Question

An enzyme is discovered that converts substrate A to product B. The molecular weight of the enzyme is 10000 Daltons. The Km and Vmax values for the enzyme are shown below:

Km (mM) 2
Vmax (mmole/min/mg) 20

What is the reaction initial velocity for this enzyme when the concentration of A is 4 mM?

What is the turnover number of this enzyme?

Answer 1

Michaelis-Menten equation describes the velocity of enzymatic reactions (v) by relating it to [S] - concentration of a substrate S.

?0 = C(Vmax) / C + Km

v0 = initial velocity

C = concentration of substrate

Vmax= maximum velocity

Km = substrate concentration when velocity is half of Vmax

according to question km = 2mM, Vmax = 20 mmole/min/mg, C= 4 mM

PUTTING THE VALUE IN EQUATION ?0 = C(Vmax) / C + Km

V0= 4X20/4+2

= 80/6

13.33 mmol/min/mg

Turnover number of an enzyme, which is the number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate. It is equal to the kinetic constant k2, which is also called kcat.

kCAT= VMAX/ET

ET is a total enzyme  

Vmax= 20mmole/min/mg

10000 dalton weight of enzyme =10000mg/mMole

putting this in the equation kCAT = VMAX/ET

-= 20mmole/min/mg/ / 10000mg/mMole

=0.002/min