ATP Production through Lipid Catabolism
Solution to the in-class example of triglyceride breakdown for ATP production:
Starting with a triglyceride consisting of three 16-C fatty acid chains:
The triglyceride can be considered as four “pieces” – a glycerol molecule and three 16-C fatty acid chains.
The glycerol molecule can be converted to pyruvate. The pyruvate can then go through pyruvate oxidation,
and the resulting acetyl-CoA can enter the TCA cycle.
Glycerol  Pyruvate       Acetyl-CoA
NAD+  NADH
(Pyruvate oxidation results in one acetyl-CoA and one NADH – the acetyl-CoA will then enter the TCA
cycle; the NADH will take the electrons to the electron transport chain)
One “turn” of the TCA cycle for the acetyl-CoA will result in the production of 3 NADH, 1 FADH2, and
one ATP (from substrate-level phosphorylation). The NADH and FADH2 will take the electrons to the
ETC).
So, for the glycerol molecule breakdown, we have the following ATP production:
4 NADH (1 from pyr. oxidation, 3 from TCA cycle) X 3 ATP/NADH = 12 ATP
1 FADH2 (from TCA cycle) X 2 ATP/FADH2 = 2 ATP
1 ATP (from TCA cycle)
Total ATP from one glycerol = 15
Now we have to deal with the three 16-C fatty acid chains. This is where the process of beta-oxidation
comes in. Each chain is broken down into 2-C fragments (acetyl-CoA). Since all three chains are the
same, it is easiest to just calculate the maximum ATP production from one chain and then multiply that by
three.
Each cleavage of the fatty acid chain will result in the production of one acetyl-CoA, one NADH, and one
FADH2. There will be 7 cleavages that occur in order to break off all the 2-carbon acetyl-CoA fragments.
This will then result in 7 NADH, 7 FADH2, and 8 acetyl-CoA (the final cleavage will split a 4-C chain into
two acetyl-CoA molecules). Each acetyl-CoA molecule will then enter the TCA cycle and give rise to the
same products listed above (3 NADH, 1 FADH2, and one ATP per acetyl-CoA). So, for one 16-C chain,
we’ll have the following ATP outcome:
8 Acetyl-CoA X 3 NADH/acetyl-CoA entering TCA = 24 NADH X 3 ATP/NADH = 72 ATP
8 Acetyl-CoA X 1 FADH2/acetyl-CoA entering TCA = 8 FADH2 X 2 ATP/FADH2 = 16 ATP
8 Acetyl-CoA X 1 ATP/acetyl-CoA entering TCA = 8 ATP
7 NADH (from the actual beta-oxidation process) X 3 ATP/NADH = 21 ATP
7 FADH2 (from the actual beta-oxidation process) X 2 ATP/FADH2 = 14 ATP
This gives us a total of 131 ATP. From this total, we have to subtract 2 ATP to account for the use of ATP
to initiate the beta-oxidation process (ATP  AMP = 2 ATP equivalent). So, for one 16-C fatty acid
chain, we have a net gain of 129 ATP. This can then be mulitplied by three to account for the ATP
production of all three fatty acid chains, giving an ATP sum of 387 ATP. To this, we add the 15 ATP
gained from the breakdown of the glycerol (pyruvate), for a final sum of 402 ATP from the original
triglyceride.