enzyme kinetic

Question

if the reaction proceeds past V₀ what would be the equation that determines the rate of product (P) formation be 

Answer 1

I would say that this really depends on the model that is most appropriate for your enzyme. Typical enzymatic reactions taught in school use the Michaelis-Menten kinetic model so that is what I will be referring to in this instance. Otherwise the elementary steps that will define your rate laws and kinetic model will be different and will result in drastically different rate of formations.

At its core, a reaction where E+ S ⇔ [ES] ⇒ E + P, the rate of formation of [P] = d[P]/dt = k₂[ES].

Using the steady-state approximation on [ES], assuming that the formation of the intermediate [ES] is slow and the consumption of [ES] is fast, allows us to set

d[ES]/dt = k₁[E][S] + k₂[ES] + k_-1[ES] = 0

so

k₁[E][S] = (k_-1 + k₂)[ES]

[ES] = k₁[E][S]/(k_-1 + k₂)

Simplifying the rate constants into one unified dissociation constant, K_M, makes the equation easier to manipulate.

K_M = (k₂ + k_-1)/k₁

[ES] = [E][S]/K_M

Because the concentration of enzyme is going to be changing throughout the reaction, you need to substitute [E] for [E₀], the total number of initial enzyme.

[E₀] = [ES] + [E]

[ES] = [E₀] - [ES])[S]/K_M

[ES] = [E₀][S]/([S] + K_M]

Therefore,

d[P]/dt = k₂[E₀][S]/([S] + K_M]

Where you would typically say V_max, the maximum rate of reaction, is

V_max = k₂[E₀]

making

d[P]/dt = V_max[S]/([S] + K_M].

Hope that helps.

-Eric