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An enzyme has a KM of 20 µM and a Vmax of 50 mmoles of product/minute/µg...

An enzyme has a KM of 20 µM and a Vmax of 50 mmoles of product/minute/µg of enzyme. After exposure to an inhibitor and analysis on a Lineweaver - Burk plot the following values are obtained: -1/KM = - 0.05 liters/µmole and 1/Vmax = 0.04 (mmoles of product/minute/µg of enzyme)-1. What kind of inhibitor was used in the experiment?

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Answer #1

From the question the Km comes out to be 1/0.05 = 20 \muM and Vmax is 1/0.04 = 25 mmoles of product/min/ug of enzyme. Hence, the Km remains constant but the Vmax decreases. Hence, this is an example of reversible noncompetitive inhibition where the inhibitor binds reversibly to a site other than the active site and causes an overall change in the three dimensional conformation of the enzyme that decreases its catalytic activity.

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