How would you create this graph Bradford protein assay data:Absorbance Readings at 595nm You will need to graph this0.104 for P1data anddetermine theexactconcentrationof0.098 for P2protein (mg/ml) insamples(fractions) allowing for thedilution factor in the assay(0.2 into 1.0 ml = 1/5 dilutiontherefore multiply by 5).the0.211 for P3unknown0.135 for P40.124 for P50.95 for P6 TUBEP1P2P3P4P5P6P7P8P9P10P11P12BSA Standard (µg/ml)Fraction (No)10203040T1T2T3T4T5T6Standard/BSAstock solution0.10.20.40.60.8(ml)Fraction (ml)0.20.20.20.20.20.2Distilled H200.80.70.60.40.20.60.60.60.60.60.6(ml)Dye reagent0.20.20.20.20.20.20.20.20.20.20.20.2(ml)

Bradford assay is extensively applied for estimating unknown protein sample concentration.…

Bradford protein assay data: Absorbance Readings at 595nm You will need to graph this 0.104 for P1 data and determine the exact concentration of 0.098 for P2 protein (mg/ml) in samples (fractions) allowing for the dilution factor in the assay the 0.211 for P3 unknown 0.135 for P4 0.124 for P5 (0.2 into 1.0 ml = 1/5 dilution therefore multiply by 5). 0.95 for P6

Bradford protein assay data: Absorbance Readings at 595nm You will need to graph this 0.104 for P1 data and determine the exact concentration of 0.098 for P2 protein (mg/ml) in samples (fractions) allowing for the dilution factor in the assay the 0.211 for P3 unknown 0.135 for P4 0.124 for P5 (0.2 into 1.0 ml = 1/5 dilution therefore multiply by 5). 0.95 for P6

Curren'S Math For Meds: Dosages & Sol

11th Edition

ISBN:9781305143531

Author:CURREN

Publisher:CURREN

Chapter14: Formula Method

Section: Chapter Questions

Problem 5SST

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This graph represents a standard curve from a BCA assay for protein concentration. If your BCA reaction with your unknown protein has an absorbance of 0.2, approximately what concentration of protein is in your unknown sample? Absorbance (562nm) 0.5 0.45 3 0.35 0.3. 0.25 0.15. 0.1- 0.05 ~0.1 μg/ml ~4.5 μg/ul less that 0.005 | 0.2 μg/μ. [Protein (BSA)) (pg/ml)
(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.
Electrophoresis A protein required 6.8 min to travel 82 cm to the detector in a 96 cm -long capillary tube with 25.4 kV between the ends. Find the apparent electrophoretic mobility.
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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I need to prepare a standard calibration curve for gamma globulin. absorbances on Y and mg of standard protein per assay on X. used 0.1mg/ml gamma stock for tubes 2-6. (Water (ml), gamma (ml), Abosrbance)--> (.036, .004, .290) (.036, .008, .358) (.024, .016, .341) (.016, .024, .520) (.008, .032, .597) - What is the math and how do you get the standard curve?
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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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…
You perform a Bradford assay to determine the concentration of isolated α-lactalbumin. You use 50 μL of a two-fold diluted solution of α-lactalbumin in the assay. You generate a standard curve with the following equation for the line: y = 0.163x + 0.082. The absorbance of your sample was 0.674 AU. What is the concentration of α-lactalbumin, in mg/mL, in your sample? Give your answer to three significant figures.