### Michaelis Menten Kinetics (Quasi Steady State Assumption) Solved Example 3

Find the product formation rate for the follow biochemical reactions, do not include [ES] in your final answer.

Start by doing a mass balance on all the enzyme-substrate intermediates. In this case, it's just [ES].

$\frac{{d[ES]}}{{dt}} = {k_1}[E][S] + {k_{ - 2}}[E][P] - {k_{ - 1}}[ES] - {k_2}[ES] = 0$

And we can develop an equation to describe the product formation rate.

$\frac{{d[P]}}{{dt}} = \nu = {k_2}[ES] - {k_{ - 2}}[E][P]$

And create equations for enzyme conservation, basically all enzyme concentrations will come from the initial enzyme concentration.

$[{E_O}] = [E] + [ES]$

$[E] = [{E_O}] - [ES]$

Now it's a lot of substituting into each equation and simplifying.

$[E] = \frac{{{k_{ - 1}} + {k_2}}}{{{k_1}[S] + {k_{ - 2}}[P]}}[ES]$

$[EO] = \left[ {\frac{{{k_1}[S] + {k_{ - 2}}[P] + {k_{ - 1}} + {k_2}}}{{{k_1}[S] + {k_{ - 2}}[P]}}} \right][ES]$

Rearrange in terms of [ES]

$[ES] = \frac{{[{E_O}]{k_1}[S] + {k_{ - 2}}[P]}}{{{k_1}[S] + {k_{ - 2}}[P] + {k_{ - 1}} + {k_2}}}$

Recall:

$\nu = {k_2}[ES] - {k_{ - 2}}[E][P]$

Now substitute in.

$\nu = \frac{{{k_1}{k_2}[{E_O}][S] + {k_2}{k_{ - 2}}[{E_O}][P]}}{{{k_{ - 1}} + {k_2} + {k_1}[S] + {k_{ - 2}}[P]}} - {k_{ - 2}}\left( {[{E_O}] - [ES]} \right)[P]$

And substitute [ES] into the right part of the equation as well and a lot will simplify.

$\nu = \frac{{{k_1}{k_2}[{E_O}][S] - {k_{ - 1}}{k_{ - 2}}[{E_O}][P]}}{{{k_{ - 1}} + {k_2} + {k_1}[S] + {k_{ - 2}}[P]}}$

Make the following substitutions:

${V_s} = {k_2}[{E_O}]$

${V_p} = {k_{ - 1}}[{E_O}]$

$\nu = \frac{{{k_1}{V_S}[S] - {k_{ - 2}}{V_P}[P]}}{{{k_{ - 1}} + {k_2} + {k_1}[S] + {k_{ - 2}}[P]}}$

Now divide both the numerator and denominator by k-1+k2.

$\nu = \frac{{\frac{{{k_1}{V_S}[S] - {k_{ - 2}}{V_P}[P]}}{{{k_{ - 1}} + {k_2}}}}}{{\frac{{{k_{ - 1}} + {k_2}}}{{{k_{ - 1}} + {k_2}}} + \frac{{{k_1}[S] + {k_{ - 2}}[P]}}{{{k_{ - 1}} + {k_2}}}}}$

$\nu = \frac{{\frac{{{k_1}{V_S}[S] - {k_{ - 2}}{V_P}[P]}}{{{k_{ - 1}} + {k_2}}}}}{{1 + \frac{{{k_1}[S] + {k_{ - 2}}[P]}}{{{k_{ - 1}} + {k_2}}}}}$

You need to do a little bit more algebra. Use the following substitutions:

${K_m} = \frac{{{k_{ - 1}} + {k_2}}}{{{k_1}}}$

${K_P} = \frac{{{k_{ - 1}} + {k_2}}}{{{k_{ - 2}}}}$

Be careful with the algebra, it's easy to make a mistake. Km and Kp will be in the denominator of both parts of the fraction. But once you do all the algebra and clean it up, you'll get a nice equation for your product formation rate:

$\nu = \frac{{\frac{{{V_S}}}{{{K_m}}}[S] - \frac{{{V_P}}}{{{K_P}}}[P]}}{{1 + \frac{{[S]}}{{{K_m}}} + \frac{{[P]}}{{{K_P}}}}}$

See other solved Michaelis-Menten examples below:

Michaelis Menten Fast Equilibrium Assumption (Ex1)

Michaelis Menten Quasi Steady State Assumption (Ex2)

Alternatively return to the Michaelis Menten hub page:

Michaelis Menten Examples