Question

The sequence of action of following enzymes of plant fatty acid synthase complex is:
(A). $\beta$-Ketoacyl-ACP reductase
(B). $\beta$-Ketoacyl-ACP synthase
(C). $\beta$-Hydroxyacyl-ACP dehydratase
(D). Enoyl-ACP reductase
Choose the correct answer from the options given below:

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

Hint:

Remember the mnemonic C-R-D-R for fatty acid elongation: Condensation $\rightarrow$ Reduction $\rightarrow$ Dehydration $\rightarrow$ Reduction. This corresponds to Synthase $\rightarrow$ Reductase $\rightarrow$ Dehydratase $\rightarrow$ Reductase.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Fatty acid synthesis in plants occurs in the plastids via a repeated four-step cycle of condensation, reduction, dehydration, and a second reduction. This cycle extends the growing acyl chain by two carbons per round.

Step 2: Detailed Explanation:
The sequential reactions catalyzed by the Fatty Acid Synthase (FAS) complex during each round of elongation are:
1. Condensation: $\beta$-Ketoacyl-ACP synthase (B) condenses acetyl-ACP (or a growing acyl-ACP) with malonyl-ACP to yield $\beta$-ketoacyl-ACP.
2. First Reduction: $\beta$-Ketoacyl-ACP reductase (A) reduces the keto group of $\beta$-ketoacyl-ACP to form $\beta$-hydroxyacyl-ACP using NADPH.
3. Dehydration: $\beta$-Hydroxyacyl-ACP dehydratase (C) removes a water molecule, introducing a trans double bond to form enoyl-ACP.
4. Second Reduction: Enoyl-ACP reductase (D) reduces the double bond of enoyl-ACP to form a fully saturated acyl-ACP.
Therefore, the correct sequence of enzyme action is (B) $\rightarrow$ (A) $\rightarrow$ (C) $\rightarrow$ (D).

Step 3: Final Answer:
The correct sequence is (B), (A), (C), (D).