Question

Calculate the number of ATP molecules produced by complete β−oxidation of the fourteen-carbon saturated fatty acid tetradecanoic acid (common name: myristic acid).

THE ANSWER IS NOT 92!!!

Answer 1

Since the Acetyl group of the acetyl coA is formed by two carbons, we should divide the number of carbons in the acyl group between two.

myristic acid (14 carbons): 14 carbons /2 = 7 Acetyl CoA

In order to be completed degraded to Acetyl CoA, the fatty acid in form of acyl CoA should experiment several “rounds” in the Beta-oxidation process. En each round, an Acetyl CoA is released and, a NADH.H+ and a FADH2 are produced.

We know already how many acetyls CoA are formed from each fatty acid: n/2, where n is the number of carbons.

Now, how many rounds are necessary for converting all the fatty acid to acetyl CoA?

Since in the last round we already obtain two Acetyl CoA, the number of necessary rounds is (n/2) -1

Observe:

myristic acid (14 carbons)

1st round:

Produce one acyl CoA of 12 carbons and one Acetyl CoA + NAD H.H+ + FADH2

2nd round:

Produce one acyl CoA of 10 carbons and one Acetyl CoA + NAD H.H+ + FADH2

3rd round:

Produce one acyl CoA of 8 carbons and one Acetyl CoA + NAD H.H+ + FADH2

4th round:

Produce one acyl CoA of 6 carbons and one Acetyl CoA + NAD H.H+ + FADH2

5th round:

Produce one acyl CoA of 4 carbons and one Acetyl CoA + NAD H.H+ + FADH2

6th round:

Produce one acyl CoA of 2 carbons and one Acetyl CoA + NAD H.H+ + FADH2

But the acyl CoA of 2 carbons is already an Acetyl CoA, so it is the last round!

So, after 4 rounds, all the myristic acid has been converted to Acetyl CoA.

Then, what we have obtained as a result of the beta-oxidation of Miristic acid?

7 acetyl CoA

And 6 NADH.H+ y 6 FADH2

In terms of ATP, the yielding depends on the kind of yielding that is used for reduced cofactors:

a) If during your course it is considered that each NADH yields 2.5 ATP and each FADH2 yields 1.5 ATP, then

7 acetyl CoA x 10 ATP/AcetylCoA in the Krebs Cycle: 70 ATP

6 NADH x 2.5 ATP/NADH = 15 ATP

6 FADH2 x 1.5 ATP/FADH2 = 9 ATP

Minus 2 ATP used in the activation (myristic acid to myristyl CoA)

70+11+9-2 = 92 ATP

b) If during your course it is considered that each NADH yields 3 ATP and each FADH2 yields 2 ATP, then

7 acetyl CoA x 12 ATP/AcetylCoA in the Krebs Cycle: 84 ATP

6 NADH x 3 ATP/NADH = 18 ATP

6 FADH2 x 2 ATP/FADH2 = 12 ATP

Minus 2 ATP used in the activation (myristic acid to myristyl coA)

84+18+12-2 = 112 ATP

YOUR ANSWER IS 112 ATP

Thank You So Much! Please Rate this answer as you wish.