Question

Given the enzyme system just described (Km = 8.20*10-5M and kcat = 40.0 s-1, [Eltot = 2.00*10-7 M), suppose we introduce a classical noncompetitive inhibitor with K1 = 1.20*10-7M. If the concentration of inhibitor introduced is 3.0 um, what will the effective kcat be? Express your answer in s-1.

Answer 1

Kcat = Vmax/[E]_tot.

Kcat = 40.0 s^-1

1M = 10⁶ uM

[E]_tot = 2.0 * 10-7 M = 0.2uM

Vmax/[E]_tot.*  Kcat

40.0 * 0.2 = 10M/s

Vmax = 10M/s

First, we need to find the Vmax upon adding the inhibitor.

Vmaxapp = Vmax / (1+I/Ki)

Ki = 1.2*10^-7M =

So Ki = 0.12uM

I= inhibitor concentration = 3.0uM.

Vmax = 10 uM/s

Vmaxapp = 10 /( 1+3/0.12)

= 10 /(1+25)

= 10/26

= 0.384 uM/s.

So, Vmax after adding inhibitor is 0.384M/s.

Kcat = Vmax/[E]_tot.

Kcat = 0.384uM/ 0.2uM/s

Kcat = 1.92

Kcat = 1.84 s^-1

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