CHM361H5 Lecture Notes - Lecture 15: Equilibrium Constant, Enzyme

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hassan_rizvi96

27 Apr 2021

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Vmax stays the same because you can flood the system with a higher concentration of substrate and throw off the enzyme. Inhibitors bind to both e and es. Or inhibitor could bind s in presence and absence of e, but this rare ! S and i binding occurs simultaneously to e at different sites. Because cannot throw off the enzyme by increasing the concentration of substrate. E + s es e + p. Ki dissociation constant for the inhibitor. This reaction decreases the concentration of es. Ki" dissociation constant for the inhibitor. No reaction of esi to form p. *** we will not talk about mixed non-competitive inhibition. Binding of s and i are independent. Affinity of e for s is unchanged. Effects of inhibitor i can not be overcome by high [s] Vmax decreases in presence of inhibitor i. When you add more inhibitor velocity decreases. If ki is bigger the velocity will increase.

Related Questions

To study the kinetics of this enzyme, a graduate student measured the velocity of the reaction at a substrate concentration of 0.1 mM substrate. He found that after 1 mM competitive inhibitor was added, the initial rate decreased to 1/4 of the original rate. KM, the Michaelis constant, is 0.5 mM. What is the KI, the dissociation constant for the inhibitor binding to the enzyme? (Hint: in the presence of the competitive inhibitor, Vo= Vmax [S]/(1+[I]/KI) KM+ [S])

The answer is 0.28 mM, please give a detailed explanation

(Biochemistry)

Given the double reciprocal plot below, what can you say about the observed form of enzyme inhibition?

A) The inhibitor is structurally similar to the substrate

B) This is mixed inhibition because Vmax decreases and Km decreases

C) This in uncompetitive inhibition because Vmax decreases and Km increases

D) Both A and C are correct statements

E) none of the above

You are investigating the effects of a new inhibitor on an enzyme and your double reciprocal plot (1/[S] on the x-axis and 1/v_o on the y-axis) looks like the figure shown. What is the most logical conclusion you can draw from these data?

A) Your new inhibitor is an uncompetitive inhibitor because both V_max and K_m change

B) Your inhibitor is likely structurally similar to the substrate because K_m increases

C) Your inhibitor can likely interact with both free enzyme (forming an EI complex) and can also interact with the ES complex (forming an ESI complex).

D) In the presence of the inhibitor -1/K_m decreases while 1/v_o increases making this a mixed inhibitor.

E) Answers C and D are correct

1. (3%) You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data are obtained: [S] V0 âInhibitor +Inhibitor 0.0001 33 17 0.0002 50 29 0.0005 71 50 0.001 83 67 0.002 91 80 0.005 96 91 0.01 98 95 0.02 99 98 0.05 100 99 0.1 100 100 0.2 100 100

a) What is the Vmax in the absence of inhibitor?

b) What is the Km in the absence of inhibitor?

c) When [S] = 0.0004, what will V0 be in the absence of inhibitor?

d) When [S] = 0.0004, what will V0 be in the presence of inhibitor?

e) What kind of inhibitor is it likely to be?

2. (2%) An enzyme is inhibited by two distinct competitive inhibitors, IA and IB. IA is shown to bind with a Ki of 13 Î¼M while IB binds with a Ki of 150 Î¼M.

a) Which is the more potent of the two inhibitors?

b) The natural substrate for the enzyme has a Km value of 0.1 mM. How does the substrate binding affinity compare to each of the two competitive inhibitors?

3. (2%) Assuming they have equal affinity for the enzyme, why would a noncompetitive inhibitor be a more effective drug than a competitive inhibitor?

CHM361H5 Lecture Notes - Lecture 15: Equilibrium Constant, Enzyme

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