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An enzyme has a Km of 4.7 x 10-5M. If the Vmax of the preparation is 22m moles liter-1 min-1, what velocity would be observed in the presence of 2.0 x 10-4M substrate and 5.0 x 10-5M of a competitive inhibitor?

Question: An enzyme has a Km of 4.7 x 10-5M. If the Vmax of the preparation is 22m moles liter-1 min-1, what velocity would be observed in the presence of 2.0 x 10-4M substrate and 5.0 x 10-5M of a competitive inhibitor?
[A].

13.54μ moles liter-1min-1

[B].

6.68μ moles liter-1min-1

[C].

7.57μ moles liter-1min-1

[D].

17.8μ moles liter-1min-1

Answer: Option A

Explanation:

No answer description available for this question.