Bioenergetics focus: During one full turn of the citric acid cycle (tricarboxylic acid cycle) per acetyl-CoA oxidized, how many ATP equivalents are produced using the conventional textbook P/O ratios?

Introduction / Context:
The citric acid cycle (also called the tricarboxylic acid cycle or Krebs cycle) harvests high-energy electrons in the form of reduced cofactors. A common exam convention asks for ATP equivalents produced per acetyl-CoA using the classic P/O ratios (3 ATP per NADH and 2 ATP per FADH2). This question uses that traditional convention to compute the total ATP yield per cycle turn.

Given Data / Assumptions:

• Accounting is per acetyl-CoA oxidized through one turn of the cycle.
• Products per turn: 3 NADH, 1 FADH2, and 1 GTP (substrate-level phosphorylation).
• Classic P/O ratios assumed: 1 NADH → 3 ATP; 1 FADH2 → 2 ATP; 1 GTP ≈ 1 ATP equivalent.

Concept / Approach:
The cycle itself produces reduced cofactors; ATP is realized when these donate electrons to the respiratory chain. Under the specified classic P/O values, we convert cofactors into ATP equivalents, then sum with the substrate-level GTP to obtain the total ATP yield for one cycle turn.

Step-by-Step Solution:

Verification / Alternative check:
Some modern texts use refined values (approximately 2.5 ATP per NADH and 1.5 ATP per FADH2). Under those values, the same cycle turn yields roughly 10 ATP equivalents (3 * 2.5 + 1 * 1.5 + 1 ≈ 10). However, the conventional exam answer under classic P/O ratios is 12, which matches the options provided here and the requested assumption set.

Why Other Options Are Wrong:

• 10: Reflects modern P/O ratios, not the classic convention explicitly used by this question.
• 13 or 8 or 9: Do not match either classic or modern accepted accounting per acetyl-CoA.

Common Pitfalls:
Mixing modern P/O ratios with classic exam conventions, or accidentally counting ATP yield per glucose (two acetyl-CoA) rather than per acetyl-CoA. Always read the assumption set and the unit of accounting.

Final Answer:
12