# User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/02

(Difference between revisions)
 Revision as of 22:34, 6 October 2012 (view source) (→Preparation of the Reactants of HRP Oxidase Assay)← Previous diff Revision as of 22:37, 6 October 2012 (view source) (→Preparation of the Reactants of HRP Oxidase Assay)Next diff → Line 9: Line 9: * The reactants listed in [http://openwetware.org/wiki/Image:3760-Protocol_HRP_AbsorbanceAssay1.pdf HRP Oxidase Assay] were weighed out. * The reactants listed in [http://openwetware.org/wiki/Image:3760-Protocol_HRP_AbsorbanceAssay1.pdf HRP Oxidase Assay] were weighed out. * The phosphate buffer was made with white granules of sodium phosphate dibasic (FW 268.07, heptahydrate) instead of potassium phosphate. The substitution is not detrimental to the procedure since the ion of interest is the phosphate group. The total volume chosen for the buffer is 60 mL. * The phosphate buffer was made with white granules of sodium phosphate dibasic (FW 268.07, heptahydrate) instead of potassium phosphate. The substitution is not detrimental to the procedure since the ion of interest is the phosphate group. The total volume chosen for the buffer is 60 mL. - * Calculations are listed below: + * Calculation for obtaining Na2HPO4 0.020 mol of Na2HPO4 × $\frac{268.07g}{1mol}$ of Na2HPO4 = 53.61 g of Na2HPO4 0.020 mol of Na2HPO4 × $\frac{268.07g}{1mol}$ of Na2HPO4 = 53.61 g of Na2HPO4 Line 15: Line 15: x60 mL = $\frac{.060 L * 53. 61 g}{1L}$ = 3.22 g of Na2HPO4 x60 mL = $\frac{.060 L * 53. 61 g}{1L}$ = 3.22 g of Na2HPO4 + + +

## Revision as of 22:37, 6 October 2012

Project name Main project page
Previous entry      Next entry

## Preparation of the Reactants of HRP Oxidase Assay

• The reactants listed in HRP Oxidase Assay were weighed out.
• The phosphate buffer was made with white granules of sodium phosphate dibasic (FW 268.07, heptahydrate) instead of potassium phosphate. The substitution is not detrimental to the procedure since the ion of interest is the phosphate group. The total volume chosen for the buffer is 60 mL.
• Calculation for obtaining Na2HPO4

0.020 mol of Na2HPO4 × $\frac{268.07g}{1mol}$ of Na2HPO4 = 53.61 g of Na2HPO4

x60 mL = $\frac{.060 L * 53. 61 g}{1L}$ = 3.22 g of Na2HPO4