User:Floriane Briere/Notebook/CHEM-496/2011/09/14: Difference between revisions
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So, the 5/6000 diluted MPB solution has a concentration of 0.0243nM. | So, the 5/6000 diluted MPB solution has a concentration of 0.0243nM. | ||
The pure MPB solution has a concentration of '''29.16nM (0.0243*(6000/5) = 29.16nM)'''. | |||
<b>What are the final concentrations (in μg/mL of your standard solutions? Are they diluted from your original calculations. Your value for the concentration of MBP seems a bit low from where it should be. Also, you needed to take a spectrum of just MBP in water/buffer with NO Bradford Assay reagent. We need that so we can calculate ε.</b> [[User:Matt Hartings|Matt Hartings]] 20:47, 20 September 2011 (EDT) | <b>What are the final concentrations (in μg/mL of your standard solutions? Are they diluted from your original calculations. Your value for the concentration of MBP seems a bit low from where it should be. Also, you needed to take a spectrum of just MBP in water/buffer with NO Bradford Assay reagent. We need that so we can calculate ε.</b> [[User:Matt Hartings|Matt Hartings]] 20:47, 20 September 2011 (EDT) |
Revision as of 11:18, 21 September 2011
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ObjectiveToday's experiment is to test the Bradford Assay technique which is going to allow us to determine the unknown proteins’ concentration of a solution. We are going to use 6 standards solutions of BSA whose concentrations are known. Then, thanks to a spectrophotometer, we’ll be able to determine the concentration of the MBP solution. Protocol
Results* Curve of the concentration (nM) as a function of the Absorbance at 595nm To draw this curve, we add to take into account that the Bradford Assay was made with 5/6 diluted solutions (because we added 1ml of solution to 200µl of Bradford reagent). Red lines represent the Absorbance at 595nm of the 5/6000 diluted MPB solution. This curve allow us to determine the concentration of the MPB solution. We can calculate the unknown concentration thanks to the line’s equation (y = 2.1105x + 0.3886): Abs(595nm) = 2.1105 * concentration + 0.3886 => concentration = (0.44 - 0.3886)/2.1105 = 0.0243nM So, the 5/6000 diluted MPB solution has a concentration of 0.0243nM. The pure MPB solution has a concentration of 29.16nM (0.0243*(6000/5) = 29.16nM). What are the final concentrations (in μg/mL of your standard solutions? Are they diluted from your original calculations. Your value for the concentration of MBP seems a bit low from where it should be. Also, you needed to take a spectrum of just MBP in water/buffer with NO Bradford Assay reagent. We need that so we can calculate ε. Matt Hartings 20:47, 20 September 2011 (EDT) * Curve of the Absorbance as a function of the Wavelength (nm) This curve is made with the standards solution and the 1/1000 diluted MPB solution (with bradford reagent)) measures. This curve is made with the 1/1000 diluted MPB solution (without Bradford reagent) measures.
According to Beer-Lambert law, Absorbance = molar absorptivity (L.mol^-1.cm^-1) * concentration (M) * length of the cuve (cm) So, molar absorptivity (ε) = Absorbance/(concentration (c)* length of the cuve (l)) We used the Absorbance of the 1/1000 MPB solution (without Bradford reagent) whose concentration is 29,16*10^-12M. And we know that l = 1cm NotesThis area is for any observations or conclusions that you would like to note.
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