Shown below is Lineweaver-Burk plot for an enzymatic reaction at different substrate concentrationsin the presence and absence of an inhibitor. The enzyme concentrationis identical in both reactions:1/v (sec/mM)4.543.532.521.510.5000.20.4y = 0.997 +3.01x (+1)0.61/[S] (mM-¹)y = 0.09999 + 3.01x0.8(-1)1(c) What is the type of inhibition mechanism?A. competitive inhibitionB. uncompetitive inhibition (substrate-dependent)C. mixed inhibition(a) Does this enzyme obey Michaelis-Menten kinetics? (yes or no) Explain:D. noncompetitive inhibition (substrate-independent)E. allosteric inhibition1.2(b) What are the apparent values of Vmax and KM for each experiment (with and without inhibitor)?d) If the concentration of the inhibitor is 0.1 mM, what is the value of KI and/or K'I (whichever isrelevant)? (e) Draw v versus [S] for this enzyme in the presence and absence of inhibitor on the plot below (youdo not need to be exact and your axes may not be scaled properly, but use thevariables you calculated in part (A) to label relevant values).v (mm/sec)[S] (MM)

Enzymes kinetics - is the study of the reaction rate of enzyme catalyzed reactions. The reaction…

Answered: Shown below is Lineweaver-Burk plot for…

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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Please note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
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 :
Proline racemase catalyzes the conversion between L-proline and D-proline. The Km and kcat for this reaction are 0.15 M and 550/sec respectively. If the enzyme concentration is 1.45 X 10-5 mmole/ml what is the Vmax of this reaction?
The Michaelis-Menten equation for the enzyme chymotrypsin is 0.14[S] v : 0.015 + [S] where v is the rate of an enzymatic reaction and [S] is the concentration of a substrate S. Calculate dv/d[S] and interpret it.
The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)
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
The rates at various substrate concentrations for an enzyme-catalysed reaction (b) are recorded as follows: 2.5 5.0 10.0 15.0 (S]/10 M 2.2 3.8 5.9 7.1 Vo/ M min Assuming the reaction obeys Michaelis-Menten kinetics, calculate the following values by using Lineweaver-Burk plot () maximum rate, Vmax- (i) Michaelis constant, Km- (ii) turnover number, kz.
In your acid phosphatase enzyme kinetics lab, you constructed a Lineweaver-Burk plot. Lefs assume that you graphicaly obtain 1/Vmax - 9.33 (umoliey and -1Km--0.012 M. Which of the following is correct? A) Vmax = 9.33 (no units) B) Km = 0.012 (no units) C) Vmax = (1/Vru" = 0.107 umolis D) Km =-14-1K = 83.3 pM E) Both C) and D) are correct
A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.
The KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affi nity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?
The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.
The effect of temperature on the hydrolysis of lactose by a ß-galactosidase is shown below in Table 1. The temperature coefficient, Q10 is the factor by which the rate increases by raising the temperature 10°C. The universal gas constant, R is 8.314 J/mol.K. (a) (b) Table 1: Data of Vmax over temperature T (°C) 20 30 35 40 45 Vmax (umoles/min.mg protein) 4.50 8.65 11.80 15.96 21.36 Plot the graph of In Vm vs 1/T using any spreadsheet software (include all appropriate labels and equation). Calculate the activation energy Ea and temperature coefficient Q10.

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Shown below is Lineweaver-Burk plot for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions: