The fractional saturation of an enzyme is 80% at a substrate concentration of 20 mM. In the presence of 5 mM of a competitive inhibitor the fractional saturation at the same substrate concentration is 40%. The Km for this reaction is mM, and the apparent Km is mM. The value of alpha is while the Ki of the inhibitor is mM.

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Q: Km V. (V) 15) + V = [S] max a' = 2 a' = 1.5 (7) -15⁰° max a Vmax 1 [S] MM a'=1 [1]

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Q: The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme:

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A:

Q: Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the…

A: Vmax : Reaction rate- when the enzyme gets fully occupied by substrate. Vmax- Maximum velocity

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A: Note : Hi ! Thank you for the question. We are authorized to answer one question at a time. Since…

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