# Michaelis-Menten derivation for 2 enzyme substrates

We know that the Michaelis-Menten derivation for the following reaction:

$E + S \rightleftharpoons ES \rightarrow E + P$

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

$E + S \rightleftharpoons ES_1 \rightarrow ES_2 \rightarrow E + P$

What would be the derived Michaelis-Menten equation now?

[ES2] is assumed to be at steady state assumption.

What would be the derived Michaelis Menten Equation?

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You may have noticed that most of your questions were edited so as to format the formulas using MathJax. You too can do that! There's a tutorial here. – Keep these mind May 6 '13 at 16:10
You've just edited in "[ES2]". Why not give it a try, and change this to "$[ES_2]$"? Thanks – Keep these mind May 6 '13 at 16:17

For us, physicists, the Menten kinetics is just an applied law of mass action.

I'm constructing this system of ODEs: $$\frac{\partial [E]}{\partial t}=-k_{f}[E][S]+k_{r}[ES_1]+k_{cat}[ES_2]$$ $$\frac{\partial [S]}{\partial t}=-k_{f}[E][S]+k_{r}[ES_1]$$ $$\frac{\partial [ES_1]}{\partial t}=k_{f}[E][S]-k_r[ES_1]-k_{tran}[ES_1]$$ $$\frac{\partial [ES_2]}{\partial t}=k_{tran}[ES_1]-k_{cat}[ES_2]$$ $$\frac{\partial [P]}{\partial t}=k_{cat}[ES_2]$$ All the information about your reaction is in the system above. You can extract conservation laws from it, by adding the equations, trying to cancel right hand side expressions. Like, by adding first, third and fourth: $$\frac{\partial}{\partial t}\left([E]+[ES_1]+[ES_2]\right)=0\quad\Rightarrow\quad[E]+[ES_1]+[ES_2]=[E]_0$$
Like if you want to assume, that substrate $[S]$ is in eqilibrium, then $$\frac{\partial [S]}{\partial t}=0 \quad\Rightarrow\quad k_{f}[E][S]=k_{r}[ES_1]$$ From here you can express $\frac{\partial [P]}{\partial t}$ as a function of $[E]$,$[S]$ and $[E]_0$.
@ngzongyi Well, by analogy: $$\frac{\partial [ES_2]}{\partial t}=0 \quad\Rightarrow\quad k_{tran}[ES_1]=k_{cat}[ES_2]$$ – Kostya May 6 '13 at 16:17