1 ptpt9146Bb 9146Bb 1031ClassEtsyE TrapsE 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 theturnover number if the enzyme had a molecular weight of 128,000 g/mol?(Enter your answer to two significant figures.)turnover number =sec-1D1 ptSubmit AnswerTry Another Version2 item attempts remainingestionstion 5on 671pt1 pt1 pt1pt1pt1pt1 pt1 ptDis the substrate concentration multiplied by thecatalytic constant. KM is equivalent to thesubstrate concentration multiplied by the ratioof rate constants for the formation anddissociation of the enzyme-substrate complex. KMis equivalent to the substrate concentration. KMis equivalent to the substrate concentrationdivided by 2A: KM is equivalent to the substrateconcentration when the reaction rate is at halfof its maximum velocity (Vmax/2).Cengage Learning | Cengage Technical SupportPreviousNew

Turnover number refers to the number of substrate molecules that is converted to the product…

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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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The purpose of the progress curve is to determine if an enzymatic reaction rate remains constant for given concentrations of enzyme and substrate, for a given assay time (20 min). How long did the rate of reaction of WGAP remain constant in your experiments? How do you know? What would cause the reaction rate to not be constant and plateau after some time?
An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-
Use correct sig figs The concentration of a purified monoclonal antibody was measured using UV280 nm. The sample was diluted (200 μL of purified antibody in 800 μL buffer) prior to analysis using a spectrophotometer. Calculate the concentration of antibody in the purified fraction if the Abs=0.95 of the diluted antibody. The molar absorptivity is known to be 191,411.6 M-1cm-1 and the molecular weight is 150 kDa. The pathlength for the cuvette is 1 cm.
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Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. QUESTION: Line (A) represents ____________. A. an enzyme-catalyzed reaction in the absence of any inhibitor. B. an enzyme-catalyzed reaction in the presence of a competitive inhibitor. C. an enzyme-catalyzed reaction in the presence of an uncompetitive inhibitor. D. an enzyme-catalyzed reaction in the presence of a noncompetitive, or mixed, inhibitor.
Table 1. Kinetic parameters of a chosen model wild-type (WT) enzyme and three catalytically-improved mutants (M1 to M3). Enzyme variants k (8¹) WT MI M2 M3 50 500 1000 50 K₁, (μM) 10 100 100 1 k/K (HM¹¹ s¹) 5 The study above was done on a wildtype enzyme and three genetically-engineered mutants. The purpose was to find a mutation to improve the catalytic efficiency of the enzyme for industrial use. Which is the most accurate statement concerning the data? M2 is the optimal choice when substrate concentrations are saturating. The wildtype enzyme is still preferable to M3 when S<
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pls explainWhy was absorbance measured at 420 nm in the enzyme kinetics experiment? To monitor thea. decrease in catechol concentrationb. increase in quinone concentrationc. decrease in polyphenol oxidase concentrationd. polyphenol oxidase-catechol complex concentrationLineweaver-Burk plot is the most used linearization method of the MichaelisMenten equation, but the main drawback is a/an:a. overemphasis to low substrate concentration and less emphasis to high substrate concentration.b. overemphasis to high substrate concentration and less emphasis to low substrate concentration.c. exaggeration to both ends of the reciprocal of the substrate concentration.d. lack of independent variables to x- and y- coordinates
I am not understanding why TA of alcohol precipitation is used as TA of purified protein. I provided 2 photos, 1 with the questions and the correct answer, and the other photo is a screenshot of the explanation.
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In a alkaline phosphatase kinetics experiments a studnet first made a standard curve with different amounts of PNP. In a reaction carried out with 10nM alkaline phosphatase and 100 uM PNPP substrate the student measured an absorbance value of 0.76 at 410nm after 30 seconds of reaction. Assuming the 30 second time point is in the lienar range of the reaction and the substrate has not been depleted at this time point, calculate the Vo at 100uM PNPP (in units of concentration per time).Beer's law formula given is y=0.0323x (y axis is Absorbance) (x Axis is PNP uM)
5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00

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