Rat heart muscle operating aerobically fills more than 90% of its ATP needs by o

Question

Rat heart muscle operating aerobically fills more than 90% of its ATP needs by oxidative phosphorylation. Each gram of tissue consumes Oxygen at the rate of 10.0 micromol/min, with glucose as the fuel source.

(c) For a steady-state concentration of ATP of 5.0 micromol/g of heart muscle tissue, calculate the time required (in seconds) to completely turn over the cellular pool of ATP. What does this result indicate about the need for tight regulation of ATP production? (Note: Concentrations are expressed as micromoles per gram of muscle tissue because the tissue is mostly water.)

Explanation / Answer

First we need to find the rate at which the heart muscle consumes glucose and produces ATP.

So,as we can see here,

Glucose oxidation requires 6 mol of O2per mol of glucose. Therefore, glucose is consumed at the rate of (10.0 micromol/min*g)/6 = 1.7micromol/min*g of tissue. If each glucose produces 32 ATP , the muscle produces ATP at the rate of (1.7micromolglucose/min*g)(32 ATP/glucose) = 54 micromol/min g, or 0.91 micromol/s*g.

(C)

It takes (5.0 micromol/g)/(0.91micromol/s*g) = 5.5 s to produce 5.0 micromol of ATP per gram, so the entire pool of ATP must be regenerated (turned over) every 5.5 s. In order to do this, the cell must regulate ATP synthesis precisely .

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