Answer to Question #101578 in Chemistry for NIBEDITA RURA

Question

Explain the mechanism of action of the enzyme aconitase on citrate in TCA cycle.
Compare the energetics of metabolic oxidation of glucose and a fatty acid.

Answer 1

Transformation of citrate to isocitrate by the enzyme aconitase is performed in two major steps. First, aconitase provides dehydration of citrate by removing -OH group and proton. The process results in the formation of a cis-aconitate intermediate. Second, aconitase mediates the 180-degree flip of cis-aconitate in the active centre of the enzyme that is accompanied by the rehydration of cis-aconitate resulting in the formation of citrate.

Metabolic oxidation of one molecule of glucose (glicolysis and TCA) results in the formation of:

+4 NADH = 10 ATP

+4 ATP

+2 Acetyl-CoA = 20 ATP

-2 ATP (used for phosphorylation)

Total yield = 32 ATP

Metabolic oxidation of one molecule of palmitate (beta-oxidation and TCA) leads to the formation of:

+7 FADH₂ = 10.5 ATP

+7 NADH = 17.5 ATP

+8 Acetyl-CoA = 80 ATP

- 2 ATP (used in the activation phase)

Total yield = 106 ATP

As a result, although glucose oxidation and fatty acid oxidation results in the formation of acetyl-CoA that enters the TCA cycle, the oxidation of one fatty acid molecule leads to the formation of more ATP molecules than the oxidation of one glucose molecule.