An Enzyme Has a K_M of 20 µM and a V_max

question 45

Multiple Choice

An enzyme has a K_M of 20 µM and a V_max 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/ K_M = - 0.03 liters/µmole and 1/ V_max = 0.02 (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 K_M has remained the same and the V_max has decreased
• B)competitive inhibitor because the K_M has remained the same and the V_max has decreased
• C)noncompetitive inhibitor because the K_M has increased and the V_max has decreased
• D)competitive inhibitor because the K_M has increased and the V_max remains the same

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