# Question #4323e

Apr 1, 2016

The β-oxidation of an 18-carbon acid produces ${\text{9 mol of Acetyl-CoA, 8 mol of NADH, and 8 mol of FADH}}_{2}$.

#### Explanation:

The β-oxidation pathway for a saturated fatty acid is a four-step cycle.

A ${\text{C}}_{18}$ acid (stearic acid) enters as $\text{stearyl-CoA}$.

Each pass through the cycle removes 2 carbon atoms and generates 1 molecule each of ${\text{acetyl-CoA, NADH, and FADH}}_{2}$.

The first seven passes through the cycle remove 14 carbon atoms and generate 7 molecules each of ${\text{acetyl-CoA, NADH, and FADH}}_{2}$.

In the eighth pass, the remaining 4 carbon atoms are converted to 2 molecules of $\text{acetyl-CoA}$ and 1 each of $\text{NADH}$ and ${\text{FADH}}_{2}$.

This makes a total of ${\text{9 acetyl-CoA, 8 NADH, and 8 FADH}}_{2}$.

The overall equation for the β-oxidation of $\text{stearyl-CoA}$ is:

${\text{CH"_3("CH"_2)_16"CO-SCoA" + "8FAD" + "8NAD"^+ + "8HSCoA" + "8H"_2"O" → color(red)(9)"CH"_3"CO-SCoA" + color(red)(8)"NADH" + color(red)(8)"FADH"_2 + "8H}}^{+}$