User:Michaela Harper/Notebook/Biological Chemistry Lab/2011/09/13
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ObjectivePerform a Bradford Assay to determine the concentration of an unknown concentration of Maltose Binding Protein (MBP). DescriptionMake standard solutions of BSA (1mL for each solution) in water of 10, 8, 6, 4, 2, and 1 μg/mL. Using the MM of BSA (66776 g/mol) convert these concentration values into M.
Assay: Data analysis: DataThis graph shows the linear relationship between concentration and absorbance for a BSA standard solution. The stock solution was 1.46 mg/mL. This was diluted by 100 to get a 0.0146 mg/mL solution. This was used to make standard solutions of 1, 2, 4, 6, 8, and 10 ug/mL. The absorbance at 595 nm is plotted against these concentrations. The R2 value for the linear regression is 0.9958. The equation for the line is shown on the graph. This equation was used to determine the concentration of the unknown MBP. We used the corrected absorbance at 595 nm as the y value and solved for x. We also incorporated the fact that we diluted the original stock solution by 1000. This graph shows the calculated molar absorptivity vs. wavelength for MBP without Bradford reagent. The molar absorptivity was calculated using the equation, ε = A / bc where ε is molar absorptivity, A is absorbance, b is the pathlength and c is concentration. The pathlength of the cell is a 1 cm cuvette. The concentration was calculated as the diluted concentration of the standard we used, determined by the information from the BSA graph and line equation. The protein in this sample is waaaaay overconcentrated. For protein spectra, we need a graph that shows features at 280nm. Matt Hartings 21:37, 20 September 2011 (EDT) NotesCalculations: To determine volume of diluted standard (0.0146 mg/mL) needed to make 1 mL samples of desired concentrations (1, 2, 4, 6, 8, and 10 ug/mL): For 10 ug/mL M_{1}V_{1} = M_{2}V_{2} (0.0146 mg/mL) V_{1} = (0.01 mg/mL) (1 mL) V_{1} = 0.685 mL Repeated for 1, 2, 4, 6 and 8 ug/mL.
Absorbance at 595 nm = 0.077 y = 0.0487x + 0.0247 0.077 = 0.0487x + 0.0247 x = 1.074 ug/mL > diluted concentration from stock solution 1.074 x 1000 = 1.074 mg/mL > original stock solution concentration
