How to calculate the energetic balance of the total oxidation of a fatty acid


The total oxidation of a fatty acid comprehends different processes:

 

1. – The activation of the fatty acid

 

2. – Beta-Oxidation

 

3. – Krebs Cycle.

 

For being metabolized, a fatty acid should experiment 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 hydrolyzed (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 reactions occur 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 oxidized to Acetyl groups in form of Acetyl CoA units.

 

Beta-Oxidation

Beta-Oxidation

 

 

 

 

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 Acetyl 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 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:

 

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 = 15 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 =  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 (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!

 

 

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27 thoughts on “How to calculate the energetic balance of the total oxidation of a fatty acid

  1. I can’t seem to figure out how the initial step costs 2 ATP. I know that fatty acyl activation to a fatty acyl-CoA costs 1 ATP and yields AMP and pyrophosphate (pyrophosphate then becomes hydrolyzed to 2 phosphates). I’ve seen the terminology 2 ATP equilavents before which often refers to the ATP hydrolyzing to AMP and PP (since PP will eventually become 2 P). Any insight would be helpful into determining how the initial priming step costs 2 ATP.

  2. A little more info about this subject (thanks) because I have seen conflicting reports. The balanced equation for the priming step involves just 1 ATP. The calculation of a complete oxidation of an even number, saturated fatty acid has to subtract 2 ATPs for the initial priming of the fatty acid to fatty acyl-CoA. Often I see the reasoning for the -2 ATP being “The equivalent of 2 molecules of ATP is consumed in the activation of a fatty acid, in which ATP is split into AMP and 2 molecules of Pi.” Although it states the equivalent of 2 ATP, the priming reaction appears to only consume 1 ATP (but has the energy of 2 due to the 2 “high energy” bonds cleavages of (1) ATP to AMP and PP; then (2) PP to 2 Pi. So I’m not sure which source is correct and I hope I posed my concern as clear as I could possible manage. Thanks in advance for you insights. I suppose I could just go with what the textbook deems as correct, but my understanding is that the textbook misunderstood that “the equivalent of 2 molecules of ATP” to actually mean 2 ATPs rather than 1 ATP with 2 high energy bonds.

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  4. This is really easy to understand and educating thanks so much aw about the calculation on odd number fatty acid carbons?

  5. hello.
    urm i just need to know
    i am not quite understand in which it said that each beta-oxidation, one molecule of ATP is required to prime the process.
    Right now i am calculating the net ATP produce from 6C-fatty acid.
    so should the answer that I got minus with 3 molecule of ATP to pay back the ATP used at first?

  6. Linoleic acid 18 carbon atom. Number acetylcoa form 18/2=9 acetyl CoA. Number rounds =9-1=8rounds FADH2=2*8=16ATP NADH+H=3*8=24ATP. acetyl coa 12*9=118ATP. TOTAL =118+16+24-2=156ATP

  7. Linoleic acid 18 carbon atom. Number acetylcoa form 18/2=9 acetyl CoA. Number rounds =9-1=8rounds FADH2=2*8=16ATP NADH+H=3*8=24ATP. acetyl coa 12*9=108ATP. TOTAL =108+16+24-2=146ATP

  8. (-2) + 32 + 90 =120 ATP
    8 NADH × 2.5 ATP/NADH = 20ATP
    8 FADH2 × 1.5 ATP/FADH2= 12ATP
    TOTAL 32

    9+ acetyl CoA × 10ATP/acetyl CoA. = 90atp
    122ATP – 2 = 120ATP

  9. Nalla825: Think… The one ATP goes to one AMP and not ADP as usual. The energy in ATP, or any other nucleotide-triphosphate, is contained in the terminal P-O bond. So, if two P-O bonds have been broken, then in effect two ATP’s have been consumed. It’s going too take a little more than that much energy to put the two phosphates back together.

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