Question

The enzyme fumarase catalyzes the hydrolysis of fumarate:

Fumarate(aq) + H₂O(liq) → L-maleate(aq)

The turnover number k₂ for this enzyme is 2.5 × 10³ s^-1, and the Michaelis constant is 4.2 × 10^-6 mol / L.

(a) What is the rate of fumarate conversion v if the total concentration of the enzyme is 1.0 × 10^-7 mol / Land the concentration of fumarate is 3.0 × 10^-4 mol / L?

(b) What is the ratio (v/v_max) in the conditions of question (a)?

(c) At what concentration of fumarate is (v/v_max) = 0.28?

Answer 1

Ans. Given,

            Turnover number = 2.5 x 10³ s^-1

            Km = 4.2 x 10^-6 mol L^-1       

            [E] = 1.0 x 10^-7 mol L^-1

            [S] = [Fumarate] = 3.0 x 10^-4 mol L^-1

#A. Turnover number, Kcat = V_max / [E]₀                                    - equation 1

Putting the values in equation 1-

            2.5 x 10³ s^-1 = V_max / (1.0 x 10^-7 mol L^-1)

            Or, V_max = (2.5 x 10³ s^-1) x (1.0 x 10^-7 mol L^-1)

            Or, Vmax = 2.5 x 10^-4 mol L^-1 s^-1

Hence, Vmax = 2.5 x 10^-4 mol L^-1 s^-1

#B. Using MM equation

            V₀ = V_max [S] / (K_m + [S])              - equation 2

            Or, V₀ = [ V_max (3.0 x 10^-4 mol L^-1)] / [(4.2 x 10^-6 mol L^-1) + (3.0 x 10^-4 mol L^-1)]

            Or, V₀ / V_max = 0.986

Therefore, the ration (V₀ / V_max) = 0.986

#C. Let the desired [Fumarate] = [S]

Putting The values in equation 2-

            V₀ / V_max = [S] / (K_m + [S])              

            Or, 0.28 = [S] / {(4.2 x 10^-6 mol L^-1) + [S] }

            Or, 0.28 x {(4.2 x 10^-6 mol L^-1) + [S] } = [S]

            Or, 1.18 x 10^-6 x 0.28 [S] = [S]

            Or, [S] – 0.28 [S] = 0.72 [S] = 1.18 x 10^-6

            Or, [S] = (1.18 x 10^-6) / 0.72 = 1.63 x 10^-6

Therefore, desired [Fumarate] = 1.63 x 10^-6 mol L^-1