Difference between revisions of "User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/17"
Mary Mendoza (talk  contribs) (→Preparation of Lysozyme stock) 
Mary Mendoza (talk  contribs) (→Preparation of Gold and Lysozyme stocks) 

(4 intermediate revisions by the same user not shown)  
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* Data collection was not possible. No signal was collected.  * Data collection was not possible. No signal was collected.  
−  ==Preparation of Lysozyme  +  ==Preparation of Gold and Lysozyme stocks== 
−  * Michael Nagle prepared a new set of Au/BSA solutions.  +  * Michael Nagle prepared a new set of Au/BSA solutions. The stock concentration of BSA was 1.49 x 10<sup>5</sup> mM/L and the stock concentration for Au was 8.24 mM. The mole ratios were kept the same as 60, 80, 100, 120, 128, 130, 132, 133, 134, 136, 138, 140, 160, and 170. Nagle finished the preparation of the set and placed the rack at 85°C for 4h. at the Thermo Scientific incubator. 
+  * Weighed out .014 g of monohydrate chloroauric acid and dissolved into 5 mL of water. The molarity of gold was calculated to be at 8240 μM.  
+  
+  
+  .014 g of gold × <math>\frac{1 mol}{339.79 g}</math> = 4.12E(5) mol ÷ .005 L = <math>\frac{8.24E(3)mol}{L}</math> = 8240 μM  
+  
+  
* Lysozyme (14,307 Da) appeared as a white granular solid provided by Sigma. Knowing that 1 Da = 1 g/ mol, the following calculations were made to obtain a stock solution of 15.24 μM.  * Lysozyme (14,307 Da) appeared as a white granular solid provided by Sigma. Knowing that 1 Da = 1 g/ mol, the following calculations were made to obtain a stock solution of 15.24 μM.  
+  
.012 L of water × <math>\frac{15E(6)mol}{1 L}</math> = <math>\frac{1.80E(7)mol}{L}</math> × <math>\frac{14,307 g}{mol}</math> = 0.00258 g of lysozyme to obtain 15 μM  .012 L of water × <math>\frac{15E(6)mol}{1 L}</math> = <math>\frac{1.80E(7)mol}{L}</math> × <math>\frac{14,307 g}{mol}</math> = 0.00258 g of lysozyme to obtain 15 μM  
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* The amount to be measured is too small to be certain that the entire solid would be dissolved in water. As a result, a multiple of the calculated amount was chosen to be dissolved in water; the amount chosen was 0.0109 g. Once dissolved, this will be diluted to 15 μM.  * The amount to be measured is too small to be certain that the entire solid would be dissolved in water. As a result, a multiple of the calculated amount was chosen to be dissolved in water; the amount chosen was 0.0109 g. Once dissolved, this will be diluted to 15 μM.  
+  
0.0109 g of lysozyme × <math>\frac{1 mol}{14,307 g}</math> = 7.62E(7) mol ÷ .050 L of water = 1.524E(5) M = 15.24 μM  0.0109 g of lysozyme × <math>\frac{1 mol}{14,307 g}</math> = 7.62E(7) mol ÷ .050 L of water = 1.524E(5) M = 15.24 μM  
+  
+  
+  * After preparing the stock, it was decided to prepare the Au/lysozyme solution next week. The clear and colorless, 15.24 μM stock solution of lysozyme was refrigerated.  
+  * In addition, the Au/lysozyme mole ratios chosen were similar to the 14 Au/BSA solutions. Likewise to the Au/BSA solutions, the lysozyme will be kept constant at a volume of .590 mL = 590 μL.  
+  
Revision as of 06:18, 7 December 2012
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Electrophoresis of the mutated DNA plasmids
UVVis for Chemiluminescence
Preparation of Gold and Lysozyme stocks
