An enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a keat value of 181 s1. At an initial enzyme concentration of 0.0100 µM, the initial reactionvelocity was found to be 1.07 x 10-0 µM/s. What was the initial concentration of the substrate, S, used in the reaction ?Express your answer in micromolar to three significant figures.> View Available Hint(s)?[S]!!µMSubmit

An enzyme is a type of biological catalyst that aids in the acceleration of chemical reactions.…

Answered: Án enzyme that follows Michaelis-Menten…

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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An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.
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If a data from an enzyme experiment is plotted as a Lineweaver - Burk plot, and the Vmax is 0.02 sec/mol, and x-intercept is -2.5 mM-1, then what is the KM value?
Lisa decides to obtain values for the Km and Vmax of an enzyme that was isolated from liver cells.. Using the Michaelis Menten plot. In what kind of measurements are needed and what would be needed to plot on a graph to estimate Km and Vmax?
If the data from an enzyme experiment is plotted as a Lineweaver-Burk plot, and the Vmax is 0.02 mol/sec, and x-intercept is –2.5 mM then what is the KM value? Show yourwork/reasoning.
For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K M
Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
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Enzyme X has a molecular weight of 48,000. It converts substrate Z into product Y. Z absorbs at 340 nm, and Y absorbs at 480 nm. A.) At what wavelength would you measure the change in absorbance to assay for enzyme X? Would the absorbance increase or decrease over time? B.)If Vmax = 60 μmol/min and you used 400 μL of a 0.1 mg/mL solution of enzyme, what is the turnover number?
The turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?

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Án enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a kat value of 181s1. At an initial enzyme concentration of 0.0100 uM, the initial reaction velocity was found to be 1.07 x 10- µM/s. What was the initial concentration of the substrate, [S), used in the reaction ?