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.