# Michaelis Menten Quasi Steady State Assumption (Example 2)

##### 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, [E$$_0$$] represents the initial concentration of the enzyme, [S] represents concentration of the substrate, and [ES$$_1$$], [ES$$_2$$] 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 [ES$$_1$$] and [ES$$_2$$].

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

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

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 [ES$$_2$$].

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

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

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

Now carefully divide by k$$_3$$ 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 K$$_m$$$$_1$$ and K$$_m$$$$_2$$ be new variables.

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

where

${K_{m1}} = \frac{{{k_2}}}{{{k_1}}}$

${K_{m2}} = \frac{{{k_4}}}{{{k_3}}}$

Try another example of Michaelis Menten Kinetics, with the pseudo or quasi steady state assumption!