For the following fatty acids determine the (i) number of ATP you would get from

Question

For the following fatty acids determine the (i) number of ATP you would get from beta oxidation followed by entry into TCA and (ii) track a label at the terminal carbon (carbon 12 for (a) and carbon 13 for (b)) and show where it would turn up in oxaloacetate in the first round of TCA. (a) 12:2c Delta 4, 8 (b) 13:2c Delta 5, 9 Consider how many ATP you can hypothetically create from the energy stored in linoleic acid (18:2c Delta 9, 12). Now count how many ATP you 'actually' achieve. What is the efficiency of ATP creation for linoleic acid?

Explanation / Answer

Ans 1:

The oxidation of a fatty acid follows:

Fatty acid + CoA + ATP --- AcylCoA + AMP + 2P

2 ATP are consumed in the activation process

The formed acyl CoA undergoes oxidation reactions and these reactions occur in the beta carbon thus known as beta-oxidation.

First round generates:

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

Second round generates:

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

Third round generates:

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

Fourth round generates:

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

Fifth round generates:

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

Sixth round generates:

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, therefore, after fourth round all the miristic acid is converted to acetyl CoA. And there is a formation of seven acetyl CoA and six NADH.H and yielding 6 FADH2

Now, ATP yielding depends on the reduced cofactors:

If considering, NADH.H+ yields 2.5 ATP and each FADH2 yields 1.5 ATP, then

7 acetyl CoA *10 ATP/AcetylCoA in the Krebs Cycle = 70 ATP

6  NADH * 2.5 ATP/NADH = 15 ATP

6 FADH2 * 1.5 ATP/FADH2 = 9 ATP

Now, substracting 2 ATP which are used in the activation, therefore,

                                          = 70 + 11 + 9 - 2 = 92 ATP

If considering, NADH yields 3 ATP and each FADH2 yields 2 ATP

Therefore, 7 acetyl CoA * 12 ATP/AcetylCoA in the Krebs Cycle = 84 ATP

6 NADH * 3 ATP/NADH = 18 ATP

6 FADH2 * 2 ATP/FADH2 = 12 ATP

Now, substracting 2 ATP which are used in the activation, therefore,

                                                     = 84 + 18 + 12 - 2 = 112 ATP

Now, calculating the energetic balance of 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 fatty acid to Acetyl Co A units: Number of Acetyl CoA minus 1 = [(n/2)-1]

Step 3 (i) If considering, NADH.H+ 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 time to 10

(ii) If considering, NADH yields 3 ATP and each FADH2 yields 2 ATP then, multiply the number of rounds times to 5 and multiply the number of Acetyl CoA times to 12.

Step 4. Two ATP that were used for the activation of the Fatty Acid

Fatty acid with 12 Carbons

Following (i):

Step 1:  12/2 = 6 Acetyl CoA

Step 2: 6-1 = 5 rounds

Step 3 (i): (5 * 4) + (6 * 10)

Step 4 = -2 ATP

Total: 78 ATP

Following (ii):

Step 1:  12/2 = 6 Acetyl CoA

Step 2: 6-1 = 5 rounds

Step 3 (ii): (5 * 5) + (6 * 12)

Step 4 = -2 ATP

Total: 95 ATP

Ans 2:

Linoleic acid has has 18 carbonwith two double bond in the structure. As a result there is a reduction in the two FADH2. Therefore, a total of 9 acetyl-CoA, 6 FADH2 and 8 NADH

Therefore, activation energy = two less ATP

Oxidatio of acetyl-CoA, 9*10= 90 ATP

Oxidation of FADH2, 6*15 = 9 ATP

Oxidation of NADH, 8*25 = 20 ATP

Thus, total ATP = 90+20+9-2 = 117 ATP.

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