Show all steps to the solutions of the following problems Ro

Show all steps to the solutions of the following problems. Round your answers no more than three significant figures.

a. What is the Vmax for the enzyme considering that it has only one substrate binding site?

½ Vmax = Km
     Vmax = 2 Km
               = 2 *0.102 * 10-3
             = 0.204 * 10-3

b. If 0.25 pmol of an irreversible or classic non-competitive inhibitor is added before starting the reaction. What is the resulting Km for the enzyme?

Solution

The K1 , K2 and K_-1 values are useful to determine the Km value.

Km = ( K_-1 + K2 )/ K1

Substitute the value and you will get,

Km = 10^-4 M

a.

The non-competitive inhibitor binds to site away from the substrate binding site. Hence it does not interfere with substrate binding and does not increase the Km. Only Vmax reduces.

Hence the answer is Km remains unchanged at 10⁴ M.

b.

Reversible inhibitor did not change the Vmax, but increased the Km. Here the inhibitor is competing with the substrate for the binding site and hence the Km increases, but the Vmax remains constant.

This means the inhibitor should be competitive inhibitor.

c.

The concentration of inhibitor is 1 x 10-4 M.

K_m^apparent = Km ( 1 + [I]/Ki)

[I] = conc of inhibitor

Ki = Ki = 1 x 10^-5 M ( From question b)

Therefore, substituting the values in the above formula,

Km ^apparent = 1.1 x 10-3 M