A mountaineer ascends to 5500 m altitude (barometric pressure 380 mmHg). After 5 days of acclimatisation, his haemoglobin is 17 g/dL and 2,3-DPG is elevated. Which of the following is the calculated PO2 of inspired air (PiO2) at this altitude, assuming normal FiO2?

• A (380 − 47) × 0.21 = 333 × 0.21 ≈ 70 mmHg ✓
• B 380 × 0.21 = 79.8 mmHg
• C (380 − 47) × 0.21 / 0.8 = 87.4 mmHg (corrected for respiratory quotient)
• D 760 × 0.21 − 47 = 112.4 mmHg

Explanation

PiO2 is calculated as (Pb − PH2O) × FiO2, where Pb is barometric pressure and PH2O at 37°C is 47 mmHg. At 5500 m: PiO2 = (380 − 47) × 0.21 = 333 × 0.21 = 69.93 ≈ 70 mmHg. The water vapour pressure of 47 mmHg must be subtracted because inspired air is humidified at body temperature in the airways, diluting the O2 partial pressure. Option B omits PH2O correction. Option C incorrectly incorporates the respiratory quotient (which is used for the alveolar gas equation to calculate PAO2, not PiO2). Option D uses sea-level barometric pressure.

Reference: Guyton & Hall, Textbook of Medical Physiology, 14th ed.

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