Biochemistry: Concepts and Connections (2nd Edition)
2nd Edition
ISBN: 9780134641621
Author: Dean R. Appling, Spencer J. Anthony-Cahill, Christopher K. Mathews
Publisher: PEARSON
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Textbook Question
Chapter 8, Problem 13P
The accompanying figure shows three Lineweaver—Burk plots for enzyme reactions that have been carried out the presence, or absence, of an inhibitor. Indicate what type of inhibition is predicted based on each Lineweaver—Burk plot. For each plot indicate which line corresponds to the reaction without inhibitor and which line corresponds to the reaction with inhibitor present.
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The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.
Please answer A & B attached with image.
By using Excel or GoogleSheets, graph the Lineweaver-Burk plots for the behavior of
an enzyme for which the following experimental data are available. What are the Km and
V
values for the inhibited and uninhibited reactions? Is the inhibitor competitive or
max
noncompetitive?
[S]
(mM)
V, No Inhibitor
(mmol min-')
V, Inhibitor Present
(mmol min-')
1× 10-4
5 × 10-4
1.5 × 10-3
2.5 × 10-3
5 × 10-3
0.026
0.092
0.136
0.150
0.010
0.040
0.086
0.120
0.165
0.142
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Chapter 8 Solutions
Biochemistry: Concepts and Connections (2nd Edition)
Ch. 8 - Prob. 1PCh. 8 - The enzyme urease catalyzes the hydrolysis of urea...Ch. 8 - An enzyme contains an active site aspartic acid...Ch. 8 - The folding and unfolding rate constants for a...Ch. 8 - In some reactions, in which a protein molecule is...Ch. 8 - Would you expect an “enzyme” designed to bind to...Ch. 8 - The initial rate for an enzyme-catalyzed reaction...Ch. 8 - a. If the total enzyme concentration in Problem 7...Ch. 8 - Prob. 9PCh. 8 - Prob. 10P
Ch. 8 - The following data describe the catalysis of...Ch. 8 - At 37 oC, the serine protease subtilisin has kcat...Ch. 8 - The accompanying figure shows three...Ch. 8 - The steady-state kinetics of an enzyme are studied...Ch. 8 - The same enzyme as in Problem 14 is studied in the...Ch. 8 - Enalapril is an anti-hypertension “pro-drug"...Ch. 8 - Initial rate data for an enzyme that obeys...Ch. 8 - Prob. 18PCh. 8 - Suggest the effects of each of the following...Ch. 8 - The inhibitory effect of an uncompetitive...Ch. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - Prob. 23PCh. 8 - In kinetics experiments, the hydrolysis of the...
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- Draw a hypothetical Michaelis-Menten plot for an enzyme reaction (i) without and (ii) with a pure noncompetitive inhibitor. Label Vax and Km on plotarrow_forwardThe Lineweaver-Burk plot, which illustrates the reciprocal of the reaction rate (1/v) versus the reciprocal of the substrate concentration (1/[S]), is a graphical representation of enzyme kinetics. This plot is typically used to determine the maximum rate, Vmax, and the Michaelis constant, Km, which can be gleaned from the intercepts and slope. Identify each intercept and the slope in terms of the constants Vmax and Km. What term is represented by C? Linewegver-Burk Pt 1/Vmax O A. В. -1/Km Km/Vmax C.arrow_forwardAn enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.arrow_forward
- The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.arrow_forwardA с Wan WWW GHEDE MAK am2 Increasing CE a-1 (no inhibitor) Slope #KY... ...." 7-15 7 a>c²=1 [no inhibitori Slope R 101 9-19 aver-Burk plots in this figure represent the activities of enzymes in the 4-5 n. 155 B D MIN -a'/KM 0.8- 1/V >~- -1/K 0 01 m(app) d'Nmax a=2 a=1.5 1[S] per mM inhibitor a=10 Slope Karrow_forwardThe Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.arrow_forward
- (b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90arrow_forwardDiscuss how the equation for enzymatic reaction (given below) was demonstrated in the experimental results in Table 1. Use the appropriate color codes (-), (+), (++), (+++) to describe the expected results in Table 1 Table 1. Enzyme Action MIXTURE ENZYME ACTIVITY Water + catechol (tt A) Enzyme + catechol (tt B) Water + enzyme (tt C)arrow_forwardThe table below lists experimental conditions that can be applied to a reaction catalyzed by a hypothetical Michaelis–Menten enzyme. For each experimental condition described, complete the table to indicate as precisely as possible the effect (no change, half as large doubles, increases, decreases) on the maximal velocity, Vmax, and Michaelis constant, KM, of the hypothetical enzyme.arrow_forward
- Under the following conditions, fill in the blanks. Then, describe why this inhibitor is the type of inhibitor you identified it as. If you were to add 5nM of a reversible inhibitor, the Km for the measured enzyme catalyzed reaction would ______ (Increase, Decrease, Stay the same) to ______µM (choose appropriate value) and Vmax would _______ (Increase, Decrease, Stay the same) to ______µMs-1. So, this inhibitor is a ______ (Competitive, Uncompetitive, Mixed) inhibitor. Conditions: kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µMarrow_forwardEnzyme X can be inhibited by two distinct inhibitors, only one of which is acting as a competitive inhibitor, while the other one is non-competitive. The three plots below (designated A, B, and Cy show Michaelis-Menten kinetics for the reactions carried out with the same concentr ation of the enzyme and either in the absence of any inhibitors or in the presence of one of the two inhibitors. Match the type of reaction (uninhibited, competitive inhibition, non-competitive inhibition) with the corresponding plot? Plot A Plot B Plot C Substrate concentration 1.Plot A v LUninhibited roaction 2. Plot C v Inhibition by a competitive inhibitor 3Plot B v Inhibition by a non-competitive inhibitor A Moving to another question will save this response. luparrow_forwardYou have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1 is much larger than ?2, which of the following statements are correct? The mutant version has a higher affinity for the substrate. The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. The reaction equilibrium is reached once there is no net change in the concentration of the substrate or the product. Based on the data table and your initial…arrow_forward
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