Question

The total degradation of a fatty acid with an odd number of carbons yields acetyl-CoA and another compound, X. Show the structure of X, and describe the pathway by which it is converted into a citric acid cycle intermediate. including where any cofactors participate.

Answer 1

Answer: X is propionyl-CoA, CH3CH2C0CoA

The structure and reaction pathway are shown in the attachments.

Explanation: In the oxidation of odd-number fatty acids, the substrate for the last pass through beta-oxidation is fatty acyl-CoA with a five-carbon fatty acid. This is oxidized and cleaved to acetyl-CoA and propionyl-CoA. Acetyl-CoA enters the citric acid cycle while the propionyl-CoA is converted in three enzymatic steps to succinyl-CoA which can then enter the citric acid cycle.

Step 1: Conversion of propionyl-CoA to D-methylmalonyl-CoA

Propionyl-CoA is carboxylated to D-methylmalonyl-CoA by the enzyme propionyl-CoA carboxylase, which contains the cofactor biotin. A molecule of ATP and Carbon (iv) oxide (in the form of hydrogen carbonate ion) is required also.

Step 2: Epimerization of D-methylmalonyl-CoA

D-methylmalonyl-CoA is epimerized by the enzyme methylmalonyl-Co epimerase to its L stereoisomer, L-methylmalonyl-CoA.

Step 3: Conversion of L-methylmalonyl-CoA to succinyl-CoA

This reaction is catalysed by methylmalonyl-CoA mutase which requires the coenzyme B12.