\/A compound was suspected of being a negative allosteric regulator of enzyme A. Which of thefollowing would be the expected Rate vs [S] in the absence and presence of the allosteric effector?For each of these figures rate is graphed on the y-axis and substrate concentration is graphed onthe x-axis. (one best answer)OAOBD torithegulatorwithoutregulator6withoutregulatorWithregulatorwithoutregulatorwithoutregulatorWithregulatorDB

Allosteric regulation is the regulation of an enzyme in the body by binding an effector molecule at…

Answered: A compound was suspected of being a…

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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Similar questions

Solve the Michaelis-Menten equation for KM when vo=Vmax/2. What does this tell you about the relationship between substrate concentration and enzymes with high KM values? With low KM values?
(i) Which graph indicates an enzymatic reaction without inhibitor?(ii) Which type of inhibitor is it? Briefly explain.(iii) Which graph indicates the highest concentration of inhibitor?(iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers.
Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
Which of these statements about enzyme-catalyzed reactions is false? The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction. The Michaelis-Menten constant Km equals the [S] at which V = 1/2 V, max: At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. The rate of a reaction decreases steadily with time as substrate is depleted. If enough substrate is added, the normal V, of max a reaction can be attained even in the presence of a competitive inhibitor.
You begin to study enzyme Z, which catalyzes a simple reversible reaction that interconverts compound S and compound P. You observe that the ∆G´° for the S to P conversion to be –6 kJ/mol, and that compound S has ∆G´° for binding to enzyme Z of –15 kJ/mol, while compound P has a ∆G´° for binding to enzyme Z of –13 kJ/mol. Please explain the effect of enzyme Z on conversion of S to P. (Your answer should include a graph qualitatively showing energy versus reaction progress; however, you still need to explain youranswer in words!) not sure how to make the correct graph.
enzyme-inhibitor complex requires 450 kJ.mol-1 to dissociate and that it displays kinetics somehow similar to non-competitive inhibition, does this is inhibitor good to use to inhibit toxanthine oxidase in the case of hyperuricemia and gout?
A purely competitive inhibitor of an enzyme has which of the following kinetic effects? decrease in Km increase in Vmax decrease in Vmax increase in Km Question 60 The Michaelis-Menten constant (KM) has the following characteristics, EXCEPT: The dimension for KM is concentration, such as molarity It is equal to ½ of Vmax It is the substrate concentration necessary to reach ½ of Vmax
The initial velocities of two different enzyme-catalyzed reactions were measured over a series of substrate concentrations. The following results were obtained: Enyme A: KM = 1.5 mM, Vmax = 10 μM s-1 Enyme B: KM = 5.0 mM, Vmax = 85 µM s-1 (a) Which enzyme binds to its substrate more tightly (assume k.1 >> k₂ in the Michaelis-Menten model)? (b) Calculate the initial velocities of each reaction when the substrate concentration is 2.5 mM. (c) Calculate the Kcat of each enzyme if the total enzyme concentration is 100 nM. (d) Which enzyme is the more efficient catalyst? Explain your answer. The enzyme carbonic anhydrase is strongly inhibited by the drug acetazolamide. A plot of the initial reaction velocity (as a percentage of Vmax) in the absence and presence of the inhibitor is shown below. What type of inhibition is taking place? Explain your reasoning. V (% of Vmax) 100 50 0.2 0.4 No inhibitor Acetazolamide [S] (MM) 0.6 0.8 1
In a bisubstrate reaction, a small amount of the fi rst product P is isotopically labeled (P*) and added to the enzyme and the fi rst substrate A. No B or Q is present. Will A (= P—X) become isotopically labeled (A*) if the reaction follows a Sequential mechanism?
For an enzyme-catalyzed reaction, the velocity was determined at two different concentrations of the substrate. Estimate the value of Vmax. [S] (MM) 10 20 88 nmol/min 79 nmol/min 67 nmol/min V(nmol/min) 51 nmol/min 27 Not enough information is given to form a reasonable estimation. 48
Select the graph that correctly illustrates the effect of a positive modifier (effector) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes. The solid blue curve represents the unmodified enzyme. The dashed green curve represents the enzyme in the presence of the effector. R is the highly active form of the enzyme and T is the less active form of the enzyme. Assume that this is a positively cooperative enzyme, meaning that the affinity for substrate increases with increasing substrate concentration.
Two experiments were performed with the enzymeribonuclease. In experiment 1 the effect of increasingsubstrate concentration on reaction velocity was measured.In experiment 2 the reaction mixtures were identical tothose in experiment 1 except that 0.1 mg of an unknowncompound was added to each tube. Plot the data accordingto the Lineweaver-Burk method. Determine the effect of theunknown compound on the enzyme’s activity. (Substrateconcentration is measured in millimoles per liter. Velocity ismeasured in the change in optical density per hour.)

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