Question

The sequence of reactions catalyzed by pyruvate dehydrogenase complex are:
(A). Decarboxylation of pyruvate
(B). Formation of Acetyl-CoA
(C). Formation of acetyl lipoamide
(D). Oxidation of dihydrolipoamide
Choose the correct answer from the options given below

• A. (A), (B), (C), (D).
• B. (A), (C), (B), (D).
• C. (B), (A), (D), (C).
• D. (C), (B), (D), (A).

Hint:

Follow the journey of the acetyl group: it starts on pyruvate (A), gets passed to lipoamide (C), and finally ends up on Coenzyme A (B). The regeneration of the enzyme complex happens last (D).

Answer 1

The Correct Option is B

Step 1: Overview:
The pyruvate dehydrogenase complex (PDC) bridges glycolysis and the citric acid cycle. It converts pyruvate to acetyl-CoA through oxidative decarboxylation, involving three enzymes (E1, E2, E3) and five cofactors.
Step 2: Reaction Steps:
The process unfolds in this order:
(A) Pyruvate Decarboxylation: Pyruvate dehydrogenase (E1) removes a carboxyl group from pyruvate, releasing CO\(_2\). The remaining two-carbon fragment binds to the thiamine pyrophosphate (TPP) cofactor.
(C) Acetyl Lipoamide Formation: The two-carbon acetyl group transfers from TPP to the oxidized lipoamide arm of dihydrolipoyl transacetylase (E2), creating an acetyl-lipoamide intermediate.
(B) Acetyl-CoA Formation: The acetyl group shifts from acetyl-lipoamide to coenzyme A, yielding acetyl-CoA. The lipoamide arm is left reduced (dihydrolipoamide).
(D) Dihydrolipoamide Oxidation: Dihydrolipoyl dehydrogenase (E3) re-oxidizes the dihydrolipoamide arm. Electrons are transferred to FAD (producing FADH\(_2\)), then to NAD\(^+\) (producing NADH), regenerating the complex.
Step 3: Conclusion:
The correct sequence is (A) \(\rightarrow\) (C) \(\rightarrow\) (B) \(\rightarrow\) (D), corresponding to option (2).