When two molecules of acetyl CoA enter the TCA cycle, net gain at the end of the cycle is

{2 N A D H}_{2} + {2 F A D H}_{2} + 1 G T P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{3 N A D H}_{2} + {2 F A D H}_{2} + 2 G T P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{6 N A D H}_{2} + {2 F A D H}_{2} + 2 G T P

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{3 N A D H}_{2} + {1 F A D H}_{2} + 4 G T P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ${6 N A D H}_{2} + {2 F A D H}_{2} + 2 G T P$
Krebs' cycle produces 2 GTP(or 2ATP) through substrate level phosphorylation, Six molecules of ${N A D H}_{2}$ and 2molecules of ${F A D H}_{2}$ for every two molecules of Acetyl CoA oxidised by it.

Suggest Corrections