In the presence of active surfactants, all of the following are expected to decrease EXCEPT?

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Question 1 of 5

In the presence of active surfactants, all of the following are expected to decrease EXCEPT?

Correct Answer: B

Rationale: Surfactant, produced by type II alveolar cells, is a phospholipid mixture that reduces surface tension in alveoli, stabilizing them and aiding lung function. Normally, high surface tension from water molecules at the air-liquid interface promotes alveolar collapse, but active surfactant lowers this tension, decreasing the tendency of lungs to collapse (atelectasis). Reduced surface tension also lessens the work of breathing by making lung expansion easier and decreases lymph flow in the lung, as less fluid is forced into the interstitium due to lower alveolar pressure gradients. However, lung compliance the ease with which lungs expand increases with active surfactant, not decreases. Compliance is inversely related to surface tension; when tension drops, the lungs become less stiff, improving their ability to stretch per unit of pressure. Thus, lung compliance is the exception, as it rises while the other factors diminish, reflecting surfactant's critical role in maintaining alveolar stability and efficient ventilation.

Question 2 of 5

Which person would be expected to have the largest PAO2-PaO2 gradient? (A stands for alveolar and a-stands for arterial)

Correct Answer: B

Rationale: The PAO2-PaO2 gradient measures the difference between alveolar oxygen (PAO2) and arterial oxygen (PaO2), normally small (~5-10 mmHg) due to efficient diffusion. In pulmonary fibrosis, thickened alveolar walls from scarring impair O2 diffusion, lowering PaO2 (e.g., 60 mmHg) while PAO2 (calculated via the alveolar gas equation, ~100 mmHg at sea level) remains closer to normal, widening the gradient (e.g., 40 mmHg). During exercise in a normal person, increased cardiac output and ventilation match perfusion, keeping the gradient minimal despite higher O2 demand. Anemia reduces oxygen-carrying capacity (low hemoglobin), not diffusion, so PaO2 approximates PAO2, maintaining a normal gradient. At 5000 meters, low atmospheric PO2 reduces both PAO2 and PaO2 proportionately (e.g., PAO2 ~50 mmHg, PaO2 ~45 mmHg), keeping the gradient small. Pulmonary fibrosis uniquely disrupts diffusion, causing the largest gradient, as fibrotic barriers hinder O2 transfer more than ventilation or perfusion issues.

Question 3 of 5

In normal individual, regarding gas exchange across pulmonary capillaries during mild exercise, which of the following statements is TRUE?

Correct Answer: A

Rationale: During mild exercise, pulmonary gas exchange adapts to increased O2 demand and CO2 production. CO2 diffuses ~20 times faster than O2 across the alveolar-capillary membrane due to its higher solubility (0.51 vs. 0.024 ml/mmHg/L), despite a slightly higher molecular weight (44 vs. 32), per Fick's law (D ∝ solubility / √MW) making it cross easier, a true statement. Diffusing capacity (DL) for O2 is less than for CO2 normally, and while exercise increases DL for both (recruiting capillaries), CO2's advantage persists, not O2's, and molecular weight is secondary to solubility. Capillary equilibrium length shortens for O2 and CO2 as blood flow rises, but this is nuanced and not uniquely true without context. Arterial blood gases (ABGs) remain normal in healthy individuals during mild exercise (e.g., PaO2 ~100 mmHg, PaCO2 ~40 mmHg), as ventilation matches perfusion. CO2's easier diffusion is the standout truth, rooted in its physicochemical properties, critical for efficient CO2 elimination.

Question 4 of 5

An experiment is conducted in two persons (subjects T and V) with identical VTs (1000 milliliters), dead space volumes (200 milliliters), and ventilation frequencies (20 breaths per minute). Subject T doubles his VT and reduces his ventilation frequency by 50%. Subject V doubles his ventilation frequency and reduces his VT by 50%. What best describes the total ventilation (also called minute ventilation) and Va of subjects T and V?

Correct Answer: B

Rationale: Total ventilation (VE) = VT × RR; alveolar ventilation (VA) = (VT - VD) × RR. Initially, T and V have VT = 1000 ml, VD = 200 ml, RR = 20/min. VE = 1000 × 20 = 20 L/min; VA = (1000 - 200) × 20 = 800 × 20 = 16 L/min. For T: VT doubles to 2000 ml, RR halves to 10/min. VE = 2000 × 10 = 20 L/min (constant); VA = (2000 - 200) × 10 = 1800 × 10 = 18 L/min (increases). For V: VT halves to 500 ml, RR doubles to 40/min. VE = 500 × 40 = 20 L/min (constant); VA = (500 - 200) × 40 = 300 × 40 = 12 L/min (decreases). T's larger VT boosts VA despite lower RR, as more air exceeds VD. V's smaller VT reduces VA, as dead space consumes a larger fraction per breath despite higher RR. Option B (T: VE constant, VA increases; V: VE constant, VA decreases) matches, reflecting how VT impacts VA efficiency at fixed VE.

Question 5 of 5

Which of the following values is above normal in-patient suffering from severe respiratory muscle weakness?

Correct Answer: B

Rationale: Severe respiratory muscle weakness (e.g., in myasthenia gravis) impairs ventilation by weakening inspiratory and expiratory muscles. Tidal volume (VT, ~500 ml normally) decreases due to limited inspiratory force, reducing breath size. Vital capacity (VC, ~4-5 L) drops as maximal inhalation and exhalation are compromised. Oxyhemoglobin saturation falls (e.g., from 95-100% to <90%) as hypoventilation lowers PaO2, causing hypoxemia. Arterial pH may decrease (acidosis) if CO2 retention raises PCO2, but this isn't specified as above normal. However, PCO2 itself (normal 35-45 mmHg) rises above normal (e.g., 50-60 mmHg) due to inadequate CO2 expulsion, a direct result of weak ventilation. Though not listed, if B intended PCO2 (a common mix-up), it fits; otherwise, none are above normal' assuming intent, PCO2's rise is the key abnormality, reflecting ventilatory failure's impact on gas exchange.

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