D. 30
E. 40

Answer 45

A is correct.

As described in the section titled “Blood-Tissue Gas Exchange,” the PO2 inside skeletal muscle cells is about 3 mm Hg. The blood in the peripheral capillaries has much higher PO2 values in order to enable the diffusion of oxygen to the mitochondria.

Question 46

A patient with chronic pulmonary disease undergoes emergency surgery. Postoperatively, the arterial PO2, PCO2, and pH are 50 mm Hg, 50 mm Hg, and 7.20, respectively. How would the acid-base status be best described?

A. Mixed respiratory and metabolic acidosis
B. Uncompensated respiratory acidosis
C. Fully compensated respiratory acidosis
D. Uncompensated metabolic acidosis
E. Fully compensated metabolic acidosis

Answer 46

A is correct.

There is a respiratory acidosis because thePCO2 is increased to 50 mm Hg and the pH is reduced to 7.20. However, there must be a metabolic component to the acidosis because as Figure 6-8A shows, a PCO2 of 50 will reduce the pH to only about 7.3 if the point moves along the normal blood buffer line. Therefore, there must be a metabolic component to reduce the pH even further. The other choices are incorrect because, as indicated above, an uncompensated respiratory acidosis would give a pH of above 7.3 for thisPCO2 . Clearly, the patient does not have a fully compensated respiratory acidosis because then the pH would be 7.4. There is not an uncompensated metabolic acidosis because the PCO 2 is increased, indicating a respiratory component. Finally, there is not a fully compensated metabolic acidosis because this would give a pH of 7.4.

Question 47

The laboratory provides the following report on arterial blood from a patient: PCO2 32 mm Hg, pH 7.25, HCO3− concentration 25 mmol·liter−1. You conclude that there is

A. Respiratory alkalosis with metabolic compensation.
B. Acute respiratory acidosis.
C. Metabolic acidosis with respiratory compensation.
D. Metabolic alkalosis with respiratory compensation.
E. A laboratory error.

Answer 47

E is correct.

A is incorrect because there is no metabolic compensation. In fact, the bicarbonate concentration is abnormally high. B is incorrect because thePCO2 is low, which is incompatible with a respiratory acidosis. C is incorrect because a metabolic acidosis requires an abnormally low bicarbonate concentration, which this patient does not have. D is incorrect because the patient has an acidosis, not an alkalosis. Therefore, the correct answer can be found by eliminating the other four. How- ever, in addition, Figure 6-8A shows that there is no way that the three given values can coexist on the diagram. Therefore, there must be a
laboratory error.

Question 48

A patient with shortness of breath is breathing air at sea level, and an arterial blood sample shows PO2 90 mm Hg, PCO2 32 mm Hg, pH 7.30. Assuming that the respiratory exchange ratio is 0.8, these data indicate

A. Primary respiratory alkalosis with metabolic compensation.
B. Normal alveolar-arterial PO2 difference.
C. Arterial O2 saturation less than 70%.
D. The sample was mistakenly taken from a vein.
E. Partially compensated metabolic acidosis.

Answer 48

E is correct.

The reduction in the pH to 7.30 with a small reduction in thePCO2 from 40 to 32 is consistent with a partially compensated metabolic acidosis. Compensation is only partial because if it was complete, the pH would be 7.4. The other choices are incorrect. This is not a respiratory alkalosis because the pH is abnormally low. When the alveolar-arterial PO2 difference is calculated using the alveolar gas equation, the alveolar PO2 is about 149 − 32/0.8, that is, 109 mm Hg giving a difference of 109 − 90, or 19 mm Hg. This is abnormally high. The arterial oxygen saturation will be greater than 70% because with a PO2 of 90 mm Hg, the saturation will be above 90% as shown in Figure 6-1. It is true that the reducedPCO2 will shift the curve slightly to the left and the increased hydrogen ion concentration will shift it slightly to the right, but the PO2 is so high that the saturation must be more than 70%. Recall that with a normal oxygen dissociation curve, an arterial PO2 of 40 gives an oxygen saturation of about 75%, so a PO2 of 90 will certainly result in a saturation of over 70%. The sample was not mistakenly taken from a vein because then the PO2 would be very much lower.

Question 49

Concerning contraction of the diaphragm,

A. The nerves that are responsible emerge from the spinal cord at the level of the lower thorax.
B. It tends to flatten the diaphragm.
C. It reduces the lateral distance between the lower rib margins.
D. It causes the anterior abdominal wall to move in.
E. It raises intrapleural pressure.

Answer 49

B is correct.

When the diaphragm contracts, it becomes flatter as shown in Figure 7-1. The other choices are incorrect. The phrenic nerves that innervate the diaphragm come from high in the neck, that is, cervical segments 3, 4, and 5. Contraction of the diaphragm causes the lateral distance between the lower rib margins to increase and anterior abdominal wall to move out as also shown in Figure 7-1. The intrapleural pressure is reduced because the larger volume of the chest cage increases the recoil pressure of the lung.

Question 50

Concerning the pressure-volume behavior of the lung,

A. Compliance decreases with age.
B. Filling an animal lung with saline decreases compliance.
C. Removing a lobe reduces total pulmonary compliance.
D. Absence of surfactant increases compliance.
E. In the upright lung at FRC, for a given change in intrapleural pressure, the
alveoli near the base of the lung expand less than those near the apex.

Answer 50

C is correct.

If there is less lung, the total change in volume per unit change in pressure will be reduced. The other choices are incorrect. Compliance increases with age, filling a lung with saline increases compliance (Figure 7-5), absence of surfactant decreases compliance, and in the upright lung at FRC, inspiration causes a larger increase in volume of the alveolar at the base of the lung compared with those near the apex (Figure 7-8).

Question 51

Two bubbles have the same surface tension, but bubble X has 3 times the diameter of bubble Y. The ratio of the pressure in bubble X to that in bubble Y is

A. 0.3:1
B. 0.9:1
C. 1:1
D. 3:1
E. 9:1

Answer 51

A is correct.

The Laplace relationship shown in Figure 7-4C states that the pressure is inversely proportional to the radius for the same surface tension. Since bubble X has three times the radius of bubble Y, the ratio of pressures will be approximately 0.3:1.

Question 52

Pulmonary surfactant is produced by

A. Alveolar macrophages. B. Goblet cells.
C. Leukocytes.
D. Type I alveolar cells.
E. Type II alveolar cells.

Answer 52

E is correct.

Surfactant is produced by type II alveolar epithelial cells as discussed in relation to Figure 7-6.

Question 53

The basal regions of the upright human lung are normally better ventilated than
the upper regions because

A. Airway resistance of the upper regions is higher than of the lower regions.
B. There is less surfactant in the upper regions.
C. The blood flow to the lower regions is higher.
D. The lower regions have a small resting volume and a relatively large increase in volume.
E. The PCO2 of the lower regions is relatively high.

Answer 53

D is correct.

As Figure7-8 shows, the lower regions of the lung have a relatively small resting volume and large increase in volume compared with those near the top of the lung. The other choices are incorrect. The airway resistance of the upper regions is probably somewhat less than that of the lower regions because the parenchyma is better expanded there. However, in any event, this is not the explanation of the difference in ventilation. There is no evidence that there is less surfactant in the upper regions of the lung. It is true that the blood flow to the lower regions is higher than to the upper regions, but this is not relevant here. It is also true that the PCO 2 of the lower regions is relatively high compared with the upper regions, but this is not the explanation of the difference in ventilation.

Question 54

Pulmonary surfactant

A. Increases the surface tension of the alveolar lining liquid.
B. Is secreted by type I alveolar epithelial cells.
C. Is a protein.
D. Increases the work required to expand the lung.
E. Helps to prevent transudation of fluid from the capillaries into the alveolar spaces.

Answer 54

E is correct.

The presence of surfactant reduces the surface tension of the alveolar lining layer and therefore the inward pull of the alveolar wall (Figure 7-4B). This in turn means that the hydrostatic pressure in the interstitium around the capillaries is less negative when surfactant is pre- sent. As a result, this helps to prevent transudation of fluid from the capillaries into the interstitium or into the alveolar spaces. The other choices are incorrect. Surfactant decreases the surface tension of the alveolar lining liquid, it is secreted by type II alveolar epithelial cells, it is a phospholipid, and it decreases the work required to expand the lung.

Question 55

Concerning normal expiration during resting conditions,

A. Expiration is generated by the expiratory muscles.
B. Alveolar pressure is less than atmospheric pressure.
C. Intrapleural pressure gradually falls (becomes more negative) during the
expiration.
D. Flow velocity of the gas (in cm·s−1) in the large airways exceeds that in the
terminal bronchioles.
E. Diaphragm moves down as expiration proceeds.

Answer 55

D is correct.

The velocity of the gas in the large airways exceeds that in the terminal bronchioles because the latter have a very large combined cross-sectional area (see Figure 1-5). The other choices are incorrect. Under resting conditions, expiration is passive, it is associated with an alveolar pressure that exceeds atmospheric pressure, intrapleural pressure gradually increases (becomes less negative) during expiration, and the diaphragm moves up as expiration proceeds.

Question 56

When a normal subject develops a spontaneous pneumothorax of his right lung, you would expect the following to occur:

A. Right lung contracts.
B. Chest wall on the right contracts.
C. Diaphragm on the right moves up.
D. Mediastinum moves to the right.
E. Blood flow to the right lung increases.

Answer 56

A is the correct answer.

Spontaneous pneumothorax of the right lung will decrease its volume because the normal expanding pressure is abolished. All the other choices are incorrect. The increase in pressure on the right will cause the chest wall on that side to expand, the diaphragm to move down, and the mediastinum to shift to the left. The blood flow to the right lung will be reduced both because its volume is small and also there is hypoxic pulmonary vasoconstriction.

Question 57

According to Poiseuille’s law, reducing the radius of an airway to one-third will
increase its resistance how many fold?

A. 1/3
B. 3
C. 9
D. 27
E. 81

Answer 57

E is correct.

Poiseuille’s law states that during laminar flow, airway resistance is inversely proportional to the 4th power of the radius, other things
Answers 189 being equal. Therefore, a reduction in the radius by a factor of 3 increases the resistance by 34, that is, 81.

Question 58

Concerning airflow in the lung,

A. Flow is more likely to be turbulent in small airways than in the trachea.
B. The lower the viscosity, the less likely is turbulence to occur.
C. In pure laminar flow, halving the radius of the airway increases its resistance
eightfold.
D. For inspiration to occur, mouth pressure must be less than alveolar pressure.
E. Airway resistance increases during scuba diving.

Answer 58

E is correct.

During scuba diving, the density of the air is increased because of the raised pressure, and therefore, airway resistance rises. The other choices are incorrect. Flow is most likely to be turbulent in large airways; the higher the viscosity, the less likely is turbulence to occur; halving the radius of the airway increases its resistance 16-fold; and during inspiration, alveolar pressure must be less than mouth pressure.

Question 59

The most important factor limiting flow rate during most of a forced expiration from total lung capacity is

A. Rate of contraction of expiratory muscles.
B. Action of diaphragm.
C. Constriction of bronchial smooth muscle.
D. Elasticity of chest wall.
E. Compression of airways.

Answer 59

E is correct.

During most of a forced expiration from TLC, dynamic compression of the airways limits flow (Figures 7-16 to 7-18). All the other choices are incorrect. In particular, flow is independent of effort.

Question 60

Which of the following factors increases the resistance of the airways?

A. Increasing lung volume above FRC
B. Increased sympathetic stimulation of airway smooth muscle
C. Going to high altitude
D. Inhaling cigarette smoke
E. Breathing a mixture of 21% O2 and 79% helium (molecular weight 4)

Answer 60

D is correct.

Inhalation of cigarette smoke causes reflex constriction of airway smooth muscle as a result of stimulation of irritant receptors in the airway wall (see Chapter 8). The other choices are incorrect. Both increasing lung volume above FRC and sympathetic stimulation of airway smooth muscle reduce airway resistance. Going to high altitude does the same because the density of the air is reduced. The density is also decreased when nitrogen is replaced by helium in the inspired gas.

Question 61

A normal subject makes an inspiratory effort against a closed airway. You would expect the following to occur:

A. Tension in the diaphragm decreases.
B. The internal intercostal muscles become active.
C. Intrapleural pressure increases (becomes less negative).
D. Alveolar pressure falls more than intrapleural pressure.
E. Pressure inside the pulmonary capillaries falls.

Answer 61

E is correct.

When an inspiratory effort is made against a closed airway, all the pressures inside the thorax fall including the pulmonary vascular pressures. The other choices are incorrect. During inspiration, the ten- sion in the diaphragm increases, external not internal intercostal muscles become active, intrapleural pressure becomes more negative, and alveolar pressure will fall equally with intrapleural pressure if lung volume does not change. If lung volume does increase slightly, intrapleural pressure will fall more than alveolar pressure.

Question 62

Concerning the respiratory centers,

A. The normal rhythmic pattern of breathing originates from neurons in the motor area of the cortex.
B. During quiet breathing, expiratory neurons fire actively.
C. Impulses from the pneumotaxic center can stimulate inspiratory activity.
D. The cortex of the brain can override the function of the respiratory centers. E. The only output from the respiratory centers is via the phrenic nerves.

Answer 62

D is correct.

The cortex can override the function of the respiratory centers, for example, during voluntary hyperventilation, or voluntary breath- holding. The other choices are incorrect. The normal rhythmic pattern of breathing originates in the brainstem, not the cortex. Expiration is passive during quiet breathing, impulses from the pneumotaxic center inhibit inspiration, and the output from the respiratory centers includes impulses from the spinal cord to the intercostal and other muscles in addition to the phrenic nerves.

Question 63

Concerning the central chemoreceptors,

A. They are located near the dorsal surface of the medulla.
B. They respond to both the PCO2 and the PO2 of the blood.
C. They are activated by changes in the pH of the surrounding extracellular fluid.
D. For a given rise in PCO2, the pH of cerebrospinal fluid falls less than that of blood.
E. The bicarbonate concentration of the CSF cannot affect their output.

Answer 63

C is correct (see Figure 8-2).

The other choices are incorrect. The central chemoreceptors are located near the ventral surface of the medulla; they do not respond to the PO2 of blood; for a given rise in PCO 2 , the CSF pH falls more than that of blood because the CSF has less buffering; and the bicarbonate concentration of the CSF can affect the output of the central chemoreceptors by buffering the changes in pH.

Question 64

Concerning the peripheral chemoreceptors,

A. They respond to changes in the arterial PO2 but not pH.
B. Under normoxic conditions, the response to changes in PO2 is very small.
C. The response to changes in PCO2 is slower than for central chemoreceptors.
D. They are the most important receptors causing an increased ventilation in
response to a rise in PCO2.
E. They have a low blood flow per gram of tissue.

Answer 64

B is correct.

The peripheral chemoreceptors are responsive to the arterialPO2 , but during normoxia, the response is small (see Figure 8-3B). The other choices are incorrect. Peripheral chemoreceptors do respond to changes in blood pH, the response to changes in PCO2 is faster than is the case for central chemoreceptors, the central chemoreceptors are more important than the peripheral chemoreceptors in the ventilatory response to increased CO2, and peripheral chemoreceptors have a very high blood flow in relation to their mass.

Question 65

Concerning the ventilatory response to carbon dioxide,

A. It is increased if the alveolar PO2 is raised.
B. It depends only on the central chemoreceptors.
C. It is increased during sleep.
D. It is increased if the work of breathing is raised.
E. It is a major factor controlling the normal level of ventilation.

Answer 65

E is correct.

The normal level of ventilation is controlled by the ventilatory response to CO2. The other choices are incorrect. The ventilatory response to CO2 is increased if the alveolar PO2 is reduced, the ventilatory response depends on the peripheral chemoreceptors in addition to the central chemoreceptors, and the ventilatory response is reduced during sleep and if the work of breathing is increased.

Question 66

Concerning the ventilatory response to hypoxia,

A. It is the major stimulus to ventilation at high altitude.
B. It is primarily brought about by the central chemoreceptors.
C. It is reduced if the PCO2 is also raised.
D. It rarely stimulates ventilation in patients with chronic lung disease.
E. It is important in mild carbon monoxide poisoning.

Answer 66

A is correct.

Ventilation increases greatly at high altitude in response to hypoxic stimulation of chemoreceptors. The other choices are incorrect. It is the peripheral chemoreceptors, not the central chemoreceptors that are responsible for the response. The response is increased if thePCO2 is also raised. Hypoxic stimulation is often important in patients with long- standing severe lung disease who have nearly normal values for the pH of the CSF and blood. Mild carbon monoxide poisoning is associated with a normal arterial PO2 , and therefore, there is no stimulation of the peripheral chemoreceptors.

Question 67

The most important stimulus controlling the level of resting ventilation is

A. PO2 on peripheral chemoreceptors.
B. PCO2 on peripheral chemoreceptors.
C. pH on peripheral chemoreceptors.
D. pH of CSF on central chemoreceptors.
E. PO2 on central chemoreceptors.

Answer 67

D is correct.

As Figure 8-2 shows, the most important stimulus comes from the pH of the CSF on the central chemoreceptors. The other choices are incorrect. The effect of PO2 on the peripheral chemoreceptors under normoxic conditions is very small. Changes inPCO2 do affect the peripheral chemoreceptors, but the magnitude is less than that for the central chemoreceptors. The effect of changes in pH on peripheral chemoreceptors under normal conditions is small, and changes in PO2 do not affect the central chemoreceptors.

Question 68

Exercise is one of the most powerful stimulants to ventilation. It primarily works by way of

A. Low arterial PO2.
B. High arterial PCO2.
C. Low PO2 in mixed venous blood.
D. Low arterial pH.
E. None of the above.

Answer 68

E is correct.

Moderate exercise does not reduce the arterial PO2 , increase the arterial PCO 2 , or reduce the arterial pH. The PO2 of mixed venous blood does fall, but there are no known chemoreceptors that are stimulated as a result.

Question 69

Concerning the Hering-Breuer inflation reflex,

A. The impulses travel to the brain via the carotid sinus nerve.
B. It results in further inspiratory efforts if the lung is maintained inflated.
C. It is seen in adults at small tidal volumes.
D. It may help to inflate the newborn lung.
E. Abolishing the reflex in many animals causes rapid, shallow breathing.

Answer 69

D is correct.

The other choices are incorrect. The impulses travel to the brain via the vagus nerve, the reflex inhibits further inspiratory efforts if the lung is maintained inflated, the reflex is not seen in adults at small tidal volumes, and abolishing the reflex by cutting the vagal nerves in experimental animals causes slow deep breathing.

Question 70

Concerning exercise,

A. It can increase the oxygen consumption more than tenfold compared with rest.
B. The measured respiratory exchange ratio cannot exceed 1.0.
C. Ventilation increases less than cardiac output.
D. At low levels of exercise, blood lactate concentrations typically rapidly
increase.
E. The change in ventilation on exercise can be fully explained by the fall in
arterial pH.

Answer 70

A is correct.

In some elite athletes, oxygen consumption can increase 15-fold or even 20-fold. The other choices are incorrect. The measured R value can exceed 1 at high levels of exercise because lactic acid is produced and there are very high levels of ventilation. Ventilation increases much more than cardiac output (Figure 9-13), and at low levels of exercise, little or no lactate is normally produced. During moderate levels of exercise, there is essentially no change in pH.

Question 71

Concerning acclimatization to high altitude,

A. Hyperventilation is of little value.
B. Polycythemia occurs rapidly.
C. There is a rightward shift of the O2 dissociation curve at extreme altitudes.
D. The number of capillaries per unit volume in skeletal muscle falls.
E. Changes in oxidative enzymes occur inside muscle cells.

Answer 71

E is correct.

There is a rise in oxidative enzymes in muscle cells that assists acclimatization. The other choices are incorrect. Hyperventilation is the most important feature of acclimatization, polycythemia occurs slowly, there is a leftward shift of the O2 dissociation curve at extreme altitude because of the respiratory alkalosis, and the number of capillaries per unit volume of skeletal muscle increases with acclimatization.

Question 72

If a small airway in a lung is blocked by mucus, the lung distal to this may become atelectatic. Which of the following statements is true?

A. Atelectasis occurs faster if the person is breathing air rather than oxygen.
B. The sum of the gas partial pressures in mixed venous blood is less than in
arterial blood during air breathing.
C. The blood flow to the atelectatic lung will rise.
D. The absorption of a spontaneous pneumothorax is explained by a different
mechanism.
E. The elastic properties of the lung strongly resist atelectasis.

Answer 72

B is correct (see Figure 9-4 for a full explanation).

The other choices are incorrect. Atelectasis occurs faster during oxygen breathing than air breathing, blood flow to an atelectatic lung is reduced because of the low lung volume and perhaps hypoxic pulmonary vasoconstriction, the absorption of a spontaneous pneumothorax can be explained by the same mechanism, and the elastic properties of the lung have little effect in resisting atelectasis caused by gas absorption.

Question 73

If helium-oxygen mixtures rather than nitrogen-oxygen mixtures (with the same oxygen concentration) are used for very deep diving,

A. Risk of decompression sickness is reduced.
B. Work of breathing is increased.
C. Airway resistance is increased.
D. Risk of O2 toxicity is reduced.
E. Risk of inert gas narcosis is increased.

Answer 73

A is correct

because decompression sickness is caused by bubbles of gas, and helium is less soluble than nitrogen. The other choices are incorrect. The work of breathing and the airway resistance are both decreased. The risk of O2 toxicity is unchanged, but the risk of inert gas narcosis is decreased.

Question 74

If a seated astronaut makes the transition from 1G to 0G, which of the following
decreases?

A. Blood flow to the apex of the lung
B. Ventilation to the apex of the lung
C. Deposition of inhaled aerosol particles
D. Thoracic blood volume
E. Pco2 in the alveoli at the apex of the lung

Answer 74

C is correct.

In zero G, the deposition of inhaled particles by sedimentation is abolished. The other choices are incorrect. Both blood flow and ventilation to the apex of the lung are increased because the normal effects of gravity are abolished (see Figures. 2-7, 4-7, and 5-8). Thoracic blood volume increases because blood no longer pools in dependent regions of the body as a result of gravity. ThePCO2 at the apex of the lung increases because the abolition of gravity results in a reduction of the VA/Q at the apex (see Figure 5-10).

Question 75

Which of the following increases by the largest percentage at maximal exercise compared with rest?

A. Heart rate
B. Alveolar ventilation
C. PCO2 of mixed venous blood
D. Cardiac output
E. Tidal volume

Answer 75

B is correct.

Alveolar ventilation like total ventilation can increase by a factor of 10 or more. The other choices are incorrect. Heart rate, cardiac output, and the PCO2 of mixed venous blood increase much less. Also, tidal volume increases much less because part of the increase in alveolar ventilation is caused by the increase in respiratory frequency.

Question 76

The transition from placental to pulmonary gas exchange is accompanied by

A. Reduced arterial PO2.
B. Rise of pulmonary vascular resistance.
C. Closure of the ductus arteriosus.
D. Increased blood flow through the foramen ovale.
E. Weak respiratory efforts.

Answer 76

C is correct.

The ductus arteriosus closes (see the discussion of Figure 9-5). There is a big increase in arterial PO2 , a large fall in pulmonary vascular resistance, a decreased blood flow through the foramen ovale, and very large inspiratory efforts.

Question 77

Concerning the 1-second forced expiratory volume,

A. The test can be used to assess the efficacy of bronchodilators.
B. It is unaffected by dynamic compression of the airways.
C. It is reduced in patients with pulmonary fibrosis but not chronic obstructive
pulmonary disease.
D. It is normal in patients with asthma.
E. The test is difficult to perform.

Answer 77

A is correct.

Bronchodilators reduce airway resistance, and their efficacy can therefore be assessed by this test. The other choices are incorrect. Dynamic compression of the airways is the main factor limiting maximal expiratory flow, the flow is greatly reduced in chronic obstructive pulmonary disease but may be normal or even increased in pulmonary fibrosis, it is reduced in patients with asthma, and it is easy to perform.

Question 78

The following may reduce the FEV1 in a patient with chronic obstructive
pulmonary disease:

A. Hypertrophy of the diaphragm.
B. Administration of a bronchodilator drug.
C. Increased expiratory effort.
D. Loss of radial traction on the airways.
E. Increased elastic recoil of the lung.

Answer 78

D is correct.

Loss of radial traction is one of the factors contributing to dynamic compression of the airways in COPD. The other choices are incorrect. The action of the diaphragm does not affect dynamic compression; if a bronchodilator drug is effective, it may increase the FEV; the flow is independent of expiratory effort; and increased elastic recoil does not occur in COPD although if it did, this could increase the FEV.

Question 79

Concerning the single-breath nitrogen test for uneven ventilation,

A. The slope of the alveolar plateau is reduced in chronic bronchitis compared with normal.
B. The slope occurs because well-ventilated units empty later in expiration than poorly ventilated units.
C. The last exhaled gas comes from the base of the lung.
D. A similar procedure can be used to measure the anatomic dead space.
E. The test is very time consuming.

Answer 79

D is correct (see discussion of Figure 2-6).

The other choices are incorrect. The slope of the alveolar plateau is increased in chronic bronchitis because poorly ventilated units empty later in expiration than well-ventilated units. The last exhaled gas comes from apex of the lung because of airway closure at the base, and the test is not very time consuming.

Question 80

In the assessment of ventilation-perfusion inequality based on measurements of PO2 and PCO2 in arterial blood and expired gas,

A. The ideal alveolar PO2 is calculated using the expired PCO2.
B. The alveolar PO is calculated from the alveolar gas equation.
C. Va / Q inequality reduces the alveolar-arterial PO2 difference.
D. Va / Q inequality reduces the physiologic shunt.
E. VA / Q inequality reduces the physiologic dead space.

Answer 80

B is correct (see the Discussion under “Measurement of Ventilation- Perfusion Inequality” in Chapter 5).

The other choices are incorrect. The ideal alveolar Po2 is calculated using the arterial PCO 2 , and VA/Q inequality increases the alveolar-arterial Po2 difference, the physiologic shunt, and the physiologic dead space.

Question 81

If a seated normal subject exhales to residual volume (RV),

A. The volume of gas remaining in the lung is more than half of the vital capacity.
B. The PCO2 of the expired gas falls just before the end of expiration.
C. If the mouthpiece is closed at RV and the subject completely relaxes, the
pressure in the airways is greater than atmospheric pressure.
D. Intrapleural pressure exceeds alveolar pressure at RV.
E. All small airways in the lung are closed at RV.

Answer 81

B is correct.

Near the end of the expiration, the expired gas comes preferentially from the apex of the lung because of airway closure at the base (see Figure 7-9). The apex of the lung has a relatively low PCO2 (see Figure 5-10). The other choices are incorrect. The residual volume is much less than half of the vital capacity; if the airway is obstructed at RV and the subject relaxes, the pressure in the airways is less than atmospheric pressure (see Figure 7-11); intrapleural pressure is always less than alveolar pressure; and only the airways near the base of the lung are closed at residual volume (see Figure 7-9).