2. The bar-headed goose (green triangles) flies at very high altitudes, sometime

Question

2. The bar-headed goose (green triangles) flies at very high altitudes, sometimes greater than 8500 meters above sea level. It has been observed flying over Mt. Everest.

a. The atmospheric pressure at 8500m is about 250 mmHg. What are the partial pressures of oxygen and carbon dioxide at this altitude? Assume 21% oxygen and 0.05% carbon dioxide in the atmosphere.

b. If this partial pressure of oxygen were found in the blood of the high altitude goose, approximately what percent of the barred goose hemoglobin binding sites would be occupied by oxygen?

c. In one or two sentences, explain why the partial pressure of oxygen in the goose’s arterial blood would actually be less than the partial pressure of oxygen you found in part a. Estimate what the P02 would be for the situation described in (a).

Explanation / Answer

a. Partial pressure of oxygen at 8500 m at 250 mm of Hg= Amount of oxygen * Atmospheric pressure

Partial pressure of oxygen=21/100 * 250= 0.21 x 250 mm= 52.5 mm Hg

Partial pressure of carbon dioxide= Amount of carbon dioxide * Atmospheric pressure= 0.05/100 * 250

= 0.125 mm Hg

b. At 52.5 mmHg partial pressure, saturation of oxygen for bar headed goose is around 85%. Oxygen saturation is the percent of Hemoglobin (Hgb) binding sites in the blood carrying oxygen.

c. PA02 in arterial blood is directly proportional to atmospheric pressure, PH20 (water vapour) and PCO2. As PH20 and PCO2 remain unaltered in lungs, despite reduced atmospheric pressure, the PAO2 in arterial

Pao2 = Fio2 · (PBPH2O)Paco2/R

PAO2 = partial pressure of O2 in the alveoli, FiO2 = fractional concentration of O2= 0.21

PB = barometric pressure= 250 mm Hg

PH20 = partial pressure of water vapor in the alveoli= 13.3 mmHg (Approximate mean value at 8500 m)

PaCO2 = partial pressure of CO2 in the arteries, R = respiratory exchange ratio (VCO2/VO2)=0.8

Hence,

PaO2= 0.21* (250-47) – 13.3/0.8= 0.21 * (203)-16.625= 26.005 mmHg

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