A) \[3\text{ }FAD{{H}_{2,}}~2\text{ }NAD{{H}_{2,}}1\text{ }ATP\]
B) \[2\text{ }FAD{{H}_{2,}}~2\text{ }NA{{D}_{2,}}~~~2\text{ }ATP\]
C) \[2\text{ }NAD{{H}_{2,}}~1\text{ }FAD{{H}_{2,}}2\text{ }ATP\]
D) \[3\text{ }NAD{{H}_{2,}}~1\text{ }FAD{{H}_{2,}}1\text{ }ATP\]
Correct Answer: D
Solution :
The total yield in one Krebs' cycle is \[3NAD{{H}_{2}},\text{ }IFAD{{H}_{2}},\text{ }1ATP\]
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