As we have seen, you can analyze the kinetic properties of a Michaelis enzyme by

Question

As we have seen, you can analyze the kinetic properties of a Michaelis enzyme by measuring its velocity of catalysis at a range of substrate concentrations and plotting [S] vs V, or 1/[S] vs 1/V. You can also analyze the properties of an enzyme inhibitor by carrying out the same experiment and graphic analysis in the presence of the inhibitor. If you do this for a noncompetitive inhibitor, what difference will you see when you compare the enzyme kinetics to an uninhibited control?

A. Vmax will decrease but KM will be unchanged.
B. Vmax will be unchanged but KM will increase.
C. Vmax and KM both remain the same with a non-competitive inhibitor.
D. Vmax will decrease and KM will increase.
E. Vmax and KM will both decrease.

Explanation / Answer

A. Vmax will decrease but Km will be unchanged.

As it is a non-competitive inhibitor it will not compete for the active site. It will just decrease the activity.

So, as it is not competing for the active site there will be no change in the Km but Vmax will definitely decrease.

If it would have been competitive inhibition then Vmax and Km both would have decreased.

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