MM

Why does the rate of product formation decrease over time?

there is less substrate to react with

Intial Velocity

linear, assume overall substrate is not changing over initial points

Why can you ignore k-2?

since we are looking at intitial reactions there is very little product so reverse reaction is negliglibel

How does v0 change as a function of [S]?

at low S, v0 is proportional to S, reaction is first order. at high S, v0 is independent of S.

Assumptions made with MM equation

the concentration of ES reaches a constant value soon after enzyme is mixed with substrate, thus es is constant, The initial substrate concentration is much greater than the total enzyme concentration, reaction velocity is measured right at the beginning when product accumulation is negligible. This allows scientists to ignore the reverse reaction

MM Equ

v0 = vmax[S]/Km+[S]

Km

Michaelis Constant, =Vmax/2

kcat

turnover number for the enzyme, =Vmax/[E], determined when enzyme is saturated, constant for any given enzyme

Axes for MM plots

x: [S] y: v0