An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence...

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An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence...

An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 microM inhibitor, Vmax is 36 nmoles substrate converted per sec with no change in Km. Determine the Ki of the inhibitor. What type of inhibitor is this?

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Answer 1

Answer #1

Since the inhibitor affects the V_max of the reaction but the K_m remains constant hence we can say that it is non-competitive inhibition.

K_i = [V_maxapp.] [I] / V_max

= 36 x 4 x 10 ^-4 / 48

= 3 x 10 ^-4 moles

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Answer 2

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An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence...

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