Why is the Krebs cycle considered a part of aerobic metabolism if molecular oxygen is not involved in any of the reactions in the cycle?

I originally thought that Krebs cycle was aerobic metabolism because it is in the process that oxygen is the final electron acceptor, but my teacher claims that glycolysis is anaerobic so what makes the Krebs cycle aerobic?


1 Answer 1


Oxygen is actually not needed in the Krebs cycle - it is needed in the electron transport chain that is downstream of the Krebs cycle to regenerate NAD+ from NADH. NAD+ is a co-enzyme and acts as an electron carrier in oxidizing reactions at various positions in the Krebs cycle. However, note that without O2, NADH accumulates and the cycle cannot continue as it needs NAD+ to run.

Krebs cycle - No O2 needed:

Krebs cycle

Electron transport chain - O2 needed to regenerate NAD+ essential for the Krebs cycle: e- transport chain

  • 2
    $\begingroup$ Why is it often the case that krebs is considered to be the stage before the electron transport chain, if the krebs can't complete without the Electron transport chain completing? And another question - since glycolysis requires NAD+ and NAD+ is produced by the electron transport chain, then if glycolysis uses the NAD+ from the electron transport chain, would that make glycolysis aerobic, and if so, then why is it considered anaerobic? $\endgroup$
    – barlop
    Jan 19, 2017 at 9:23
  • $\begingroup$ "... needed in the electron transport chain that is upstream of the Krebs cycle..." - "upstream" makes me think of the glycolytic pathway - which is not what the second figure shows, correct? Why isn''t it "downstream"? Are there two contexts of "aerobic"? A. See answer: O2 is essential for metabolisum, which is not the case in glycolysis? ATP in glycolysis is not produced by breaking down O2? B. in context of Warburg/cancer "aerobic" means that O2 does exist, is there (but not metabolized by the metabolism referred to by "aerobic" (glycolytic cancer cells). $\endgroup$ Feb 2, 2021 at 17:31
  • $\begingroup$ @PeterBernhard - it should be downstream for sure thanks. Does that answer your question as well? $\endgroup$
    – AliceD
    Feb 2, 2021 at 21:03

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