1.8 Calculations for the Purification of a Recombinant Protein The purification of arecombinant protein is carried out starting with 100 liters of a clarified cell lysate (i.e., thecells have been lysed, and the cell debris has been removed to give a clarified solution),which has a total protein concentration of 0.36 mg/ml and a recombinant protein con-centration of 2.2 U/ml, where U denotes units of biological activity of the recombinantprotein. It is known that the completely pure recombinant protein has a specific activityof 40.0 U/mg. Purification is continued until a chromatography step that yields 2.0 litersof a fraction containing the protein, with a total protein concentration of 1.11 mg/mland a recombinant protein concentration of 43.2 U/ml.For the recombinant protein,calculate the starting and ending purity, the starting and ending specific activity, and thepercentage yield and fold purification through the chromatography step.

Purification of a Recombinant Protein:This process involves the following steps-Cel harvesting- In…

Answered: starting and ending purity, t

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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Calculations for the Purification of a Recombi- nant Protein The purification of a recombinant protein is carried out starting with 100 liters of a clarified cell lysate (i.e., the cells have been lysed, and the cell debris has been removed to give a clar- ified solution), which has a total protein concentra- tion of 0.36 mg/ml and a recombinant protein con- centration of 2.2 U/ml, where U denotes units of biological activity of the recombinant protein. It is known that the completely pure recombinant pro- tein has a specific activity of 40.0 U/mg. Purifica- tion is continued until a chromatography step that yields 2.0 liters of a fraction containing the protein, with a total protein concentration of 1.11 mg/ml and a recombinant protein concentration of 43.2 U/ml.. For the recombinant protein, calculate the. starting and ending purity, the starting and ending specific activity, and the percentage yield and fold purification through the chromatography step. :
2.7 Problem in the Determination of Molecular Welght by Gel Electrophoresis A puri- fied recombinant protein is analyzed for molecular weight by SDS-PAGE at pH 8.5. From the protein sequence deduced from the gene that was expressed in bacteria, the protein is expected to have a molecular weight of 44,000. However, the molecular weight of the pro- tein is found by SDS-PAGE to be 52.000. Explain the reason or reasons for this difference in molecular weight. What calculation could you make to help explain this discrepancy?
Each group will be provided with 20 g double-stranded DNA oligomer in STE buffer (0.1M NaCl/ Tris/ 10 mM EDTA, pH 7.4). The sequence of one strand of the oligomer is: 5’-AATGGTACGTATACTTTAT-3’ - show a table of % denaturation of DNA at different temperatures, using the equation provided in “Calculation” - plot a graph of % denaturation of DNA vs different temperatures with clear label and title, and hence determine the experimental Tm of the DNA sample - calculate the theoretical Tm of the sample using Marmur and Doty equation, compare the experimental Tm with theoretical Tm of the DNA sample
33. Gel filtration chromatography shows that the molecular weight of a protein is 240,000 Daltons. However, when the protein was subjected to SDS-PAGE in the absence of ß- mercaptoethanol, two polypeptides (100,000 Daltons, 140,000 Daltons) were identified. Furthermore, when the protein was subjected to SDS-PAGE in the presence of ß- mercaptoethanol, three polypeptides (100,000 Daltons, 90,000 Daltons, and 50,000 Daltons) were identified. Which of the following statements is true of the protein? A. The protein is composed of three polypeptides. B. The three-dimensional structure of the protein is stabilized by both covalent and non- covalent bonds. C. Both A and B. D. Neither A nor B. 34. What is purpose of SDS in SDS-PAGE? E. To selectively bind the target protein F. To maintain buffer pH in the gel G. To cause the separation of proteins to be on the basis of molecular weight only H. To initiate polymerization of acrylamide to form gel
A novel protein runs to give a single band on SDS PAGE (45kDa) and three bands on native PAGE. Explain..
Show the calculations for the preparation of 0.01%, 0.001%, 0.0001%, 0.00001% standard DNA solutions via serial dilution from a 0.1% stock solution. Prepare 10.0 mL of 0.10% standard DNA solution using acetate buffer pH 4.6 as solvent. Prepare four different concentrations of the 0.10% standard DNA solution from step 1 via serial dilution to obtain 0.01%, 0.001%, 0.0001%, 0.00001% standard DNA solutions. Obtain seven clean and dry test tubes. Place 1.50 mL solutions in each tube as follows: Table 2.1. Set-up for the diphenylamine assay. Test tube # Content 1 (blank) acetate buffer pH 4.6 2 0.10% standard DNA 3 0.01% standard DNA 4 0.001% standard DNA 5 0.0001% standard DNA 6 0.00001% standard DNA 7 DNA extract from Part I Add 3.50 mL diphenylamine reagent to each tube, swirl each tube to thoroughly mix the contents and heat for 10 mins in a boiling water bath. Cool immediately under tap water. Read absorbances at 595 nm.…
Purification of a protein of unknown structure has been achieved. The natural protein has a molecular weight of 240,000, according to size-exclusion chromatography. Using a concentration of 6 M guanidine hydrochloride in the chromatography, a single peak can be identified as the molecular weight (MW) 60,000 of a protein. B-mercaptoethanol (BME) and guanidine hydrochloride (GHC) are used in tandem to produce proteins with mass masses of 34,000 and 26,000, respectively. The structure of this protein can be inferred from these facts.
Low-resolution X-ray diffraction analysis of a protein composed of long stretches of the sequence (-Gly-Ser-Gly-Ala-Gly-Ala-)n, where n indicates any number of repeats, shows an extended structure of stacked layers, with a repeat distance between layers that alternates between 3.5 Å and 5.7 Å. Propose a model that explains this scenario.
(b) Both laboratories used 10 micrograms of protein each in their kinetic assays. Protein concentrations weredetermined by the Bradford protein assay. Assay conditions employed in the two labs (pH, temperature,etc.) were also identical. What would be the most plausible cause for the discrepancy in the Vmax valuesfor the compound I? Explain.Recall that the Bradford assay measures total protein amounts in sample solution based on complexformation between a dye and proteins. Also, the assay solution used in both labs does not contain anyinhibitors.
The A280 of a protein sample loaded onto a gel was determined to be 0.767 (1.00 cm path length, after subtracting the blank). The total volume of this sample was 428 µL. 19.0 µL of this protein sample was mixed with 19.0 µL of 2X laemalli sample buffer and then 12.0 µL of the entire sample was loaded into the gel and electrophoresed. Calculate the amount of protein that was loaded into the gel (in µg).
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Approximate molecular weight for an unknown protein from gel-filtration experiment is 130 kDa. Thirty six mg of this pure protein was treated with excess of fluorodinitrobenzene. After the reaction and complete acid hydrolysis the mixture was found to contain 356 µg of dinitrobenzene derivative of methionine (free acid) and no other amino acid-dinitrobenzene derivatives. Is this protein consisting of single polypeptide chain or multiple subunits? If it consists of multiple subunits, then how many? From these data, can you calculate the molecular weight of the protein more precisely?

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