How much does the inspired Oâ‚‚ concentration have to be raised to abolish the hypoxaemia?

Correct Answer: B

Rationale: To abolish hypoxemia (PaO₂ < 60 mmHg, here ≈ 50 mmHg), inspired O₂ (FiO₂) must raise PAO₂ to ≈100 mmHg. From the alveolar gas equation: PAO₂ = FiO₂ × (760 - 47) - (80 / 0.8). Set PAO₂ = 100: 100 = FiO₂ × 713 - 100; 200 = FiO₂ × 713; FiO₂ ≈ 0.28. Normal FiO₂ = 0.21 (21%), so increase = 0.28 - 0.21 = 0.07 (7%, choice B). Choice A (5%) yields PAO₂ ≈ 85 mmHg, insufficient; C (10%) overshoots to 121 mmHg; D (15%) is excessive (157 mmHg). A 7% rise (to 28% O₂) restores normoxemia without overcompensation, matching physiological needs under hypoventilation, making B correct.

Correct Answer: D

Rationale: decreased pO₂ doesn't shift the Hb dissociation curve right it's a condition, not a modifier. Choices A (2,3-DPG), B (pCO₂), and C (temperature) shift it right, reducing Hb-O₂ affinity (Bohr effect), enhancing O₂ unloading in exercise. 2,3-DPG rises in active tissues, stabilizing deoxy-Hb. Elevated pCO₂ and H⁺ (CO₂ hydration) and heat (muscle work) favor release. Low pO₂ moves along the curve (P₅₀ = 26 mmHg), not shifting it affinity stays unless pH, CO₂, or temperature change. Exercise boosts O₂ delivery; right shifts (P₅₀ to ≈30 mmHg) aid this, while pO₂ reflects demand, not curve position. D is the exception.

Correct Answer: D

Rationale: diffusion is inversely proportional to the square root of molecular weight (Graham's law), though less dominant in lungs. Choice A is false; diffusion is proportional to the diffusion constant (Fick's law: V = D·A·ΔP/T). Choice B is wrong; it's proportional to area. ' it's proportional to solubility. Choice E is false; it's proportional to partial pressure difference. In lungs, solubility and gradient drive O₂/CO₂ diffusion, but molecular weight (√MW) affects gas speed in free states. D's inverse relationship holds in diffusion theory, making it correct.

Correct Answer: D

Rationale: RQ = 1.0 for carbohydrate metabolism (VCOâ‚‚ = VOâ‚‚). Choice A is true; high V/Q (apex) lowers RQ (less COâ‚‚ relative to Oâ‚‚). Choice B is false; base RQ is lower (higher perfusion). Choice C is wrong; RQ = VCOâ‚‚ / VOâ‚‚, not reverse. D reflects metabolic stoichiometry, a standard definition.

Correct Answer: C

Rationale: PCO₂ = 600 mmHg (extreme hypercapnia) with HCO₃⁻ = 28 mEq/L (near normal) indicates acute respiratory acidosis. Using Henderson-Hasselbalch (pH = 6.1 + log(HCO₃⁻ / (0.03 · PCO₂))), pH ≈ 6.1 + log(28 / 18) ≈ 6.3, severely acidic due to high PCO₂. Choice A (metabolic) requires low HCO₃⁻. Choice B needs higher HCO₃⁻ (e.g., 40). Choice D implies compensation (HCO₃⁻ rise, e.g., 40-50), not seen here. Choice E (alkalosis) contradicts PCO₂. Acute hypoventilation (e.g., obstruction) causes this uncompensated state, making C accurate.