2Two version of the same enzyme were isolated, a wild type and a mutant differing at a single amino acid. After a thorough study the following kinetic parameters wereaccurately measured. For the second blank, use a number with two significant figures.Wild-typeMutantMaximum velocity90 micromol/min1 micromol/minKm10 mM0.1 mMAssuming a two-step reaction where k-1 is much larger than k2, thechoose your answer...form has the higher affinity for the substrate. The initial velocity of the reaction catalyzed by the wild-type enzyme at a substate concentration of 10 mM istype your answer...micromol/min. choose your answer...enzyme alters the equilibrium more in the direction of the product.

Enzymes are protein molecules that increase the rate of reaction by decreasing the activation…

Answered: 2 Two version of the same enzyme were…

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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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?
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.
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=
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
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.
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
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.
1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
Vmax [S] Km + [S] Vo = Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km [S] Km 1 Vm Vm [S] 1 The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For uncompetitive inhibition The Michaelis-Menten equation becomes Vo-Vmax(S)/(Km+a'[S)) Put the Minchaelis-Menten equation for uncompetitive inhibition in the Lineweaver-Burk format. The slope of this plot will be
The glucose oxidase used in this experiment has a concentration of 27 U/mg. Calculate the mass (in g) of solid glucose oxidase required to react with all the glucose in 4.00 µL of10.0- mM glucose sample assuming the reaction proceeds for 30 minutes.
An enzyme catalyzes a reaction at a velocity of 20 micromole/min when the concentration of substrate (S) is 0.01 M. The Km for this 1 X 10^-5 M. Assuming that Michaelis-Menten Kinetics are followed, what will the reaction velocity be when the concentration of S is 1.0 X 10^-6 M?
A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).

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