1. For enzyme catalyzed reaction:kiS + E=X (fast)k2X -2> P + E (slow)derive the rate law and answer the following:a) What is meant by Michaelis-Menten constant km?(b) Prove that order of reaction in substrate changes from unity to zero athigher [S].(c) What is the rate and value of km, if k2 >> k1.(d) Draw the potential energy diagram for enzyme catalyzed reaction.

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Answered: 1. For enzyme catalyzed reaction: ki S…

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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k_1 E + SES E + P +0 k₂ S ↓↑ KIS ESS Based on this model, please answer the following questions: i) This is the model for what? Uncompet tive ii) Show the rate equation for an enzyme reaction of this type. iii) Show the Lineweaver-Burk plot for this case. A
The rate constants of an enzyme-catalyzed reaction, obeying the Michaelis-Menten kinetics, have been determined : E + S K₁ = 2 x 108 M-¹ S-¹ -1 -1 -1 K-₁= 1 x 10³ S K₂ = 5 x 10³ S-1 K₁ 1 K-1 ES K₂ E +P 1- Determine the Michaelis constant Km of the enzyme. 2- Determine the catalytic constant (kcat) of the enzyme. 3- Determine the catalytic efficiency of the enzyme.
Assume that an enzyme-catalyzed reaction follows the scheme shown: E + S ES E + P k₁ = 1 x 10%/M-s k-1 = 2.5 x 10%/s k2= 3.4 x 107s What is the dissociation constant for the enzyme-substrate, Ks? What is the Michaelis constant, Km, for this enzyme? What is the turnover number, Kcat, for this enzyme? What is the catalytic efficiency for the enzyme? If the initial Et concentration is 0.25mM, what is Vmax?
1) Is the ratio of the forward rate constant and reverse rate constant changed by the presence of an enzyme catalyst? 2) 4) What is the simplest mathematical relationship between substrate concentrations [S], initial velocity (Vo) of an enzyme catalyzed reaction, the maximal velocity of the reaction (Vmax), and the ½ maximal Vo (i.e. Km) ? (Hint: what is the Michaelis-Menten Equation? 3) Graphically illustrate the most useful derivation of the Michaelis-Menton equation (that darn linearized, double-reciprocal)
The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate ?0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<?m[S]<<Km, [S]=?m[S]=Km, and [S]>>?m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity ?0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.
For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K M
The initial rate for an enzyme-catalyzed reaction has been determined at a number of substrate concentrations. Data are as follows: [S(u mol/L) v[(u mol/L) min-'] 5 22 10 39 20 65 50 102 100 120 200 135 (a) Estimate Vmax and K, from a direct graph of v versus [S]. Do you find difficulties in getting clear answers? (b) Now use a Lineweaver-Burk plot to analyze the same data. Does this work better?
For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
(a) At what substrate concentration would an enzyme with a kcat of 30.0s−1 and a Km of 0.0050 M operate at one-quarter of its maximum rate?(b) Determine the fraction of Vmaxthat would be obtained at the following substratecon centrations [S]: ½Km, 2Km, and 10Km.(c) An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species.The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme?
For the following reaction: k1 E + S ES P+E (a) What is the rate equation for the formation of ES? (b) If all the enzyme is bound to substrate, what is the effect of adding more substrate on the forward rate of the reaction? (c) What is meant by the 'steady state assumption"?
Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+S ES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.

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