1. The answer is E [I A 4, 5, B 2,3, 5]. Residual volume (RV) cannot be measured by spirometry. Therefore, any lung volume or capacity that includes the RV cannot be measured by spirometry. Measurements that include R V are functional residual capacity (FRC) and total lung capacity (TLC). Vital capacity (VC) does not include RV and is, therefore, measurable by spirometry. Physiologic dead space is not measurable by spirometry and requires sampling of arterial Pco2 and expired C02.
2. The answer is B [II D 2], Neonatal respiratory distress syndrome is caused by lack of adequate surfactant in the immature lung. Surfactant appears between the twenty-fourth and the thirty-fifth gestational week. In the absence of surfactant, the surface tension of the small alveoli is too high. When the pressure on the small alveoli is too high (P = 2T/r), the small alveoli collapse into larger alveoli. There is decreased gas exchange with the larger, collapsed alveoli, and ventilation/ perfusion (V/Q) mismatch, hypoxemia, and cyanosis occur. The lack of surfactant also decreases lung compliance, making it harder to inflate the lungs, increasing the work of breathing, and producing dyspnea (shortness of breath). Generally, lecithin:sphingomyelin ratios greater than 2:1 signify mature levels of surfactant.
3. The answer is B [VI C]. Pulmonary blood flow is controlled locally by the Po2 of alveolar air. Hypoxia causes pulmonary vasoconstriction and thereby shunts blood away from unventilated areas of the lung, where it would be wasted. In the coronary circulation, hypoxemia causes vasodilation. The cerebral, muscle, and skin circulations are not controlled directly by Po2.
4. The answer is D [VIII B 2 a]. The patient's arterial Pco2 is lower than the normal value of 40 mm Hg because hypoxemia has stimulated peripheral chemoreceptors to increase his breathing rate; hyperventilation causes the patient to blow off extra C02 and results in respiratory alkalosis. In an obstructive disease, such as asthma, both forced expiratory volume (FEV,) and forced vital capacity (FVC) are decreased, with the larger decrease occurring in FEVi. Therefore, the FEV/ FVC ratio is decreased. Poor ventilation of the affected areas decreases the ventilation/perfusion (V/Q) ratio and causes hypoxemia. The patient's residual volume (RV) is increased because he is breathing at a higher lung volume to offset the increased resistance of his airways.
5. The answer is C [II E 3 a (2)]. The cause of airway obstruction in asthma is bronchiolar constriction. (32-adrenergic stimulation ((^-adrenergic agonists) produces relaxation of the bronchioles.
6. The answer is E [II F 2]. During inspiration, intrapleural pressure becomes more negative than it is at rest or during expiration (when it returns to its less negative resting value). During inspiration, air flows into the lungs when alveolar pressure becomes lower (due to contraction of the diaphragm) than atmospheric pressure; if alveolar pressure were not lower than atmospheric pressure, air would not flow inward. The volume in the lungs during inspiration is the functional residual capacity (FRC) plus one tidal volume (TV).
7. The answer is D [IX B; Table 4-8]. At high altitudes, the Po2 of alveolar air is decreased because barometric pressure is decreased. As a result, arterial Po2 is decreased(< 100 mm Hg), and hypoxemia occurs and causes hyperventilation by an effect on peripheral chemoreceptors. Hyperventilation leads to respiratory alkalosis. 2,3-Diphosphoglycerate (DPG) levels increase adap-tively; 2,3-DPG binds to hemoglobin and causes the hemoglobin-02 dissociation curve to shift to the right to improve unloading of 02 in the tissues. The pulmonary vasculature vasoconstricts in response to hypoxia, resulting in increased pulmonary arterial pressure and hypertrophy of the right ventricle (not the left ventricle).
8. The answer is G [I A 3; Figure 4-1], Expiratory reserve volume (ERV) equals vital capacity (VC) minus inspiratory capacity. [Inspiratory capacity includes tidal volume (TV) and inspiratory reserve volume (IRV)]. Residual volume (RV) is not needed for this calculation.
9. The answer is C [VI B], The distribution of blood flow in the lungs is affected by gravitational effects on arterial hydrostatic pressure. Thus, blood flow is highest at the base, where arterial hydrostatic pressure is greatest and the difference between arterial and venous pressure is also greatest. This pressure difference drives the blood flow.
10. The answer is D [II C 2; Figure 4-3], By convention, when airway pressure is equal to atmospheric pressure, it is designated as zero pressure. Under these equilibrium conditions, there is no airflow because there is no pressure gradient between the atmosphere and the alveoli, and the volume in the lungs is the ftmctional residual capacity (FRC). The slope of each curve is compliance, not resistance; the steeper the slope, the greater the volume change for a given pressure change, or the greater compliance. The compliance of the lungs alone or the chest wall alone is greater than that of the combined lung-chest wall system (the slopes of the individual curves are steeper than the slope of the combined curve, which means higher compliance). When airway pressure is zero (equilibrium conditions), intrapleural pressure is negative because of the opposing tendencies of the chest wall to spring out and the lungs to collapse.
11. The answer is C [II E 4], The medium-sized bronchi actually constitute the site of highest resistance along the bronchial tree. Although the small radii of the alveoli might predict that they would have the highest resistance, they do not because of their parallel arrangement. In fact, early changes in resistance in the small airways may be "silent" and go undetected because of their small overall contribution to resistance.
12. The answer is D [VII B 2]. Alveolar Po2 in the left lung will equal the Po2 in inspired air. Because there is no blood flow to the left lung, there can be no gas exchange between the alveolar air and the pulmonary capillary blood. Consequently, 02 is not added to the capillary blood. The ventilation/perfusion (V/Q) ratio in the left lung will be infinite (not zero or lower than that in the normal right lung) because Q (the denominator) is zero. Systemic arterial Po2 will, of course, be decreased because the left lung has no gas exchange. Alveolar Po2 in the right lung is unaffected.
13. The answer is C [IV C 1; Figure 4-7]. Strenuous exercise increases the temperature and decreases the pH of skeletal muscle; both effects would cause the hemoglobin-02 dissociation curve to shift to the right, making it easier to unload 02 in the tissues to meet the high demand of the exercising muscle. 2,3-Diphosphoglycerate (DPG) binds to the [} chains of adult hemoglobin and reduces its affinity for 02, shifting the curve to the right. In fetal hemoglobin, the p chains are replaced by y chains, which do not bind 2,3-DPG, so the curve is shifted to the left. Because carbon monoxide (CO) increases the affinity of the remaining binding sites for 02, the curve is shifted to the left.
14. The answer is A [IV C 1; Figure 4-7]. A shift to the right of the hemoglobin-02 dissociation curve represents decreased affinity of hemoglobin for 02. At any given Po2, the percent saturation is decreased, the P-„ is increased (read the Po2 from the graph at 50% hemoglobin saturation), and unloading of 02 in the tissues is facilitated. The 02-carrying capacity of hemoglobin is determined by the hemoglobin concentration and is unaffected by the shift from curve A to curve B.
15. The answer is D [V B], In venous blood, C02 combines with H20 and produces the weak acid H2C03, catalyzed by carbonic anhydrase. The resulting H+ is buffered by deoxyhemo-globin, which is such an effective buffer for H+ (meaning that the pK is within 1.0 unit of the pH of blood) that the pH of venous blood is only slightly more acid than the pH of arterial blood. Oxyhemoglobin is a less effective buffer than deoxyhemoglobin.
16. The answer is B [VI A]. Blood flow (or cardiac output) in the systemic and pulmonary circulations is nearly equal; pulmonary flow is slightly less than systemic flow because about 2% of the systemic cardiac output bypasses the lungs. The pulmonary circulation is characterized by both lower pressure and lower resistance than the systemic circulation, so flows through the two circulations are approximately equal (flow = pressure/resistance).
17. The answer is D [I A 5 b, 6 b]. Alveolar ventilation is the difference between tidal volume (TV) and dead space multiplied by breathing frequency. TV and breathing frequency are given, but dead space must be calculated. Dead space is TV multiplied by the difference between arterial Pco2 and expired Pco2 divided by arterial Pco2. Thus: dead space = 0.45 x (41 - 35/41) = 0.066 L. Alveolar ventilation is then calculated as: (0.45 L - 0.066 L) x 16 breaths/min = 6.14 L/min.
18. The answer is B [VII C; Figure 4-10; Table 4-5]. Ventilation and perfusion of the lung are not distributed uniformly. Both are lowest at the apex and highest at the base. However, the differences for ventilation are not as great as for perfusion, making the ventilation/perfusion (V/Q) ratios higher at the apex and lower at the base. As a result, gas exchange is more efficient at the apex and less efficient at the base. Therefore, blood leaving the apex will have a higher Po2 and a lower Pco2 because it is better equilibrated with alveolar air.
19. The answer is E [VIII B 2], Hypoxemia stimulates breathing by a direct effect on the peripheral chemoreceptors in the carotid and aortic bodies. Central (medullary) chemoreceptors are stimulated by C02 (or H+). The J receptors and lung stretch receptors are not chemoreceptors. The phrenic nerve innervates the diaphragm, and its activity is determined by the output of the brain stem breathing center.
20. The answer is A [IX A]. During exercise, the ventilation rate increases to match the increased 02 consumption and C02 production. This matching is accomplished without a change in mean arterial Po2 or Pco2. Venous Pco2 increases because extra C02 is being produced by the exercising muscle. Because this C02 will be blown off by the hyperventilating lungs, it does not increase the arterial Pco2. Pulmonary blood flow (cardiac output) increases manyfold during strenuous exercise.
21. The answer is B [VII B 1]. If an area of lung is not ventilated, there can be no gas exchange in that region. The pulmonary capillary blood serving that region will not equilibrate with alveolar Po2, but will have a Po2 equal to that of mixed venous blood.
22. The answer is A [V B; Figure 4-9], C02 generated in the tissues is hydrated to form H+ and HCO:-f in red blood cells (RBCs). H+ is buffered inside the RBCs by deoxyhemoglobin, which acidifies the RBCs. HC03" leaves the RBCs in exchange for CP and is carried to the lungs in the plasma. A small amount of C02 (not HC03~) binds directly to hemoglobin (carbaminohemoglobin).
23. The answer is E [I B 2]. During normal breathing, the volume inspired and then expired is a tidal volume (TV). The volume remaining in the lungs after expiration of a TV is the functional residual capacity (FRC).
24. The answer is D [I A 4], During a forced maximal expiration, the volume expired is a tidal volume (TV) plus the expiratory reserve volume (ERV). The volume remaining in the lungs is the residual volume (RV).
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