Question

At the end of Krebs cycle, but before the electron transport chain, the oxidation of glucose has produced a net gain of -

• A. $3C{O}_2,5NAD{H}_2,1FAD{H}_2,2ATP$
• B. $6C{O}_2,10NAD{H}_2,2FAD{H}_2,4ATP$
• C. $6C{O}_2,10NAD{H}_2,2FAD{H}_2,38ATP$
• D. None of these

Answer 1

Answer: (B)

$6C{O}_2,10NAD{H}_2,2FAD{H}_2,4ATP$

One molecule of the glucose that enters in the glycolysis produces 2 molecules pyruvate and 2 molecules of NADH${}_2$ and 2 ATP. 2 molecules of pyruvate will form 2 molecules of Acetyl Co-A and this will release 2 molecules of CO${}_2$ and 2 NADH${}_2$. These 2 Acetyl Co-A will enter into Kreb's cycle and will release 4CO${}_2$, 6NADH${}_2$, 2FADH${}_2$ and 2ATP.

so, the total gain before the ETC comes out to be 6CO${}_2$, 10NADH${}_2$, 2FADH${}_2$ and 4ATP.

Hence, the correct answer is '6CO${}_2$, 10NADH${}_2$, 2FADH${}_2$, 4ATP'.