User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/08/29: Difference between revisions
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* Using calibrated micropipets, the appropriate amounts of Au, BSA, and water were distributed to their respective mole ratio containers. The test tubes were capped with metal tops and wrapped in foil. | * Using calibrated micropipets, the appropriate amounts of Au, BSA, and water were distributed to their respective mole ratio containers. The test tubes were capped with metal tops and wrapped in foil. | ||
* Upon placing the Au/BSA solutions into the oven, the solutions appeared as clear and colorless liquid solutions. | |||
* The rack containing the 14 mole ratios was placed inside an oven set to 85 °C for 4 hours. | * The rack containing the 14 mole ratios was placed inside an oven set to 85 °C for 4 hours. |
Revision as of 20:13, 14 September 2012
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Preparation of Gold and Bovine Serum Albumin Stock Solutions
.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
.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
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)
1. First obtained the mole concentration needed for Au in correspondence to its mole ratio using the following factors: X= mole concentration needed for Au 1.5 μM is the mole concentration of BSA used since this is held constant XAu= 1.5 μM × 133 (mole ratio) = 199.50 μM of Au XAu= 1.5 μM × 134 (mole ratio) = 201 μM of Au XAu= 1.5 μM × 136 (mole ratio) = 204 μM of Au XAu= 1.5 μM × 138 (mole ratio) = 207 μM of Au XAu= 1.5 μM × 140 (mole ratio) = 210 μM of Au 2. After obtaining the mole concentrations of Au, these mole concentrations were applied into the dilution equation used earlier to obtain the volume of Au needed. The following volumes for Au are: VAu 133= (199.5 × 6 mL) ÷ 7722 μM of Au = 0.155 mL of Au VAu 134= (201 × 6 mL) ÷ 7722 μM of Au = 0.156 mL of Au VAu 136= (204 × 6 mL) ÷ 7722 μM of Au = 0.158 mL of Au VAu 138= (207 × 6 mL) ÷ 7722 μM of Au = 0.160 mL of Au VAu 140= (210 × 6 mL) ÷ 7722 μM of Au = 0.163 mL of Au
Volumes of water needed V133 = 6 mL − .155 mL of Au − 0.6 mL of BSA = 5.245 mL V134 = 6 mL − .156 mL of Au − 0.6 mL of BSA = 5.244 mL V136 = 6 mL − .158 mL of Au − 0.6 mL of BSA = 5.242 mL V138 = 6 mL − .160 mL of Au − 0.6 mL of BSA = 5.240 mL V140 = 6 mL − .163 mL of Au − 0.6 mL of BSA = 5.237 mL
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