Question

An enzyme follows Michaelis-Menten kinetics. How does the following things affect the Km and the Vmax. Explain how they are affected and why.

1. A competitive inhibitor

2. A Mixed inhibitor

3. 6 M urea

4. doubling [S]

Answer 1

V_max = Vo*(K_m+[S])/[S]

1. Competitive inhibitors competes with the substrate for the active sites of the enzyme molecule.

Competitive inhibitors raise the value of K_m while the value of V_max remains unchanged.

2. A mixed inhibitor binds to both of the free enzyme and the ES complex.

Mixed inhibitors have no effect on the value of K_m while they decrease the value of V_max because some enzyme molecules will always be out of commission.

3. Urea is a competitive inhibitor.

So it will raise the value of K_m while the value of V_max remains unchanged.

4. Km is equal to the substrate concentration at which the reaction rate is half its maximum value.

Doubling [S] will increase the value of Km while the value of V_max will remain unaltered.