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Question

(D) When one mole of pyruvate enters the mitochondria, it undergoes oxidative decarboxylation to form one mole of $Acetyl-CoA$, producing $1$ $NADH +

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) When one mole of pyruvate enters the mitochondria, it undergoes oxidative decarboxylation to form one mole of $Acetyl-CoA$, producing $1$ $NADH + H^+$.In the $TCA$ cycle, one mole of $Acetyl-CoA$ produces $3$ $NADH + H^+$, $1$ $FADH_2$, and $1$ $GTP$ (equivalent to $1$ $ATP$).Total yield per pyruvate: $4$ $NADH + H^+$, $1$ $FADH_2$, and $1$ $ATP$.Using the standard $P/O$ ratios ($1$ $NADH = 2.5$ $ATP$ and $1$ $FADH_2 = 1.5$ $ATP$):$4$ $NADH 	imes 2.5 = 10$ $ATP$.$1$ $FADH_2 	imes 1.5 = 1.5$ $ATP$.$1$ $ATP$ (substrate-level phosphorylation) = $1$ $ATP$.Total = $10 + 1.5 + 1 = 12.5$ $ATP$.However, in older textbook conventions where $1$ $NADH = 3$ $ATP$ and $1$ $FADH_2 = 2$ $ATP$:$4 	imes 3 = 12$ $ATP$, $1 	imes 2 = 2$ $ATP$, and $1$ $ATP$.Total = $12 + 2 + 1 = 15$ $ATP$. Thus, $15$ is the standard answer based on traditional calculations.

Upon the oxidation of one mole of pyruvate by mitochondrial respiration, the moles of $ATP$ generated are

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