Total number of ATP produced through ETS only | vedclass

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Question

(A) One molecule of $3$-phosphoglycerate ($3$-$PGA$) is converted into one molecule of pyruvic acid,producing $1$ $NADH + H^+$ and $1$ $ATP$ (substrat

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) One molecule of $3$-phosphoglycerate ($3$-$PGA$) is converted into one molecule of pyruvic acid,producing $1$ $NADH + H^+$ and $1$ $ATP$ (substrate-level phosphorylation).Then,the pyruvic acid enters the link reaction and the Krebs cycle.In the link reaction,$1$ $NADH + H^+$ is produced.In the Krebs cycle,$3$ $NADH + H^+$,$1$ $FADH_2$,and $1$ $GTP$ (equivalent to $ATP$) are produced.Total reduced coenzymes produced from $1$ molecule of $3$-$PGA$ are:$1$ ($3$-$PGA$ to Pyruvate) + $1$ (Link reaction) + $3$ (Krebs cycle) = $5$ $NADH + H^+$.$1$ $FADH_2$ (Krebs cycle).Through $ETS$,$1$ $NADH + H^+$ produces $3$ $ATP$ and $1$ $FADH_2$ produces $2$ $ATP$.Total $ATP$ from $ETS$ = $(5 	imes 3) + (1 	imes 2) = 15 + 2 = 17$ $ATP$.However,the provided image calculation uses $4$ $NADH$ (likely considering only the Krebs cycle and link reaction) and $1$ $FADH_2$,resulting in $14$ $ATP$. Based on the standard metabolic pathway,$1$ $3$-$PGA$ yields $5$ $NADH$ and $1$ $FADH_2$,totaling $17$ $ATP$. Given the options and the provided image logic,the intended answer is $14$ $ATP$.

Total number of $ATP$ produced through $ETS$ only from one molecule of $3$-phosphoglycerate in aerobic respiration is

Similar Questions

How many times more efficient is aerobic respiration compared to anaerobic respiration?

Complete oxidation of $1$ gram molecule of glucose gives rise to:

$A-$ $A$ total of $10$ $NADH + H^+$ and $2$ $FADH_2$ molecules are formed in aerobic respiration.
$R-$ One $NADH + H^+$ and $1$ $FADH_2$ are helpful in the formation of $3$ and $2$ $ATP$ molecules respectively.

The energy yield as a result of the total oxidation of one glucose molecule during cellular respiration is to convert:

Aerobic respiration of glucose produces energy of approximately: (in $kcal$)

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