Michaelis Menten Kinetics (Quasi Steady State Assumption) Solved Example 2

We have two different assumptions when finding the product formation rate in Michaelis-Menten kinetics, either the fast equilibrium approach or the pesudo steady state approach.

Let's solve a series of biochemical reactions using the quasi steady state assumption. Here, [E0] represents the initial concentration of the enzyme, [S] represents concentration of the substrate, and [ES1], [ES2] represent concentrations of the enzyme-substrate complexes.

Find the product formation rate of:

Using the quasi steady state, we need to perform mass balances on [ES1] and [ES2].

\[\frac{{d[E{S_1}]}}{{dt}} = {k_1}[E][S] - {k_2}[E{S_1}] + {k_4}[E{S_2}] - {k_3}[E{S_1}] = 0\]

\[\frac{{d[E{S_2}]}}{{dt}} = {k_3}[E{S_1}] - {k_4}[E{S_2}] - {k_5}[E{S_2}] = 0\]

Rearrange the equations a little. Remember, when we develop an equation for the product formation rate, we don't want enzyme-substrate intermediates to appear in our final equation.

\[[E{S_1}] = \frac{{{k_4} + {k_5}}}{{{k_3}}}[E{S_2}]\]

\[[E] = \left[ {\frac{{{k_2}{k_4} + {k_2}{k_5} + {k_3}{k_5}}}{{{k_1}{k_3}[S]}}} \right][E{S_2}]\]

And we can also develop an equation for enzyme conservation (basically all enzyme concentrations will equal the concentration of initial enzyme you added).

\[[{E_O}] = [E] + [E{S_1}] + [E{S_2}]\]

\[[{E_O}] = \left[ {\frac{{{k_2}{k_4} + {k_2}{k_5} + {k_3}{k_5} + {k_1}{k_4}[S] + {k_1}{k_5}[S] + {k_1}{k_3}[S]}}{{{k_1}{k_{}}3[S]}}} \right][E{S_2}]\]

Now rewrite that in terms of [ES2].

\[[E{S_2}] = \frac{{{k_1}{k_3}[{E_O}][S]}}{{{k_2}{k_4} + {k_2}{k_5} + {k_3}{k_5} + {k_1}{k_4}[S] + {k_1}{k_5}[S] + {k_1}{k_3}[S]}}\]

\[\frac{{d[P]}}{{dt}} = \nu = {k_5}[E{S_2}]\]

\[\nu = \frac{{{k_3}{k_5}[{E_O}][S]}}{{{k_2}{k_4} + {k_2}{k_5} + {k_3}{k_5} + {k_1}{k_4}[S] + {k_1}{k_5}[S] + {k_1}{k_3}[S]}}\]

Now carefully divide by k3 and simplify.

\[\nu = \frac{{{k_5}[{E_O}][S]}}{{\frac{{{k_2}{k_4}}}{{{k_3}}} + \frac{{{k_2}{k_5}}}{{{k_3}}} + \frac{{{k_3}{k_5}}}{{{k_3}}} + \frac{{{k_1}{k_4}}}{{{k_3}}}[S] + \frac{{{k_1}{k_5}}}{{{k_3}}}[S] + \frac{{{k_1}{k_3}}}{{{k_3}}}[S]}}\]

\[\nu = \frac{{{k_5}[{E_O}][S]}}{{\frac{{{k_2}{k_4}}}{{{k_3}}} + \frac{{{k_2}{k_5}}}{{{k_3}}} + {k_5} + \frac{{{k_1}{k_4}}}{{{k_3}}}[S] + \frac{{{k_1}{k_5}}}{{{k_3}}}[S] + {k_1}[S]}}\]

And let Km1 and Km2 be new variables.

\[\nu = \frac{{{k_5}[{E_O}][S]}}{{\left( {{K_{m2}} + \frac{{{k_5}}}{{{k_3}}}} \right)({K_{m1}} + [S]) + \frac{{{k_5}}}{{{k_1}}} + [S]}}\]


\[{K_{m1}} = \frac{{{k_2}}}{{{k_1}}}\]

\[{K_{m2}} = \frac{{{k_4}}}{{{k_3}}}\]

See other solved Michaelis-Menten examples below:

Michaelis Menten Fast Equilibrium Assumption (Ex1)

Michaelis Menten Quasi Steady State Assumption (Ex3)

Alternatively return to the Michaelis Menten hub page:

Michaelis Menten Examples

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