5Kinetic data was collected for a new enzyme and its substrate. The same data has been plotted in two ways, vvs. [S] and 1/v vs. 1/[S]. Use these plots to answer the following questions about its enzymatic properties.6v(micromole/sec)41/v (sec/micromole)2.42.221.81.61.41.210.80.60.40.20050100150[S] (MM)200250300-0.3 -0.2-0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11/[S] (1/mM)a.b.Determine Vmax and Kм for enzyme and its substrate. Don't forget to justify your work and include units!Vmax 5 umol/secKm = 10 mMYou duplicated the above experiment, but this time at a higher initial enzyme concentration. Draw onBOTH of the plots above what you expect the new data will look like at the higher enzyme concentration.Explain how this affects your Vmax and KM values, if at all.Vmax increases, Km stays the same as you increase enzyme concentration

Detailed explanation: a) Let's break down the problem step by step and determine the Michaelis…

Answered: 5 Kinetic data was collected for a new…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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The initial rates of an enzyme-catalyzed reaction have been determined for five different [S] (see table below). [S]o (mol.L-¹) 1.0 x 10 1.5 x 10-4 2.0 x 10-4 -4 5.0 x 10 7.5 x 10-4 -1 Vo (μmol.L-¹ min-¹) 28 35 42 63 75 1- Determine graphically the value of Km (first with the Michaelis-Menten representation and then with the Lineweaver-Burk representation, then compare by reporting on the Michaelis-Menten representation the values found by the Lineweaver-Burk representation) 2- Calculate the concentration of a competitive inhibitor 1, with a K₁ value of 2.4.104 M, whose action would quadruple the apparent value of Km.
-Inhibitor +Inhibitor [S] (mM). V&νβσπ:(μmol/sec) V0&νβσπ&ν βσπ;(μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITHOUT iinhibitor?
-Inhibitor +Inhibitor [S] (mM) V0 &ν βσπ:(μmol/sec) V0&νβσπ: &ν βσπ:(μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Km of this enzyme WITHOUT iinhibitor?
You were asked to determine the mode of inhibition exerted by Inhibitor DEDS to a newly discovered enzyme known as BILISTASE. The kinetics data are shown below: [Substrate, uM) Vo with DEDS (uM/min) Vo without DEDS (uM/min) 1.667 1.600 3.544 2.500 1.923 4.259 3.333 2.139 4.737 14.000 2.268 5.019 5.000 2.410 5.336 Using linear regression analysis, determine the values of Vmax and KM of the enzyme in the ABSENCE of inhibitor: (Express your answer in 3 decimal places, do NOT include the units) Vmax KM
The following data were collected for an enzyme that can be described by Michaelis Menton kinetics. Use graphical analysis to determine the Km and Vmax for this exzyme / substrate combination. [S] mM Product formed (umol/min) 0.21 1.5 2.0 0.24 3.0 0.28 4.0 0.33 8.0 0.40 16.0 0.45
-Inhibitor +Inhibitor [S] (mM) V0&νβσπ; (μmol/sec). V0&νβσπ:&νβ σπ: (μmollsec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Km of this enzyme WITH iinhibitor?
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?
The following reactions were catalyzed by an enzyme that follows the Michaelis-Menten mechanism in the absence and presence of inhibitor (10mM). Assume [E]T is the same for each reaction. Determine K, and/or K'. [S] (mM) Uninhibited v (um/s) Inhibited v. (um/s) 1 3 1.67 2 4.5 2.6 6.8 4.5 10 8.1 6.2 20 9.5 7.7
1). The initial velocity of an enzyme-catalyzed reaction has been determined at several substrate concentrations with and without an inhibitor I. The total enzyme concentration [E]ot present in each experiment is 1.20 x 10 M. [1] = 5.00 x10 M. (Note: v initial rate or velocity.) [S], M 1.25E-3 2.50E-3 5.00E-3 10.0E-3 v, M/min, no inhibitor 0.480E-3 0.660E-3 0.810E-3 0.920E-3 v, M/min, 5.00x10 M 9.43E-5 1.41E-4 1.70E-4 2.10E-4 b). What kind of inhibition occurred. c). Calculate the value of Ki = a). Use the Lineweaver-Burk plot to determine the values of Vmax and KM at [1] = 0.00 mM, and the values of Vmax and KM at 0.500 mM of I. it in 'How many molecules of
1. Consider an enzyme-catalyzed reaction giving the following results at a fixed enzyme concentration of 1 x 10-6 M in the presence or absence of 5 mM or 10 mM concentrations of an inhibitor I [S] (mol/L) 0.0010 0.0025 0.0050 0.0100 0.0500 0.1000 a) Control 0.00167 0.00333 0.00500 0.00667 0.00909 0.00952 v (m/s) [I] = 5 mM 0.00090 0.00200 - 0.00333 0.00500 0.00833 0.00909 [I] = 10 mM 0.00062 0.00142 0.00250 0.00400 0.00769 0.00869 Estimate Vmax and KM for all cases. What type inhibitor is [I]? Estimate K₁ and/or Kı' depending on the type inhibitor. Estimate Kcat. Estimate Kcat/KM.
-Inhibitor +Inhibitor [S] (mM) V0&νβσπ;(μmol/sec). νο&νβσπ: &ν βσπ (μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITH iinhibitor?
The steady-state kinetics of an enzyme are studied in the absence and presence of an inhibitor (inhibitor A). The initial rate is given as a function of substrate concentration in the following table: v[(mmol/L)min- [S] (mmol/L) No inhibitor Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38 (a) What kind of inhibition (competitive, uncompetitive, or mixed) is involved? (b) Determine Vmax and Ky in the absence and presence of inhibitor.

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