# Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state?

Suppose an enzyme $\ce{E}$ can catalyze two reactions:

\begin{align} \ce{S1 + E &<=> S1E -> P1 + E} \tag{R1} \\ \ce{S2 + E &<=> S2E -> P2 + E} \tag{R2} \end{align}

I want to derive a rate law. Can I assume that

\begin{align} \frac{d[\ce{S1E}]}{dt} &= 0 \tag{1}\\ \frac{d[\ce{S2E}]}{dt} &= 0 \tag{2} \end{align}

like in the derivation of the Michaelis-Menten rate law?

• Why do you think this assumption might be different in the two reaction case versus the one reaction case? (Note that the steady-state approximation relies on particular conditions, and isn't necessarily universally valid.)
– R.M.
Nov 16, 2017 at 16:47
• @R.M. I want the same conditions as given when the Michaelis Menten rate law is valid: high amount of substrate, only initial velocities etc. The assumption in the one reaction case relies on "the concentration of the intermediate complex does not change on the time-scale of product formation " (citing wikipedia) I do not know how to estimate if this still holds true for the two reactions case. Nov 17, 2017 at 11:35

In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P.

\begin{align} \frac{d[\ce{S1E}]}{dt} &= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}] - \ce{k2}[\ce{S1E}] = 0\\ \end{align}

Therfore:

\begin{align} [\ce{S1E}] &= \frac{\ce{k1}}{k_{-1}+\ce{k2}}[\ce{S1}][\ce{E}] = K_a[\ce{S1}][\ce{E}] \\ \end{align}

Similarly for the second reaction:

\begin{align} [\ce{S2E}] &= \frac{\ce{k3}}{k_{-3}+\ce{k4}}[\ce{S2}][\ce{E}] = K_b[\ce{S2}][\ce{E}] \\ \end{align}

The enzyme E is involved in both reactions and its total concentration(bound and unbound) is constant. This total concentration of enzyme $$[E]_{0}$$ is equivalent to the concentration of the free enzyme before adding the substrates. The concentration of the free enzyme at a certain time t is [E]:

\begin{align} [E]_{0} &= [E] + [\ce{S1E}] + [\ce{S2E}] \\ \end{align}

if you substitute $$[\ce{S1E}]$$ and $$[\ce{S2E}]$$ with the previous expressions then:

\begin{align} [E] &= \frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]} \\ \end{align}

Eventually: \begin{align} \frac{d[P_{1}]}{dt} = k_2[\ce{S1E}] = \ce{k2}(K_a[\ce{S1}][\ce{E}]) = K_p[\ce{S1}][\ce{E}] = K_p[\ce{S1}]\frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]} \\ \end{align}

and similarly for the rate of production of $$P_2$$