4. Volumetric blood flow through a 1 g section of muscle is 0.025 mL/min. Arteri

Question

4. Volumetric blood flow through a 1 g section of muscle is 0.025 mL/min. Arterial O2 content (oxygen in blood entering the muscle) is 20 mL of O2 per dl of blood, and venous O2 content (oxygen in blood leaving the muscle) is 14 mL of O2 per dl of blood. The density of blood is constant, at 1.05 g/mL. (a) What is the rate of O2 taken up by the muscle, in mL/min? Now suppose that the blood flow to the muscle increases from 0.025 to 0.1 mL/min. (b) Ifarteriolar and venous O2 content is unchanged, what happens to O2 consumption in the muscle? (c) If O2 consumption of the muscle does not change, what is the venous O2 content? (in mL O2 per dl of blood)

Explanation / Answer

a) The appropriate equation here is
QO2 = Qv x [ [O2]arteriole - [O2]venule]
Where QO2 is the rate of O2 consumption, Qv is the cardiac output, and the concentrations are total
The O2 content of the blood. Here we solve for QO2:
QO2 = 0.025 mL min-1 x [0.20 mL O2 mL blood-1 - 0.14 mL O2 mL blood-1 ]
= 0.0015 mL O2 min -1g-1

b) Nothing happens to O2 consumption. Increasing flow by itself does not alter O2 consumption. Rather typically
increased O2 consumption increases blood flow.

c) We solve QO2 = Qv x [ [O2]arteriole - [O2]venule] For [O2]venule
0.0015 mL O2 min-1 g-1 = 0.1 mL m in-1g-1 x [ 0.20 mL O2 ml blood-1 - x]
[O2]venule = 0.185 mL mL blood-1 = 18.5 %

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.