When 100% glucose is oxidized to CO_2, the ef | vedclass

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(A) In the process of aerobic respiration, the complete oxidation of one molecule of glucose $(C_6H_{12}O_6)$ yields $36$ to $38$ $ATP$ molecules.Taki

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) In the process of aerobic respiration, the complete oxidation of one molecule of glucose $(C_6H_{12}O_6)$ yields $36$ to $38$ $ATP$ molecules.Taking the standard yield of $36$ $ATP$ molecules, where each $ATP$ molecule stores approximately $7.6 	ext{ kcal}$ of energy, the total energy captured in $ATP$ is $36 	imes 7.6 = 273.6 	ext{ kcal}$.The total energy released by the complete combustion of one mole of glucose is $686 	ext{ kcal}$.Respiratory efficiency is calculated as the ratio of energy captured in $ATP$ to the total energy released: $	ext{Efficiency} = \frac{273.6}{686} 	imes 100 \approx 40\%$.In decimal form, $40\%$ is equivalent to $0.4$.

When $100\%$ glucose is oxidized to $CO_2$, the efficiency of such a respiration is

Similar Questions

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