Question

A eighteen-carbon-long fatty acid would yield ____ B acetyl CoA molecules. If each of these can enter a Krebs cycle and produce 10 ATP per turn of the cycle, producing B x 10: ____ ATP. Including the ATP formed from the oxidation of the end carbon of the fatty acid chain, this fatty acid could yield a grand total of ____ ATP.

Answer 1

An 18-carbon fatty acid yields 9 acetyl CoA molecules that enter the Krebs cycle, producing a grand total of 120 ATP after accounting for NADH, FADH₂, and energy investment for activation.

Step-by-step explanation:

An eighteen-carbon-long fatty acid would yield 9 acetyl CoA molecules. This is because the number of ß-oxidation cycles is equal to n/2 - 1, where n is the number of carbon atoms in the fatty acid, so for an 18-carbon fatty acid, there would be 18/2 - 1 = 8 + 1 (from the last cycle which yields two acetyl CoA) cycles.

Each acetyl CoA molecule can enter the Krebs cycle and generate 10 ATP per cycle, so the total ATP production from the Krebs cycle would be 9 x 10 = 90 ATP.

Including the ATP produced from the oxidation of the end carbon of the fatty acid chain, which involves the reduced coenzymes NADH and FADH₂, as well as the two ATP used for activation (2 ATP converted to 2 ADP), the formula for the total ATP yield would be ((0.5 x n - 1) x 14) + 10 - 2.

For an 18-carbon fatty acid, this amounts to (0.5 x 18 - 1) x 14) + 10 - 2 = (9 - 1) x 14 + 10 - 2 = 8 x 14 + 8 = 112 + 8 = 120 ATP.

Therefore, the grand total of ATP produced by the complete oxidation of an eighteen-carbon-long fatty acid is 120 ATP.