### Michaelis Menten Kinetics (Fast Equilibrium Assumption) Solved Example

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 fast equilibrium 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: Let's use the equilibrium assumption first. Do a mass balance on the left reaction, a mass balance on the right reaction, and a mass balance on all enzyme-containing chemicals, and we obtain (remember, fast equilibrium assumption):

Remember, when we find the product formation rate, we want it to be a function of only [S] and [E0], we don't want any of the enzyme-substrate intermediates to appear in our equation. Now substitute and rearrange to acquire:

$[E{S_1}] = \frac{{{k_1}}}{{{k_2}}}[S][E]$

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

And substitute into the equation for [E].

$[E] = \frac{{{k_2}{k_4}}}{{{k_1}{k_3}}} \cdot \frac{{[E{S_2}]}}{{[S]}}$

Plug these into the equation for [E0] and factor out the [ES2] term.

$[{E_0}] = \left[ {\frac{{{k_2}{k_4} + {k_1}{k_4}[S] + {k_1}{k_3}[S]}}{{{k_1}{k_3}[S]}}} \right][E{S_2}]$

Now isolate for [ES2] by rearranging.

$[E{S_2}] = \frac{{{k_1}{k_3}[{E_0}][S]}}{{{k_2}{k_4} + {k_1}{k_4}[S] + {k_1}{k_3}[S]}}$

The product formation rate (or reaction velocity) is:

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

Now substitute in what we found for [ES2] earlier.

$\nu = \frac{{{k_1}{k_3}{k_5}[{E_0}][S]}}{{{k_2}{k_4} + {k_1}{k_4}[S] + {k_1}{k_3}[S]}}$

Now, if you carefully do algebra, you can show this follows Michaelis-Menten kinetics. Divide the numerator and denominator terms by k1k3.

$\nu = \frac{{{k_5}[{E_0}][S]}}{{\frac{{{k_2}{k_4}}}{{{k_1}{k_3}}} + \frac{{{k_1}{k_4}}}{{{k_1}{k_3}}}[S] + \frac{{{k_1}{k_3}}}{{{k_1}{k_3}}}[S]}}$

$\nu = \frac{{{k_5}[{E_0}][S]}}{{\frac{{{k_2}{k_4}}}{{{k_1}{k_3}}} + \frac{{{k_4}}}{{{k_3}}}[S] + [S]}}$

$\nu = \frac{{{V_m}[S]}}{{{K_{m2}}({K_{m1}} + [S]) + [S]}}$

Where $${K_{m1}} = \frac{{{k_2}}}{{{k_1}}}$$ and $${K_{m2}} = \frac{{{k_4}}}{{{k_3}}}$$

See other solved Michaelis-Menten examples below:

Michaelis Menten Quasi Steady State Assumption (Ex2)

Michaelis Menten Quasi Steady State Assumption (Ex3)