For the $\beta$-oxidation of a fatty acid with notation of 18:3c$\Delta$6,9,12, how many NADPH would be required and how many FADH₂ would be produced?

Question

Quizplus · Accepted Answer

The fatty acid 18:3cΔ6,9,12 has three double bonds. Two of these double bonds are in positions that require reduction by NADPH during β-oxidation. The complete oxidation of an 18-carbon fatty acid involves 8 cycles of β-oxidation, producing 8 FADH2. However, the presence of double bonds affects the number of FADH2 produced. For each double bond that bypasses the acyl-CoA dehydrogenase step, one less FADH2 is produced. Thus, 7 FADH2 are produced, and 2 NADPH are required for the reduction of the double bonds.

For The β\betaβ -Oxidation of a Fatty Acid with Notation of 18:3c

For The $\beta$ -Oxidation of a Fatty Acid with Notation of 18:3c