22. The graph to the right shows the rates of a chemical reaction and the effects of inhibition.
Remember that [S] is concentration of substrate. A. What is Vmax? (1 pt) B. Predict the effect of
adding more substrate to enzyme VMax catalyzed reaction under normal conditions (black line
in fig). Will Vmax increase, decrease or stay the same? Explain your answer. (2 pts) Normal
Enzyme Competitive inhibitor C. Explain what would happer Noncompetitive inhibitor if more
substrate were added to the reaction with competitive inhibitor. (2 pts) D. Explain the effects of
noncompetitive inhibition on the Vmax of the reaction. Why does the graph show a decrease in
rate versus the competitive inhibitor reaction? (2 pts)
Solution
Answer A] Vmax is the maximum rate of reaction ,when enzyme is saturated with substrate.
B] The effect of adding more substrate to enzyme catalysed reaction under normal conditions are
as follows.
The Vmax will remain constant.
A]At low concentration of substrate ,there is a steep increase in the rate of reaction with
increasing substrate concentratio n.The catlytic site of the enzyme is empty,waiting for substrate
to bind,for much of time,& the rate at which product can be formed is limited by the
concentration of sustrate which is available.
B]As the conc. of substrate increases ,the enzyme becomes saturated with substrate.As soon as
catalytic site is empty,more substrate is available to bind and undergo reaction.
The rate of formation now will depends on the activity of the activity of the enzyme itself,and
adding more substrate will not affect the rate of reaction to any significant effect.
hence Vmax does not change it remains same
C] If more substrate is added to the recation with competitive inhibitor
competitive inhibitors can only bind to E and not to ES.
they do not affect Vmax because the inhibitor does not change the catalysis in ES because it can
not bind to ES.
D]The effect of noncompetitive inhibitor on the Vmax is ,the Vmax decreases because some of
the enzyme molecules will always be out of commission.
Remember that [S] is concentration of substrate. A. What is Vmax? (1 pt) B. Predict the effect of
adding more substrate to enzyme VMax catalyzed reaction under normal conditions (black line
in fig). Will Vmax increase, decrease or stay the same? Explain your answer. (2 pts) Normal
Enzyme Competitive inhibitor C. Explain what would happer Noncompetitive inhibitor if more
substrate were added to the reaction with competitive inhibitor. (2 pts) D. Explain the effects of
noncompetitive inhibition on the Vmax of the reaction. Why does the graph show a decrease in
rate versus the competitive inhibitor reaction? (2 pts)
Solution
Answer A] Vmax is the maximum rate of reaction ,when enzyme is saturated with substrate.
B] The effect of adding more substrate to enzyme catalysed reaction under normal conditions are
as follows.
The Vmax will remain constant.
A]At low concentration of substrate ,there is a steep increase in the rate of reaction with
increasing substrate concentratio n.The catlytic site of the enzyme is empty,waiting for substrate
to bind,for much of time,& the rate at which product can be formed is limited by the
concentration of sustrate which is available.
B]As the conc. of substrate increases ,the enzyme becomes saturated with substrate.As soon as
catalytic site is empty,more substrate is available to bind and undergo reaction.
The rate of formation now will depends on the activity of the activity of the enzyme itself,and
adding more substrate will not affect the rate of reaction to any significant effect.
hence Vmax does not change it remains same
C] If more substrate is added to the recation with competitive inhibitor
competitive inhibitors can only bind to E and not to ES.
they do not affect Vmax because the inhibitor does not change the catalysis in ES because it can
not bind to ES.
D]The effect of noncompetitive inhibitor on the Vmax is ,the Vmax decreases because some of
the enzyme molecules will always be out of commission.
