Question

Given a saturated fatty acid, be able to calculate the ATP that would be produced upon complete oxidation. Be able to explain the lower ATP yield obtained with unsaturated fatty acids. Calculate how many ATPs would be produced from oxidation of a 14 carbon fatty acid.

Answer 1

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 Miristic acid has been converted to Acetyl CoA.

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

7 acetyl CoA

And 6 NADH.H+ y 6 FADH2

The acetyl CoA units are oxidized up to CO2 and H2O in the Krebs Cycle.

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

a) If during your course it is considered that each NADH.H+ 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

Beta-oxidation of unsaturated fatty acids changes the ATP yield due to the requirement of two possible additional enzymes.

?-Oxidation of unsaturated fatty acids poses a problem since the location of a cis bond can prevent the formation of a trans-?2 bond. These situations are handled by an additional two enzymes, Enoyl CoA isomerase or 2,4 Dienoyl CoA reductase.

Beta oxidation (Unsaturated fatty acid)

Whatever the conformation of the hydrocarbon chain, ?-oxidation occurs normally until the acyl CoA (because of the presence of a double bond) is not an appropriate substrate for acyl CoA dehydrogenase, or enoyl CoA hydratase:

•             If the acyl CoA contains a cis-?3 bond, then cis-?3-Enoyl CoA isomerase will convert the bond to a trans-?2 bond, which is a regular substrate.

•             If the acyl CoA contains a cis-?4 double bond, then its dehydrogenation yields a 2,4-dienoyl intermediate, which is not a substrate for enoyl CoA hydratase. However, the enzyme 2,4 Dienoyl CoA reductase reduces the intermediate, using NADPH, into trans-?3-enoyl CoA. As in the above case, this compound is converted into a suitable intermediate by 3,2-Enoyl CoA isomerase.