2. Two different enzymes are able to catalyze the same reaction, A→ B. They both have the same Vmax, butdiffer in their Km for the substrate A. For enzyme 1, the Km is 1.0 mM; for enzyme 2, the Km is 10 mM. Whenenzyme 1 was incubated with 0.1 mM A, it was observed that B was produced at a rate of 0.0020mmoles/minute.a. What is the value of the Vmax of the enzymes?b.What will be the rate of production of B when enzyme 2 is incubated with 0.1 mM A?C. What will be the rate of production of B when enzyme 1 is incubated with 1 M (i.e., 1000 mM) A?d. Which enzyme is a more efficient catalyst?

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Answered: 2. Two different enzymes are able to…

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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2. A 500µl reaction containing 0.5 pmol of an enzyme following Michaelis -Menten kinetics with saturating substrate concentration is studied. For this enzyme kl = 1 x 10° M-1 s', k-1 = 2 x 10's', and k2 = 1 x 10° s'. a. What is the Vmax for the enzyme considering that it has only one substrate binding site? b. What are the values of Km and Ks? Is this a rapid equilibrium enzyme or does it follow steady state kinetics? Explain why. c. What is the substrate concentration needed to achieve half maximal velocity? Explain the basis for your answer. d. If the substrate concentration in part e is doubled, how many fold will the initial velocity increase? e. What is the Km is the concentration of the enzyme is doubled? f. What is the efficiency of this enzyme? g. Would this enzyme be considered a perfect enzyme? Explain. h. If 0.25 pmol of an irreversible or classic non-competitive inhibitor is added before starting the reaction. What is the resulting Km for the enzyme? i. A reversible…
1. The initial rate of an enzymatic reaction was determined at different substrate concentrations. The data is below: [S] (μmoles/L) v [(μmol/L) min -¹] 20 50 100 200 65 102 120 135 A. Graph Michaelis and Menten and use it to estimate Vmax and Km. B. Make the Lineweaver-Burk plot and use it to estimate Vmax and Km. C. What would you expect to find in the values of Vmax and Km if you do the same experiment but with twice the amount of enzyme used here. Draw your answer on the MM graph. 2. The Table below shows the data collected by an undergraduate student of the enzyme rate at different substrate concentrations. In the presence and in the absence of an inhibitor.
3. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained: [S] (M) 2.5 x 10-6 4.0 x 106 1х 105 2 x 105 4х 10-5 1 x 104 2 x 10-3 1 x 102 Vo (µM / min) 28 40 70 95 112 128 139 140 A. Vmax = 140 and Km=1 x 10-5 B. Vmax = 70 and Km = 1 x 103 C. Vmax = 140 and Km=1 x 10-2 D. Vmax = 70 and Km = 1 x 10-² %3D
The figure below shows the dependence of the enzyme's rate, v (in µM/min). as a function of substrate concentration. S (in mM). Also shown is the dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. 1/v 1- 05 -2 -1 3 1/IS] The KM of the enzyme is approximately: O A. 0.25 mM O B. 0.5 mM OC. 1.0 mM D. 2 mM O E. 0.3 mM
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
An enzyme catalyzes the reaction A --> B. The enzyme is present at a concentration of 2 nM, and the Vmax is 10 uM/s. The Km for substrate A is 10mM Calculate the catalytic efficiency of this enzyme
3. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained: [S] (M) Vo (uM / min) 2.5 x 106 4.0 x 106 28 40 1 x 105 2 x 105 4 x 105 1 x 104 2х 103 1x 10-2 70 95 112 128 139 140 A. Vmax = 140 and Km =1 x 10s B. Vmax = 70 and Km = 1 x 105 C. Vmax = 140 and Km =1 x 102 D. Vmax = 70 and Km = 1 x 102 %3D %3!
2. Below is experimental data regarding mode of inhibition of "“Enzyme K" by sucralose. This is a double- reciprocal plot of enzyme velocity (Vo) versus substrate concentration, and each line represents a series of data collected at a different sucralose concentration. This is preliminary experimental data, so for a first- approximation of the experimental data, velocity data without inhibitor present can be overlooked (ignored). 0.04 00 mM sucralose 0.035 O 2.5 mM sucralose 0.03 O 10 mM sucralose 0.025 0.02 0.015 0.01 0.005 -6 2 4 -0.005 -0.01 1/ [substrate] (mM) A) Propose a mechanism of inhibition of this enzyme by sucralose. B) Explain your reasoning to part A). 1/V. (mM1 min)
1. Consider an enzyme-catalyzed reaction giving the following results at a fixed enzyme concentration of 1 x 10-6 M in the presence or absence of 5 mM or 10 mM concentrations of an inhibitor I [S] (mol/L) 0.0010 0.0025 0.0050 0.0100 0.0500 0.1000 a) Control 0.00167 0.00333 0.00500 0.00667 0.00909 0.00952 v (m/s) [I] = 5 mM 0.00090 0.00200 - 0.00333 0.00500 0.00833 0.00909 [I] = 10 mM 0.00062 0.00142 0.00250 0.00400 0.00769 0.00869 Estimate Vmax and KM for all cases. What type inhibitor is [I]? Estimate K₁ and/or Kı' depending on the type inhibitor. Estimate Kcat. Estimate Kcat/KM.
6. Examine the plot of the enzyme-catalyzed reaction below: Wheat-germ acid phosphatase shows Michaelis-Menten kinetics when acting on para-nitrophenol phosphate 16 14 12 10 8 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 (S] / mM a) Is this reaction better characterized by Michaelis-Menten kinetics or simple mass action? b) What is the approximate value of Vmax? Km? c) Assuming that this plot was made at 100 nM enzyme, draw the curve, including axis labels, for 300 nM enzyme. d) Imagine that this enzyme is inhibited by a competitive inhibitor with Ki (for inhibitor) = Km (for substrate). Note that v = (kcat [E] tot [S]/Km) / (1 + [S]/Km + [I]/Ki). Draw the curve of reaction velocity (v) versus [S] for [I] = 1 mM. e) Now consider an experiment where you simultaneously introduce substrate and competitive inhibitor at equal concentrations. Make a plot of v versus concentration of [S] = [I], i.e. the X-axis should be the concentration of both I and S, which are equal. v/ uM min-
1. Explain what happens to the velocity in the following scenarios. a. More substrate is added when the substrate concentration is low. b. More substrate is added when the enzyme is saturated with substrate. 2. If an enzyme has a K = 9.00mM and V max = = 1.85mM s', what would the velocity be if you had a substrate concentration of 3.0 mM? b. What would happen to the velocity if K, decreased to 3.0mM? 3. A researcher recorded the following velocity (R) data with an enzyme whose substrate is ATP. Find V and K, and show work. CMI 4. How would each of the following effect the rate of an enzymatic reaction? a. Increase of temperature (from 0C-100C) b. Increase in pH c. Increase in substrate concentration d. Increase in enzyme concentration
1. Provided in the Table below kinetic data for an enzymatic reaction that was carried out in the presence (last two columns) and absence (second column = control) of enzyme inhibitor. Both inhibitors were added in eachreaction at a concentration of 2 mM. The enzyme concentration was similar in all and was approximately 0.001 µM. a. Calculate both Vmax and KM for the control using Lineweaver-Burk curve. b. Provide the type of inhibition for both? c. Find, KI, for the inhibitor binding to the enzyme, for experiments (2) and (3). d. Calculate the reaction Kcat for the Control in experiment (1). e. Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and KM for the enzyme.

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