# Michaelis Menten derivation for 2 enzyme substrates

We know the Michaelis Menten derivation for the following reaction:

$$\ce{E + S <=> ES -> E + P}$$

However, what if the reaction took place in a different scenario whereby:

$$\ce{E + S <=> ES1 -> ES2 -> E + P}$$

What would the derived Michaelis Menten equation be now?

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1. Set up the reaction with rate constants, assuming $k_{-2}\approx k_{-3}\approx0$: $$\ce{E + S <=>[k_1][k_{-1}] ES_1->[k_2] ES_2 ->[k_3] E + P}$$
2. Set up the differential equations describing the reaction, i.e. the rate of change for each component with time. The rate of substrate change, for example, will be $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$.
4. Make the pseudo-steady-state assumption (PSSA): assume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e. $\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0$.
5. Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.