Total Oxidation of a 17 carbon fatty acid, including oxidation of the resulting Propionyl CoA


Let’s use the basic calculations described in previous posts about this issue:

 

Beta-oxidation of  fatty acids with an odd number of carbons.

 

Energetic balance of the total (and I mean total) oxidation of a fatty acid with an odd number of carbons.

  

Oxidation of a fatty acid with 17 atoms of carbon.

 

 

Activation of the fatty acid to Acyl CoA = -2 ATP

 

Number of rounds in the Beta oxidation

(17/2) -1.5 = 8.5-1.5 = 7

7 rounds x  5 ATP/ round = 35 ATP

 

Number of units of Acetyl CoA produced = 7 Acetyl CoA

7 Acetyl CoA x 12 ATP/Acetyl CoA  = 84 ATP

 

Propionyl CoA up to Succinyl CoA = -1ATP

 

Succinyl CoA up to Malate = 3 ATP

 

Malate up to Pyruvate (1 NADPH.H+)

 

Pyruvate up to Acetyl Co A = 3 ATP

 

Acetyl CoA oxidation in the Krebs Cycle = 12 ATP

 

Total of ATP (considering the total oxidation of Propionyl CoA converted to Malate and then from Malate to Pyruvate and then from Pyruvate to Acetyl CoA = -2 + 35 +84 -1 + 3 +3 +12 =  134 ATP

 

In summary  (following the equivalence of  1 NADH.H+ yielding 3 ATP in the Respiratory Chain and 1 FADH2 yielding 2 ATP):

 

-Calculate the number of rounds of the fatty acid in the Beta-oxidation:

  Number of rounds  = (Number of carbons/2) -1.5

 

-The number of Acetyl CoA is the same as the number of rounds

 

-Subtract 2 ATP that were used in the initial activation of the fatty acid.

 

-Multiply the number of rounds x 5 ATP/round.

 

-Multiply the number of Acetyl CoA x 12 ATP/Acetyl CoA.

 

-Add 17 ATP produced in the total oxidation of Propionyl CoA to CO2

 

 

To practice this kind of exercise, I suggest that you do the calculations using now  the criteria that considers that each NADH.H+ oxidized in the Respiratory Chain yields 2.5 ATP and each FADH2 yields 1.5 ATP

 

 

I am looking forward to see your answers and comments!!!

 

 

Energetic Balance of the Total (and I mean Total) Oxidation of a Fatty Acid with an Odd number of Carbons.


In previous posts we have discussed how the fatty acids with an odd number of carbons chain, release 1 unit of Acetyl CoA and 1 unit of Propionyl CoA, instead of the two Acetyl CoA units released when a fatty acid with an even number of carbons is beta-oxidized.

 

Let’s use some examples, (we represent here just the carbons in the chains):

 

Example 1: a fatty acid with 6 carbons (Hexanoic acid)

 

C-C-C-C-C-C-C

 

During the Beta-oxidation, three units of Acetyl CoA are released (two carbons each):

 

C-C/C-C/C-C

 

Example 2: A fatty acid with 7 carbons (Heptanoic acid):

 

C-C-C-C-C-C-C

 

During the Beta-oxidation two units of Acetyl CoA and one unit of Propionyl CoA are released (two units of two carbons and one unit of three carbons):

 

C-C-C/C-C/C-C

 

As discussed previously in another post, the Acetyl CoA are oxidized in the Krebs Cycle, but the Propionyl CoA is used in the formation of Succinyl CoA, in a process that consumes 1 ATP (-1 ATP).

 

The Succinyl CoA can continue in the Krebs Cycle and form Oxalacetate. Oxalacetate will react with Acetyl CoA (Citrate Synthase reaction) to form Citrate, following the reactions in the Krebs Cycle. If it happens, we can consider that the atoms of carbons of the Propionyl CoA have followed an anaplerotic pathway (to be used in the Kreb’s Cycle without being consumed). 

 

BUT these carbons could also be completely oxidized if they follow this sequence of reactions:

 

Succinyl CoA + GDP + (P) – -> Succinate + CoA + GTP (it is equivalent to 1 ATP)

 

Succinate + FAD – – – – >Fumarate + FADH2 (It generates 2 ATP in the Respiratory Chain)

 

Fumarate + H2O—–> Malate

 

But the Malate can now diffuse from the matrix through the mitochondrial membranes and be decarboxylated (under the action of the cytoplasmatic malic enzyme) to Pyruvate (and production of 1 CO2).

 

Pyruvate can return to the interior of the mitochondria,  where another decarboxylation occurs, this time under the action of the  Pyruvate dehydrogenase complex, (with production of another CO2) and the formation of Acetyl CoA, whose Acetyl group will be oxidized in the Cycle producing other two molecules of CO2.

 

We can see that through this sequence of reactions it is possible the total oxidation of the three original carbons of the Propionate to 3 molecules of CO2! (To avoid confusions, observe that, yes, there are 4 decarboxylations, but one of the CO2 does not come originally from the Propionyl CoA, but from the carboxylation process in the conversion of Propionyl CoA to Succinyl CoA)

 

Which would be the energetic balance of the total oxidation of an odd chain fatty acid considering this sequence of reactions?

 

Let’s see:

 

Propionyl CoA to Succinyl Co A = -1 ATP

 

In the mitochondria, (following the reactions of the Kreb’s Cycle up to Malate):

 

Succinyl CoA + GDP + (P) –> Succinate +CoA + GTP (it is equivalent to 1 ATP)

 

Succinate + FAD ——– – – – > Fumarate + FADH2 (It generates 2 ATP in the Respiratory Chain)

 

Fumarate + H2O————–>Malate

 

In the cytoplasm:

 

Malate + NADP+ – – – >Pyruvate + NADPH.H+ + CO2 (we will not consider this  reduced cofactor in the balance since NADPH.H+ is not a source of energy, but a source of reduction equivalents for synthetic reactions)

 

In the mitochondria again:

 

Pyruvate + CoA + NAD+ —-> Acetyl CoA + CO2 + NADH.H+ (Observe that this NADH.H+ is generated inside the mitochondria, so it yields 3 ATP)

 

The Acetyl CoA produced in the previous reaction, when oxidized in the Krebs Cycle: 12 ATP

 

Therefore, considering this metabolic way,

 

-1 +1 +2 +3 + 12 = 17 ATP as a result of  the total oxidation of the Propionyl CoA generated by the beta-oxidation of a fatty acid of odd number of carbons.

 

 

Therefore, for calculating the energetic balance we should add 17 ATPs from the oxidation of the Propionyl CoA, to the ATPs generated in the Beta-oxidation, and the ATPs generated as a result of the oxidation in the Krebs Cycle  of the Acetyl CoA units formed during the Beta-oxidation of the odd chain fatty acid. ( We should recall also that 2 ATPs are consumed in the initial activation of the fatty acid)

 

In our next post we will analyze the oxidation of the heptadecanoic acid (17 carbons) as an example of the application of these calculations.

 

Oxidation of a fatty acid with 17 atoms of carbon


(This post analise the energetic balance considering that the Propionyl CoA follows an anaplerotic fate)

 

Apply the equations described in the previous post:

 

N= Number of Carbons

 

(N/2) -1.5 = Number of rounds in Beta-oxidation

 

(N/2) -1.5 = Number of acetyl CoA produced in Beta-oxidation

 

So, in terms of production and consumption of ATP of ATPs, the oxidation of a 17-carbons fatty acid will show the following energetic balance:

 

Activation of a fatty acid to Acyl CoA = -2 ATP

 

Number of rounds in Beta-Oxidation:

 

(17/2) – 1.5 = 8.5 -1.5 = 7

7 rounds x 5 ATP/round = 35 ATP

 

Number of produced Acetyl CoA: 7 Acetyl CoA

 

7 Acetyl CoA x 12 ATP/Acetyl CoA = 84 ATP

 

Additionally, the Beta-oxidation has produced 1 Propionyl CoA. The conversion of Propionyl CoA to Succinyl CoA, as described in a former post, will consume 1 ATP (Consider -1 ATP).

 

As described in a previous post:

 

We can consider the conversion of Propionyl CoA to Succinyl CoA as an anaplerotic pathway, in which case, the molecule of Succinyl CoA continue in the Krebs Cycle generating Oxalacetate in the following sequence of reactions:

 

Succinyl CoA + GDP + (P) —> Succinate + CoA + GTP (equivalent to 1 ATP)

 

Succinate   +  FAD ———-> Fumarate +  FADH2 (Generates 2 ATPs in the Respiratory Chain)

 

Fumarate +     H2O————– > Malate

 

Malate +   NAD+ —————->  Oxalacetate + NADH.H+ (Generates 3 ATPs in the Respiratory Chain) “

 

Total = (-1+1+2+3) = 5 ATPs

 

Total of ATPs produced (considering the anaplerotic fate of the Propionyl CoA turned into Succinyl CoA) = -2+35+84-1+6 = 122 ATP                        

 

BUT…

 

We may also consider a total oxidation that would include the carbon atoms of the Propionyl CoA!

 

In our next post, we will consider what happens if, instead of the Succinyl CoA following an anaplerotic pathway, it follows a path that allow the carbon atoms of Propionyl CoA end up being oxidized to CO2. It would allow a real  total oxidation of all the original carbons of the heptadecanoic acid or any other fatty acid with an odd number of carbons.

 

 

Oxidation of fatty acids with an odd number of carbons


 

Some readers have asked about the oxidation of odd chain fatty acids.

 

Reviewing the subject, I found a lack of detailed information in the literature available on the Internet, and even in texts of Biochemistry that are frequently used in the Schools of Medicine and Biochemistry courses in other academic careers.  This lack of information is the result of the fact that the bulk of fatty acids in our bodies, (and in the diet we consume) are generally fatty acids with an even number of carbons.

 

Referring to the oxidation of odd-chain fatty acids, texts and articles usually are limited to report that in the last round of the Beta-oxidation of fatty acids of this type, one Propionyl CoA and one Acetyl CoA are produced, and then these texts generally describe the conversion of Propionyl CoA to Succinyl CoA, but without specifically mentioning the ATP balance in these reactions.

 

Therefore, and answering your questions, I have included in this post the energetic considerations to take into account when analyzing the ATP production in the oxidation of fatty acids of an odd number of carbons:  

 

 

N= number of carbons 

 

(N/2) –1.5 =  Number of rounds in the Beta-oxidation

 

(N/2) –1.5 = Number of acetyl CoA produced in the Beta-oxidation

 

Additionally, 1 Propionyl CoA (3-carbon Acyl CoA) is produced in the last round.

 

Based on a yield of 3 ATP per NADH.H+ and 2 ATP per FADH2 that are oxidized in the respiratory chain:

 

-Multiply the number of turns in Beta-oxidation x 5 ATP / turn

 

Multiply the number of Acetyl CoA x 12 ATP / Acetyl CoA (since each Acetyl CoA yields 12 ATP when oxidized in the Krebs Cycle)

 

Subtract now two ATP (-2 ATP) consumed in the initial activation of the fatty acid (see the related post for explanation)

 

But also, in this process, as was written before,  1 Propionyl CoA is released, because in the last round of the Beta-oxidation, instead of obtaining two Acetyl CoA (as with the even chain fatty acids), the odd fatty acids now yields 1 Acetyl CoA and 1 Propionyl CoA.

 

What happens with this Propionyl CoA?

 

The Propionyl CoA must undergo a carboxylation in a sequence of reactions requiring Biotin and Vitamin B12. These reactions produce Succinyl CoA.

 

 

Note also that this sequence of reaction requires the consumption of 1 ATP (then consider it as  -1 ATP )

 

What happens to the carbon atoms of the Succinyl CoA?

 

Let’s discuss an anaplerotic fate for the Succinyl CoA:

 

We can consider the conversion of propionyl CoA to Succinyl CoA as an anaplerotic pathway, in which case, the molecule of Succinyl CoA continue in the Krebs Cycle generating Oxalacetate in the following sequence of reactions:

 

Succinyl CoA + GDP + (P) —> Succinate + CoA + GTP (equivalent to 1 ATP)

 

Succinate   +  FAD ———-> Fumarate +  FADH2 (Generates 2 ATP in the Respiratory Chain)

 

Fumarate +     H2O————– > Malate

 

Malate +   NAD+ —————->  Oxalacetate + NADH.H+ (Generates 3 ATP in the Respiratory Chain)

 

Remember that by definition the products of the anaplerotic reactions are incorporated into the Krebs Cycle, increasing its activity, but without being oxidized. In this case, because of the sequence of reactions experienced by the Succinyl CoA up to Oxalacetate, we can consider that the incorporation of the Succinyl CoA originated from the Propionyl CoA, has generated 6 additional ATPs.

 

Therefore, considering the “anaplerotic” fate of the Propionyl CoA:

 

– Add (-1 +1 +2 +3) = 5 ATP to the previous calculations.

 

In our next post, we will use an example. We will apply this information in the calculation of the energetic balance of the Beta-oxidation of the decaheptanoic acid (17  Carbons), assuming that the propionyl CoA has followed the anaplerotic pathway up to oxalacetate, as described in this post.

 

 

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


 

The basis for the calculation and how to do it have been explained in detail in other post.

 

Anyway, since time is an issue in exams, some readers are interested in knowing a method of doing the calculation in a real fast way.

 

For obtaining a quick answer when calculating the energetic balance of the total oxidation of a fatty acid (and also for verifying your answers if you use some other method of calculation) you can apply these formulas:

 

a) If your course use for the reduced cofactors these energetic yields

NADH.H+ = 2.5 ATP

FADH2 = 1.5 ATP

 

Use this formula:

 

[(n/2) -1] x (14) +8 = Number of ATP formed

 

Where n= Number of carbons

 

Example:

 

Balance of the total oxidation of palmitic acid (16 carbons):

(16/2)-1] x (14) +8 = 106 ATP

 

b) If the equivalence used for the energetic yields in your course are:

NADH.H+ = 3 ATP

FADH2 = 2 ATP

 

Then apply this formula:

 

[(n/2) -1] x 17 +10 = Number of ATP formed

 

For the total oxidation of palmitic acid, following the former energetic yielding criteria:

 

[(16/2)-1] x 17 + 10 = 129 ATP

 

Thanks for the questions and comments!

 

 

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!

 

 

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!