User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/08/29: Difference between revisions
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==Preparation of Gold and Bovine Serum Albumin Stock Solutions== | ==Preparation of Gold and Bovine Serum Albumin Stock Solutions== | ||
* 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. | * 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: | * The amount of .0249 g of BSA was acquired from the following calculation: | ||
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3.75×10<sup>-7</sup> mM of BSA × 66430 g (formula weight of BSA) = .0249 g of BSA | 3.75×10<sup>-7</sup> mM of BSA × 66430 g (formula weight of BSA) = .0249 g of BSA | ||
* Gold (Au) stock solution was made by dissolving .085 g of hydrogen tetrachloroaurate (HAuCl<sub>4</sub>) in .025 L of water to obtain 10 mM HAuCl<sub>4</sub>. | * Using a metal spatula, Gold (Au) stock solution was made by dissolving .085 g of hydrogen tetrachloroaurate (HAuCl<sub>4</sub>) in .025 L of water to obtain 10 mM HAuCl<sub>4</sub>. | ||
* The amount of .085 g of HAuCl<sub>4</sub> was acquired from the following calculation: | * The amount of .085 g of HAuCl<sub>4</sub> was acquired from the following calculation: | ||
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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 |