Simple squamous epithelium is found in:

Correct Answer: A

Rationale: Simple squamous epithelium, a single layer of flat cells, facilitates diffusion and filtration. It lines alveoli (A) in the lungs, enabling gas exchange due to its thinness. The gastrointestinal tract (GIT, B) uses simple columnar epithelium for absorption, with taller cells and microvilli, not squamous. Skin (C) has stratified squamous epithelium for protection, not a single layer. The trachea (D) has pseudostratified columnar epithelium with cilia, not squamous. A is correct alveoli rely on simple squamous for efficient oxygen and carbon dioxide transfer, unlike the other structures' distinct functional needs.

Correct Answer: C

Rationale: Most CO₂ diffusing into blood in systemic capillaries converts to bicarbonate ions in red blood cells (RBCs), making choice C correct. Only about 5-10% remains dissolved as CO₂ (choice A) due to low solubility. Around 20-25% forms carbamino compounds (choice B) by binding to hemoglobin's amino groups, but this is less dominant. The majority (65-70%) enters RBCs, where carbonic anhydrase catalyzes its reaction with H₂O to form carbonic acid (H₂CO₃), which dissociates into H⁺ and HCO₃⁻ (bicarbonate). ' CO₂ doesn't combine directly with hemoglobin significantly carbamino formation is specific. Choice E occurs minimally in plasma (slow without enzyme) compared to RBCs. Bicarbonate then exits RBCs via the chloride shift, maintaining pH balance. This process's efficiency in RBCs, driven by enzymatic speed and volume, ensures C is the primary fate of CO₂ in systemic capillaries.

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: C

Rationale: In the upright lung, the ventilation/perfusion (V̇/Q̇) ratio is highest at the apex (choice C), due to gravity's effect. Ventilation is greater at the base (choice A is false), as dependent alveoli expand more. Perfusion is also greater at the base (choice B is false), with blood flow higher in lower zones (e.g., 4:1 base-to-apex ratio). V̇/Q̇ at the apex (≈3) exceeds the ideal 0.8, being ‘abnormally high' relative to perfect matching, while the base is ≈0.6. PO₂ is highest at the apex (choice D is false), as high V̇/Q̇ over-ventilates relative to perfusion; bases have lower PO₂ due to lower V̇/Q̇. pH follows PO₂ trends, not bases (choice E false). C's high V̇/Q̇ reflects physiological mismatch, driving regional gas exchange differences.

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.