If inhibitor X changes lactase activity to a Vo of 0.111111111111 mM per minute when [S] = 0.40 mM, and a Vo of 0.166666666667 mM per minute when [S] = 1.0 mM, calculate the new Vmax. A. 0.10 mM per minute B. 0.25 mM per minute C. 0.50 mM per minute D. 1.0 mM per minute E. 2.0 mM per minute

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If inhibitor X changes lactase activity to a Vo of 0.111111111111 mM per minute when [S] = 0.40
mM, and a Vo of 0.166666666667 mM per minute when [S] = 1.0 mM, calculate the new Vmax.
A. 0.10 mM per minute
B. 0.25 mM per minute
C. 0.50 mM per minute
D. 1.0 mM per minute
E. 2.0 mM per minute

Expert Solution
Step 1

Vmax, also known as the maximum velocity, is the maximum rate of an enzymatic reaction under optimal conditions when the enzyme is saturated with substrate. At Vmax, all enzyme molecules are bound to substrate, and any further increase in substrate concentration does not increase the rate of reaction.   

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Follow-up Question

Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X.

A. 0.10 per minute
B. 0.25 per minute
C. 0.50 per minute
D. 1.0 per minute
E. 2.0 per minute

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Follow-up Question

Calculate the new Km of the lactase reaction with inhibitor X.
A. 0.10 mM
B. 0.25 mM
C. 0.50 mM
D. 1.0 mM
E. 2.0 mM

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