Question

Oxaloacetate uniformly labeled with radioactive 14C (i.e., with equal amounts of 14C in each of its carbon atoms) is mixed with unlabeled acetyl-CoA and citrate synthase. The resulting molecule then takes a single pass through the citric acid cycle in order to reform oxaloacetate. What percentage of the original radioactivity will be found in the newly formed oxaloacetate

Answer 1

50% i.e
`(1)/(2)`

Step-by-step explanation:

Oxaloacetic acid is a dicarboxylic acid, i.e. a molecule comprising two carboxylic groups. Oxaloacetate conjugate base is an essential intermediary in several metabolic pathways, and in particular in the Krebs cycle (worsening to become Oxaloacetate). During the formation of acetate ion, oxaloacetic acid undergoes two successive deprotonations.

Since Oxaloacetate has 4 Carbons, suppose we label them with C14 it will be 100% radiolabelled. At each cycle of the Krebs cycle(citric acid cycle), two molecules of CO₂ are being eliminated from Oxaloacetate.

Thus, the addition of Acetyl-CoA to the radiolabelled Oxaloacetate leads to the formation of Citrate. Still, here, the Carbons belongs to Oxaloacetate only and lose Carbons in the form of CO₂.

Therefore, the Oxaloacetate formed from this Citrate has 50% of the total radiolabelled carbons.