Transcription for Educational Content:---### Enzyme Kinetics: Lineweaver-Burk PlotsThis section explores the behavior of an enzyme through Lineweaver-Burk plots using the given experimental data. The objective is to determine the Km and Vmax values for reactions with and without inhibitors and identify whether the inhibition is competitive or noncompetitive.**Experimental Data:**| [S] (M) | V, No Inhibitor (Arbitrary Units) | V, Inhibitor Present (Arbitrary Units) ||--------------|-----------------------------------|----------------------------------------|| 10,000 - 1 x 10⁴ | 0.026 | 0.010 || 2,000 - 5 x 10⁴ | 0.092 | 0.040 || 666.67 - 1.5 x 10⁻³ | 0.136 | 0.086 || 400 - 2.5 x 10⁻³ | 0.150 | 0.120 || 200 - 5 x 10⁻³ | 0.165 | 0.142 |**Analysis for No Inhibitor:**- **Slope:** 0.033- **Y-intercept:** 5.057- **Vmax:** 0.167 units- **Km:** 6.57 x 10⁻⁴ M**Analysis for Inhibitor Present:**- **Slope:** 0.009- **Y-intercept:** 5.21- **Vmax:** Close to 0.142 units- **Km:** 1.73 x 10⁻³ MBased on this analysis and data on slope and intercept changes, the type of inhibition can be classified as *noncompetitive*.---### Graph Explanation (Hypothetical):The Lineweaver-Burk plot graphs the reciprocal of the substrate concentration [1/[S]] against the reciprocal of the reaction velocity [1/V]. - **Plot Characteristics**: - A steeper slope indicates slower reaction velocity. - The y-intercept gives the inverse of Vmax; a higher value suggests lower maximal velocity. - Changes in Km and Vmax with an inhibitor present typically indicate the nature of the inhibition.This understanding assists in comprehensively analyzing enzyme behavior under different conditions, a crucial aspect of enzyme

For a one-substrate enzyme-catalyzed reaction, the Michaelis-Menton equation shows the quantitative…

Answered: 10,000 2.000 66667 Draw Lineweaver-Burk…

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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A3
A particular enzyme has a ΔΔG‡ of -22.1 kJ mol-1 at 37.0 °C. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)
28. If 20 mM solution of Acetyl-phosphate is transformed to acetate by incubating with a catalytic amount of acetate kinase until equilibrium is reached. Acetyl-phosphate -> Acetate + Pi At equilibrium, the concentration of Acetyl-phosphate is 8 mM. Determine Keq and AG" for the hydrolysis of Acetyl-phosphate reaction. (R = 0.008314 kJ/mol K) .
Suppose the biochemical pathway represented in Equation 5-10a is coupled at its first step to the hydrolysis of ATP to ADP. How would this change the expres- sion for AG of the pathway?
What are the benefits of measuring the initial rate of a reaction Vå for use in kinetic studies? (This is a multi-select question). [ES] can be measured accurately. changes in [S] are negligible, so the value of [S] is known. changes in Km are negligible, so Km can be treated as a constant. V₁ = Vmax. --> A negligible amount of product has formed, so that the back reaction P -- need not be considered. ES
What is the likely keq for an enzyme that has a deltaG =-18.3kj/mol?
The following kinetic data were obtained for an enzyme in the absence of an inhibitor (1) and in the presence of two different inhibitors (2) and (3) at 5mM concentration. [S], mM (mmol/mL/sec) (mmol/mL/sec) (mmol/mL/sec) (1) (2) (3) 1 12 4.3 5.5 2 20 8 9 3 29 14 13 8 35 21 16 12 40 26 18 Determine the Vmax and KM of the enzyme Determine the type of inhibition.
The kcat and KM for chymotrypsin-catalysed cleavage of a synthetic substrate, S, were determined to be 60 s-1 and 0.5 mM, respectively, using steady-state kinetics. The concentration of enzyme in the assay was 5 x 10-8 M. ii) Draw a graph on the axes shown in Figure 1 below showing how the initial rate of the reaction varies with [S], making use of the values for KM and Vmax. Label the axes with appropriate titles and units.
From a kinetics experiment, Kcat was determined to be 55sec^-1. For the kinetic assay, 0.05mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 30,000g/mole. Assume a reaction volume of 3mL. Calculate Vmax (um*min^-1) for the enzyme and catalytic efficiency (in M^-1sec^-1) for the enzyme. The Km for the enzyme was determined to be 8.3*10^-2M.
5) In an experiment to investigate the inhibition of the enzyme-glucosidase the following data for the rates of reaction with glucopyranoside for various substrate concentrations was obtained. By constructing a Leaver-Burk plot, determine the value of the Michaelis constant. [S]/ (10-6 mol dm-3) v/ (10-3 mol dm-3 s-1) 1.00 2.00 3.00 4.00 16.7 33.3 41.1 49.8
Use the relationships revealed by a Lineweaver-Burk plot and the table of enzyme performance to calculate the Vmax and Km of the enzyme with no inhibitor. with inhibitor A, and with inhibitor B. [S] (uM) Vo (umol/min); w/ no inhib. Vo (umol/min); w/ inhib. A Vo (umol/min); w/ inhib. B 10 6.3 5.1 4.0 40 18.4 15.8 11.8 100 29.9 27.0 19.1 150 34.7 32.0 22.2 No inhib. Vmax= Km= Inhib. A Vmax= Km= Inhib. B Vmax= Km=
The ΔG°′ for hydrolytically removing a phosphoryl group from ATP is about twice as large as the ΔG°′ for hydrolytically removing a phosphoryl group from AMP (−14 kJ · mol−1). Explain the discrepancy.

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