During strenuous exercise, the muscle interstitial fluid $\mathrm{pO}_2$ falls to 20 mm Hg . The oxy

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During strenuous exercise, the muscle interstitial fluid $\mathrm{pO}_2$ falls to 20 mm Hg . The oxygen delivered by blood that passes through the exercising muscle tissues will be

Professor PC Thomas Classes · Answer

(c) : The given graph shows the relationship between partial pressure of oxygen with the amount of oxygen that can bind with haemoglobin. The $\mathrm{pO}_{2}$ of the air within alveoli is about 100 mm Hg . As the blood travels through the systemic blood capillaries, oxygen leaves the blood and diffuses into the tissues. Consequently, the blood that leaves the tissues in the veins has a $\mathrm{pO}_{2}$ that is decreasing (in a resting person) to about 40 mm Hg . Therefore, A depicts oxygenated blood and $B$ depicts systemic blood.\ The steep portion of the curve from 60 down to 20 mm Hg is ideal for unloading oxygen in the tissue, i.e., for a small decrease in $\mathrm{pO}_{2}$ (due to diffusion of oxygen) a large quantity of oxygen can be unloaded to the peripheral tissue capillary. During strenuous exercise, the muscles accelerated metabolism uses more oxygen from the capillary blood and thus decreases the venous blood $\mathrm{pO}_{2^{\prime}}$ for example the $\mathrm{pO}_{2}$ of the venous blood could drop to 20 mm Hg . Hence, $\mathrm{D}, \mathrm{E}$ and F shows venous blood in exercise, normal venous blood and normal arterial blood respectively.\ As we know, in the resting person, the $\mathrm{pO}_{2}$ of tissues is 40 mm Hg , which is favourable for oxygen diffusion from arterial blood to the tissues. Oxygen unloaded to the tissues by arterial blood is $5 \mathrm{~mL} / 100 \mathrm{~mL}$ of blood $(20 \mathrm{~mL}-15 \mathrm{~mL}=5 \mathrm{~mL} / 100 \mathrm{~mL}$, from the graph). During strenuous exercise, the muscle interstitial fluid $\mathrm{pO}_{2}$ falls to 20 mm Hg , so the oxygen that unload to the tissue is $15 \mathrm{~mL} / 100 \mathrm{~mL}$ of blood, $(20-5) \mathrm{mL} / 100 \mathrm{~mL}=15 \mathrm{~mL} / 100 \mathrm{~mL}$, from the graph i.e., three times as much as normal.\