User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/08/29: Difference between revisions

From OpenWetWare
Jump to navigationJump to search
Line 39: Line 39:


* After making this calculation, it was decided by our group to keep the BSA concentration constant. As a result, 0.6 mL of 15 μM BSA stock solution was distributed to all other remaining mole ratios in their respective test tube containers.  
* After making this calculation, it was decided by our group to keep the BSA concentration constant. As a result, 0.6 mL of 15 μM BSA stock solution was distributed to all other remaining mole ratios in their respective test tube containers.  
* Calculation of the required Au concentration for all other mole ratios was divided amongst other group members, Michael Nagle and Puja Moody. Calculations for mole ratios 133, 134, 136, 138, and 140 are provided below:


<!-- ##### DO NOT edit below this line unless you know what you are doing. ##### -->
<!-- ##### DO NOT edit below this line unless you know what you are doing. ##### -->

Revision as of 12:57, 14 September 2012

Project name <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page
Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html>

Preparation of Gold and Bovine Serum Albumin Stock Solutions

  • Using a metal spatula, Bovine Serum Albumin (BSA) was prepared by dissolving .0249 g of white crystalline powder of BSA into .025 L of water to obtain 15 μM of BSA.
  • The amount of .0249 g of BSA was acquired from the following calculation:

.025 L of water × (15×10-6 mM of BSA ÷ 1 L of water) = 3.75×10-7 mM of BSA

3.75×10-7 mM of BSA × 66430 g (formula weight of BSA) = .0249 g of BSA

  • Using a metal spatula, Gold (Au) stock solution was made by dissolving .085 g of hydrogen tetrachloroaurate (HAuCl4) in .025 L of water to obtain 10 mM HAuCl4.
  • The amount of .085 g of HAuCl4 was acquired from the following calculation:

.025 L of water × (10×10-3 mM of HAuCl4 ÷ 1L of water)= .00025 mM of HAuCl4

.00025 mM of HAuCl4 × 339. 79 g (formula weight of HAuCl4)= .085 g of HAuCl4

Preparation of different mole ratios of Au/BSA

  • It was decided to prepare solutions of mole ratios; 60, 80, 100, 120, 128, 130, 132, 133, 134, 136, 138, 140, 160, and 170. The total volume to be prepared for each mole ratio solution is 6 mL.
  • After allowing some time for the stock solutions of Au and BSA to completely dissolve, calculations were made to verify the volume needed of each constituent, Au, BSA, and water, to obtain the appropriate mole ratio.
  • Using the dilution equation, M1V1 = M2V2, the following calculations were made:
  • For mole ratio 60

M1 = 15 μM

V1 = Volume needed from 15 μM BSA stock solution to complete ratio

M2 = 1.5 μM (concentration of BSA for mole ratio 60)

V2 = 6 mL (total volume of final solution)

  • V2 for 60 = (1.5 μM × 6 mL) ÷ 15 μM = 0.6 mL of 15 μM BSA stock solution
  • After making this calculation, it was decided by our group to keep the BSA concentration constant. As a result, 0.6 mL of 15 μM BSA stock solution was distributed to all other remaining mole ratios in their respective test tube containers.
  • Calculation of the required Au concentration for all other mole ratios was divided amongst other group members, Michael Nagle and Puja Moody. Calculations for mole ratios 133, 134, 136, 138, and 140 are provided below:




Retrieved from "https://openwetware.org/mediawiki/index.php?title=User:Mary_Mendoza/Notebook/CHEM_571_Experimental_Biological_Chemistry_I/2012/08/29&oldid=626324"