User:Floriane Briere/Notebook/CHEM-496/2011/09/14

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Objective

Today's experiment consists on performing a 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

  • Preparation of the stock solution (4ml - 10µg/ml)
  1. Add 0.027ml of BSA (1.46mg/ml - 66776 g/mol) to 3.973ml of water
  2. Vortex the solution
  • Preparation of the 6 standard solutions (1ml)

Image:14sept - Tableau SSolution.png

  • Dilution to 1/1000 of the MPB solution
  1. Add 0.1ml of pure MPB to 0.9ml of water (dilution to 1/10 of pure solution)
  2. Add 0.1ml of 1/10 MPB diluted solution to 0.9ml of water (dilution to 1/100 of pure solution)
  3. Add 0.1ml of 1/100 MPB diluted solution to 0.9ml of water (dilution to 1/1000 of pure solution)
  • Bradford Assay Technique
  1. Add 1ml of standard solution (1,2,3,4,5 and 6) to 200µl of Bradford reagent and measure absorbance at 595nm
  2. Add 1ml of MPB diluted solution (1/100 and 1/1000) to 200µl of Bradford reagent and measure absorbance at 595nm
  3. Add 1ml of water to 200µl of Bradford reagent and measure absorbance at 595nm (blank)
  4. Measure absorbance at 595nm of only MPB diluted solution (1/100 and 1/1000)

Image:14sept - Tableau SSolutionN2.png

Results

  • Curve of the concentration (nM) as a function of the Absorbance at 595nm:

Image:14sept - Abs(595)=f(concentration).png

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 - 0.0204):

Abs(595nm) = 2.1105 * concentration - 0.0204 => concentration = (0.44 + 0.0204)/2.1105 = 0.218nM

So, the 5/6 diluted MPB solution has a concentration of 0.218nM.

The pure MPB solution has a concentration of 0.261nM (0.218*(6/5) = 0.261nM).

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):

Image:14sept - Absorbance = f(wavelength).png

This curve is made with the standards solution and the 1/1000 diluted MPB solution (with bradford reagent)) measures.

Image:14sept - Absorbance = f(wavelength) v2.png

This curve is made with the 1/1000 diluted MPB solution (without Bradford reagent) measures.


  • Curve of the molar absorptivity (L.mol^-1.cm^-1) as a function of the Wavelength (nm):

Image:14sept - Molar absorbitivity= f(wavelength).png

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



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