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$(C)$ The net gain of $ATP$ during aerobic respiration is calculated as follows:$1$. Glycolysis produces $2$ $ATP$ and $2 NADH + H^+$.$2$. Pyruvate ox

Vedclass · Accepted Answer

$36$

Vedclass · Answer

$(C)$ The net gain of $ATP$ during aerobic respiration is calculated as follows:$1$. Glycolysis produces $2$ $ATP$ and $2 NADH + H^+$.$2$. Pyruvate oxidation (Link reaction) produces $2 NADH + H^+$.$3$. Krebs' cycle produces $2$ $GTP$ (equivalent to $2$ $ATP$), $6 NADH + H^+$, and $2 FADH_2$.$4$. In the Electron Transport System $(ETS)$, each $NADH + H^+$ typically yields $3$ $ATP$ and each $FADH_2$ yields $2$ $ATP$.Total $ATP$ calculation: $2$ ($ATP$ from glycolysis) + $2$ ($GTP$ from Krebs' cycle) + $10 NADH + H^+$ ($2$ from glycolysis + $2$ from link reaction + $6$ from Krebs' cycle) $	imes 3$ $ATP$ = $30$ $ATP$ + $2 FADH_2$ $	imes 2$ $ATP$ = $4$ $ATP$.Total = $2 + 2 + 30 + 4 = 38$ $ATP$ (in prokaryotes) or $36$ $ATP$ (in eukaryotes due to the shuttle cost of $2$ $ATP$).Given the options, $36$ is the standard accepted value for eukaryotic aerobic respiration.

Net gain of $ATP$ molecules per hexose during aerobic respiration is

Similar Questions

One molecule of glucose + $6$ molecules of $O_2$ + $38$ molecules of $ADP$ combine to form $6$ $H_2O$,$6$ $CO_2$,and .........

Describe the respiratory balance sheet.

If the complete oxidation of one mole of glucose yields $CO_2$ and $H_2O$ with a release of $686 \, Kcal$ of energy,and the energy stored in one mole of $ATP$ as a phosphate bond is $12 \, Kcal$,what is the maximum number of $ATP$ molecules that can be produced from one glucose molecule?

$A-$ $A$ 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 products of aerobic respiration of glucose are ......... .

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