62.7k views
0 votes
Stage 2 of fatty acid synthesis consists of two reactions. What are they?

User Grambo
by
8.7k points

1 Answer

3 votes

Final answer:

Stage 2 of fatty acid synthesis consists of the reduction of acetoaceyl-S-ACP to 3-hydroxyacyl-S-ACP and the dehydration of 3-hydroxyacyl-S-ACP to form trans-Δ2-enoyl-S-ACP. These reactions are catalyzed by 3-ketoacyl-ACP reductase and 3-hydroxyacyl-ACP dehydratase, respectively.

Step-by-step explanation:

Stage 2 of fatty acid synthesis involves two critical reactions. Firstly, the acetoaceyl-S-ACP is reduced to form 3-hydroxyacyl-S-ACP, a step catalyzed by the enzyme 3-ketoacyl-ACP reductase using NADPH as a reducing agent. Secondly, the 3-hydroxyacyl-S-ACP is dehydrated to form trans-Δ2-enoyl-S-ACP through the action of 3-hydroxyacyl-ACP dehydratase.

These reactions repeatedly add two-carbon units to the growing fatty acid chain, which ultimately releases a 16-carbon palmitoyl-ACP. Other systems, not the fatty acid synthase (FAS) complex, perform further elongation and the addition of double bonds. The released palmitate quickly reacts with coenzyme-A to form an acyl-CoA in the cell, with the synthesis using NADPH predominantly from the HMP pathway.

User Shackleton
by
7.9k points

No related questions found