User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/02: Difference between revisions

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* 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 Na<sub>2</sub>HPO<sub>4</sub>


0.020 mol of Na<sub>2</sub>HPO<sub>4</sub> × <math>\frac{268.07g}{1mol}</math> of Na<sub>2</sub>HPO<sub>4</sub> = 53.61 g of Na<sub>2</sub>HPO<sub>4</sub>
0.020 mol of Na<sub>2</sub>HPO<sub>4</sub> × <math>\frac{268.07g}{1mol}</math> of Na<sub>2</sub>HPO<sub>4</sub> = 53.61 g of Na<sub>2</sub>HPO<sub>4</sub>
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x<sub>60 mL</sub> = <math>\frac{.060 L * 53. 61 g}{1L}</math> = 3.22 g of Na<sub>2</sub>HPO<sub>4</sub>
x<sub>60 mL</sub> = <math>\frac{.060 L * 53. 61 g}{1L}</math> = 3.22 g of Na<sub>2</sub>HPO<sub>4</sub>


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Revision as of 19:37, 6 October 2012

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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 × [math]\displaystyle{ \frac{268.07g}{1mol} }[/math] of Na2HPO4 = 53.61 g of Na2HPO4


x60 mL = [math]\displaystyle{ \frac{.060 L * 53. 61 g}{1L} }[/math] = 3.22 g of Na2HPO4




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