Dihydrofolate reductase (DHFR) reduces dihydrobiopterin (BH2) to tetrahydrobiopterin (BH4). Someone told me that this reaction needs only one NADPH molecule (I am not sure if this is correct), namely, NADPH provides one hydride atom and becomes NADP+.

Could someone explain to me how is this possible? How can BH2 -> BH4 conversion be done with one NADPH molecule, while I am expecting two (to give two hydrogen atoms)?

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    – David
    Feb 17 '18 at 10:37
  • $\begingroup$ @xusr, David is right, your comments belongs to an answer. Please rewrite them in an answer, I do not want that someone deletes them. $\endgroup$ Feb 17 '18 at 12:17
  • $\begingroup$ @xusr, I did not request to delete them, I just said that they would be more organized in an answer. There were valuable info in them, it would be great to rewrite them in an answer please. Additionally, I would be able to vote you up and accept the answer if it really answers me. $\endgroup$ Feb 17 '18 at 12:22
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    $\begingroup$ This is a very interesting question. It is impossible to ever prove a chemical mechanism, but there can be evidence gathered to support the mechanism. I found this article pnas.org/content/111/51/18225. So the hydride, being a nucleophile, will attack at the N5 atom (the one between the hydroxyls and nitrogen). This then pushes the electrons in the double bond onto an electrophilic nitrogen. These lone pairs, being a nucleophile can abstract the second hydrogen atom. The second hydrogen atom that shows up on the nitrogen could be from water or from a residue on the enzyme itself. $\endgroup$
    – user10370
    Feb 18 '18 at 4:30
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    $\begingroup$ Oh sorry, chemist brain went to mechanism first. But what I was trying to get was that NADPH is a source of hydride (the anion of hydrogen), so it will behave as a nucleophile. The other hydrogen on the nitrogen should be an electrophile (or H+). This H+ could either be from water or from the enzyme DHFR. It shouldn't be from another NADPH because a hydride already has fulfilled its 2 electron shell and it would be extremely hard to put in more electrons. And water undergoes self ionization, so some exists as H30+ and OH-. The H-O-H that had it's H+ abstracted becomes OH-. $\endgroup$
    – user10370
    Feb 18 '18 at 17:09

The short answer is that in reduction of BH2 to BH4 by dihydofolate reductase not only is a hydride ion (a proton plus an electron) transferred from NADPH to BH2, but the product also picks up a proton from the solvent: N5 of BH4 is protonated.

$$\ce{BH2 + 2e- + 2H+ <=> BH4\tag{1}}$$

Enzymically (see EC

$$\ce{NADPH + BH2 + H+ <=> BH4 + NADP+\tag{2}}$$

Thus the reduction of BH2 to BH4 involves 2 electrons and 2 protons. NADPH donates 2 electrons and 1 proton (as a hydride), forming a new carbon-hydrogen bond. The solvent accounts for the other proton. The situation is very similar to the alcohol dehydrogenase reaction (discussed below).

The reduction of biopterin all the way to tetrahydrobiopterin by dihydrofolate reductase is a 4-electron reduction, and requires 2 NAD(P)H.


There a couple of points about NADP-linked enzymic oxido-reduction, and about oxido-reduction in general, that possibly need clarification.

NAD(P)-linked enzymes, almost invariably (and perhaps always), catalyze the direct, stereospecific transfer of a hydride ion (a proton plus two electrons) as a single entity between the nicotinamide cofactor and acceptor.

This statement probably needs some explaining. The fact that NADP-linked oxidoreductase reactions are consistent with the transfer of a hydride ion is one of the great contributions of Westheimer to enzymology (see here). In many cases, this results in the formation of a carbon-hydrogen bond.

Secondly, hydride transfer is stereospecific. Amazingly, NADP-linked oxido-reductases distinguish between the two hydrogens at C4 of NAD(P)H. This is an example of prochirality (the Ogston effect in the Krebs cycle is another).

For simplicity, the two hydrogens at C4 are called the HA and HB. The work of Westheimer, Vennesland and co-workers showed that NAD(P)-linked enzymes can be divided into two classes: those that transfer HA from cofactor to substrate (A-sterospecific dehydrogenases) and those that transfer HB (B-stereospecific dehydrogenases).

It is now known that sterespecificity of hydride transfer is one of the most evolutionary conserved traits of NAD(P)-linked oxido-reductases. Enzymes that may be shown to be descended from a common ancestor by amino acid sequence analysis invariably display the same stereospecificity of hydride transfer (Yeast and horse liver alcohol dehydrogenase is an example) 1†.

In addition hydride transfer displays apparent absolute stereochemistry fidelity: the 'wrong' hyride is never transferred. Transfer of HB by lactate dehydrogenase, which is A-stereospecific, could not be detected experimentally, for example (see here).

A great review of this aspect of dehydrogenase biochemistry has been published by You (1985), available as a a pdf from here. (An earlier review may be found here)

Finally, a brief word on oxido-reduction in general. Oxidation is the loss of electrons, reduction is the gain of electrons. An ionization is neither an oxidation or a reduction, and neither is hydration/dehydration. This, of course, is 'basic stuff', but seems to sometimes cause confusion. When dealing with oxido-redution reactions we can add in as many protons as we like, or take out as many protons as we like, and this may be important stoichiometrically, but ionization does not change the oxidation state of a molecule.

An Illustrative Example - Alcohol dehydrogenase

Alcohol dehydrogenase (EC catalyzes the following reaction, the 2-electron reduction of acetaldehyde to ethanol with NADH as hydride donor:

$$\ce{Ethanol + NAD+ <=> Acetaldehyde + NADH + H+ \tag{3}}$$

This very nice diagram, taken from Cornforth's nobel lecture, Asymmetry and Enzyme Action, illustrates all the points made above.


  • The hydrogen labelled with the asterisk it HA. Stereospecific transfer of hydride to acetaldehyde results in the formation of a new carbon-hydrogen bond. If the asterisk represents deuterium or tritium, the result is deuterated or radiolabelled ethanol.

  • If, instead, HB is deuterated or tritiated, the result is deuterated or tritiated NAD+, with no transfer of label to product.

  • Formation of the alcohol involves 'picking up a proton' (from the solvent).

  • Transfer of hydride acetaldehyde is also stereospecific. The hydride is stereospecifically transferred to one of the faces of the planar carbonyl group. I am ignoring this aspect here (but see Cornforth's lecture) 2†.

  • To be very pedantic, HA is more correctly called the pro-4R hydrogen, and HB is the pro-4S hydrogen.

(In Cornforth's original diagram the H+ was left out).

The Specific Question Posed by the OP

How can BH2 -> BH4 conversion be done with one NADPH molecule, while I am expecting two (to give two hydrogen atoms)?

As stated above, BH4 (tetrahydrobiopterin) contains one more pair of electrons than does BH2 (dihydobiopterin), but (like the alcohol dehydrogenase reaction) there are two protons involved. The fact that reduction involves hydride transfer accounts for one. The other is picked up from the solvent.

This is illustrated in the diagram below, where the key point is that the carbon atom at C6 now has a new carbon-hydrogen bond. This bond accounts for 2 electrons plus 1 proton. But the nitrogen atom at C5 has also picked up a proton.

(We must be very careful when dealing with reduction of the pteridine ring because of the quininoid form (see here). This is why I first picked alcohol dehydrogenase as an illustrative example, and I am ignoring such complications here).


The OP chooses an interesting substrate for DHFR, an unconjugated pteridine. The 'more usual' substrate, of course, is the conjugated pteridine, dihydrofolate (FH2). Although DHFR does catalyze the transformation of BH2 to BH4 , the enzyme usually associated with this transformation is dihydropteridine reductase (EC

One final point. Both chromosomal and plasmid dihydrofolate reductase, although unrelated from an evolutionary point of view, are both A-stereospecific with respect to hydride transfer from NAD(P)H.

(The above diagram was modified from one found on the internet. The original may be found here)

Four-Electron Oxidoreductases

Most NAD(P)-linked oxidoreductases catalyze a two-electron reduction that involves just one molecule of nicotinamide cofactor.

A few, such as dihydrofolate reductase can catalyze two sequential 2-electron reductions, where the first 2-electron product is released into solution before it is reduced to the final product by a second encounter with enzyme. (In the case of DHFR, the reduction of folate to dihydrofolate is a two-electron reduction, and the reduction of dihydrofolate to tetrahydrofolate is another 2-electron reduction).

Some oxido-reductases, however, catalyze a four-electron oxido-reduction where the intermediate product is not necessarily released into solution. Such enzymes are called four-electron oxido-reductases, and are very rare. They have been reviewed by Feingold & Franzen, (1982).

One example is histidinol dehydrogenase (EC, that catalyzes the reduction of an alcohol to an acid (the aldehyde being the intermediate 2-electron reduction product).

$$\ce{L-histidinol + 2 NAD+ + H2O = L-histidine + 2 NADH + 3 H+ \tag{4}}$$

Another example is HMG-CoA reductase (EC, a key enzyme of cholesterol metabolism (that is inhibited by statins). In this case, the enzme reduces a substrate formally at the oxidation level of a carboxylic acid (but 'activated' by thiol ester linkage to coenzyme A) all the way to the alcohol.

$$\ce{Mevalonate + CoA + 2 NADP+ = 3-hydroxy-3-methylglutaryl-CoA + 2 NADPH + 2 H+ \tag{5}}$$

A third example is UDP-D-glucose 6-dehydrogenase (EC, which catalyzes the reduction of the C6 alcohol group of UDP-glucose all the way to the acid.

$$\ce{UDP-D-glucose + 2 NAD+ + H2O = UDP-D-glucuronate + 2 NADH + 2 H+}$$

So, a four electron NAD(P)-linked oxido-reduction is possible. But such reactions are relatively rare. The key to the biochemistry of the NAD(P)-linked enzymes, IMO, is to realize that a hydride is transferred as a single entity (and that hydride transfer is stereospecific).


  1. Alcohol dehydrogenase from Drosophila is unrelated to the yeast and horse liver enzymes from a evolutionary perspective, and is B-stereospecific.
  2. The same Cornforth that reputedly said that if the β-lactam structure of penicillin was correct he would give up chemistry and grow mushrooms (see here).

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