Final answer:
Stage 3 of fatty acid synthesis consists of reduction, dehydration, and saturation reactions catalyzed by specific enzymes using NADPH as an electron donor, which ultimately results in the formation of a 16-carbon fatty acid, palmitoyl-ACP.
Step-by-step explanation:
Stage 3 of fatty acid synthesis involves a set of chemical reactions which are the reverse of the ß-oxidation pathway used in fatty acid degradation. However, the enzymes involved in fatty acid synthesis are different from those of the degradation pathway.
- The first step is the reduction of acetoacetyl-S-ACP to 3-hydroxyacyl-S-ACP, catalyzed by 3-ketoacyl-ACP reductase, using NADPH as an electron donor.
- This is followed by dehydration of 3-hydroxyacyl-S-ACP by 3-hydroxyacyl-ACP dehydratase to form trans-A²-enoyl-S-ACP.
- Finally, enoyl-S-ACP is converted to saturated acyl-S-ACP by the action of enoyl-ACP reductase, using another molecule of NADPH.
The process is cyclic, repeating until palmitoyl-ACP, a 16-carbon fatty acid, is synthesized, after which further elongation and addition of double bonds are carried out by different enzyme systems.
Stage 3 of fatty acid synthesis consists of a series of reactions that are essentially the opposite of the reactions involved in fatty acid degradation. The reactions in stage 3 are as follows:
Reduction of acetoaceyl-S-ACP to form 3-hydroxyacyl-S-ACP, catalyzed by 3-ketoacyl-ACP reductase.
Dehydration of 3-hydroxyacyl-S-ACP to form trans-A²-enoyl-S-ACP, catalyzed by 3-hydroxyacyl-ACP dehydratase.
Conversion of enoyl-S-ACP to saturated acyl-S-ACP, catalyzed by enoyl-ACP reductase.
These reactions are necessary for the synthesis of fatty acids.