In the whole chain of oxidative phosphorylation, does the reduction of oxygen step limit the speed at which NADH can be processed?
Suppose you had 2 OXPHOS systems side by side - one in which O2 was always, 100% completely reduced (producing no free radicals) and one in which O2 was only reduced halfway (producing lots of free radicals). Which system would run faster, and by how much?
Hi King,
Firstly, oxidative phosphorylation is the process whereby F0F1ATPase uses the Gibbs free energy generated through the transfer of electrons by the Complexes of the Electron Transport Chain to phophorylate ADP with Pi thereby producing ATP. Although ETC and OP are married together by a coupling process but they are two different processes.
The electron transport chain Complex I also known as NADH dehydrogenase. I think inorganic phosphorus would regulate NADH better. I think!
I think that an electron transfer from 2Fe-2S to heme c1(or f6) is a rate limiting step in Oxidative phosphorylation (photophosphorylation).
A link of our animation of this process:
https://m.youtube.com/watch?v=CeZxSyeDBwk
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