5. 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 Vmax values for the inhibited and uninhibited reactions? Is the inhibitor competitive or noncompetitive? [S] (mM) V, No Inhibitor (mmol min-1) V, Inhibitor Present (mmol min-1) 1x 10-4 5 × 10-4 1.5 × 10-3 2.5 × 10-3 5 × 10-3 0.026 0.010 0.092 0.136 0.040 0.086 0.120 0.150 0.165 0.142

5. 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 Vmax values for the inhibited and uninhibited reactions? Is the inhibitor competitive or noncompetitive? [S] (mM) V, No Inhibitor (mmol min-1) V, Inhibitor Present (mmol min-1) 1x 10-4 5 × 10-4 1.5 × 10-3 2.5 × 10-3 5 × 10-3 0.026 0.010 0.092 0.136 0.040 0.086 0.120 0.150 0.165 0.142

Biochemistry

9th Edition

ISBN:9781305961135

Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal

Publisher:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal

Chapter6: The Behavior Of Proteins: Enzymes

Section: Chapter Questions

Problem 60RE

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Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?
a. Calculate both Vmax and KM for the control using Lineweaver-Burk curve. b. Provide the type of inhibition for both? 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 Ky for the enzyme. c. (1) V. [(umol/(ml.s)] 7.6 (2) V- Τ (μmol/ (ml.s)] [S] (mM) (3) V. [(umol/(ml.s)] 6.6 2 4 14.6 26.6 45.8 4.4 8.6 16.4 29.8 11.4 17.8 24.6 28.2 16 24 60 40.8
Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
a. What is the Vmax of this enzyme WITHOUT inhibitor? Please show your work. b. What is the Km of this enzyme WITHOUT inhibitor? Please show your work. c. The specificity constant of enzyme X is 8 x 10^7 /(M * seconds) What is the kcat of enzyme X WITHOUT inhibitor? Please show your work d. What was the concentration of enzyme used for measuring the kinetics of enzyme X WITHOUT inhibitor? Please show your work
Draw the isomers for each aldotetrose and ketopentose in the figure below and designate each isomer as a D or L sugar and designate also the R and S in every chiral centers. Label the enantiomers and diastereomers respectively. CH,OH C=0 Н-С—ОН Н-С—ОН Н-С —ОН Н-С—ОН ČH,OH ČH,OH
8) plot of enzyme activity with and without an inhibitor present gave the following plot. What type of inhibitor is present? How does this inhibitor function? What changes are seen in Vmax and KM? Draw a line that approximates the result from addition of twice as much inhibitor to the reaction. 1/v (v in mM/min) 0.8 0.6 0.4 0.2 0 0.2 0.4 06 1/[s] ([S] in mM) 0.8
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 Activate
1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
columns) and absence (second column = control) of enzyme inhibitor. Both inhibitors were added in each reaction at a concentration of 2 mM. The enzyme concentration was similar in all and was approximately 0.001 Им. Calculate both Vmax and KM for the control using Lineweaver-Burk curve. b. Provide the type of inhibition for both? 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). Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and Ky for the enzyme a. C. e. [S] (mM) (1) V. {(umol/(ml.s)] (2) V. [[umol/(ml.s)]| 4.4 (3) V. umol/{ml.s] 6.6 11.4 76 14.6 8.6 26.6 16.4 29 8 17.8 24.6 28.2 16 45.8 24 60 40.8
An enzyme has a V of 1.2 uM s The Km for its substrate is 10 µM. max Calculate the initial reaction velocity, Vo, for each substrate concentration, [S]. Calculate Vo when [S] is 2 µM. Vo = µM s Calculate Vo when [S] is 10 µM. Vo = µM s-1 Calculate Vo when [S] is 30 µM. Vo = µM s-1
A. Lineweaver-Burk plot of the enzyme with increasing amounts of substrate in the absence or the presence of the inhibitor is shown below. Graph A : x-intercept Graph B : x-intercept = - 0.012, y-intercept = 0.8 Graph C : x-intercept = - 0.027, y-intercept = 0.8 Graph D : x-intercept = - 0.039, y-intercept = 0.8 - 0.007, y-intercept = 0.8 Graph A 4 Graph B Graph C Graph D 1 -0,04 -0,02 0,00 0,02 0,04 1/[Substrate] (uM) (i) Which graph indicates an enzymatic reaction without inhibitor? (ii) Which type of inhibitor is it? Briefly explain. (iii) Which graph indicates the highest concentration of inhibitor? (iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers. 1/Rate (umol/min)
Table below shows the kinetic data obtained for an enzyme in the absence of any inhibitor (1), and in the presence of an inhibitor (2), at a concentration of 7.0 mM. Assume the total enzyme concentration, [E]T, is the same for each experiment. Table 1 [S] (mM) Vo Vo {umol/(mL•s)} {umol/(mL•s)} Without Inhibitor With Inhibitor (1) 3.8 (2) 2.2 1 2 7.3 4.3 4 13.3 8.2 8 22.9 14.9 12 30.0 20.4 a); Construct a Lineweaver -Burk plot using the kinetic data shown in Table 1. Determine Vmax and Km for the uninhibited enzyme and inhibited enzyme.