The Michaelis-Menten rate equation for reversible mixed inhibition is written asVo=Vmax [S]aKm + a' [S]where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitorbound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is theMichaelis constant that represents the [S] at which the reaction reaches/VmVmax2α'Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation.Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer.=Apparent, or observed, Km is equivalent to the [S] at which Vomax.apparent Km =

Hence solved.Km=[S][αV0Vmax−1]Explanation:Determine the rate constant's expression, r.First, in…

Answered: The Michaelis-Menten rate equation for…

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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e) Two 20th century scientists, Leonor Michaelis and Maud Leonora Menten, proposed the model known as Michaelis-Menten Kinetics to account enzymatic dynamics. The model serves to explain how an enzyme can cause kinetic rate enhancement of a reaction and explains how reaction rates depends on the concentration of enzyme and substrate. k1 k2 E + S= ES → E + P k-1 Equation 1 i. Prove Equation 2 by referring to Equation 1. Vmax[S] V = [S] + KM Equation 2 ii. Give three characteristic features of catalyst.
Using equilibrium argument, why does Km apparently increase, decrease or stay the same in uncompetitive inhibition?
☐ Consider the Lineweaver-Burk graph below. The "+ inhibitor" and "no inhibitor present" have the same y intercept. Which of the following accurately describe the effects of the inhibitor on the enzyme's kinetics? (Pick two.) + inhibitor 1/Vo -No inhibitor present 1/[S] The inhibitor doesn't change the Vmax The inhibitor decreases the Vmax The inhibitor increases the Km The inhibitor doesn't change the Km The inhibitor decreases the Km
what concentration of a competitive inhibitor is required to yield 60% inhibition of an enzyme at a substance concentration of 1.0x10^-4 M. The Km for the enzyme is 22 uM and the Ki for the inhibitor is 59 uM. To what concentration must the substrate be increased to reestablish velocity at the uninhibited value?
In pure noncompetitive inhibition: a. Where on the enzyme does the inhibitor bind? b. Does the inhibitor bind to E, ES or both? c. What is the effect of I on Vmax? d. What is the effect of I on Km?
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?
An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min−110 µmol⋅min−1 when the substrate concentration is five times greater than the ?M.KM. What is the ?maxVmax of this enzyme?
In the same kinetics experiment above without the inhibitor, what is the fraction of occupied enzyme at substrate concentration 1microM ? How about when it's 1mM ?
Noncompetitive inhibition can often be explained by which of the following models? the induced fit model the lock and key model both (a) and (b) neither (a) nor (b)
-Inhibitor +Inhibitor [S] (mM) V0&νβσπ;(μmol/sec). νο&νβσπ: &ν βσπ (μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITH iinhibitor?
Consider the Michaelis-Menten equation, below: Vmaz (S V. k + [S] %3D What is the relationship between changes in substrate concentration and velocity when the concentration of substrate, [S), is well below k7 The S terms cancel out in this equation, so there is no effect of changing substrate concentration. The S] term in the numerator is negligible, so there is no impact of changing substrate concentration. Because the enzyme has reached Vr there is no effect of changing substrate concentrations on enzyme velocity. OThe [S term in the denominator is negligible compared to k There is a linear relationship between substrate concentration and velocity.
Drug D reversibly binds to enzyme E and inhibits its activity toward substrate S. D binds equally well whether or not S is bound. Sketch a graph of the expected 1/v vs. 1/[S] relationship for: A) The enzyme reacting with S in the absence of drug D, B) The enzyme reacting with S in the presence of a small amount of drug D, and C) The enzyme reacting with S in the presence of a large amount of drug D.

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