Explain how the Krebs cycle contributes to ATP production during aerobic cell respiration.

IB DP Biology Standard Level (2023 syllabus) — C1.2 Cell respiration · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

Acetyl coenzyme A (acetyl CoA) enters the Krebs cycle by combining with oxaloacetate to form citrate, a 6‑carbon compound. During the cycle carbon dioxide is released and the oxidation reactions reduce the coenzymes NAD⁺ to NADH and FAD to FADH₂, carrying hydrogen ions. A small amount of ATP is produced directly in the cycle by substrate‑level phosphorylation of ADP to ATP. The reduced coenzymes NADH and FADH₂ then transfer their electrons to the electron transport chain, where the flow of electrons drives oxidative phosphorylation and generates the majority of ATP.

Examiner tips

Mention acetyl CoA entry, CO₂ release, NADH/FADH₂ formation, substrate‑level ATP, and link to ETC/oxidative phosphorylation
Use exact terms: acetyl CoA, oxaloacetate, citrate, NADH, FADH₂, CO₂, substrate‑level phosphorylation, electron transport chain, oxidative phosphorylation

Common mistakes

Confusing the order of steps (e.g., saying CO₂ is released before NADH formation)
Failing to state that ATP from the cycle is substrate‑level, not oxidative
Omitting the role of NADH/FADH₂ in the electron transport chain

Mark scheme (4 marks)

Acetyl coenzyme A (acetyl CoA) enters the Krebs cycle by combining with oxaloacetate to form citrate (a 6-carbon compound)
Carbon dioxide is released and coenzymes (NAD and FAD) are reduced / hydrogen carriers are reduced during oxidation reactions in the cycle
A small amount of ATP is produced directly in the Krebs cycle by substrate-level phosphorylation
The reduced coenzymes (NADH/FADH₂) pass electrons to the electron transport chain, driving oxidative phosphorylation and producing the majority of ATP

Key terms in this question

Krebs cycle