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.- 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!

 

 

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

  1. Pingback: How to calculate QUICKLY the energetic balance of the total oxidation of a fatty acid « The Biochemistry Questions Site

  2. i dont understand why we hav to consider 2 ATP has been consumed instead of 1 ATP? Could you explain in detail? Thank you.

    • When 1 molecule of ATP is hydrolyzed to ADP + (P), it is considered that from the energetic point of view, it has been expended 1 rich in energy bond (equivalent to 1 ATP)
      When 1 molecule of ATP is hydrolyzed to AMP + 2 (P), it is considered that from the energetic point of view 2 rich in energy bonds have been consumed (2 ATPs)

      In the activation of a fatty acid:
      Fatty acid +CoA +ATP —> Acyl CoA + AMP + 2 (P)
      It explains that when calculating the energetic balance, it is considered that 2 rich in energy bonds (2 ATP) have been consumed, even when from the molecular point of view, only 1 molecule of ATP was used.
      This kind of reasoning is found in other energetic balances, like in urea cycle.

  3. Solution:
    . step 1 18/2 = 9 acetyl-CoA
    . step 2 9-1 = 8rounds
    . step 3 (8x 5) + (9×12)= 148
    . step 4 148-2

    tOTAL: 146 ATP

    Thanks!
    Your method is easier than the one i’ve learnt from text book. I’m going to hav a metabolism test on this coming wed. :p

  4. thaaaaaaaaanxxxxxxxxx alot this really helped me 2 b able to calculate the total amount of ATP when given any numbar of carbons… this is well explaned and written in an easy simple way… It defently will help me in my bio exam nxt week!!! thanks again!

  5. Does anyone know how to calculate the energy given off by the oxidation of butter fat?

    Here is what I’m given:
    butter = 5g
    water in calorimeter = 1 kg
    Initial Temperature = 23 C
    Final Temperature = 68 C

    I just need help approaching this. Thanks =)

  6. Howdy just wanted to give you a brief heads up and let you
    know a few of the images aren’t loading properly. I’m not sure why but I think its a linking issue.

    I’ve tried it in two different web browsers and both show the same results.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s