Question

For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B. For substrate A, she determined 30 min that Km 3.0 HM and kcat Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows Michaelis- Menten kinetics In experiment A, where [A] 4 mM, she found that the initial velocity, Vo, was 394 nM min. What was the [E] in experiment A? nM EJ In experiment B, where [E4] 0.5 HM, she found that Vo 5 HM min. What was the [A] in experiment B? [A] им The compound Z was found to be a very strong competitive inhibitor of the enzyme, with an a of 10. In experiment C, in which the same [E4] was used as in experiment A but a different [A] was used, an amount of Z was added that reduced the rate Vo to 197 nM min. What was the [A] in experiment C?

Answer 1

Keat CA LE. Va Km 3.0 kab 30 min Experiment A V, 3q4 nMmin CAQ- 4M M Keat CA3E V = LADKn 30 x 4i x Ee 3 94 x10-9 x10 334 -8 . 314 10 13.14 x 10- - 3.1 M Expeniment Et 0.SMM Vo 5 X106M 2 o.SxI M Keat Cn3CE CA+Kn V CA3 Vxkm -Keat CA3 CEE V Km keaCA) CEE] - V CA CA KeatE) CA) VKM SK10-6x 3 x 10 30 x0.5x10~ Sxio E AQ S M

Experiment c Et 1-31 x 10- Vo 14 nM m Km 3 x1o-6 197 x1o- KCat 20 m Kzat CECAJ ox Kmt CA Vkno CAJ= 197x10 3x10-6 x10 30x 1-314 x1 0~ 8- 197x10-9 2.9x 10 ~S = 29.97x1o-6 CAJ 29.9 M 20 10 Keat -3x 10-6 Km 10
I have given the result and also have shown the calculations