Krebs cycle yields per turn – which product is made in the greatest number? Considering one turn of the Krebs (TCA) cycle starting from one acetyl-CoA, which listed product is generated in the highest number of molecules?

Introduction / Context:The Krebs cycle generates reduced electron carriers and a small amount of ATP/GTP per acetyl unit oxidized. Recognizing stoichiometry per cycle helps estimate ATP yield via oxidative phosphorylation and compare substrates.

Given Data / Assumptions:

• Input: one acetyl-CoA condenses with oxaloacetate.
• Classic stoichiometry per turn: 3 NADH, 1 FADH₂, 1 ATP (or GTP), and 2 CO₂.
• Counts refer to one cycle turn, not per glucose (which yields two turns).

Concept / Approach:NADH is produced at three dehydrogenase steps (isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase). FADH₂ is produced once (succinate dehydrogenase). Substrate-level phosphorylation yields one ATP (or GTP) at the succinyl-CoA synthetase step. Therefore, the most numerous reduced carrier per turn is NADH.

Step-by-Step Solution:List outputs per acetyl-CoA: 3 NADH, 1 FADH₂, 1 ATP/GTP, 2 CO₂.Compare counts: 3 (NADH) > 2 (CO₂) > 1 (FADH₂, ATP/GTP).Select NADH as produced in the greatest number.

Verification / Alternative check:Per glucose, the cycle turns twice, giving 6 NADH, 2 FADH₂, and 2 ATP/GTP—again confirming NADH predominance.

Why Other Options Are Wrong:

• Acetyl-CoA: consumed, not produced.
• FADH₂/ATP: only one each per turn.
• CO₂: two per turn, fewer than the three NADH.

Common Pitfalls:Mixing per-turn counts with per-glucose totals; forgetting that some textbooks denote GTP interchangeably with ATP at the succinyl-CoA step.

Final Answer:NADH.