Question

The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. Graph a Lineweaver-Burk plot. What are the apparent values of vmax and km for each experiment? what is the inhibition mechanism If the concentration of inhibitor is 0.5 mM, what is the value of K1?

Answer 1

From the given chart firstly we will find out the reciprocal values to graph a lineweaver-burk plot. As:

(a) Now we graph the plot as uploaded here:

(b) The apparent values of K_M and V_max are as can calculated:

   - 1/K_M = x intercept

    so from the graph       - 1/K_M = - 0.5   (when inhibitor is absent)

                                       K_M   = -1/-0.5 = 2mM

                                           -1/K_M = - 0.2     (inhibitor present)

                                       K_M = -1/-0.2 = 5mM

        1 / V_max = y intercept

                             1/ V_max = 1/0.10   = 10 molecules/min

In a competetive enzyme inhibition V_max does not change with the presence of inhibitor while K_M increases as shown above.

(c) In a competetive inhibition,at a given moment, the enzyme may be bound to the inhibitor, the substrate, or neither, but it can bind both at the same time. A competetive inhibitor could bind to an allosteric site of the free enzyme to prevent the substrate binding, as long as it does not bind to the allosteric site when the substrate is bound.