Question

An enzyme that follows simple michaelis–menten kinetics has an initial reaction velocity of 10 μmol⋅min−1 when the substrate concentration is five times greater than the km. What is the vmax of this enzyme?.

Answer 1

The Vmax of the enzyme is 12 μmol⋅min−1

We use Michaelis-menten equation,  which is given by;

$V=\frac{Vmax(S)}{Km+(S)}$

Where v is velocity of reaction

Vmax = maximum rate achieved by the system

[S] = concentration of a substrate

K$_{m}$= Michaelis constant

Given V$_{0}$ = 10$umolmin^{-1}$

[S] = 5km

Now, we calculate Vmax:

10$umolmin^{-1}$ = $\frac{Vmax*(5Km)}{Km+5Km}$

Km = 1

$10=\frac{5KmVmax}{6Km}$

$\frac{10*6}{5} =Vmax$

$Vmax$ =12 μmol⋅min−1

Learn more about Michaelis-menten kinetics here:

brainly.com/question/15180587

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