If his R=0.8 how much will his arterial pOâ‚‚ fall?

Correct Answer: D

Rationale: With ventilation halved, pCO₂ rising to 80 mmHg, and R = 0.8, arterial pO₂ falls by 50 mmHg (choice D). Using the alveolar gas equation: PAO₂ = FiO₂ × (P_atm - PH₂O) - (PaCO₂ / R), at sea level (760 mmHg), normal PAO₂ = 0.21 × (760 - 47) - (40 / 0.8) ≈ 100 mmHg. Post-overdose, PAO₂ = 0.21 × (760 - 47) - (80 / 0.8) = 149.7 - 100 = 49.7 mmHg. Normal PaO₂ ≈ 100 mmHg, so it falls to ≈50 mmHg, a drop of 50 mmHg. Choice A (85) implies PaO₂ = 15 mmHg, too low; B (75) suggests 25 mmHg, insufficient; C (60) miscalculates R's effect. Hypoventilation raises pCO₂, reducing PAO₂ proportionally, and R adjusts the CO₂-O₂ exchange ratio, confirming D's accuracy.

Correct Answer: A

Rationale: carbon monoxide (CO) poisoning doesn't stimulate carotid bodies effectively, as they sense arterial pO₂, not O₂ content. CO binds hemoglobin (COHb), reducing O₂ delivery, but PaO₂ stays normal (≈100 mmHg), masking hypoxia. Choice B (cyanide) triggers them via metabolic acidosis/hypoxia signals. Choice C (hypoxia, low pO₂) directly activates them (<60 mmHg). Choice D (hypercapnia) stimulates via pCO₂ and pH changes. Choices E (H⁺) and F (nicotine) also activate them. Located at carotid bifurcations, these chemoreceptors drive ventilation in hypoxia or acidosis. CO's failure to lower PaO₂ distinguishes A as the non-stimulant, despite tissue hypoxia.

Correct Answer: C

Rationale: the Bohr effect shifts the O₂ dissociation curve right due to pCO₂ increasing H⁺, reducing Hb affinity. Choice A is false; left shift increases affinity. Choice B is wrong; decreased 2,3-DPG shifts left, not right. ' temperature shifts the curve (right with increase). Choice E is false; 2,3-DPG rises at altitude. The Bohr effect, driven by CO₂'s pH impact, aids O₂ unloading in tissues (P₅₀ rises), a key adaptation. C accurately describes this mechanism.

Correct Answer: D

Rationale: Choice D is correct (replacing E); helium (low density) reduces turbulence (lower Reynolds number, Re = ρvD/μ). Choice A is false; large airways (higher velocity) are more turbulent than small (laminar). Choice B is wrong; turbulent flow scales less than linearly with pressure (not doubled). ' turbulence depends on density, not viscosity (laminar does). Low-density gases decrease Re, easing flow in obstructive disease, making D true.

Correct Answer: D

Rationale: the chloride shift moves HCO₃⁻ out of the RBC (not in) and Cl⁻ in, maintaining neutrality as CO₂ forms HCO₃⁻. Choice A is true (Fick's law: V ∝ A/T). Choice B is correct (Bohr effect: PCO₂ → H⁺ → right shift). Choice C is accurate (Henry's law). Choice E is true (Haldane effect). D reverses ion directions; HCO₃⁻ exits via band 3 exchanger, Cl⁻ enters, supporting CO₂ transport, making it the error.