Home»Math Guides»Michaelis Menten Kinetics, Pseudo/Quasi Steady State Equilibrium 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!