The total oxidation of a fatty acid comprehends different processes:
1.- The activation of the fatty acid
2.- Beta-Oxidation
3.- Kreb’s Cycle.
For being metabolized, a fatty acid should experiment an activation:
Fatty acid + CoA + ATP —-à Acyl CoA + AMP + 2(P)
The activation of the fatty acid requires 1 molecule of ATP, but since two energy rich bonds are hydrolized (the ATP is hydrolyzed to AMP and 2 (P) ) for energetic balance purposes it is considered that 2 ATP have been consumed in this activation process)
The formed acyl CoA will experiment different oxidation reactions. These reaction occurs in the Beta-carbon. That is why the process is called Beta-oxidation
Recalling:
CH3 -……………………………….. -CH2 – CH2 – CH2 – CH2 -COOH
(Omega carbon) -Delta-Gamma-Beta-Alpha-Carboxyl
In Beta-oxidation (a mitochondrial process) the acyl CoA is totally oxidazed to Acetyl groups in form of Acetyl CoA units.
How many Acetyl CoA units will be formed?
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.
Miristic acid (14 carbons): 14 carbons /2 = 7 Acetil CoA
Palmitic acid (16 carbons): 16 carbons/2 = 8 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 acetyl 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:
Miristic 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 Miristic 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 yielding 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 = 11 ATP
6 FADH2 x 1.5 ATP/FADH2 = 9 ATP
Minus 2 ATP used in the activation (Miristic acid to miristyl CoA)
70+11+9-2 = 88 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 NADHx 3 ATP/NADH = 18 ATP
6 FADH2 x 2 ATP/FADH2 = 12 ATP
Minus 2 ATP used in the activation (Miristic acid to Miristyl coA)
84+18+12-2 = 112 ATP
How to calculate the energetic balance of any fatty acid?
Step 1.- Number of Carbons/2 = Number of Acetyl CoA formed.
Step 2.- Number of rounds in the Beta-oxidation necessary for converting the whole fatty acid to Acetyl Co A units: Number of Acetyl CoA minus 1 [(n/2)-1]
Step 3 (Option a) If you consider that each NADH yields 2.5 ATP and each FADH2 yields 1.5 ATP then multiply the number of rounds times 4 and multiply the number of Acetyl CoA times 1o
OR
Step 3 (Option b).- If you consider that each NADH yields 3 ATP and each FADH2 yields 2 ATP then multiply the number of rounds times 5 and multiply the number of Acetyl CoA times 12.
Step 4.- Take two ATP that were used for the activation of the Fatty Acid
Example:
Fatty acid with 12 Carbons
– Option (a):
Step 1: 12/2 = 6 Acetyl CoA
Step 2: 6-1 = 5 rounds
Step 3 (Option a): (5 x 4) + (6 x10)
Step 4 = -2
Total: 78 ATP
– Option (b):
Step 1: 12/2 = 6 Acetyl CoA
Step 2: 6-1 = 5 rounds
Step 3 (Option b): (5 x 5) + (6 x12)
Step 4 = -2
Total: 95 ATP
I would like now that the reader calculates the energetic balance of the total oxidation of a fatty acid with 18 carbons. Assume that the oxidation of each NADH.H+ yields 3 ATP and that each FADH2 yields 2 ATP.
I am looking forward to read your answers in the comments!