PROBLEM SET 6.2: GAS TRANSPORT AND pH DISTURBANCES
ANSW ER 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 O
...
PROBLEM SET 6.2: GAS TRANSPORT AND pH DISTURBANCES
ANSW ER 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 Hg6.PS2.2
B.2. W hat would the oxygen content of her mixed venous blood be?
At 40 mm Hg the oxygen content of venous blood would be calculated as in A.1:
0.003 mL O
2 dL-1 mm Hg-1 x 40 mm Hg + 0.75 x 1.35 mL O2 g Hb-1 x 10 g Hb dL-1 =
0.12 mL dL-1 + 10.12 mL dL-1 = 10.24 mL dL-1
C. If arterial blood has P
aO2 = 95 mm Hg as in part A and mixed venous blood has SO2 = 75%
(part B), how much O2 would be extracted by the tissues per L of blood?
The O
2 extracted per L of blood is just the A-V difference. At 95 mm Hg we calculated the arterial
content as 135.2 mL L-1 and at 75% saturation we calculated 102.4 mL L-1. The difference is
32.8 mL L-1
D. Given the [O2]a - [O2]v from part C, what must Qa be for tissue extraction to match her QO2
of 225 m l min-1?
Now the equation for O2 extraction is
225 m L m in-1 = Q
a x 32.8 mL L-1
Solving for Qa, we get Qa = 6.86 L min-1
E. According to your calculations, the response to anem ia in principle could be to keep cardiac
output, Qa, constant and let PvO2 fall, or to raise Qa to keep PvO2 constant. With your new
knowledge of respiratory and circulatory physiology, which do you think happens?
In circulatory physiology, you learned that hypoxia of tissues cause local vasodilation. This should
decrease TPR. The circulatory system works to keep blood pressure constant. With a constant MAP
and a reduced TPR, the cardiac output should increase. However, it is likely that some interm ediate
situation will prevail; PvO2 will fall, but not as low as in the calculation, and cardiac output will rise, but
not as much as in the calculation. Despite this prediction, what actually appears to happen is that the
viscosity of blood is lowered by the smaller hematocrit, and that the larger cardiac output occurs from
a reduction in TPR because of th
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