2. The kinetic effect of purely competitive inhibitor of an enzyme:(A) Increases K without affecting Vmax (B) Decreases Km without affecting Vmax(C) Increases Vmax Without affecting K (D) Decreases Vmax without affecting Km3. In reversible non-competitive enzyme activity inhibition:(A) Vma is increased(C) Vmax is decreased4. In enzyme kinetics Vmax reflects:(A) The amount of an active enzyme(C) Half the substrate concentration5. In non-competitive enzyme activity inhibition, inhibitor:(A) Increases Km(C) Does not affect Km(B) K is increased(D) Concentration of active enzyme is reduced(B) Substrate concentration(D) Enzyme substrate complex(B) Decreases Km(D) Increases Km

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Answered: 2. The kinetic effect of purely…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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4. An enzyme-substrate complex has a Kd = 100 nM. A competitive inhibitor with which of the following Ki values will be most effective at inhibiting this enzyme? A) K₁ = 10 nM B) K₁ = 100 nM C) K₁ = 1000 nM D) None of the above - only the KM is important Briefly explain:
(a) You are given the following experimental measurements for the effects of an inhibitor on the initial velocity of an enzymatic reaction. Plot the data and determine what type of inhibition is being observed. Label all axes. [S] (MM) Vo (mM • Vo with I present (mM ⚫s-¹) 1 2.8 3 6.0 1.0 2.5 4 7.0 3.1 8 9.9 12 11.5 4.8 5.7 (b) Briefly explain how varying [S] can be used to determine whether an inhibitor is uncompetitive or competitive. Be sure to include in your explanation why this method would work.
Graph a double reciprocal plot that satisfy the following: a. Michaelis-Menten kinetics enzyme, b. an inhibitor that binds only free enzyme (competitive), c. an inhibitor that binds only enzyme-substrate complex.
2. An enzyme-catalyzed reaction is carried out at several different substrate concentrations (listed below), either (1) with no inhibitor, or (2) in the presence of an inhibitor, I, at 0.63 g/L. Given the following data, determine: A. what type of inhibitor I is B. the value of Ki No Inhibitor: With Inhibitor: Substrate Concentration (g/L) 0.86 Reaction Velocity (g/L-s) Reaction Velocity (g/L-s) 0.179 0.161 2.35 3.80 4.90 6.10 18.4 0.398 0.567 0.666 0.726 0.950 0.322 0.407 0.443 0.495 0.639
2. Enzyme-catalyzed reactions. Answer the following with true or false. If false, explain why. (a) The initial rate of an enzyme-catalyzed reaction is independent of substrate concentration. (b) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. (c) The Michaelis constant Km equals the substrate concentration at which velocity (v) = Vmax/2. (d) The Km for a regulatory enzyme varies with enzyme concentration. (e) If enough substrate is added, the normal Vmax of an enzyme-catalyzed reaction can be attained even in the presence of a noncompetitive inhibitor. (f) The Km of some enzymes may be altered by the presence of metabolites structurally unrelated to the substrate. (g) The rate of an enzyme-catalyzed reaction in the presence of a rate-limiting concentration of substrate decreases with time. (h) The sigmoidal shape of the v versus [S] curve for some regulatory enzymes indicates that affinity of the enzyme for the…
V 23. The graph below is a graph of Vmax (a) Label the graph clearly with both the Vmax and the Km. Estimate the Km from this graph giving the correct units. v/Vmax 1.0 0.5 0.0 0.0 vs [S] for an enzyme. Enzyme Activity vs [Substrate] :0.2 0.4 Substrate (HM): 0.6 (b) If the Vmax = 25 mmoles per minute per µmole of enzyme calculate the Keat and the specificity constant."
Please explain your reasoning for every answer
1. ( . tion can be represented by the so-called Michaelis equation ) The velocity of an enzyme catalyzed reac- Vmax [S] Vo = Km Vo = max kcat [Eo] [So] (Км + [Sol) V max where [So] is the initial substrate concentration, [Eo] is the total enzyme concentration, kcat is the cataly- tic rate constant or turnover number, and KM repre- sents the Michaelis constant. The graph on the right illustrates the dependence of the velocity of the reaction as a function of substrate concentration. Ex- Km [S] (mm) plain why the maximal velocity can be represented by the expression Vmax = Kcat [Eo] at very high substrate concentrations. Vo (µm/min)
1. Consider the following parameters related to an enzyme that follows Michaelis-Menten kinetics for the reaction: k(1) k(2) S ----> ES ----> P k(-1)
6. An enzyme catalyzed reaction has a KM of 1 mM and a Vmax of 5 nM/s. What is the reaction velocity when the substrate concentration is (a) 0.25 mM (b) 1.5 mM, and (c) 10 mM?
1. The concentration of substrate X is high. What happens to the rate of the enzyme-catalyzed reaction if the concentration of substrate X is reduced? Explain. 2. An enzyme has an optimum pH of 7.2. What is most likely to happen to the activity of the enzyme if the pH drops to 6.2? Explain
Explain how running a reaction with the enzyme aldolase at a pH of 14 affects its activity Enzyme Activity e Activity 765432TO Factors Affecting Enzyme Activity 0 6543 2 6 Solution pH 8 a) It significantly diminishes activity because enzyme is denatured b) It significantly diminishes activity because the enzyme becomes too reactive Oc) It will quadruple its activity because enzyme is denatured and free Question 9 (5 points) Estimate the optimum pH required for maximum enzyme activity 10 d) It will double its activity because enzyme now has more active sites 12 Factors Affecting Enzyme Activity

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