I was preparing for the Canadian Chemistry Olympiad and I ran into the following question on one of the past exams:

If enzyme A binds to the substrate 25 times stronger than enzyme B, what is the ratio of the catalytic rate between enzyme A and enzyme B if the energy of the two transition states is identical? What is the difference in activation energy between the two reactions?

I feel as if I can make a qualitative prediction, but cannot derive a quantitative prediction. I suspect, since enzyme A bind more strongly to the substrate than enzyme B, it would have a greater activation energy, and thus result in a slower reaction.


I suspect that it is necessary to use the equation $$ \begin{align} k = \frac{kT}{h} e^{-\Delta G^\ddagger/RT} \end{align} $$

But the relationship between the two activation energies does not seem simple

$$\begin{cases} \Delta G^\ddagger_A = \Delta G_{E + S \rightarrow T} - 25\Delta G_\text{bind} \\ \Delta G^\ddagger_B = \Delta G_{E + S \rightarrow T} - \Delta G_\text{bind} \end{cases}$$

Plugging this into the the original equation, and solving for $k_B/k_A$, I get the following:

$$ \frac{k_B}{k_A} = \frac{e^{\Delta G_\text{bind}/RT}}{e^{25\Delta G_\text{bind}/RT}} = e^{-24\Delta G_\text{bind}/RT}$$

Since $\Delta G_\text{bind} < 0$, $k_B/k_A > 1$, supporting my qualitative prediction. Yet this does not give me a useful ratio, since it is dependent on the binding energies of enzymes A and B. This makes me suspect I am missing/overlooking something, any insight would be appreciated.


The question as written is a little unclear but if we assume that the two enzymes are initially at the same energy and that the transition state from species enzyme + substrate, [ES] to its product is also at the same energy, say $E_{TS}$, then possibly an answer can be found. ( I assume that by 'catalytic rate', 'catalytic rate constant ' is meant).

An equilibrium constant in a reaction only informs us about the ratio of rate constants and free energy change $\Delta G^0 $. It does not on its own help us find activation energies or individual rate constants. In this question the transition states are of equal energy, which additional information we can use. I have assumed that the schematic of the energy levels shown below represents the levels in the question. The dotted lines are only schematic. The activation energy is from ES to TS in each case.
energy levels
Let $K_B$ be the binding equilibrium constant for species B, then $25K_B$ is that for A. The species [ES] for A is then at free energy $\Delta G^0_A=-RTln(25K_B)$ and B at $\Delta G^0_B=-RTln(K_B)$, which means that the activation energy (from [ES] to its transition state towards products) is greater for A by the difference between these values which is $RTln(25)$. The ratio of rate constants is the ratio of Arrhenius equations each of which we assume has the form $k=A_0exp(-(E_{TS}+\Delta G)/RT)$ (where $A_0$ is the pre-exponential term assumed equal). Most terms cancel giving the ratio of rate constants as 25 with B being faster.


Your intuition that the enzyme binding more strongly to substrate will be slower is correct, this is called ground-state stabilization. I think where you might be making an error is how you handle the factor of 25. The factor of 25 probably refers to the ratio of equilibrium constants between E and ES for the two enzymes, not the ratio of binding energies. So you need to start with delta G = - RT ln Keq, and compare for 2 different Keq's. Then use your equation to estimate the difference in rate constants.


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