Two different enzymes are able to catalyze the same reaction, A B. They both hav

Question

Two different enzymes are able to catalyze the same reaction, A B. They both have the same Vmax, but differ in their Km for the substrate A. For enzyme X, the Km is 1 mM; for enzyme Y, the Km is 10 mM. When enzyme X was incubated with 0.1 mM A, it was observed that B was produced at a rate of 0.002 mM/min.
a) What is the value Vmax of each enzymes? b) What will be the rate of production of B when enzyme Y is incubated with 0.1 mM A? c) What will be the rate of production of B when enzyme X is incubated with 1 M A? 7. Two different enzymes are able to catalyze the same reaction, A B. They both have the Y, th produced at a rate of 0.002 mM/min. (6 marks) max, but differ in their Km for the substrate A. For enzyme X, the Km is mM; for enzyme e Km is 10 mM. When enzyme X was incubated with 0.1 mM A, it was observed that B was a) What is the value Vmax of each enzymes? b) What will be the rate of production of B when enzyme Y is incubated with 0.1 mM A? c) What will be the rate of production of B when enzyme X is incubated with 1 M A?

Explanation / Answer

answer)

a: The value of Vmax for the enzymes can be calculated from the values given for the first enzyme. In particular wii be:
  Vmax = Vo ([S] + Km) / [S] = 0.0020 (0.1 + 1) / 0.1 =  0.022 mmol/min

B: the rate ofproduction of B when enzyme 2 is incubated with 0.1 mM A will be:

  Vo = ([S] Vmax) / ([S] + Km) = (0.1 * 0.022) / (0.1 + 10) = 0.00022 mmol/min

c: the rate of productionof B when enzyme 1 is incubated with 1 M (i.e., 1000 mM) A will be:

  Vo = ([S] Vmax) / ([S] + Km) = (1000 * 0.022) / (1000 + 1) = 0.022 mmol/min

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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