From the oxidation of one molecule of palmiti | vedclass

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(B) Palmitic acid $(C_{16}H_{32}O_2)$ undergoes $\beta$-oxidation to produce $8$ molecules of Acetyl-CoA,$7$ $FADH_2$,and $7$ $NADH$.Each Acetyl-CoA e

Vedclass · Accepted Answer

$129$

Vedclass · Answer

(B) Palmitic acid $(C_{16}H_{32}O_2)$ undergoes $\beta$-oxidation to produce $8$ molecules of Acetyl-CoA,$7$ $FADH_2$,and $7$ $NADH$.Each Acetyl-CoA enters the Krebs cycle to produce $3$ $NADH$,$1$ $FADH_2$,and $1$ $ATP$ (or $GTP$).For $8$ Acetyl-CoA,the total yield is $24$ $NADH$,$8$ $FADH_2$,and $8$ $ATP$.Adding the products from $\beta$-oxidation ($7$ $FADH_2$ and $7$ $NADH$),the total yield is $31$ $NADH$,$15$ $FADH_2$,and $8$ $ATP$.Using standard conversion factors ($1$ $NADH = 2.5$ $ATP$ and $1$ $FADH_2 = 1.5$ $ATP$): $31 	imes 2.5 = 77.5$ $ATP$ and $15 	imes 1.5 = 22.5$ $ATP$.Total $ATP = 77.5 + 22.5 + 8 = 108$ $ATP$.However,considering the older convention ($1$ $NADH = 3$ $ATP$ and $1$ $FADH_2 = 2$ $ATP$): $31 	imes 3 = 93$ $ATP$ and $15 	imes 2 = 30$ $ATP$.Total gross $ATP = 93 + 30 + 8 = 131$ $ATP$.Since $2$ $ATP$ equivalents are consumed for the activation of fatty acid into Acyl-CoA,the net gain is $131 - 2 = 129$ $ATP$.

From the oxidation of one molecule of palmitic acid (fatty acid),the number of $ATP$ molecules gained as net are

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