Oxaloacetate uniformly labeled with 14c (i. e., with equal amounts of 14c in each of its carbon atoms) is condensed with unlabeled acetyl-coa. after a single pass through the citric acid cycle back to oxaloacetate, what fraction of the original radioactivity will be found in the oxaloacetate?

answer is 1/2. explain how and why that is correct.

Your answer is correct as ½ fraction of 14C signal would be recovered.

The original oxaloacetate scaffold has 4 labeled carbon atom of which 2 labeled carbon atom are deposited as the other 2 labeled carbon atom are lost during the Krebs cycle resulting in ½ fraction of 14C signal to be recovered.

In the Krebs cycle, the oxaloacetate-contributed carbons gives the 2 carbon atoms involved in the releasing of two molecules of CO2

solvent

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the answer would be d : p