User:Javier Vinals Camallonga/Notebook/Javier Vinals notebook/2013/09/02: Difference between revisions

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This procedure was taken from the following reference and has been used by our previous two Experimental Biological Chemistry groups.
This procedure was taken from the following reference and has been used by our previous two Experimental Biological Chemistry groups.
#Add 1mL of the (~2.5mM -note the exact concentration) gold (HAuCl4) solution to a 10mL volumetric flask
#Add an appropriate amount of BSA solution so that the final concentration of gold is 90X that of BSA
#Add 1mL of the (~2.5mM -note the exact concentration) gold (HAuCl4) solution to a 10mL volumetric flask
#Add an appropriate amount of BSA solution so that the final concentration of gold is 90X that of BSA
                     - Through calculation, we obtained that 0.001809 L was the necessary amount of BSA solution needed to add, so that the final concentration of gold is 90X that of BSA. Calculations done: First we find the amount of moles of BSA necessary for the reaction. [(2.54*10^-3 moles/L of HAuCL4)*(1L/1000mL)*1mL]/90=2.82 *10^-8 moles of BSA. Now we find the volume of BSA necessary to add to the reaction: (2.82 *10^-8 moles BSA) *(1L/15.6 *10^-6 moles)= 0.001809 L of BSA or 1.8 mL of BSA
                     - Through calculation, we obtained that 0.001809 L was the necessary amount of BSA solution needed to add, so that the final concentration of gold is 90X that of BSA. Calculations done: First we find the amount of moles of BSA necessary for the reaction. [(2.54*10^-3 moles/L of HAuCL4)*(1L/1000mL)*1mL]/90=2.82 *10^-8 moles of BSA. Now we find the volume of BSA necessary to add to the reaction: (2.82 *10^-8 moles BSA) *(1L/15.6 *10^-6 moles)= 0.001809 L of BSA or 1.8 mL of BSA
   
   
#Add deionized water up to 10mL
#Add deionized water up to 10mL
#Transfer solution to a test tube and cap with aluminum foil
#Transfer solution to a test tube and cap with aluminum foil
#Heat in oven at 80C for 3 hours
#Heat in oven at 80C for 3 hours
                   - After we took the solution out of the oven, we observed that some nanoparticles had formed, due to an error during the process fo the experiment. Thus it was discarded away.  
                   - After we took the solution out of the oven, we observed that some nanoparticles had formed, due to an error during the process fo the experiment. Thus it was discarded away.  
#Transfer solution to a plastic falcon tube (with blue cap)
#Transfer solution to a plastic falcon tube (with blue cap)
Stock solutions made
Stock solutions made
Gold solution (HAuCl4·3H2O) 0.0100g in 0.0100mL water → 2.54mM
Gold solution (HAuCl4·3H2O) 0.0100g in 0.0100mL water → 2.54mM
BSA solution 0.0104g BSA (MW = 66776g/mol) in 0.0100mL water → 15.6μM
BSA solution 0.0104g BSA (MW = 66776g/mol) in 0.0100mL water → 15.6μM

Revision as of 07:52, 2 September 2013

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Objective

Synthesize two different sets of gold nanoparticles. In one set, Au3+ is reduced by a protein (bovine serum albumin, BSA) and the synthesized nanoparticle is also surrounded and stabilized by BSA. In the second set, Au3+ is reduced by citrate, and the AuNP is stabilized by citrate in solution. The BSA-AuNPs are purple in color and the citrate-AuNPs are more of a burgundy (reddish) color.

BSA-AuNP

This procedure was taken from the following reference and has been used by our previous two Experimental Biological Chemistry groups.

  1. Add 1mL of the (~2.5mM -note the exact concentration) gold (HAuCl4) solution to a 10mL volumetric flask
  2. Add an appropriate amount of BSA solution so that the final concentration of gold is 90X that of BSA
                   - Through calculation, we obtained that 0.001809 L was the necessary amount of BSA solution needed to add, so that the final concentration of gold is 90X that of BSA. Calculations done: First we find the amount of moles of BSA necessary for the reaction. [(2.54*10^-3 moles/L of HAuCL4)*(1L/1000mL)*1mL]/90=2.82 *10^-8 moles of BSA. Now we find the volume of BSA necessary to add to the reaction: (2.82 *10^-8 moles BSA) *(1L/15.6 *10^-6 moles)= 0.001809 L of BSA or 1.8 mL of BSA
  1. Add deionized water up to 10mL
  2. Transfer solution to a test tube and cap with aluminum foil
  3. Heat in oven at 80C for 3 hours
                  - After we took the solution out of the oven, we observed that some nanoparticles had formed, due to an error during the process fo the experiment. Thus it was discarded away.
  1. Transfer solution to a plastic falcon tube (with blue cap)

Stock solutions made Gold solution (HAuCl4·3H2O) 0.0100g in 0.0100mL water → 2.54mM BSA solution 0.0104g BSA (MW = 66776g/mol) in 0.0100mL water → 15.6μM

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