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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?
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 $P_2$
