When 100% carbon is oxidized to CO $_{2}$ , the efficiency of such respiration is

40%

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60%

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90%

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100%

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Solution

The correct option is A 40%
Oxidation of carbon to CO $_{2}$ takes place in aerobic respiration.
The amount of energy released by complete oxidation of 1 mol of glucose = 686 kcal
No. of ATPs produced during respiration = 36
Energy required to produce 1 mol ATP = 7.6 kcal
Total energy used to produce ATPs = 7.6 x 36 = 273.6 kcal
Percentage of total energy conserved = (273.6 / 686) x 100 = 39.88 %
Hence, Efficiency of aerobic respiration is 39.88% i.e. 40%
So, the answer is A.

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