2) Below are kinetic data at three different inhibitor concentrations. Graph the data in anappropriate form to determine whether the inhibitor is competitive or uncompetitive.a (μm)0.40.671.002.00v (μM/min)(no inhibitor)0.220.290.320.40v (μM/min)(10 nM inhibitor)0.210.260.300.36v (μM/min)(20 nM inhibitor)0.200.240.280.32

Enzyme inhibitors are the compounds that can interfere with enzyme activity and inhibit the action…

Answered: 2) Below are kinetic data at three…

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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. The following data, presented by G. Bowes and W. L. Ogre in J. Biol. Chem. (1972) 247:2171–2176, describe the relative rates of incorpo- ration of CO, by rubisco under N, and under pure O,. Decide whether O, is a competitive or uncompetitive inhibitor. [CO,] (mM) Under N2 Under O, 0.20 16.7 10 0.10 12.5 5.6 0.067 8.3 4.2 0.050 7.1 3.2
Km value should be more than 10% apart, use non-competitive inhibitor
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (uM) 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Km of the inhibited enzyme reaction?
15. G + C Content: Describe how the G+C content of an organism can be measured. In your response, be sure to address the following: b. Explain the principles behind a melting curve and how this helps determine G + C content. Include an explanation of the components of the curve and how one would interpret a melting curve.
What is the initial velocity measurement for the hydrolysis of 50 μM amoxicillin by the enzyme?
8L.3.4
Calculate KI' of the inhibitor from the information given. All information may not be needed to calculate. K'm = (29Ki+1.45x10^-10)/Ki Vmax = 11.7 µMs-1 Kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM Inhibitor Concentration = 5x10^-12
CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
D 1. 0.25 Short answer questions, write answers on blank paper. A plot of enzyme activity with and without an inhibitor present gave the following plot. 0.2 0.15 1/V v in mM/s 0.1 0.05 0 0 0.05 0.1 y=0.67x+0.02 0.15 y=0.2x+0.02 0.2 0.25 0.3 1/[S]. [S] in mM Without the inhibitor, what are the values of Vmax and Km? In the presence of the inhibitor, which of these items has changed and what is (are) the new value(s)? What type of inhibitor is present (explain why you have arrived at this conclusion)? Be sure to include units as necessary--answers that do not contain required units will be marked incorrect
Please explain your reasoning for every answer
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
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM

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