Answers to Quiz 105 · Alveolar Ventilation and Gas Exchange in Lungs
  1. In the single-breath nitrogen test, the subject breathes out after a breath of pure O₂. Why does the first N₂ in the expired air mark the end of the dead space?

    Answer: The dead space holds only O₂, so any N₂ must be alveolar gas

    The breath of pure O₂ fills the conducting airways with O₂, and N₂ is left only in the alveoli. So N₂ appearing in the curve means alveolar air has arrived. Revise dead space

  2. On the single-breath nitrogen curve, the change from dead-space air to alveolar air is S-shaped rather than sharp because:

    Answer: Some alveolar gas mixes with the dead-space gas

    Mixing blurs the boundary, so an equivalent sharp boundary is drawn instead, placed so that the N₂ on its two sides balances. Revise dead space

  3. When the nitrogen curve is analysed with a planimeter, the dead space is found from:

    Answer: The dead-space area as a fraction of the total area, applied to the volume expired

    The planimeter measures the dead-space area and the whole area, which stands for the volume expired (VE). Dead space then follows by simple proportion. Revise dead space

  4. Tidal volume is 500 mL, dead space 150 mL and respiratory rate 12/min. What is the alveolar ventilation?

    Answer: 4200 mL/min

    (500 − 150) × 12 = 4200 mL/min. 6000 mL/min is the minute ventilation, which still includes the dead-space air, and 1800 mL/min is the dead-space ventilation. Revise alveolar ventilation

  5. Two people have the same minute ventilation. One breathes rapidly and shallowly, the other slowly and deeply. The rapid, shallow breather has:

    Answer: Lower alveolar ventilation, with hypoxia and hypercapnia

    The dead space takes the same volume out of every breath, so small breaths leave little for the alveoli. Your book says rapid, shallow breathing causes hypoxia and hypercapnia. Revise alveolar ventilation

  6. In the pulmonary function laboratory, alveolar ventilation is calculated as:

    Answer: CO₂ expired per minute ÷ fractional CO₂ in alveolar gas

    All expired CO₂ comes from the alveoli, so V̇ECO₂ = VA × FACO₂, which gives VA = V̇ECO₂ ÷ FACO₂. FACO₂ is sampled from end-tidal air. Revise alveolar ventilation

  7. If alveolar ventilation is halved, the alveolar PCO₂:

    Answer: Doubles

    Alveolar ventilation and PCO₂ are inversely related, so halving VA doubles the PCO₂. A rise in PaCO₂ is the sign of hypoventilation. Revise alveolar ventilation

  8. Which of these causes peripheral rather than central hypoventilation?

    Answer: Myasthenia gravis

    Peripheral hypoventilation is failure of the ventilatory apparatus, here weak respiratory muscles. The other three depress the respiratory centres, which is central hypoventilation. Revise alveolar ventilation

  9. The normal thickness of the alveolar-capillary membrane is:

    Answer: 0.2–0.5 µ

    0.2–0.5 µ. This thinness is what lets O₂ and CO₂ diffuse across it so easily. Revise gas exchange in the lungs

  10. Which of these is NOT one of the six layers of the alveolar-capillary membrane itself?

    Answer: Red cell membrane

    The six run from surfactant to capillary endothelium. Plasma, the red cell membrane, the fluid inside the red cell and Hb are the four extra layers O₂ crosses to reach Hb. Revise gas exchange in the lungs

  11. In your book's table of partial pressures, venous blood has a PO₂ and PCO₂ of about:

    Answer: 40 and 46 mmHg

    Venous blood: PO₂ 40, PCO₂ 46 mmHg. 95 and 40 are arterial blood, 100 and 40 alveolar air, and 116 and 32 expired air. Revise gas exchange in the lungs

  12. The diffusion gradient for O₂ across the alveolar-capillary membrane is about:

    Answer: 60 mmHg

    About 60 mmHg: alveolar PO₂ (100) minus the PO₂ of venous blood entering the capillaries (40). 6 mmHg is the gradient for CO₂. Revise gas exchange in the lungs

  13. A patient with heart failure develops interstitial edema of the lungs. Which blood-gas picture does this diffusion defect typically produce?

    Answer: Low PO₂ with little change in PCO₂

    CO₂ is 20 times more soluble in water than O₂, so the extra fluid slows O₂ but hardly slows CO₂. The result is hypoxemia without significant change in PCO₂. Revise gas exchange in the lungs

  14. Why is carbon monoxide the gas used to measure the lung's diffusing capacity?

    Answer: Its uptake is diffusion-limited, and its capillary partial pressure stays near zero

    Hb binds CO about 210 times more strongly than O₂, which keeps capillary PCO near zero and makes uptake depend on diffusion alone. Venous blood has essentially no CO. Revise gas exchange in the lungs

  15. The uptake of nitrous oxide (N₂O) from the alveoli is flow-limited because:

    Answer: Blood equilibrates with alveolar N₂O within 0.1 s

    N₂O crosses so fast that the blood is equilibrated within 0.1 s, so only more blood flow can raise its uptake. Never equilibrating within 0.75 s describes CO, the diffusion-limited gas. Revise gas exchange in the lungs

15 questions on chapter 106. Pick one answer for each, then save. Your answers stay in this browser, and the right ones are shown at the top of the next quiz.

1. The mean pressure in the pulmonary artery is about:
2. Which of these is NOT true of the pulmonary circulation?
3. Pulmonary vascular resistance is lowest at:
4. When cardiac output rises, pulmonary vascular resistance falls. The main mechanism is:
5. In the middle zone (Zone 2) of an upright lung, blood flow is determined by:
6. Alveolar hypoxia causes:
7. Which of these is true of the bronchial circulation?
8. Fluid is normally absorbed rather than filtered at the pulmonary capillary because:
9. A patient with long-standing mitral stenosis develops pulmonary edema. The mechanism is:
10. In fresh-water drowning, death is usually due to:
11. In salt-water drowning:
12. The normal ventilation-perfusion ratio for the whole lung at rest is:
13. Going from apex to base of an upright lung:
14. Pulmonary tuberculosis is commonest at the apex of the lung because there:
15. The normal physiological shunt of the lung is due to: