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>. | * Finally, the molarity of the BSA stock solution was accounted for to be 1.5 × 10<sup>-8</sup> m/L | ||
* Using a metal spatula, Gold (Au) stock solution was made by dissolving .085 g of orange, granular solid 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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.00025 mM of HAuCl<sub>4</sub> × 339. 79 g (formula weight of HAuCl<sub>4</sub>)= .085 g of HAuCl<sub>4</sub> | .00025 mM of HAuCl<sub>4</sub> × 339. 79 g (formula weight of HAuCl<sub>4</sub>)= .085 g of HAuCl<sub>4</sub> | ||
* Not all weighed HAuCl<sub>4</sub> was dissolved into solution. The remaining HAuCl<sub>4</sub> material from the weighing paper amounted to .0194 g. | |||
* The amount of .0194 g HAuCl<sub>4</sub> residue was subtracted from the theoretical value .085 g of HAuCl<sub>4</sub> to obtain the amount of HAuCl<sub>4</sub> dissolved in solution. | |||
* The actual HAuCl<sub>4</sub> present in solution is .0656 g. The molarity of the HAuCl<sub>4</sub> stock solution was .007722417 m/L or 7722 μM. | |||
==Preparation of different mole ratios of Au/BSA== | ==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, M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub>, the following calculations were made: | |||
*For mole ratio 60 | |||
M<sub>1</sub> = 15 μM | |||
V<sub>1</sub> = Volume needed from 15 μM BSA stock solution to complete ratio | |||
M<sub>2</sub> = 1.5 μM (concentration of BSA for mole ratio 60) | |||
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 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, [http://openwetware.org/wiki/User:Michael_F._Nagle/Notebook/Chem_571/2012/08/29 Michael Nagle] and [http://openwetware.org/wiki/User:Puja_Mody/Notebook/Chem_571:_Gold_Nanoparticles/2012/08/29 Puja Moody](see their calculation section). Calculations for mole ratios 133, 134, 136, 138, and 140 are provided below: | |||
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 | |||
X<sub>Au</sub>= 1.5 μM × 133 (mole ratio) = 199.50 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 134 (mole ratio) = 201 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 136 (mole ratio) = 204 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 138 (mole ratio) = 207 μM of Au | |||
X<sub>Au</sub>= 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: | |||
V<sub>Au 133</sub>= (199.5 × 6 mL) ÷ 7722 μM of Au = 0.155 mL of Au | |||
V<sub>Au 134</sub>= (201 × 6 mL) ÷ 7722 μM of Au = 0.156 mL of Au | |||
V<sub>Au 136</sub>= (204 × 6 mL) ÷ 7722 μM of Au = 0.158 mL of Au | |||
V<sub>Au 138</sub>= (207 × 6 mL) ÷ 7722 μM of Au = 0.160 mL of Au | |||
V<sub>Au 140</sub>= (210 × 6 mL) ÷ 7722 μM of Au = 0.163 mL of Au | |||
* To obtain the exact volume of water needed for each mole ratio, the corresponding volume of Au for the given mole ratio and 0.6 mL of BSA was subtracted from 6 mL. The volume of water needed for mole ratios 133, 134, 136, 138, and 140 are listed below: | |||
Volumes of water needed | |||
V<sub>133</sub> = 6 mL − .155 mL of Au − 0.6 mL of BSA = 5.245 mL | |||
V<sub>134</sub> = 6 mL − .156 mL of Au − 0.6 mL of BSA = 5.244 mL | |||
V<sub>136</sub> = 6 mL − .158 mL of Au − 0.6 mL of BSA = 5.242 mL | |||
V<sub>138</sub> = 6 mL − .160 mL of Au − 0.6 mL of BSA = 5.240 mL | |||
V<sub>140</sub> = 6 mL − .163 mL of Au − 0.6 mL of BSA = 5.237 mL | |||
* 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. | |||
* Lab area was cleaned and the hygroscopic chloroauric acid was returned to its dessicator. BSA was returned back to the refrigerator for proper storage. | |||
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Revision as of 13:57, 18 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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