2x10 1 pH 8.5 1x10 1 pH 7.4 8x10-7 6x 10-7 4x10-7 2x 10-7 1.5x10- 1x10-5 5x10-6

Question

2x10 1 pH 8.5 1x10 1 pH 7.4 8x10-7 6x 10-7 4x10-7 2x 10-7 1.5x10- 1x10-5 5x10-6 0 0 Substrate concentration (M) In both experiments, the enzyme concentration was 5.5x10-8 M From the graph on the left, estimate the values of Vmax and Km, with units of uM/s and M, respectively, in the pH 7.4 buffer From the graph on the right, estimate the values of Vmax and Km, with units of ??/s and ??, respectively, in the pH 8.5 buffer The enzyme appears to be substantially more active in the pH 8.5 buffer than in the pH 7.4 buffer. From the data, do you think that this difference is due primarily to a difference in the binding of the substrate in the two buffers or to a difference in the rate of catalysis? Briefly explain your answer and any assumptions you have made At pH 8.5, what value of Vmx would you expect when the enzyme concen tration is 1?1? Calculate the expected reaction rate at pH 8.5 if the enzyme concentration is 1 ?? and the substrate concentration is 50?? Under the conditions described in part (e) above, what fraction of the enzyme will have substrate bound to it?

Explanation / Answer

(a) From graph on the left, the estimated value of Vm = 8 x 10-7 M/s = 8 x 10-1 µM/s or 0.8 µM/s.

We know Km is the substrate concentration at V = Vm/2. Again from graph Vm/2 = 4 x 10-7 M/s and corresponding substrate concentration is 5 x 10-5 M. Therefore, Km = 5 x 10-5 M

(b) From graph on the right, the estimated value of Vm = 1.5 x 10-5 M/s = 15 x 10-6 µM/s or 15 µM/s.

We know Km is the substrate concentration at V = Vm/2. Again from graph Vm/2 = 0.75 x 10-5 M/s and corresponding substrate concentration is 2.5 x 10-5 M. Therefore, Km = 2.5 x 10-5 M

(c) The enzyme is more active in pH 8.5 buffer than pH 7.4 buffer because of the higher rate of catalysis in buffer with pH 8.5. From above calculations we see that Km at pH 8.5 is lower than Km at pH 7.4, which means that the enzyme has higher affinity for substrate at pH 8.5 than at pH 7.4

(d) We know that Vm ? [E], where [E] is enzyme concentration.

Therefore, we have [Vm1]/[E1] = [Vm2]/[E2] Or Vm2 = Vm1*[E2]/[E1]

W have Vm1 = 15 µM/s, [E1] = 5.5 X 10-8 M = 5.5 x 10-2 µM = 0.05 µM, and [E2] = 1 µM

Therefore, Vm2 = 15*1/0.05 = 300 µM/s

Thanks!

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.