For an enzyme following Michaelis-Menten kinetics, the following Vo vs. S data wereobtained in steady state experiments using 2 micromolar enzyme for all measurements:[S]Vo1 mM4.0 mM/min2 mM4.8 mM/min4 mM4.9 mm/min8 mM4.95 mM/minEstimate the value of kcat for the reaction. Be sure to include the appropriate units.

To estimate the value of kcat for an enzyme following Michaelis-Menten kinetics, we first need to…

Answered: For an enzyme following…

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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From a kinetics experiment, kcat was determined to be 295sec-1. For the kinetic assay, 0.3mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 3mL. Calculate Vmax (µM∙min-1) for the enzyme and catalytic efficiency ( in M-1sec-1) for the enzyme. The Km for the enzyme was determined to be 2.55 x 10-2M.
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.
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?
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.
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 reaction velocity 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] !! µM Submit
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
Table below shows the kinetic data obtained for an enzyme in the absence of any inhibitor (1), and in the presence of an inhibitor (2), at a concentration of 7.0 mM. Assume the total enzyme concentration, [E]T, is the same for each experiment. Table 1 [S] (mM) Vo Vo {umol/(mL•s)} {umol/(mL•s)} Without Inhibitor With Inhibitor (1) 3.8 (2) 2.2 1 2 7.3 4.3 4 13.3 8.2 8 22.9 14.9 12 30.0 20.4 a); Construct a Lineweaver -Burk plot using the kinetic data shown in Table 1. Determine Vmax and Km for the uninhibited enzyme and inhibited enzyme.
You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. ?0=V0=
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?
An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 μM and a kcat value of 206 s−1 . At an initial enzyme concentration of 0.0100 μM , the initial reaction velocity was found to be 1.07×10−6 μM/s . What was the initial concentration of the substrate, [S] , used in the reaction ? Express your answer in micromolar 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
Given the following set of kinetic data, what is the KM of the enzyme? Numerical answer, only (assuming the same units as seen in the table). [S] (micro M) vo (Micro mol/sec) 1 49 2 96 8 249 50 498 100 656 1000 778 5000 999

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