15. Dodgerase is an enzyme that is exquisitely inhibited by a class of cheaters(Astrozolamide and Redsoxin). Dodgerase plays an important role in the civic pride ofone of the greatest and most diverse cities in the world. The experimental curve ofinitial reaction velocity (as percentage of Vmax) versus [S] for the Dodgerase reaction isshown (upper curve). When the experiment was run in 2017, in the presence ofinhibitor, the lower curve was obtained. From an inspection of the curves and yourknowledge of the kinetic properties of competitive, uncompetitive and mixed inhibitors,determine the nature of the inhibition by Astrozolamide and explain your reasoning!100No inhibitor50Astrozolamide0.20.40.60.8[S] (mm)pee"A JO %) A

Enzymes are globular protein molecules that catalyze a given reaction in a time-dependent manner.…

knowledge of the kinetic properties of competitive, uncompetitive and mixed inhibitors, determine the nature of the inhibition by Astrozolamide and explain your reasoning! 100 No inhibitor Astrozolamide 0.2 0.4 0.6 0.8 (S) (mm) pre"A JO %) A

knowledge of the kinetic properties of competitive, uncompetitive and mixed inhibitors, determine the nature of the inhibition by Astrozolamide and explain your reasoning! 100 No inhibitor Astrozolamide 0.2 0.4 0.6 0.8 (S) (mm) pre"A JO %) A

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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Case Study: Enzyme Kinetics Data for new statin drug (inhibits HMG CoA reductase): ● ● I Substrate (UM) 0.5 1.0 1.5 2.5 3.5 || Rate of reaction (mm/min) 23.5 32.2 36.9 41.8 44 What are the Kms of the uninhibited and inhibited reactions? Rate of reaction in presence of 5.0 nM statin Using the above date, create a double-reciprocal (Lineweaver Burk) plot. What type of inhibition is shown by the new statin? What are the Vmax of the uninhibited and inhibited reactions? 16.67 25.25 30.49 37.04 38.91
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
4. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. (8) mg V₂ μmol s1 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00 0 100 200 300 [s], mM 400 500 WT (a) Estimate the KM and Vmax for the wild-type and mutant enzyme from the graph. V105A (b) Calculate the keat and keat/KM for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
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
AutoSave 301-Enzyme Kinetics and Inhibition, Part 3 - Compatibility Mode - Word Search ff steve M SM File Home Insert Draw Design Layout References Mailings Review View Help E Share O Comments prepared five different concentrations of substrate (sucrose), and five different concentrations of inhibitor C (plus the control, with zero mM of inhibitor C). The following Table lists the inhibitor C concentrations [I], substrate concentrations [S], and resulting enzyme velocities (V.) for all six of these experiments: [I] O mM O mM O mM O mM O mM [S] Vo 1/[S] 1/ Vo 0.1 mM 0.333333333333 mM per minute 0.2 mM 0.50 0.3 mM 0.60 0.4 mM 0.666666666667 0.5 mM 0.714285714286 0.1 mM 0.1 mM 0.166666666667 0.1 mM 0.2 mM 0.25 0.1 mM 0.3 mM 0.30 0.1 mM 0.4 mM 0.333333333333 0.1 mM 0.5 mM 0.357142857143 0.20 mM 0.1 mM 0.111111111111 0.20 mM 0.2 mM 0.166666666667 0.20 mM 0.3 mM 0.20 0.20 mM 0.4 mM 0.222222222222 0.20 mM 0.5 mM 0.238095238095 0.3 mM 0.1 mM 0.083333333333 0.3 mM 0.2 mM 0.125 0.3 mM 0.3 mM…
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
3. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. -1 v, μmol s¹ mg-¹ 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00+ 0 100 → WT 200 300 [s], mM 400 500 → V105A (a) Estimate the Km and Vmax for the wild-type and mutant enzyme from the graph. (b) Calculate the Keat and Keat/Km for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
The effect of an inhibitor I on the rate of a single-substrate, enzyme-catalyzed reaction was investigated and gave the following results (See table below). a. What is the mode of action (pure competitive, mixed competitive, uncompetitive or noncompetitive) of the inhibitor? Briefly explain your answer. b. Estimate the value of Km and Vmax for uninhibited reaction.
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
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (µM). 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 Vmax of the inhibited enzyme reaction?
USSE EUSS reaction rate substrate concentration Blue line - Enzyme alone Red line - Enzyme + unknown compound The 4 graphs above represent the change in enzyme kinetics with the individual addition of different compounds that could be categorized as either: allosteric inhibitors, allosteric activators, competitive inhibitors, activators or non-competitive inhibitors. Review the graphs above. Each graph represents the activity of an enzyme and the enzyme + the addition of an unknown compound. By comparing the kinetics of the enzyme alone to the enzyme + unknown, determine what type of compound was added to each of the 4 different solutions to elicit the observed change.