Question

Suppose an enzyme has Km = 0.1mM and kcat = 10^3 s-1, and you perform an experiment in which [E]tot = 10-7M. Suppose that adding 0.1 µM of an inhibitor results in a Kmeff= 0.5 mM, and has no effect on kcat. Determine the substrate concentation that would be necessary to achieve a reaction velocity v of 5*10^-5Ms-1

Answer 1

Ans. Given, Kcat = 10³ s^-1

            [E]_tot = 0.1 uM = 1.0 x10^-7 M                                   ; [1 uM = 10^-6 M]

# Now,

            Kcat = Vmax / [E]_tot

            Or, 10³ s^-1 x (1.0 x 10^-7 M) = Vmax

            Hence, Vmax = 1.0 x 10^-4 M s^-1

# In presence of Inhibitor-

            Given, Kmeff = Km,app = 0.5 mM = 5.0 x 10^-4 M                        ; [1 mM = 10^-3 M]

            We have, Vmax = 1.0 x 10^-4 M s^-1                        - remains unaffected of inhibitor

            Vo = 5.0 x 10^-5 M s^-1

Now, Using Michaelis- Menten equation- Vo = Vmax [S] / (Km + [S])                   

            (5.0 x 10^-5 M s^-1) = (1.0 x 10^-4 M s^-1) x [S] / (5.0 x 10^-4 M + [S])

            Or, (5.0 x 10^-5 M s^-1) / (1.0 x 10^-4 M s^-1) =[S] / (5.0 x 10^-4 M + [S])

            Or, 0.1 x (5.0 x 10^-4 M + [S]) = [S]

            Or, 5.0 x 10^-5 M + 0.1[S] = [S]

            Or, 5.0 x 10^-5 M = [S] – 0.1[S]

            Or, [S] = 5.0 x 10^-5 M / 0.9

            Hence, [S] = 5.56 x 10^-5 M

Therefore, required [S] = 5.56 x 10^-5 M