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
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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.Calculate the Activity of an amylase enzyme which is diluted 1:100 times with phosphate buffer and incubated for 10 minutes at 37 degree Celsius. Given are the amount of maltose [mg] = 5.05 and the volume of enzyme used for the assay as 0.5ml.Various concentrations of recombinant human insulin were prepared for use standards for an HPLC method. To verify the prepared concentrations, the samples were analyzed by measuring the absorbance at 280 nm in a short path length (5 mm) cuvette. The molar absorption coefficient for human insulin is approximately 5.875 x 10³ M-¹cm-¹. a. Calculate the extinction coefficient in mL mg-¹cm-¹. b. Calculate the concentrations of the following human insulin standards if the measured absorbances and dilutions used are: Standard 1 Standard 2 Abs. (at 280 nm) of Diluted Sample 0.305 0.685 Dilution 145.0 µL sample, 25.0 μL buffer 130.0 µL sample, 40.0 μL buffer
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- What are the advantages of gel filtration as a technique for protein purification? Identify two types of gels (sieving matrix) that may be used for gel filtration chromatography and discuss the types of samples which may be used for each. Discuss the principle behind affinity chromatography and differentiate it from gel filtration chromatography.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.