Question

A class conducts an enzyme kinetics experiment using the digestive enzyme, trypsin and an unknown inhibior. They generate the two curves below. According this data, what type of inhibitor did the students use?

1. Competitive 2. Non- Competitive

In the same Lineweaver-Burke plot from question above, what is the approximate value of Vmax? 1/4 100 0.01 Vmax cannot be determined from this plot.

Answer 1

From the given plots, it can be inferred that it is competitive inhibition.

A competitive inhibitor competes with the substrate for the active site of the enzyme. As a result there will be a reduced rate of reation under the influence of a competitive inhibitor.

Inorder to decrease the chance of inhibitor binding to the enzyme more amount of substrate has to be added i.e substrate concentration should be increased. As a result the Km of the enzyme will be higher.

In the presence of a competitive inhibitor, it takes a higher substrate concentration to achieve the same velocities that were reached in the absence of inhibitor. So while V_max can still be reached if sufficient substrate is available, one-half V_max requires a higher substrate concentration than before and thus K_m is larger.

As we can see Km is higher in the given graph we can infer that it is competitive inhibition.

In the given Lineweaver Burk plot, x intercept (1/[S]) is -4 and y intercept (1/V) is 0.01.

The Lineweaver-Burk plot essentially relies on the following equation:

1/V = (Km/Vmax) * 1/[S] +1/V

In this plot Vmax is given by the inverse of y intercept.

Y intercept = 0.01

Inverse of y intercept = 1/0.01

= 100

So the approximate value of Vmax form the given plot is 100

Option C is correct.