User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/08/29: Difference between revisions
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V<sub>2</sub> = 6 mL (total volume of final solution) | V<sub>2</sub> = 6 mL (total volume of final solution) | ||
*V<sub>2</sub> for 60 = (1.5 μM × 6 mL) ÷ 15 μM = 0.6 mL | *V<sub>2</sub> 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. | |||
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Revision as of 12:52, 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)
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