FADH2 is produced in the conversion of Succinate to Fumarate in the tricarboxylic acid cycle. Why is this so? Why not NADH?
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3$\begingroup$ Just a guess: FAD reduction potential is higher than that of NAD; so it should be a better oxidizing agent than NAD. NAD however would be a better reducing agent. Source $\endgroup$– WYSIWYGCommented Sep 6, 2013 at 18:50
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1$\begingroup$ @TalhaIrfan — I have reverted your edit. There is absolutely no reason to place the abbreviations in italics and a different typeface — and every reason not to — which is what your markup did. The poster or who ever edited it previously was quite right in formating with HTML 'sub'. If she hadn't rendered the subscript — OK, but she had. Leave well alone. $\endgroup$– DavidCommented May 8, 2018 at 15:00
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$\begingroup$ @David Latex has been the preferred mode for both Mathematical as well as Chemical equations/formulae on SE; and its italic by default. Just the answer below is an example of its use! $\endgroup$– Failed ScientistCommented May 8, 2018 at 16:28
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2$\begingroup$ @TalhaIrfan — preferred by whom? This is not a mathematical construct it is a biochemical construct that would be set in a biochemistry text book (e.g. Berg et al.) in the same weight and typeface as other words in the text. Your Latex rendering is further ugly and makes reading difficult. And as I said, is in italic for no good reason. I have not changed the answer below, just as I would regard it rude to change US spelling to UK spelling. Please do not change answers just because you personally prefer a particular style. $\endgroup$– DavidCommented May 8, 2018 at 16:49
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$\begingroup$ @TalhaIrfan — I think that this question should be discussed further on Meta. I'll post a question when I have a moment. That way we can get clarification. $\endgroup$– DavidCommented May 8, 2018 at 18:14
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3 Answers
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In general, $\ce{NADH}$ and $\ce{FADH2}$ are coenzymes. The structure of the main part of an enzyme determines which coenzyme or which prosthetic group will work with the concerned enzyme. Unlike most other TCA cycle enzymes, Succinic Dehydrogenase involves the participation of $\ce{FAD}$ rather than $\ce{NAD}$ and that is a consequence of its specific structure.
Another possible reason might have to do with the energy gap available. The conversion of succinate to fumarate liberates less free energy as compared to other oxidation reactions. Ususally, the steps which are coupled with NADH reduction have a free energy change of about $\Delta G=-100$ to $-150~ \pu{kcal/mol}$ but the conversion catalysed by succinic dehydrogenase has a free energy change of $ -80~ \pu{kcal/mol}$ (rough figures) and hence releases less energy. It therefore might not be feasible to couple with $\ce{NADH}$ reduction but it would be favourable (thermodynamically) to couple with an easy-to-reduce (since the reduction potential is higher and hence reduction is more favourable) and hence less energetic molecule $\ce{FADH2}$.
Apart from structural and thermodynamic requirement, one last possible reason might be that succinic dehydrogenase is a part of the mitochondrial membrane and involved in the Electron Transport Chain. Here, flavin molecules are excellent electron transporters (in comparison to nicotinamide based reductants) and therefore it might be more profitable to use a flavin based reductant to later facilitate easy electron transport and proton transfer.
Again all these are probable reasons and I dont know for sure.
answered Sep 7, 2013 at 3:06
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There is a simple straightforward answer to this question, which I present here because the poster of the accepted answer seems unsure which of his three suggestions is correct, and therefore there is a danger of the reader thinking that an explanation in terms of flexibility or membrane location is correct. It is not.
It’s all to do with thermodynamics — the free energy changes in the different redox reactions. Thus, in their chapter on the tricarboxylic acid cycle Berg et al. write:
Succinate is oxidized to fumarate by succinate dehydrogenase. The hydrogen acceptor is FAD rather than NAD+, which is used in the other three oxidation reactions in the cycle. In succinate dehydrogenase, the isoalloxazine ring of FAD is covalently attached to a histidine side chain of the enzyme (denoted E-FAD).
E-FAD + succinate ⇋ E-FADH2 + fumarate
FAD is the hydrogen acceptor in this reaction because the free-energy change is insufficient to reduce NAD+. FAD is nearly always the electron acceptor in oxidations that remove two hydrogen atoms from a substrate.
An internet search brings up any many sets of university lecture notes that provide numbers to back this up. For example notes from Bryant Miles at Texas A & M (slightly reformatted by me):
Why is FAD the electron acceptor rather than NAD+?
The oxidation of an alkane is not sufficiently exergonic to reduce NAD+ — E˚' = –0.315 V
The oxidation of an alkane is sufficiently exergonic to reduce FAD. — E˚' = –0.031 V
Fumarate + 2e- → 2H+ Succinate — E˚' = 0.031 V
∆E˚' = –0.0002 V ∴ ∆G˚' = +0.04 kJ/mol
The change in free energy based on the steady state concentrations of all the reactants and products in mitochondria isolated from pig hearts is approximately zero, a near equilibrium reaction.
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I think that flavoproteins are more flexible than NAD. NAD carries exactly 2 electrons, while FAD can carry either 1 or 2 electrons. This is important, as succinate dehydrogenase is at the crossroad between the Krebs cycle and the electron transport chain. Best I can tell, the Krebs cycles is knocking out 2 electrons per enzyme, while the electron transport chains is knocking out 1 electron per complex. So maybe FAD makes the translation between the two systems work?
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1$\begingroup$ Please add some references to your answer. $\endgroup$ Commented Feb 1, 2017 at 11:03
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$\begingroup$ But in the succinate dehydrogenase reaction FAD accepts two electrons to give FADH2, so the fact that FAD is chemically capable of accepting one electron in a different reaction is quite irrelevant. The answer can be found in most biochemical texts, and comes down to the energetics of the reactions involved. $\endgroup$– DavidCommented Feb 1, 2017 at 15:27