Using the following kinetic mechanism. An enzyme catalyzed reaction obeying Mich

Question

Using the following kinetic mechanism. An enzyme catalyzed reaction obeying Michealis-Menten kinetics has the following rate constants: k_1 = 3.2 times 10^7 M^-1 s^-1 k_2 = 7.3 times 10^2 s^-1 k_3 = 5.8 times 10^5 s^-1 (A) Calculate K_ for this enzyme. (B) Calculate K_m for this enzyme. (C) Calculate k_ for this enzyme. (D) Calculate the specificity constant for this enzyme. (E) If the enzyme concentration is 2.3 times 10^-9 M, calculate V_max for this enzyme. (F) If the enzyme concentration is 2.3 times 10^-9 M, What concentration of substrate would generate a velocity equal to 0.25 V_max? (G) If the enzyme concentration was increased to 6.4 times 10^-9 M, what would be the values of V_max and K_m?

Explanation / Answer

Answer:

28. (A) Ks = k2/k1 = 7.3 x 102 / 3.2 x 107 = 2.28 x 10-5 M

(B) Km = (k2 + k3)/ k1 = (7.3 x 102) + (5.8 x 105) / 3.2 x 107 = (730 + 580000) / 32000000

= 0.018 M

(C) kcat = k3 = 5.8 x 105 s-1

(D) Specificity constant = kcat/ Km = 5.8 x 105 s-1/ 0.018 M = 32222222.22 M-1s-1 = 0.32 x 108 M-1s-1

(Since there are multiple subparts, the first 4 subparts have been answered according to the rules of Chegg)

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