the type of epithelium in the urinary bladder is:

Correct Answer: B

Rationale: The urinary bladder's epithelium is transitional (B), a stratified type unique to the urinary tract, allowing stretch as the bladder fills. Its surface cells shift from cuboidal to flattened, adapting to volume changes without rupturing. Stratified squamous (A) lines the skin and mouth, resisting abrasion, not stretching. Simple squamous (C), a single thin layer, lines alveoli and capillaries for diffusion, unsuitable for the bladder's mechanical demands. Pseudostratified columnar (D) is in the trachea, with cilia, irrelevant here. Transitional epithelium's ability to transition shapes under tension makes B the correct choice for the bladder's dynamic environment.

Correct Answer: C

Rationale: Choice C is false. Pulmonary resistance decreases as lung volume increases because extra-alveolar vessels (arteries and veins) are pulled open by radial traction, reducing resistance, not increasing it. Choice A is true; increased pulmonary arterial pressure recruits and distends vessels, lowering resistance. ' pulmonary resistance is about 1/10 of systemic due to shorter, wider vessels. Choice D is true; acetylcholine, via parasympathetic stimulation, relaxes bronchiolar smooth muscle, though its effect is less pronounced than adrenaline's. Choice E is also true; at large lung volumes, pulmonary capillaries are compressed, increasing resistance. The error in C stems from misunderstanding lung volume effects: as lungs expand, airway resistance drops (bronchioles widen), and extra-alveolar vessel resistance decreases due to mechanical stretching, not increases. This aligns with physiological principles of pulmonary circulation, making C the false statement.

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: Hypoventilation (choice A) doesn't occur at high altitudes; it's the exception. Low pOâ‚‚ triggers hyperventilation via peripheral chemoreceptors, increasing ventilation to raise PaOâ‚‚. Polycythemia (choice B) increases RBCs/Hb, boosting Oâ‚‚ capacity after days. Capillary density rises (choice C) in tissues, enhancing Oâ‚‚ delivery over weeks. The Oâ‚‚ dissociation curve shifts right (choice D) due to increased 2,3-DPG, aiding Oâ‚‚ unloading despite lower PaOâ‚‚. Pulmonary vasoconstriction (choice E) occurs acutely, shunting blood to better-ventilated areas. Hypoventilation would worsen hypoxemia, countering acclimatization's goal of optimizing Oâ‚‚ availability. Hyperventilation lowers PaCOâ‚‚, causing alkalosis (later compensated renally), making A the process that doesn't aid high-altitude adaptation.

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.