The following reactions were catalyzed by an enzyme that follows the Michaelis-Mentenmechanism in the absence and presence of inhibitor (10mM). Assume [E]T is the same for eachreaction. Determine K, and/or K'.[S] (mM)Uninhibited v (um/s)Inhibited v. (um/s)131.6724.52.66.84.5108.16.2209.57.7

The given enzyme kinetics is said to follow Michaelis-Menten mechanism. Here Substrate concentration…

Answered: The following reactions were catalyzed…

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 questions deal with a fundamental understanding of enzyme catalysis.a. Why is the rate of an enzyme-catalyzed reaction proportional to the amount of (ES) complex?b. What do you think is meant by saturation of the enzyme?c. What do you think is meant by the term “saturation kinetics”?d. How does the Michaelis-Menten equation explain why the rate of an enzyme-catalyzed reaction reachesa maximum value at high [S]?
The enzyme, fumarate, has the following kinetic constants: k 1 k 2 k -1 where k 1 = 10 9 M -1 s -1 k -1 =4.4 x 10 4 s -1 k 2 = 10 3 s -1 a. What is the value of the Michaelis constant for this enzyme? b. At an enzyme concentration of 10 -6 M, what will be the initial rate of product"
If the higher value of KM resulting in the new plot ( red curb ) is due to the presence of an enzyme inhibitor is inhibitor reversible or irreversible? And why?
For the following aspartase reaction in the presence of the inhibitor hydroxymethylaspartate, determine Km and whether the inhibition is competitive or noncompetitive. You have to plot thegraph on the graph paper and also by using excel.[S] V, No Inhibitor V, Inhibitor Present(molarity) (arbitrary units) (same arbitrary units) 1 x 10-4 0.026 0.0105 x 10-4 0.092 0.0401.5 x 10-3 0.136 0.0862.5 x 10-3 0.150 0.1205 x 10-3 0.165 0.142
Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:
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 µM
The following data, presented by G. Bowes and W. L. Ogre in J. Biol. Chem. (1972) 247:2171–2176, describe the relative rates of incorporation of CO2 by Rubisco under N2 and under pure O2. Decide whether O2 is a competitive or uncompetitive inhibitor.
You obtain a calculated Vmax of 4.26uM/s and a Km of 122.5uM from a kinetics experiment performed using 0.5uM enzyme. What is the catalytic efficiency of this enzyme?
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.
The following question focuses on how the parameters regulating enzyme function might change, and how these might appear graphically on a Michaelis-Menten plot and a Lineweaver-Burke plot. Carbonic anhydrase is an enzyme that will convert CO2 and water into HCO3. CO2 + H20 > H+ + HCO3 There are many different isoforms of this enzyme. (see for instance http://en.wikipedia.org/wiki/Carbonic_anhydrase . Assume that one variant has a Km of 10 µM and a different variant has a Km of 100 µM. Draw on the same graph a typical Michaelis-Menton plot showing the alteration in the rate of carbonic anhydrase as the CO2 level is varied for the two different variants of enzyme, assuming the concentration of the enzyme (10 mM) in the test tube is kept constant. Assume that you have equal amounts of the two different variants of carbonic anhydrase in a number of test tubes and that the Vmax for both enzymes are the same. Be sure to label the axes. For the same conditions as above, draw a…
Using equilibrium argument, why does Km apparently increase, decrease or stay the same in uncompetitive inhibition?
The following question focuses on how the parameters regulating enzyme function might change, and how these might appear graphically on a Michaelis-Menten plot and a Lineweaver-Burke plot. Carbonic anhydrase is an enzyme that will convert CO2 and water into HCO3. CO2 + H20 > H+ + HCO3 There are many different isoforms of this enzyme. (see for instance http://en.wikipedia.org/wiki/Carbonic_anhydrase 1 Assume that one variant has a Km of 10 µM and a different variant has a Km of 100 µM. Draw on the same graph a typical Michaelis-Menton plot showing the alteration in the rate of carbonic anhydrase as the CO2 level is varied for the two different variants of enzyme, assuming the concentration of the enzyme (10 mM) in the test tube is kept constant. Assume that you have equal amounts of the two different variants of carbonic anhydrase in a number of test tubes and that the Vmax for both enzymes are the same. Be sure to label the axes. For the same conditions as above, draw a…

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