Question

Propose a cycle for the β-degradation of a hypothetical saturated fatty acid with 20 carbon
atoms and calculate the number of ATPs regenerated by degradation scheme you have
proposed.

Answer 1

Explanation:

Total ATPs =129. The odd chain, unsaturated fatty acid will be able to go two rounds through the β oxidation pathway before it is unable to do

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Answer 2

Answer:

A 20 carbon saturated fatty acid would undergo 9 rounds of β-oxidation. In every cycle of β-oxidation, one FADHA2 is produce at fatty acyl-CoA dehydrogenase step and one NADH in 3-hydroxyacyl-CoA dehydrogenase step. So, 9 rounds of β-oxidation of a 20 carbon fatty acid produces 9 NADH, 9 FADH2, and 10 acetyl-COA.

Explanation:

Fatty acid oxidation takes place in four stages - dehydrogenation, hydration, oxidation, and thiolysis. These four stages keep repeating until the whole molecule is oxidized. Each of these four stages is catalyzed by a different enzyme.

The net reaction is thio-lysis of the carbon-carbon bond. The carbonyl carbon in β-ketoacyl-CoA is electrophilic. Active-site thiolate acts as a nucleophile and releases acetyl-CoA. Terminal sulfur in CoA-SH acts as a nucleophile and picks up the fatty acid chain from the enzyme.

Beta-oxidation is a four-step process, which repeats until the fatty acid has been completely broken down. The four steps are dehydrogenation, hydration, oxidation, and thiolysis. Dehydrogenation is catalyzed by Acyl-CoA-dehydrogenase and coverts FAD to FADH2 to form a double bond between C2 and C3.

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