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

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* 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  +  .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.  
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+  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  
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Revision as of 09:51, 27 October 2012
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Electrophoresis of the mutated DNA plasmids
UVVis for Chemiluminescence
Preparation of Lysozyme stock
.012 L of water × = × = 0.00258 g of lysozyme to obtain 15 μM
0.0109 g of lysozyme × = 7.62E(7) mol ÷ .050 L of water = 1.524E(5) M = 15.24 μM
