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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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    $\begingroup$ Welcome to Chemistry.SE. This is a good question, but it seems like it is phrased like a textbook or homework question. It can become a great question if you shared with us what you have tried and where you are stuck. You will get a better answer that way. I am tagging it as homework, which means that answers will tell you how to solve the problem, but not just give you the answer. See our faq for more information. $\endgroup$ – Ben Norris May 7 '13 at 14:36
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I will propose an outline for determining an expression for the rate of substrate conversion in the given case:

  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 $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$.
  3. Choose initial conditions and set up two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES1 and ES2.
  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.
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