Q. 18.74 - An Introduction to General, Organic, and Biological Chemistry

Question

Arachidic acid is a ${\mathrm{C}}_{20}$ fatty acid found in peanut and fish oils.

a. How many cycles of $\mathrm{\beta }$ oxidation are needed for the complete oxidation of arachidic acid?
b. How many acetyl CoA are produced from the complete oxidation of arachidic acid?
c. Calculate the total ATP yield from the complete $\mathrm{\beta }$ oxidation of arachidic acid by completing the following:

Step-by-Step Solution

Answer

Part(a) Nine cycles of $\mathrm{\beta }$ oxidation are needed.

Part(b) Ten acetyl CoA are produced. 

Part(c)  

Activation		$\to -2 \mathrm{ATP}$
10 acetyl CoA	$\times 10 \mathrm{ATP}/\mathrm{acetyl} \mathrm{CoA}$	$\to 100 \mathrm{ATP}$
9 NADH	$\times 2.5 \mathrm{ATP}/\mathrm{NADH}$	$\to 22.5 \mathrm{ATP}$
9 FADH2	$\times 1.5 \mathrm{ATP}/{\mathrm{FADH}}_2$	$\to 13.5 \mathrm{ATP}$
total		$134 \mathrm{ATP}$

Part(a) Step 1 : Given information

We are given arachidic acid. We need to find cycles of $\mathrm{\beta }$ oxidation that are needed for the complete oxidation of acid.

Part(a) Step 2 : Simplify

As we know arachidic is ${\mathrm{C}}_{20}$ fatty acid .

so, there will be nine oxidation cycles are needed for complete oxidation of this fatty acid.

Part(b) Step 1 : Given information

We are given arachidic acid. We need to find number of acetyl CoA  produced from the complete oxidation of acid.

Part(b) Step 2 : Simplify

As we know arachidic is ${\mathrm{C}}_{20}$ fatty acid.

Therefore, ten acetyl CoA are produced from the complete oxidation of arachidic acid.

Part(c) Step 1 : Given information

We are given Arachidic acid. We need to calculate the total ATP yield from the complete $\mathrm{\beta }$ oxidation of acid.

Part(c) Step 2 : Simplify

The total ATP  yeild will be fom acetyl CoA , NADH, and FADH_{2 .}

Required table is : 

Activation		$\to -2 \mathrm{ATP}$
10 acetyl CoA	$\times 10 \mathrm{ATP}/\mathrm{acetyl} \mathrm{CoA}$	$\to 100 \mathrm{ATP}$
9 NADH	$\times 2.5 \mathrm{ATP}/\mathrm{NADH}$	$\to 22.5 \mathrm{ATP}$
9 FADH2	$\times 1.5 \mathrm{ATP}/{\mathrm{FADH}}_2$	$\to 13.5 \mathrm{ATP}$
Total		$134 \mathrm{ATP}$