Home»Math Guides»Michaelis Menten Kinetics, Pseudo/Quasi Steady State Equilibrium Assumption, Example 3
Michaelis Menten Kinetics (Quasi Steady State Assumption) (Solved Example 3)
Find the product formation rate for the following 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\(_-\)\(_1\)+k\(_2\).
\[\nu = \frac{{\frac{{{k_1}{V_S}[S] – {k_{ – 2}}{V_P}[P]}}{{{k_{ – 1}} + {k_2}}}}}{{\frac{{{k_{ – 1}} + {k_2}}}{{{k_{ – 1}} + {k_2}}} + \frac{{{k_1}[S] + {k_{ – 2}}[P]}}{{{k_{ – 1}} + {k_2}}}}}\]
\[\nu = \frac{{\frac{{{k_1}{V_S}[S] – {k_{ – 2}}{V_P}[P]}}{{{k_{ – 1}} + {k_2}}}}}{{1 + \frac{{{k_1}[S] + {k_{ – 2}}[P]}}{{{k_{ – 1}} + {k_2}}}}}\]
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.
K\(_m\) and K\(_p\) 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:
\[\nu = \frac{{\frac{{{V_S}}}{{{K_m}}}[S] – \frac{{{V_P}}}{{{K_P}}}[P]}}{{1 + \frac{{[S]}}{{{K_m}}} + \frac{{[P]}}{{{K_P}}}}}\]
Try another topic, such as The First Law of Thermodynamics, Reversible Expansion!