User:Mary Mendoza/Notebook/CHEM572 Exp. Biological Chemistry II/2013/01/23: Difference between revisions
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.030 <math>\frac{mol}{L}</math> mol of adenosine × <math>\frac{267.24 g}{1 mol}</math> = <math>\frac{8.0172 g}{L}</math> × .010 L = 0.0802 g of adenosine | .030 <math>\frac{mol}{L}</math> mol of adenosine × <math>\frac{267.24 g}{1 mol}</math> = <math>\frac{8.0172 g}{L}</math> × .010 L = 0.0802 g of adenosine | ||
The calculations on the bottom are courtesy of Dhea Patel. | |||
[[Image:ADA_kinetics_table.png|px80]] | [[Image:ADA_kinetics_table.png|px80]] | ||
The calculation of for the buffer was formulated by [[User:Catherine Koenigsknecht/Notebook/Experimental Biological Chemistry/2013/01/23|Catherine Koenigsknecht]] | |||
* Procedure: Buffer | |||
#Protocol for 0.1 M Sodium Phosphate Buffer (pH 7.4) | |||
#Add 3.1 g of NaH2PO4•H2O and 10.9 g of Na2HPO4 (anhydrous) to distilled H2O to make a volume of 1 L. The pH of the final solution will be 7.4. This buffer can be stored for up to 1 mo at 4°C. | |||
#Dilute to 0.05M: (take 50 mL of 1M solution and dilute in 1 Liter H2O) | |||
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Revision as of 13:31, 23 January 2013
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Calculation of reagents for ADA kinetic assay
.030 [math]\displaystyle{ \frac{mol}{L} }[/math] mol of adenosine × [math]\displaystyle{ \frac{267.24 g}{1 mol} }[/math] = [math]\displaystyle{ \frac{8.0172 g}{L} }[/math] × .010 L = 0.0802 g of adenosine
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