User:Puja Mody/Notebook/Chem 571: Gold Nanoparticles/2012/08/29: Difference between revisions

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.376 μM x66430 g/mol (MW of BSA)= .0249g BSA in 25mL of water.  
.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.  
* 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 M<sub>1</sub>V<sub>1</sub>= M<sub>2</sub>V<sub>2</sub> 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.


==Results/Conclusions==
==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 HAuCl<sub>4</sub>.  
* 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 HAuCl<sub>4</sub>.  
.0656g x 1mol/339.79 (MW  HAuCl4)= 1.93 x10<sup>-4</sup> / .025L= .00772 M/L.
.0656g x 1mol/339.79 (MW  HAuCl4)= 1.93 x10<sup>-4</sup> / .025L= .00772 M/L.
*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.
 




Revision as of 19:15, 9 September 2012

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Day 1

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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.

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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