Lipid catabolism energetics — Using the classical P/O assumptions, how many ATP are generated by the complete oxidation of palmitate (C16:0) to CO2 and H2O, net of activation?

Introduction:
Quantifying ATP yield from fatty acid oxidation is a cornerstone of bioenergetics. Palmitate (C16:0) is a standard example used to illustrate β-oxidation cycles, acetyl-CoA production, and coupling to the TCA cycle and oxidative phosphorylation. This question adopts the classical textbook P/O ratios (NADH ≈ 3 ATP; FADH2 ≈ 2 ATP) and subtracts activation cost.

Given Data / Assumptions:

• Palmitate undergoes 7 β-oxidation cycles to yield 8 acetyl-CoA.
• β-Oxidation directly produces 7 NADH and 7 FADH2.
• Each acetyl-CoA in the TCA cycle yields 3 NADH, 1 FADH2, and 1 GTP (≈ 1 ATP).
• Activation of palmitate to palmitoyl-CoA costs the equivalent of 2 ATP (ATP → AMP + PPi).

Concept / Approach:
Compute ATP from β-oxidation reducing equivalents plus ATP from acetyl-CoA oxidation, then subtract the activation cost. Classical P/O: NADH → 3 ATP; FADH2 → 2 ATP; GTP → 1 ATP. Modern estimates (NADH ≈ 2.5; FADH2 ≈ 1.5) give ~106 ATP, but this problem’s canonical answer uses the older convention.

Step-by-Step Solution:

Verification / Alternative check:
Using modern P/O (2.5 and 1.5) gives ≈ 10 ATP per acetyl-CoA cycle and ≈ 106 net ATP, but many exam keys still reference the classical 129 number for palmitate under “old P/O” assumptions.

Why Other Options Are Wrong:

• 35: accounts only for β-oxidation NADH/FADH2, ignores TCA.
• 96: counts only TCA yield from acetyl-CoA, omitting β-oxidation and activation.
• 131: gross yield before subtracting activation cost.
• 106: close to modern estimates but not under the stated classical assumptions.

Common Pitfalls:
Forgetting the 2-ATP activation penalty or mixing classical and modern P/O ratios in one calculation. Choose one framework and apply it consistently.

Final Answer:
129.