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Step 7 of the glycolysis pathway is the conversion of 1,3-bisphosphoglycerate into 3-phosphoglycerate by the action of the enzyme phosphoglycerate kinase, resulting in the production of 2 ATP molecules (per glucose).

This reaction has a large negative value of ΔG (-18.5 kJ/mol). If reactions having large negative value of ΔG are classed as being irreversible, then why is Step 7 of Glycolysis reversible?

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    $\begingroup$ I like to think of it as saying "there's enough 3-PG around and so little 1,3-BPG around that we do actually have to consider the reverse reaction." As @AlanBoyd points out, in erythrocytes this reaction is actually pretty close to equilibrium. So we can't call this "irreversible" because the simplifying assumption applied to "irreversible" reactions is that there is no equilibrium (K is infinitely large). $\endgroup$
    – stords
    Apr 9, 2014 at 10:06

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The free energy change that you quote for the phosphoglycerate kinase (PGK) forward reaction is, of course, the standard free energy change (ΔG0') for the overall reaction. The standard free energy change is defined for all reactants at a concentration of 1M. Note that this value includes the formation of ATP - the free energy of hydrolysis of 1,3-BPG would be much larger, and some of that energy is 'captured' in the ATP product.

The actual free energy change for the PGK reaction, ΔG, will differ from this value because of the fact that the actual reactant concentrations are very far from the idealised 1M. You can find a Table here which compares, for each step in glycolysis, the standard free energy change with a true free energy change, calculated on the basis of cellular conditions in erythrocytes (also shown). You will see that for the PGK step the actual free energy change is 0.09 kJ mol-1 rather than the -18.9 kJ mol-1 that you quote in your question.

You can also see from the Table that there are only two reactions in the glycolytic pathway with large negative values for ΔG: those catalysed by phosphofructokinase and by pyruvate kinase. These are of course the two steps that are bypassed in gluconeogenesis because they are essentially irreversible.

So - your reasoning was correct, but it was based on the wrong ΔG values.

added much later

As pointed out in the comments, the 1st step in glycolysis also has a large negative value for ΔG and is also bypassed. Not sure how I forgot to include that!

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  • $\begingroup$ "You can also see from the table that there are only two reactions...." Isn't 1st step of glycolysis having very large -ve ∆G? So there are 3 irreversible steps, not 2. $\endgroup$
    – JM97
    Jul 16, 2017 at 15:57
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You have to look at the complete reaction including the cofactors. In general, you can drive a chemical reaction into directions which are not favorable by:

  • removing products from the environment (if they are gaseous for example or react further)
  • having a huge excess of substrates (and thus making the back reaction less likely to happen)
  • and by coupling the main reaction to a second, which provides the necessary energy for the main reaction to happen

If you look at the reaction of 1,3-BPG to 3-PG this looks like the following (picture from Wikipedia):

enter image description here

The back reaction to 1,3-BPG needs ATP as a co-factor, so that the enzyme has the necessary phosphate group to add. The lysis of the phosphor diester bond from ATP to ADP delivers the energy to make the reaction possible.

This reaction is the control point in either generating ATP (glycolysis), when there is little ATP and much ADP present or generating glucose (gluconeogenesis) when there is much ATP and little ADP. This makes sense when you consider that ATP is a rather instable molecule with its high energy diester bonds, while ADP is more stable.

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    $\begingroup$ ATP is not hydrolyzed in the reverse reaction. Instead, a phosphate carboxylic anhydride is formed. If you're trying to do a Hess' law type analysis then you need to consider the hydrolysis of 1,3-BPG, which is much more favorable than -18.5 kJ/mol (in fact 30.5 kJ/mol more favorable). $\endgroup$
    – stords
    Apr 9, 2014 at 9:27
  • $\begingroup$ I just looked it up like this. Where does the phosphate group come from then? $\endgroup$
    – Chris ♦
    Apr 9, 2014 at 9:30
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    $\begingroup$ ATP transfers a phosphate to the 3-PG to form ADP and 1,3-BPG. Hydrolysis is when water cleaves the phosphodiester, forming ADP and inorganic phosphate. $\endgroup$
    – stords
    Apr 9, 2014 at 9:43
  • $\begingroup$ Ahh, got your point. $\endgroup$
    – Chris ♦
    Apr 9, 2014 at 10:27
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    $\begingroup$ Not sure why you are calling it "phosphodiester" bond (which I think is what you mean by 'phosphor diester')...there is no ester here! Its just a bond between two phosphates, not a carbon-linked phosphate and a carbon-linked hydroxide :/ $\endgroup$
    – another 'Homo sapien'
    Jul 16, 2017 at 17:05

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