1) For the simplified representation of an enzyme-catalyzed reaction shown below, the statement "ES is in steady-state" means that: a

1) For the simplified representation of an enzyme-catalyzed reaction shown below, the statement "ES is in steady-state" means that:

a. k₂is very slow.

b. k₁=k₂.

c. k₁=k_-1.

d. k₁[E][S] =k_-1[ES] +k₂[ES].

e. k₁[E][S] =k_-1[ES].

2. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the K_mfor that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 µmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 µmol) of product to be formed?

a. 1.5 min

b. 13.5 min

c. 27 min

d. 3 min

e. 6 min

3. The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by

1/V0= KM/(Vmax[S]) + 1/Vmax

To determine Km from a double-reciprocal plot, you would:

a. multiply the reciprocal of the x-axis intercept by -1.

b. multiply the reciprocal of the y-axis intercept by -1.

c. take the reciprocal of the x-axis intercept.

d. take the reciprocal of the y-axis intercept.

e. take the x-axis intercept, where V0= 1/2 Vmax.

4. Which of the following statements about enzyme-catalyzed reactions is FALSE?

a.   The rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration when the enzyme is saturated with substrate.

b.   The normal Vmaxof a reaction can be reached even in the presence of a competitive inhibitor if enough substrate is added.

c.    The rate of a reaction decreases steadily with time as substrate is depleted.

d.   The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction.

e. The Michaelis-Menten constant Kmequals the [S] at which V0= 1/2 Vmax.