Paramesonephric ducts will give:

Correct Answer: A

Rationale: Paramesonephric (Müllerian) ducts, in females, develop into the uterus and upper vagina (A), forming under estrogen influence as the mesonephric ducts regress. Ovaries (B) arise from gonadal ridges (mesoderm), not paramesonephric ducts, which are separate from gamete-forming tissues. Testes (C) and prostate (D) develop in males from mesonephric (Wolffian) ducts under testosterone, while paramesonephric ducts degenerate. A is correct paramesonephric ducts' role in female reproductive tract formation contrasts with the others' distinct embryological pathways.

Correct Answer: D

Rationale: The Oâ‚‚ dissociation curve shifts right (decreased Hb-Oâ‚‚ affinity) with increased [H+] (choice A), pCOâ‚‚ (choice B), temperature (choice C), and 2,3-DPG (choice E), but not carbon monoxide (CO, choice D). Increased [H+] (acidosis) and pCOâ‚‚ (Bohr effect) reduce Hb affinity, unloading Oâ‚‚ to tissues. Higher temperature similarly favors Oâ‚‚ release, aiding exercising muscles. 2,3-DPG, upregulated in hypoxia, stabilizes deoxy-Hb, shifting the curve right. CO, however, binds Hb 200-250 times more strongly than Oâ‚‚, forming carboxyhemoglobin (COHb), which doesn't release Oâ‚‚ easily. This left-shifts the curve for remaining unbound Hb, increasing affinity and impairing tissue delivery, opposite to a right shift. CO's effect is thus distinct, making D the exception that doesn't shift the curve rightward.

Correct Answer: B

Rationale: The blood gas (pH = 7.52, pCO₂ = 20 mmHg, pO₂ = 120 mmHg, HCO₃⁻ = 16 mmol/L) indicates respiratory alkalosis with partial renal compensation (choice B). pH > 7.45 shows alkalosis; low pCO₂ (20 mmHg, normal 35-45) suggests hyperventilation as the cause (respiratory). HCO₃⁻ (16 mmol/L, normal 22-26) is reduced, indicating renal compensation kidneys excrete bicarbonate to offset pH rise, though not fully (partial). Choice A (metabolic alkalosis) requires high HCO₃⁻ (e.g., vomiting), not low. Choice C (metabolic acidosis) fits low pH and HCO₃⁻, not here. Choice D (respiratory acidosis) needs high pCO₂. High pO₂ (120 mmHg) aligns with hyperventilation. This pattern low pCO₂ driving alkalosis, kidneys adjusting HCO₃⁻ confirms B as the diagnosis.

Correct Answer: C

Rationale: strong inspiration produces intrapleural pressures of -20 to -30 mmHg, not -6 mmHg, which suits quiet breathing. Choice A is true; at rest (end-expiration), pressure is subatmospheric (≈-4 mmHg or -5 cmH₂O) due to lung-chest wall recoil. ' inspiration lowers it further (e.g., -7.5 cmH₂O) as the thorax expands. Choice D is plausible; -2.5 mmHg at bases reflects gravitational gradients (less negative than apex). Strong inspiration, via maximal diaphragmatic and intercostal effort, significantly drops pressure to drive large airflow (e.g., vital capacity). -6 mmHg underestimates this, fitting tidal breathing, not ‘strong' effort. This error misaligns with pleural dynamics, where forceful inspiration amplifies negative pressure, making C the incorrect statement.

Correct Answer: C

Rationale: O₂ is perfusion-limited at rest (equilibrates in 0.25 s vs. 0.75 s transit), intermediate between N₂O (perfusion-limited) and CO (diffusion-limited). Choice A is false; N₂O, highly soluble, is perfusion-limited uptake depends on flow. Choice B is wrong; CO, low solubility, is diffusion-limited. ' diffusing capacity (Dₗ) is proportional to area, inversely to thickness (Fick's law). Choice E is true; transit time is ≈0.25 s at rest. O₂'s rapid equilibration at rest shifts to diffusion limitation in exercise or disease, but C accurately reflects its resting state.