Hi Hannah – It will be a very long explanation if I tell you exactly how it happens, but look closely at where the electron transport chain in oxidative phosphorylation. On the inner membrane of the mitochondrion, Succinate dehydrogenase will be competitively inhibited by malonate. Now, the isomerisation action of succinate dehydrogenase is coupled with a Coenzyme 6Q – 6QH2 synthesis reaction at site II of the electron transport chain (on the succinate area, not the reduced nicotinamide adenine dinucelotide side). Now I’m this far, you’d might as well hear it! So, Coenzyme Q will release hydrogen cations intot he inter membrane space.Now, the reason that it can only stop oxygen consumption here is because the structure of succinate is C4O4 with double bonds to hydrogen to form succinic acid. That is released into site 2. Then. Fumarase is actually not just having a bit of an isomerisation change, it’s being bonded to another oxygen (which is obviously needed in the fumarae catabolsim process, to later form reduced nicotinamide adenine dinucleotide. Now, it can stop the oxygen consumption at succinate dehydrogenase because O2 is incorporated at the end of the electron transport chain, obviously much after the coupled coenzyme Q process, a totally different substrate by then, formed after many catabolism processes.So, when the new oxygen atom is bonded to form fumarate by fumarase, the structure changes in the cycle, so as it is a muxh similar structure to succinate and only one step after, compared to the oxygen consumption of succinate at the end of the electron transport chain, malonate is able to competitively inhibit succinate dehydrogenase. And it works the same for pyruvate and malate.

sorry, but I have no idea, what are you saying, billyfisher100.

hannahzarzar:
the reason is because succinate provides electrons into ETC from another "branch" of ETC than pyruvate or malate. The complexes I and II are not in a row, but rather different entering branches. This is a good picture

as you can see, electrons from both complexes I and II are directly passed onto complex III (rather than I -> II -> III).
Now where do the electrons come from? In complex I the electrons are from pyruvate being oxidized in tricarboxylic acid (TCA) cycle, while complex II is actually part of the TCA cycle being succinate dehydrogenase. Thus if you inhibit complex II, the electrons from complex I can be still transported onto complex III and oxygen can be consumed by complex IV 😉

I apologise for the long, confusing explanation! The essence of what I was trying to say is very similar to JackBean’s post, just worded in an entirely differently, confusing way!