Question

Some enzymes are inhibited by high concentratins of their own substrates. (a) show that when substrate inhibition is important the reaction rate v is given by

v= Vmax/ ( 1+km/[S}o + [S]o/K1

Where Ka is the equilibrium constant for dissociation of the inhibited enzyme substrate complex. (b) What effect does the substrate inhibition have on a plot of 1/v against 1/[S]o?

Answer 1

b-Many enzymes are inhibited by their own substrates, leading to velocity curves that rise to a maximum and then descend as the substrate concentration increases.

a-

In typical enzyme-catalyzed reactions, reactant and product concentrations are usually hundreds or thousands of times greater than the enzyme concentration. Consequently, each enzyme molecule catalyzes the conversion to product of many reactant molecules. In biochemical reactions, reactants are commonly known as substrates. The catalytic event that converts substrate to product involves the formation of a transition state, and it occurs most easily at a specific binding site on the enzyme. This site, called the catalytic site of the enzyme, has been evolutionarily structured to provide specific, high-affinity binding of substrate(s) and to provide an environment that favors the catalytic events. The complex that forms, when substrate(s) and enzyme combine, is called the enzyme substrate (ES) complex. Reaction products arise when the ES complex breaks down releasing free enzyme.

Between the binding of substrate to enzyme, and the reappearance of free enzyme and product, a series of complex events must take place. At a minimum an ES complex must be formed; this complex must pass to the transition state (ES*); and the transition state complex must advance to an enzyme product complex (EP). The latter is finally competent to dissociate to product and free enzyme. The series of events can be shown thus:

E + S <