algae/notes/mm_derivation.html
For the enzyme catalyzed reaction
E + S | --k1--> <--k2-- |
ES | --k3--> <--k4-- |
E + P |
v = k3*[ES]
Rate of formation of ES = k1* [E]*[S]
Rate of breakdown of ES = (k2 + k3) * [ES]
At steady state, the formation and breakdown are equal. This steady state would be temporary in any enzyme assay or nutrient uptake experiment.
k1* [E]*[S] = (k2 + k3) * [ES]
rearranging:
[ES] = [E]*[S]/((k2 + k3)/k1)
We can lump these kinetic constants to make a new constant called Km = (k2 + k3)/k1)
[ES] = [E]*[S]/Km
The total enzyme is the amount bound and unbound to the substrate
[Etot] = [E] + [ES]
Substituting [Etot] - [ES] for [E]
[ES] = ([Etot] - [ES]) *[S]/Km
Solving for [ES]
[ES] = ([Etot] *(( [S]/Km)/(1+[S]/Km))
Which simplifies to
[ES] = [Etot] * [S] / ([S] + Km)
Multiplying both sides by k3 gives the velocity of the reaction
v = k3*[ES] = k3* [Etot] * [S] / ([S] + Km)
substituting Vmax for k3* [Etot] gives the familar form of the MM equation
v = Vmax * [S] / ([S] + Km)