Michaelis Menten Quasi Steady State Assumption (Example 2)

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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.

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


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

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

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

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