Question

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Applying the Michaelis-Menten Equation IV An enzyme is found that catalyzes the reaction X= Y Researchers find that the Kyn for the substrate A is 4 pm, and the keat is 20 min . (a) In an experiment, (X) = 6 mm, and the initial velocity, V, was 480 nm min!. What was the (E) used in the experiment? (b) In another experiment, (E) = 0.5 mm, and the measured V = 5 xem min!. What was the [X] used in the experiment? The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an a of 10. In an experiment with the same [EJ as in part (a), but a different (X), an amount of Z is added that reduces the rate V, to 240 nm min . What is the (X) in this experiment? (d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.

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Below is BoX 6-1 if needed

Answer 1

Vo - Vmax [s] Umar & keat [Ex] kmi (s) may a keat LEJ Griven km = 4MM keat = 20 min (a) [x] = 6mm - [s] Vo - 480 NM min- V. - Vmax [3] km + [s] Vmax - V. (km + [3]). [3] Vmax = 480 x 10-9 M mini 6x10 3 M (4x100 M + 6x 103 m] - 450 YIO 9 M min - 16.00 4x10-3 M) Exp-3 м = 480.32 x 10-4 M mini - 0.48 x 10-6 M mint Vmax. = 0.48 MM mint Vmax = keat [EE] =0.48 MM mint - 0.024 AM [Fa] - 0.48 MM mint 0.48 MM mint keat (b) [Ex] = 0. 5MM, Vo = 5 MM mint [x] - ? 20 min Vo = Keat [Fe] [s] - Vokm + V. [s] = keat [E] [s] km + [s] - Vokm = [s] [keat (E+] – Vo] [s] = Vokm Vo km = 5x10-6 M mint x 4x100 M 1) 14 (20min xo 5x10-6 M) - 5xp-bmmin = 20x10-12 M² mint - 20x10-12 M² min 10x10-6 M min-5x 10-6 M mino 5x10-6 M mint 4x10-6 M = 4 M