Below is the mechanism for the reactions of the pyruvate dehydrogen complex, which oxidatively decarboxylates pyruvate and transfers the acetyl group to coenyzme A for further metabolism in the Krebs cycle.

pyruvate dehydrogenase complex reactions

The key point is that after $\ce{FAD}$ has reoxidized the dihydrolipoamide, the resulting $\ce{FADH2}$ is reoxidized by $\ce{NAD+}$, which is reduced to $\ce{NADH}$.

I find the chemical energetics of this difficult to understand. According to standard tables of redox-potential for half reactions, $\ce{FAD}$ is a stronger oxidising agent, and a weaker reducing agent, than $\ce{NAD+}$, which is why it used in the succinate dehydrogenase reaction, for example. How, then, can it reduce $\ce{NAD+}$?

If $\ce{FADH2}$ is really able to reduce $\ce{NAD+}$ it should generate as much ATP in oxidative phosphorylation after oxidation of succinate to fumarate, despite the relatively low free energy change for that reaction. But that is absurd.

  • 5
    $\begingroup$ This is the most perceptive question on a biochemical topic that I have seen in the last twelve months, and deserves far more votes than it has received. A student, apparently new to biochemistry, has brought his chemical knowledge to bear and pointed to an apparent anomaly which most text-books ignore. The beauty of this question to me is that the explanation for this apparent anomaly illustrates the application of chemistry in a biological context. I have learnt from finding the answer to the question, and I hope the poster's interest in biological chemistry has been stimulated by it. $\endgroup$
    – David
    Apr 11, 2021 at 16:58
  • $\begingroup$ [1-2] IMO, and as I pointed out (politely) in the comments to the accepted answer, there is no thermodynamic problem here whatsoever. Why shouldn't FADH2 reduce NAD(+)? The reaction is reversible. It may be inferred from the standard redox potentials that the position of equilibrium greatly favours FAD reduction by NADH, but this does not mean that the reaction cannot (thermodynamically) proceed in the opposite direction. All that is required is a negative delta G. The position of equilibrium in ethanol dehydrogenase reaction,for example, greatly favours acetaldehyde reduction, and NADH ... $\endgroup$
    – user338907
    Nov 21, 2023 at 20:57
  • $\begingroup$ [2-2] can reduce acetaldehyde (to produce ethanol, as happens in yeast), BUT NAD(+) can oxidize ethanol (to give acetaldehyde, as happens in liver). When dealing with redox potentials and Gibbs free energy, one needs to sharply distinguish between standard conditions and actual conditions. For a reaction to proceed spontaneously (only) requires a negative delta G (or a positive redox potential), not a negative standard delta G (or a positive standard redox potential) $\endgroup$
    – user338907
    Nov 21, 2023 at 21:06
  • $\begingroup$ IMO a great reference to this problem is Calculating apparent equilibrium constants of enzyme-catalyzed reactions at pH 7 by RA Alberty. The pdf is freely available $\endgroup$
    – user338907
    Nov 21, 2023 at 21:48

2 Answers 2


The redox potential widely quoted in text books for the half-reaction reduction of FAD to FADH2 is indeed inconsistent with the observed direction of several reactions involving this coenzyme, including its reduction of NAD+ in the dihydrolipoamide dehydrogenase reaction. The reason for this is that the value quoted is determined for free FAD in aqueous solution, which does not equate to the situation within the enzyme. In its tightly bound state in flavoproteins, the immediate environment (and, in some cases, covalent linkage) alters the affinity of the flavin ring for electrons, hence causing the redox potential to deviate from that in solution. The extent and direcction of this deviation differs in different flavoproteins.

Restating the problem
The basic idea of chemical theromodynamics is that reactions are energetic favourable if the (Gibbs) free energy change is negative (see e.g. Section 14.1.1 of Berg et al.). Although the actual free energy change, ΔG, is influenced by the relative concentration of the reactants, the usual starting point for assessing this in biochemistry is the ‘standard’ free energy change at pH7 and 25˚C — ΔG0′. In the case of oxidation/reduction reactions the standard oxidation–reduction potentials E0′ of each of the two half reactions are usually known (as they can be measured in solution), and one calculates the ΔG0′ from the oxidation–reduction potentials of the two half reactions, using the formula:

ΔG0 = –nFΔE0

What is important for this discussion is not so much calculation of actual values, but determining their sign, as for a negative ΔG (reaction energetically favourable) a positive ΔE is required. Let us consider an example using the much abbreviated version of Table 18.1 of Berg et al. Biochemistry, below: Oxidant Reductant n E´o (V)

Succinate   α-Ketoglutarate 2   – 0.67
NAD+        NADH            2   – 0.32
FAD         FADH2           2   – 0.22
Fumarate    Succinate       2   – 0.03

Consider one of the energy-generating oxidation reactions of the Krebs tricarboxylic acid cycle:
α-ketoglutarate + NAD+ → succinate + CO2 + NADH
This is made up of two half reactions:
(+ 0.67 V) α-ketoglutarate → succinate
(– 0.32 V) NAD+ → NADH
which gives an overall ΔE0′ of + 0.35 V, i.e. a negative ΔG0′, which is consistent with the actual direction of the reaction and the production of energy. (Actual worked numeric examples can be found in Berg et al.)
Now if we consider the dihydrolipoamide dehydrogenase reaction
The two half reactions are:
(+ 0.22 V) FADH2 → FAD
(– 0.32 V) NAD+ → NADH
This gives an overall ΔE0′ of – 0.1V, i.e. a positive ΔG0′, which, as the poster astutely points out, is not consistent with the direction of the reaction.

Resolving the apparent anomaly
Berg et al. do, actually, remark on this in section 17.1.1 on the pyruvate dehydrogenase complex:

This electron transfer to FAD is unusual, because the common role for FAD is to receive electrons from NADH. The electron transfer potential of FAD is altered by its association with the enzyme and enables it to transfer electrons to NAD+.

So, how is “the electron transfer potential of FAD altered by its association with the enzyme”? The quotation from Ghia and Massey1 provides a general summary:

The pyrimidine nucleus of the three-membered ring system is electron-deficient and can be viewed as an ‘electron sink’. In thermodynamic terms, any interaction which tends to lower its electron or negative charge density will increase the redox potential. The reduced flavin consists roughly of an electron-rich phenylenediamine moiety fused with a (4,5-diamino)-uracil. The latter is the moiety in which the negative charge (i.e. two electrons taken up by the oxidized flavin) is localized and, most importantly, stabilized. The degree to which this negative charge is stabilized or destabilized is an important factor governing the redox potential. Thus while a positive charge in the protein around the pyrimidine ring will contribute to increase the redox potential, the presence of negative charge or of a hydrophobic environment will lower it.

Flavin ring

In the specific case under consideration — dihydrolipoamide dehydrogenase — it has been established2 that a positively charged lysine-57 sidechain (numbering for the A. Vinelandi enzyme) that increases the redox potential (with, perhaps, a small contribution from the backbone N of alanine-326):

FAD environment in dihydrolipoamide dehydrogenase
[FAD environment in dihydrolipoamide dehydrogenase from A. Vinelandi. NAD is shown in light blue, except for the O and N atoms at the end of the flavin ring interacting with the protein. A section of the latter is shown as a wireframe in salmon pink without sidechains, except for lys57 and ala326. (Own diagram from 3LAD.pdb)]

A second apparent anomaly and its resolution
In drawing attention to the apparent anomaly in the reduction of NAD+ by FADH2, the poster suggested that this would imply one should by able to obtain enough energy to produce 3ATPs from reducing FADH2. The problem with this is that one actually needs FAD to have a lower redox potential to perform the reduction of fumarate that occurs in the Krebs cycle:

succinate oxidation

This can be seen by considering the redox potentials in the table above:
succinate + FAD + → fumarate + FADH2
The two half reactions are
(–0.22 V) FAD → FADH2
(+0.03 V) succinate → fumarate
Thus, there is an overall ΔE0′ of – 0.19V, i.e. a positive ΔG0′ — another anomaly!

Berg et al. are rather coy on this one. They do not comment, but in their Table 17.2 of free energy changes in the steps of the TCA cycle — where all the other values are given as exact figures — they quote “~0” for this step.

This implies a higher (more positive) redox potential for FAD than the –0.22 quoted in the table — certainly not the lower one of dihydrolipoamide dehydrogenase.

I turns out that this is, in fact, the case. In succinate dehydrogenase (aka as fumarate reductase) FAD is covalently attached to a histidine residue (his44 in E. coli) at the other side (position 8) of the flavin ring3.

FAD environment in succinate dehydrogenase
[FAD environment in succinate dehydrogenase from E. coli. NAD is represented as previously. A section of the latter is shown as a wireframe in salmon pink without sidechains, except for his44 and a backbone N. (Own diagram from 1L0V.pdb)]

An extensive review4 of the redox potentials of covalently and non-covalently bound flavoproteins shows that covalent attachment increases the redox potential.

Final observations

  • Chemistry is a powerful tool for understanding biology at the molecular level. Biology is subject to the laws of chemistry, but the chemistry of the special environments encountered (protein interiors, spaces compartmented by membranes) need taking into consideration, something that the classically trained chemist, new to the subject, may not fully realized.
  • Elementary textbooks often have to walk a fine line between comprehensibility and truth. The student of biological chemistry should realize that the initial simplifications that enabled him to get an overall grasp may mask greater complexities. One of the intellectual satisfactions in science can be found by resolving these complexities in a rationale manner.

1Ghisla and Massey (1989) Mechanisms of flavoprotein-catalysed reactions
2Maeda-Yorita et al. Biochemistry (1994) 33, 6213–6220
3Kim and Winge (2013) BBA 1827, 627–636
4Heuts et al. (2009) FEBS Journal 276, 3405–3427

  • 1
    $\begingroup$ Your answer is gold! I really enjoyed it. The concept you used is called symphoria or Neighbouring group participation or anchimeric assistance in organic chemistry texts. I was familiar with this concept in the field of organic chemistry and how it was used to determine the mechanism of reactions But I'd never seen the actual use of this in biochemical studies. Great job! $\endgroup$ Apr 11, 2021 at 17:18
  • 1
    $\begingroup$ (+1) [1-3] Although I don't like criticizing such an excellent answer, I think the following point is relevant: to the sentence "... which, as the poster astutely points out, is not consistent with the direction of the reaction." must be added the qualification "when the reactants are in the standard state". But what determines spontaneity is the actual reduction potential (or Gibbs free energy), not the standard reduction potential (unless standard conditions happen to apply). ... $\endgroup$
    – user338907
    Apr 12, 2021 at 16:43
  • $\begingroup$ [2-3] That is, the conditions within the active site may well favour oxidation FADH(2) by NAD(+) without the need to postulate any change in oxidation/reduction potential of cofactor by binding to the enzyme. To take an example given by Berg: the LDH reaction. When reactants are in the standard states (1M), the reaction greatly favours pyruvate reduction. But in the Cori cycle, depending on in vivo conditions, the enzyme may act either as a pyruvate reductase OR as a lactate dehydrogenase. $\endgroup$
    – user338907
    Apr 12, 2021 at 16:44
  • $\begingroup$ [3-3] If we were to consider only standard reduction potentials we would conclude that the enzyme always acts a pyruvate reductase. Similarly, although the point about isocitrate dehydrogenase is a good one, but MDH is surely a counter-example? The position of equilibrium of the MDH reaction greatly favours oxaloacetate reduction (when reactants are in the standard state, oxaloacetate reduction is the 'spontaneous' reaction), but in the TCA cycle MDH functions as a malate dehydogenase, contrary to the above analysis. $\endgroup$
    – user338907
    Apr 12, 2021 at 16:44
  • $\begingroup$ @Sam Thank you for your kind words. This sort of effect is fundamental to other aspects of protein chemistry, especially catalysis, where the environment of an amino acid residue can alter its properties. $\endgroup$
    – David
    Apr 12, 2021 at 22:46

In the example of dihydrolipoamide dehydrogenase from A. vinelandi, the interaction of the FADH2 with the protein should ideally decrease the reduction potential, (not increase it). If the reduction potential is decreased, then the protein-bound FADH2 will be better able to donate e- to NAD+. Thus, it doesn't make sense to position positively charged aa residues next to the FADH2 in this particular example provided from A. vinelandi.


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