# 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 '17 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. – PascalIv Nov 17 '17 at 11:35