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conversion of G3P to 1,3-bisphosphoglycerate (from Fundamentals of Biochemistry by Voet, 5th ed.)

In this step of glycolysis, I'm not seeing where the $\ce{H+}$ ion on the product side is coming from. It seems to me that the G3P's aldehydic H is replaced by phosphate, and that H is given to $\ce{NAD+}$ to make NADH. So where is the extra $\ce{H+}$ coming from?

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  • $\begingroup$ Look at what you are substituting OPO for $\endgroup$ – rhill45 Nov 19 '16 at 2:33
  • $\begingroup$ I've reverted the question to its original. Clearly the answer to the question needs to take into account the ambiguous nature of the equation, which does not make it clear whether it is meant to be a stoichiometric representation of the reaction occuring on the enzyme (which must involve discrete species) or adherence to a convention (which I assume, like Berg et al. is not explained) in which Pi is used to represent the fact that the reactive species (orthophosphate?) is in equilibrium with other phosphate species. Thanks @tomd for putting me right here. $\endgroup$ – David Nov 21 '16 at 11:31
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Although texts such as Berg et al. tend to refer to inorganic phosphate, $\ce{P_i}$, as orthophosphate ($\ce{PO4^{3-}}$), the term inorganic phosphate is used because in aqueous solution at pH 7.6 several phosphate species exist, the predominant one being $\ce{HPO4^{2-}}$. If this is regarded as $\ce{P_i}$, then it is the source of the $\ce{H^{+}}$, and the equation balances.

Note added by David:

On checking I find, in contrast to Berg, Lehninger’s book defines $\ce{P_i}$ as monohydrogen phosphate ($\ce{HPO4^{2-}}$), and Fersht actually writes the equation of the reaction (16-1) with $\ce{HPO4^{2-}}$ rather than $\ce{P_i}$.

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  • $\begingroup$ Ok, so do most sources assume $\ce{P_i}$ as being $\ce{HPO4^{2-}}$ or as being $\ce{H2PO4^-}$? It seems like the source of my image assumes $\ce{P_i}$ is $\ce{HPO4^{2-}}$. $\endgroup$ – carbenoid Nov 19 '16 at 4:00
  • $\begingroup$ Take a look at this link: statemaster.com/encyclopedia/ATP-hydrolysis . It is about ATP hydrolysis which also releases Pi (the phosphate ion) but notice the resonance stabilisation. I think this is what anongoodnurse is meaning $\endgroup$ – KingBoomie Nov 19 '16 at 8:40
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    $\begingroup$ This is very muddled. You are mixing up solubility and ionization, and if you want to discuss this you have to put the charges on. $\endgroup$ – David Nov 19 '16 at 12:52
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    $\begingroup$ @RickBeeloo No, the question is about the stoichiometry of the equation, it has nothing to do with resonance stabilization. In the link you cite the equation is balanced. $\endgroup$ – David Nov 19 '16 at 12:53
  • $\begingroup$ @David - I am not tech savvy enough to include both the superscript charge and the subscript. However, in truth, in cells, Pi does not exist as PO4-3. And ultimately, the H+ does, indeed, come from the solution, which has a number if weak acids and bases. $\endgroup$ – anongoodnurse Nov 19 '16 at 13:26
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Although an answer that I have contributed an edit to has been, rightly, accepted, I decided to add my own answer to clarify a couple of points.

  1. The equation in the question is what is known as a biochemical equation, not a chemical equation. As this IUPAC document notes:

Chemical equations are written in terms of specific ionic and elemental species and balance elements and charge, whereas biochemical equations are written in terms of reactants that often consist of species in equilibrium with each other and do not balance elements that are assumed fixed, such as hydrogen at constant pH.

One should not, therefore, expect Voet’s equation to balance.

  1. The 1974 paper of Buehner et al. on the crystallization of glyceraldehyde 3-phosphate dehydrogenase includes the following (net reactants in yellow, hydrogens in orange): GAPDH reaction

…which answers the question.

It is interesting that the IUPAC document defines Pi as orthophosphate, although this is irrelevant to the question at hand as none of the old papers I have looked at use the abbreviation Pi.

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