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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?

13.54μ moles liter-1min-1
6.68μ moles liter-1min-1
7.57μ moles liter-1min-1
17.8μ moles liter-1min-1

Correct Answer: 13.54μ moles liter-1min-1

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