Question

The following reaction rate data are collected upon analysis of an enzyme catalyzed hydrolysis. The reaction is carried out with two different inhibitors. Determine the modes of inhibition for each inhibitor. (The rates provided are relative to the rate measured in the absence of inhibitor. The enzyme concentration for each set of data is equal.)
No inhibitor
[S]0 (10−2 M) 1.25 3.84 5.81 7.13
v0 0.398 0.669 0.859 1.000
[I1]– –2mmol/L
[S]0 (10−2 M) 1.25 2.5 4.00 5.50
v0 0.179 0.313 0.435 0.529
[I2]– –50mmol/L
[S]0 (10−2 M) 1.75 2.50 5.00 10.00
v0 0.202 0.218 0.240 0.253

Answer 1

Ans. Determination of Vmax and Km using LB Plot”

Lineweaver-Burk plot gives an equation in from of y = mx + c

where, y = 1/ Vo, x = 1/ [S],

Intercept, c = 1/ Vmax ,

Slope, m = Km/ Vmax

#a. Enzyme kinetics in absence of inhibitor [y = 2.1747x + 0.7965]:

Vmax = 1 / c = 1 / 0.7965 = 1.2555

Hence, Vmax = 1.2555

Now,

            Km= m x Vmax = 2.1747 x 1.2555 = 2.73033585

            Hence, Km= 2.730 M

#b. Enzyme kinetics in presence of inhibitor 1 [y = 5.9786x + 0.8037]:

Vmax = 1 / 0.8037 = 1.244

Now,

            Km= 5.9789 x 1.244 = 7.44

#c. Enzyme kinetics in presence of inhibitor 2 [y = 2.1144x + 3.7422]:

Vmax = 1 / 3.7422 = 0.267

Now,

            Km= 2.1144 x 0.267 = 0.565

# Type of Inhibitor:

Inhibitor 1: Vmax remains almost the same whereas Km is increased. It is the characteristic of competitive inhibitor. Therefore, Inhibitor 1 is a competitive inhibitor.

Inhibitor 2: Both the Vmax and Km is decreased in presence of inhibitor when compared to that of un-inhibited enzyme. So, inhibitor 2 is an un-competitive inhibitor.