Please explain the answer! Thanks! An enzyme has a Vmax Value of 720 nM per seco

Question

Please explain the answer! Thanks!

An enzyme has a Vmax Value of 720 nM per second and a Vmax/Km value of 0.36 per second. When a series of experiments with an inhibitor is run the apparent value of Vmax/Km decreases to 0.09 per second and the apparent Vmax value decreases to 180 nM per second. Given this information, what type of inhibition is occurring? Choose the answer that best, most complete description of the type of inhibition that is occurring. Competitive Uncompetitive Mixed Pure Noncompetitive Irreversible

Explanation / Answer

Ans is non competitive

As value of Km remains same amd value of Vmax is decrease d.

Vmax/km =0.36 => km =720/0.36 = 2000 and

when Vmax = 180 => km = 180/ 0.09 = 2000

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