In some books the graph of the change in free energy during an enzyme-catalysed reaction is depicted as shown below, where S = substrate, E = enzyme, P = product, and T* represents the transition state.

Free Energy changes in an enzyme-catalysed reaction

In this depiction the free energy of the enzyme–substrate complex (ES) is shown as having a slightly lower free energy than the substrate (S).

What is the explanation of this?

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    – David
    Feb 28, 2023 at 9:27

1 Answer 1


The diagram presented in the question is only one of many variations found in text books and the like. It is unsatisfactory to the extent that it shows one of the intermediates (ES) in the catalysis, but not the other (EP), and that it does not label the species present at each stage. Such energy diagrams are considered in a paper by Aledo et al. in Biochemical and Molecular Biology Education, from which I reproduce their favoured example, below.

Energy diagram for enzyme catalysed reaction

Explanation: A simple first order reaction in which a single substrate (S) is converted to a product (P) is shown with intermediates, ES and EP. The transition states between the various conversions in the sub-reactions are shown in red, and that for the uncatalysed reaction (E + S) in dark blue. (I am not sure why the E is included in this, except to allow use of a unified diagram.) ΔG is the overall free energy change of the reaction and ΔGUC and ΔGC are the free energy changes to reach the transition state in the uncatalysed and enzyme-catalysed reactions, respectively.

This diagram reflects the key point made by the authors of the paper:

“For an enzyme-catalyzed reaction, it should be stressed that, under conditions where the overall reaction is spontaneous, each elementary step must exhibit a negative free energy change, and this must be properly reflected in the progression profile of the reaction.”

So, to answer the poster’s specific question, for the reaction to proceed not only must the overall free energy change from E+S → E+P be negative, but that of the intermediate stage, E+S → ES must also be negative, i.e. the free energy of ES is lower than that of S — or E+S, to be precise.

(And the free energy of E+P must be less than that of EP, the violation of which principle is apparently a common failing of many diagrams.)


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