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 scientific method.

hydrogen is a compound or a single element and a element is a pure substence that cannot be separated into simpler subtances

clorine is a compound its a chemical substance that forms when two or more different elements combine

so both hydrogen and clorine are compounds