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 and what data interpretation supports this conclusion?
A) noncompetitive inhibitor because the KM has remained the same and the Vmax has decreased
B) competitive inhibitor because the KM has remained the same and the Vmax has decreased
C) noncompetitive inhibitor because the KM has increased and the Vmax has decreased
D) competitive inhibitor because the KM has increased and the Vmax has decreased
Correct Answer:
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