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.

(a) Calculate the rate at which the heart muscle consumes glucose and produces ATP.

(b) Consider an alternate scenario – what would be the rate of consumption if the energy source were triglycerides whose fatty acyl chains were each 12-C in length and saturated. (assume the same rate of 10.0 micromol/min of oxygen consumption).

(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.)

a) Consumption of glucose is 1.7 μmol/min.g of tissue.

    The muscle produces ATP at the rate of 54.4 μmol /min.g or 0.91 μmol /sec/g.

b) The rate of consumption is s13.9 μmol /min.g or 0.23 μmol /sec/g

c)  The time required (in seconds) to completely turn over the cellular pool of ATP 5.5 secs

   To do this, the cell must regulate ATP synthesis very precisely

Explanation:

a) For the oxidation of glucose we need 6 mol of O2 per mol of glucose.

Therefore, we can say consumption of glucose is at the rate of 10/6 = 1.7 μmol/min.g of tissue.

Each glucose molecule produces 32 ATP.

Then the muscle produces ATP at the rate of = 1.7x32=54.4 μmol /min.g or 0.91 μmol /sec/g.

b) For the oxidation of triglycerides whose fatty acyl chains we need 23 mol of O2 per mol of glucose.

Therefore, we can say consumption of glucose is at the rate of 10/23=0.435 μmol/min.g of tissue.

Each glucose molecule produces 32 ATP.

Then the muscle produces ATP at the rate of = 0.435x32

   =13.9 μmol /min.g or 0.23 μmol /sec/g.

c) At 5 μmol /g level, it will take 5.5s to produce ATP.

This shows that the entire pool of ATP in muscle must be regenerated for every 5.5 s.

d

explanation:

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