Question

An enzyme-catalyzed reaction is run, and its Vmax and KM value are recorded. An inhibitor is added at a concentration of 0.675 mM. The value for KM did not change between the two reactions. However, Vmax decreased by 70%.
a. Where does this inhibitor bind to the enzyme? Briefly explain.

b. What is the value of Ki for the inhibited reaction?

Answer 1

We have that the value of Ki for the inhibited reaction

• The place the enzyme inhibitor binds solely to the complicated fashioned between the enzyme and the substrate.
• KI = 3.551mM

Chemical Reaction

a)

Generally, if the fee of Vmax binds, the place the enzyme inhibitor binds solely to the complicated fashioned between the enzyme and the substrate.

b) V diminished through 70%

Thereofore

Alternate = 1 - 0.7 = 0.3V

Hence

0.3 =1/ ( 1 + [I]/KI)

1= 0.3(1 + 0.657/KI )

0.3 + 0.1971KI = 1

KI = 3.551mM

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