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…

Answered: つく 146 Bb 9146 Bb 10311 Class E Etsy…

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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A16 Compound A is the substrate for two enzymes, E, and E, involved in the synthe- sis of an amino acid and an antibiotic, respectively. Their reaction rates, r, and r, at an arbitrary concentration of E, and E, were measured at different concen- trations of A. However, the data were not labeled properly (shown in Table P.4.1), and the student who did the measurement needs help to sort out which dataset is for E,. Determine the K, and rmax tration of A. Which set of data is more likely to be for E, and which for E,, and why? for both enzymes, with respect to the concen- Table P.4.1 Reaction kinetics of a common substrate A for two enzymes. Concentration of 0.2 0.6 1.2 3 4 6. 8 9. 12 15 A (mM) Reaction rate (r.) (mmol/L'min) 3.33 4.29 4.62 4.76 4.84 .88 4.9 4.92 4.94 4.95 4.96 4.97 Reaction rate (r,) (mmol/L*min) 0.09 0.23 0.38 0.5 0.6 0.67 0.71 0.75 0.8 0.82 0.86 0.80
A16 Compound A is the substrate for two enzymes, E, and E2, involved in the synthe- sis of an amino acid and an antibiotic, respectively. Their reaction rates, r, and r2, at an arbitrary concentration of E, and E, were measured at different concen- trations of A. However, the data were not labeled properly (shown in Table P.4.1), and the student who did the measurement needs help to sort out which dataset is for E,. Determine the K and rmax for both enzymes, with respect to the concen- tration of A. Which set of data is more likely to be for E, and which for E, and why? Table P.4.1 Reaction kinetics of a common substrate A for two enzymes. Concentration of 0.2 0.6 1.2 3. 4 5 8 9 12 15 A (mM) Reaction rate (r.) (mmol/L'min) Reaction rate (r,) (mmol/L'min) 3.33 4.29 4.62 4.76 4.84 4.88 4.9 4.92 4.94 4.95 4.96 4.97 0.09 0.23 0.38 0.5 0.6 0.67 0.71 0.75 0.8 0.82 0.86 0.80
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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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CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
Bradford technique makes use of the Coomassie blue dye that binds to the protein, with the complex absorbing strongly at 465 nm. O True O False
Two-dimensional gel electrophoresis of proteins in a cell extract provides a qualitative way to compare proteins with respect to intra- cellular abundance. Describe a quantitative approach to the deter- mination of number of molecules of an enzyme per cell.
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Deepika_Biology From a kinetics experiment, Kcat was determined to be 55sec^-1. For the kinetic assay, 0.05mL of a 0.05mg/mL solution of the 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.
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