Enzyme-Catalyzed Reaction With Inhibitors: Determining Kn And Vmax Using Direct Linear Plot

The following data were obtained for an enzyme-catalyzed reaction in the absence of inhibitor and with inhibitor A or inhibitor B. In each case, the inhibitor is present a 10 pM. 

From the data above, using the direct linear plot (it is Eisenthal-Cornish-Bowden direct plots), determine:

A. kn,

B. vmax

C. The type of inhibition by A.

D. The type of inhibition by B. 

The aim of the question is to obtain K_m and V_max and also to determine the type of inhibition of A and also the type of inhibition of B. To answer the question of the type of inhibition, we will plot the graph on inhibitor A and inhibitor B together with the graph of S. We will use the graph to determine the competitive inhibitor and non-competitive inhibitor. According to the first graph enzymes inhibitor A is the competitive inhibitor since it raises Km only. According to the graph, enzyme inhibitor B is a non- competitive inhibitor since it lowers Vmax only keeping Km constant. We use the following data:

 We obtain the following graph

We will draw another graph to determine the values of K_m and V_max too.

We achieve this, we will have to transpose the S and V data to make it easier to create series. We create the series and then graph it. Later, we will have to format the graph to make the lines extrapolated. From the extrapolated lines, the x – intercept is the value of K_m and the y-intercept is our V_max.

From the graph the x – intercept is 0.7 and that’s the value of Km and the y – intercept is 5 and that’s the value of Vmax