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PROBLEM SET 6.2: GAS TRANSPORT AND pH DISTURBANCES ANSWER KEY 1. A. An elderly woman has a hemoglobin concentration of 10 g dL-1. Her O2 dissociation curve is normal (when expressed as SO2). Assume... that her resting O2 consum ption (QO2) is within normal limits, 225 mL min-1, and that Qa, the cardiac output, is 4.5 L min-1 ; PaO2 = 95 mm Hg is normal. A.1. W hat is the total oxygen content (in mL dL-1) of her arterial blood? The total oxygen content is the sum of the dissolved O2 and O2 bound to hem oglobin. The dissolved O 2 is [O2]dissolved = 0.003 mL O2 dL-1 mm Hg-1 x 95 mm Hg = 0.285 mL dL-1 The O 2 content of Hb is given as [HbAO2] = 0.98 x 1.35 mL O2 g Hb-1 x 10 g Hb dL-1 = 13.23 mL dL-1 So the total O 2 content is 13.23 mL dL-1 + 0.29 mL dL-1 = 13.52 mL dL-1 A.2. W hat is the total oxygen content of her mixed venous blood? The QO2 is 225 mL m in-1. This is equal to the cardiac output times the A-V difference in O2 content. So we have 225 m L m in-1 = 4.5 L min-1 x [ 135.2 mL L-1 - X mL L-1] Solving for X, the venous O2 content, we get X = 8.52 mL dL-1 A.3. W hat is the S O2 of venous blood? This is a difficult problem to solve analytically, because the solution requires the exact form of the Hb saturation curve and the linear relation between dissolved O 2 and PO2. We can approximate the solution by assuming that dissolved O2 is about 0.1 mL dL-1, meaning that the Hb@O2 is 8.52 -0.1 = 8.42 mL dL-1. At an Hb@O 2 capacity of 1.35 mL g Hb-1 and 10 g dL-1, the percent saturation is S O2 = 8.42 mL dL-1/ 13.5 mL dL-1 = 0.62 or 62% saturation This corresponds to a PaO2 of about 31 mm Hg; thus the dissolved O2 would be 0.09 mL dL-1. W e could plug this back in an re-calculate the SO2 until the answers converge by iteration. The answer would differ only a little from the answer given above. B. Assume that the S O2 of venous blood in this woman is the normal 75% . B.1. W hat would be the P vO2? A normal 75% saturation means that PvO2 = 40 mm Hg [Show More]

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