1). The initial velocity of an enzyme-catalyzed reaction has been determined at several substrateconcentrations with and without an inhibitor I. The total enzyme concentration [E]ot present in eachexperiment is 1.20 x 10 M. [1] = 5.00 x10 M.(Note: v initial rate or velocity.)[S], M1.25E-32.50E-35.00E-310.0E-3v, M/min, no inhibitor0.480E-30.660E-30.810E-30.920E-3v, M/min, 5.00x10 M9.43E-51.41E-41.70E-42.10E-4b). What kind of inhibition occurred.c). Calculate the value of Ki=a). Use the Lineweaver-Burk plot to determine the values of Vmax and KM at [1] = 0.00 mM, and thevalues of Vmax and KM at 0.500 mM of I.it in 'How many molecules of

For single substrate enzymes, the mathematical relation between Km, vmax, vo and [S] is described by…

Answered: 1). The initial velocity of an…

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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(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.
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
15) The graph at right shows the results of reaction rate vs. substrate concentration for a Michaelis-Menten type enzyme 16) Th a. True b. False* reaction rate substrate concentration
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)
5. 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 equilibrium or the steady state? Justify your answer. b) What is kcat for this reaction? Justify your answer. c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1, and each enzyme has two active sites. d) What substrate concentration would be required for the reaction in (c) to reach half of Vmax. Justify your answer mathematically. e) A second Michaelis-Menten enzyme has k1 = 4.2 ⅹ 107 M-1 s-1, k-1 = 6.1 ⅹ 104 s-1, and k2 = 5.3 ⅹ 102 s-1. Which enzyme is most efficient? 6. A pharmaceutical company is trying to develop a
The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →

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