CHEM237 Chapter Notes - Chapter 12: Drug Design, Cytochrome P450, Reaction Mechanism

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coffeeherring584

20 Jul 2016

School

University of Waterloo

Department

Chemistry

Course

CHEM237

Professor

Guy Guillemette

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Document Summary

Fructofuranosidase was studied, where sucrose was cut into glucose and fructose. Steady-state assumption assumes that the es stays constant throughout a reaction due to the rate of synthesis equaling the rate of consumption of es. Concentration of s must be much higher than enzyme concentration. The use of initial velocity minizes effects of reversible reactions, inhibition of the enzyme by its products, and progress inactivation of the enzyme. Maximal velocity occurs at high substrate concentration when the enzyme is saturated (in es form) As ks decreases, the enzyme affinity for the substrate decreases. Km is therefore a measure of affinity of the enzyme for its substrate since ks is apart of the equation. Best done [s] ranges from 0. 5km to 5km: finding what actually happens within a transition state is very difficult. You have to look at nature of mechanism too.

Related Questions

1. For a Michaelis-Menten hydrolase, k1 = 6.2 x 108 M -1 s -1 , k-1 = 3.1 x 105 s -1 , and k2 = 6.9 x 102 s -1 .

a. Calculate Ks and Km for this enzyme. Does this enzyme follow rapid equilibrium kinetics or simply steady state? Justify your answer.

b. What is kcat for this enzyme? Justify your answer.

c. What is the efficiency of this enzyme? Would this enzyme be considered a perfect enzyme? Justify your answer.

d. In a 100Âµl reaction containing 50 pmol of the enzyme and saturating levels of substrate, what would be the Vmax value expressed in units of M s-1 . Assume 2 active sites per enzyme molecule.

e. To what concentration must the substrate be reduced in order to achieve half this velocity? Justify your answer mathematically.

f. To what must the enzyme concentration be reduced to achieve half the velocity in subpart d? Justify your answer mathematically.

g. Assume you set up a reaction with the hydrolase in buffer containing 3.0 mM substrate and want to add enough of a competitive inhibitor with Ki = 2.2 x 10-5 M to achieve a quarter of maximal velocity. What concentration of the inhibitor must be present in the reaction?

The zero order rate is seen in an enzyme catalyzed reaction when:
A. There is too much enzyme and too little substrate
B. Enzyme reactions are always first order
C. The Michaelis constant is very high
D. The V_max of the reaction is twice that which is observed
E. The enzyme substrate comples is at its maximum concentration

What happens when you add more enzyme to a catalyzed reaction that already is proceeding at V_max?
A. The reaction is inhibited because of competition for substrate
B. The plot of velocity vs. [S] will show a sigmoidal (an S-curve) shape
C. The reaction goes faster
D. The K_m of the reaction will change
E. There is no change in the rate of the reaction

Given two enzymes with the same V_max but with very different K_m values
A. The enzyme with the higher K_m will approach V_max at smaller [S]
B. V_max values cannot be the same with different K_m values
C. The steady state concentration of the enzyme-substrate complex is the same for each enzyme at every substrate concentration
D. The enzyme with a lower K_m value approaches V_max at lower [S]
E. Both values change when more enzyme is sdded

I think first one is E, second one is E, and last one is C. What are the correct answers for above?

CHEM237 Chapter Notes - Chapter 12: Drug Design, Cytochrome P450, Reaction Mechanism

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