# How do enzymes affect a reaction's equilibrium?

I am not sure to understand something I read in a educational journal (1)

# Introduction

Lets consider the following reversible enzyme-catalysed reversible reaction;

$$E + S \leftrightharpoons ES \leftrightharpoons E + P$$

with forward rate constants $$k_1$$ and $$k_2$$ and backward rate constants $$k_{-1}$$ and $$k_{-2}$$ respectively.

The equilibrium constant of this reaction can be expressed in terms of reaction quotient at equilibrium, in terms of thermodynamic potential and also in terms of rate constants;

$$K_{eq} = (\frac{[P]}{[S]})_{eq} = e^{-\frac{\Delta_r\mu^\circ}{RT}} = \frac{k_1 \cdot k_2}{k_{-1} \cdot k_{-2}}$$

It is possible to experimentally determine the Michaelis constants of this reaction for S and P by studying it in conditions where $$[S]>>[P]$$ and $$[S]<<[P]$$ respectively. The Michaelis constants are in fact only combinations of the rate constants;

$$K_M^S = \frac{k_{-1} + k_2}{k_1}$$

$$K_M^P = \frac{k_{-1} + k_2}{k_{-2}}$$

and the maximum forward and backward rate are respectively

$$v^+_{max} = [E] \cdot k_2$$

$$v^-_{max} = [E] \cdot k_{-1}$$

where $$[E]$$ is the total enzyme concentration (including $$[ES]$$). Which leads to the Haldane relationship by rewriting $$\frac{k_1 \cdot k_2}{k_{-1} \cdot k_{-2}}$$;

$$K_{eq} = \frac{v^+_{max} \cdot K_M^P}{v^-_{max} \cdot K_M^S}$$

# Problem

I was thinking that enzymes did not affect the position of the equilibrium of a reaction; that position would always be determined by thermodynamics, and the presence of an enzyme would simply affect the speed at which that equilibrium is reached (by lowering the activation energy). However, I stumbled upon a document (1) suggesting that both forward and backward rate expressions ($$v^+ = v^+_{max} \frac{[S]}{K_M^S + [S]}$$ and $$v^- = v^-_{max} \frac{[P]}{K_M^P + [P]}$$ respectively) apply on the whole reaction advancement axis and that the point where $$v^+ = v^-$$ then determines the apparent equilibrium of the reaction. The point of this document was to show that different values of enzyme parameters could correspond to the same $$K_{eq}$$, therefore an enzyme can set the apparent equilibrium of a reaction to a value totally different from its thermodynamic equilibrium.

I find that point confusing; if the equilibrium of a reaction changes in the presence of an enzyme, to the point that two different enzymes can change the direction of a given reaction (which is also the subject of another article of the same journal (2)), then the equilibrium constant of that chemical reaction cannot be predicted from thermodynamics (ie based on the chemical potential of its reagents). Then, can it be said that an enzyme modifies the chemical potential change of a reaction?

(1) A. Mellors, "The Haldane relationship, enzymes & equilibrium", Biochemical education 1976 vol 4 no. 4

(2) M. L. Uhr, "The influence of an enzyme on the direction of a reaction", Biochemical education 1979 vol 7 no. 1

• Are you sure the equations are correct? I might be getting confused by so many variables, but it seems to me that substituting all the variables in the last equation from the two expressions above it always gives $\mathrm{K_{eq}=1}$, which obviously doesn't make any sense. Mar 26, 2021 at 19:33
• @Shoubhik R Maiti: If OP is following Michaelis-Menten kinetics, the last equation should be $\ce{ES* -> E + P}$, which is usually the RDS. It would simplify this a lot. Mar 26, 2021 at 20:56
• @Mathew Yes, the last step is generally considered irreversible in Michaelis-Menten. But there are lots of enzymes which are completely reversible, like isomerases. So, the OP's equation isn't impossible. Mar 26, 2021 at 21:00
• With respect to the equations you do need to add the $\ce{ES <=> EP}$ step in order to derive the equations as you are doing Mar 26, 2021 at 22:48
• Set it up as $\ce{E + S <=>[k_1][k_{-1}] ES <=>[k_2][k_{-2}] EP <=>[k_3][k_{-3}] E + P}$ with Vmax substrate = $k_2$ and Vmax product = $k_{-2}$ and you should be able to derive the proper Haldane relationship which is the ratio of the respective Vmax/Km's is equal to the Keq, so you can vary Vmax and Km in tandem without affecting Keq. I'll expand on this in my answer if I have time later. Mar 27, 2021 at 0:30

I believe you have misread the first referenced paper. What it says is that the $$V_{max}$$ and $$K_m$$ values for the forward and reverse reactions can be varied to optimize a property such as the maximum velocity (represented by $$V_{max}$$) in one direction, but are always constrained by the fixed value of $$K_{eq}$$.
The reason that the $$K_m$$ values (which are pseudo equilibrium constants) can be adjusted is that they represent a reaction in which the enzyme is a reactant. For the substrate, this reaction is approximately $$\ce{E + S <=> ES}$$.
Likewise, the second paper also describes how the $$V_{max}$$ for one direction or the other can be maximized, but again subject to the constraint that the respective $$K_m$$ values must vary such that the overall $$K_{eq}$$ maintains its fixed value.