Glucose labeled with 14c in c-3 and c-4 is completely converted to acetyl-coa via glycolysis and the pyruvate dehydrogenase complex. what percentage of the acetyl-coa molecules formed will be labeled with 14c, and in which position of the acetyl moiety will the 14c label be found?
a) 100% of the acetyl-coa will be labeled at c-1 (carboxyl).
b) 100% of the acetyl-coa will be labeled at c-2
c) 50% of the acetyl-coawill be labeled, all at c-2 (methyl).
d) no label will be found in the acetyl-coa molecule
e) not enough information is given to answer this question.

The correct answer is option d.

Explanation:

The production of Acetyl-CoA takes place by the dissociation of both carbohydrates and lipids in the process of glycolysis and beta-oxidation. It then moves into the TCA cycle in the mitochondria and combines with oxaloacetate to give rise to citrate.  

In the given case, no labeling will be found in the acetyl-CoA. The labeled C3 and C4 carbon of glucose signify the carboxyl carbon of pyruvate. In the succeeding reactions of the transformation of pyruvate to acetyl-CoA, the carboxyl carbon gets lost in the form of carbon dioxide. Thus, acetyl-CoA does not comprise any labeled C3 and C4 of glucose.  

the binding of the substrate is dictated by the shape of each molecule. side chains on the enzyme interact with the substrate in a specific way, resulting in the making and breaking of bonds.

i believe the answer would be b. hope it : )