Question

BIOCHEMISTRY: Vmax and Km values

For the following problem

Answer 1

In order to calculate Vmax and Km values, for both with and without the inhibitor present follow the below mentioned steps

Step-1: Find 1/S, 1/V(w/o inhibitor), and 1/V(with inhibitor) values for all substrate concentrations.

Step-2: To calculate Vmax and Km value for enzyme action with inhibitor, plot a graph of 1/S Vs 1/V(with inhibitor)

           The graph will be a straight line touching both X- and Y- axis and of the following format

              y = mx + C    ------------- (1)

           where y = 1/V(with inhibitor), x =1/S, Y- intercept, C = 1/Vmax and m = Km / Vmax.

            also X - intercept = - 1/Km

    Now we can obtain any 2 equations by putting the values given in data to find Vmax and Km values. Hence

1/0.1 = m *(1/2) + C -----------(4)

1/0.12 = m * (1/3) + C -----------(5)

By solving (4) and (5) we get

C = 1/Vmax = 5

=> Vmax = 1/5 = 0.2 (answer)

m = Km / Vmax = 10 = Km /0.2

=> Km = 10 * 0.2 = 2 (answer)

Step-3: To calculate Vmax and Km value for enzyme action w/o inhibitor, plot a graph of 1/S Vs 1/V(w/o inhibitor)

           The graph will be a straight line touching both X- and Y- axis and of the following format

              y = mx + C    ------------- (1)

           where y = 1/V(w/o inhibitor), x =1/S, Y- intercept, C = 1/Vmax and m = Km / Vmax.

            also X - intercept = - 1/Km

    Now we can obtain any 2 equations by putting the values given in data to find Vmax and Km values. Hence

1/0.25 = m *(1/2) + C -----------(2)

1/0.28 = m * (1/3) + C -----------(3)

By solving (2) and (3) we get

C = 1/Vmax = 2.714

=> Vmax = 1/2.714 = 0.36846 (answer)

m = Km / Vmax = 2.571 = Km / 0.36846

=> Km = 2.571 * 0.36846 = 0.9473 (answer)

we can determine the type of inhibition by plotting a a graph of 1/S Vs 1/V(with inhibitor) and  1/S Vs 1/V(w/o inhibitor).