4. Using a fluorescent model substrate, you study the kinetics of an enzyme-catalyzed reaction. You observe thebelow data. Write an equation describing reaction velocity (v) versus [S]. Define and provide a numerical value forany constants that you include in your equation.161412420.00.51.01.52.02.53.03.5[S] / mM`N'at 50 mM and observe the followingIn pilot experiments, you add the putative inhibitor imidazoleDoesn't affect anything1reaction velocities:[S] = 1 mM5 µM min-1[S] = 2 mM9 9 µM min-1[S] = 4 mM 911 µM min-1Is imidazole a competitive or non-competitive inhibitor? Provide a brief explanation.>9 00v/ µM min-

Enzymes are the biocatalysts that catalyse the biochemical reactions by lowering the activation…

Answered: 4. Using a fluorescent model substrate,…

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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You are working on an enzyme that obeys standard Michaelis-Menten kinetics. You have determined the Vmax to be 0.1 mol/sec and the Km to be 2.5 mM. What would the rate of the reaction be when the substrate concentration is 20 mM? 0.09 MS-1 O 0.133 Ms-1 O 0.18 Ms ¹ 9 Ms-1 O 0.018 Ms-1 0.2 MS-1
An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s
Suppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number
7. You obtain the following set of data for two enzymes that follow Michaelis-Menten kinetics: Enzyme A Initial velocity (µM/min) 38 Enzyme B Initial velocity (пМ/min) 140 Substrate concentration (µM) 233 350 2 73 4 133 8 229 467 20 400 583 150 706 682 300 750 691 786 798 697 699 1000 3000 6000 800 700 (a) What are the approximate Km values for each enzyme based on these data? Briefly justify your answer. (b) In the above experiment, 0.1 µM of enzyme A or enzyme B were used. Calculate kçat and specificity constant for each enzyme. Which enzyme is more efficient? (c) The kinetics of Enzyme B were re-measured in the presence of a small molecule inhibitor. The Km increased while the Vmax remained unchanged. What type of inhibition is this and why does the Km increase?
An enzyme catalyzes a reaction with a Km of 6.50 mM and a Vmax of 2.00 mM s¹. Calculate the reaction velocity, Vo, for each substrate concentration. Vo : Vo: Vo : [S] = 4.00 mM [S] = 6.50 mM [S] 14.0 mM = n. mm. s-1 mm. sl mm. s-1
3.An enzyme catalyzed reaction is studied and the following kinetic analysis is obtained: |[S), mM 0.050 0.075 v, (µM min") 0.93 1.264 1.77 2.14 3.7 0.125 0.175 0.935 a. Using Excel, make a fully labeled Lineweaver-Burk plot and determine the Km and Vmax for the enzyme b. The reactions were set up dissolving 1 mg of the enzyme (MW= 100000 Da) in 100 ml of final reaction buffer. Determine the turnover number for the enzyme assuming 1 active site exists per enzyme molecule? c. Determine the catalytic efficiency for the enzyme d. The same reactions are performed in presence of an inhibitor A and the resulting velocities determined: v plus inhibitor, (µM min) 0.272 0.37 0.518 0.626 |1.08 |[S), mM 0.050 0.075 0.125 0.175 0.935 Plot these data on the same graph as above and determine the new Km and Vmax and the type of inhibitor (competitive, non-competitive). e. Can the effects of the inhibitor be over-ridden by adding more substrate? Why?
An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-
Q is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound Q
An enzyme catalyzes a reaction at a velocity of 20 micromole/min when the concentration of substrate (S) is 0.01 M. The Km for this 1 X 10^-5 M. Assuming that Michaelis-Menten Kinetics are followed, what will the reaction velocity be when the concentration of S is 1.0 X 10^-6 M?
Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of Km tor this enzyme? Substrate added, umol/L Vo uM/sec 48 97 175 187 192 197 2 16 100 1000 198 OA 1 uM O8.2 pM OC 4 uM. OD. 8 uM OE 16 IM OF 1000 uM OG 97 uM/sec OH 175 UM/sec

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