Day 1
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- Abigail E. Miller 11:59, 17 September 2012 (EDT):huh? what do you need to say here?
Goals
- Make stock solutions of gold nanoparticles (HAuCl4 in water and BSA.)
- Determine how much of each solution will be used in order to make the 14 different concentrations needed.
Procedure
- A stock solution with the AuNP was first made with 25mL of water along with .085 grams of the gold nano- particles. This resulted in a stock solution concentration of 10mM.
- Next, a 25mL stock solution of BSA was made. The desired concentration of the BSA stock solution was 15μM. so .025×(15×10-6)= 3.75 x10-7μM.
.376 μM x66430 g/mol (MW of BSA)= .0249g BSA in 25mL of water.
- These concentrations were chosen based on the concentration ranges used in the experimentation by Bakshi.
- A wide range of concentrations were then determined in order to best test the optimal range for the unfolding of the BSA protein and the insertion of the AuNP into the BSA. The concentrations chosen were 60, 80, 100, 120, 128, 130, 132, 133, 134, 136, 138, 140, 160, 170. This wide range was taken in order to determine exactly where optimal absorption of the AuNP into the BSA occurs.
- The formula M1V1= M2V2 was used in order to calculate how much of each stock solution was necessary in order to create the desired concentrations. See excel spreadsheet for full break down of measurements.
- The amount of BSA used was kept constant at a concentration of 1.5 μM and .6mL of the BSA stock solution was added into each of the 14 test tubes. This measurement was calculated with the M1V1= M2V2 equation where M2= 1.5μM V2= 6mL (because we wanted the total amount of solution in each test tube to equal 6mL. A similar calculation was done in order to determine the amount of the AuNP stock that would go into each test tube.
M1V1= M2V2
7720μM · X= (1.5 × (y)) ×6mL Where y= the desired concentration of each test tube (60, 80, 100 ....etc)
- Once the appropriate concentrations of the solutions were mixed, the test tubes were wrapped in foil and placed in an oven at a temperature of 80°C for 4 hours in order to unfold and refold the protein around the AuNP.
Calculations
- m1v1=m2v2
- 15μM*(volume BSA stock
solution) = (1.5*6)
- 0.6mL BSA stock
solution in each tube
- 7720μM*x mL=6mL*(1.5*60μM)
- 0.069mL HAuCl4
stock solution
- 5.331 mL H2O
- 7720μM*x mL=6mL*(1.5*80μM)
- 0.093mL HAuCl4
stock solution
- 5.307mL H2O
- 7720μM*x mL=6mL*(1.5*100μM)
- 0.117mL HAuCl4
stock solution
- 5.283mL H2O
- 7720μM*x mL=6mL*(1.5*120μM)
- 0.134mL HAuCl4
stock solution
- 5.266mL H2O
- 7720μM*x mL=6mL*(1.5*128μM)
- 0.149mL HAuCl4
stock solution
- 5.251mL H2O
- 7720μM*x mL=6mL*(1.5*130μM)
- .152mL HAuCl4
stock solution
- 5.248mL H2O
- 7720μM*x mL=6mL*(1.5*132μM)
- 0.154mL HAuCl4
stock solution
- 5.246mL H2O
- 7720μM*x mL=6mL*(1.5*133μM)
- 0.155mL HAuCl4
stock solution
- 5.245mL H2O
- 7720μM*x mL=6mL*(1.5*134μM)
- 0.156mL HAuCl4
stock solution
- 5.254mL H2O
- 7720μM*x mL=6mL*(1.5*136μM)
- 0.159mL HAuCl4
stock solution
- 5.244mL H2O
- 7720μM*x mL=6mL*(1.5*138μM)
- 0.161mL HAuCl4
stock solution
- 5.242mL H2O
- 7720μM*x mL=6mL*(1.5*140μM)
- 0.163mL HAuCl4
stock solution
- 5.237mL H2O
- 7720μM*x mL=6mL*(1.5*160μM)
- 0.187mL HAuCl4
stock solution
- 5.213mL H2O
- 7720μM*x mL=6mL*(1.5*170μM)
- 0.198mL HAuCl4
stock solution
- 5.202mL H2O
- The tubes were wrapped in tin foil and heated at 80°C for 4 hours.
Results/Conclusions
- The actual concentration of the AuNP stock solution ended up being .00772 M/L due to the fact that some of the AuNP compound was lost on weigh paper when it was being weighed out and was therefore not transferred into the solution. The actual amount added into the solution was .0656 grams HAuCl4.
.0656g x 1mol/339.79 (MW HAuCl4)= 1.93 x10-4 / .025L= .00772 M/L.
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