User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/16

(Difference between revisions)

Revision as of 01:54, 26 October 2012

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In reference to the PCR Mutation protocol, 100 ng/μL of the primer was needed for the reaction. The weight of the primer in the provided container was 0.46 mg. A ratio of the weight over volume was equated to the required concentration of the primer:

0.46 mg = 0.46E6 ng

$\frac{0.46E6 ng}{x \mu L}$ = $\frac{100 ng}{1 \mu L}$ of primer in water = 4600 μL of water

There is limited space in the plastic container (1 mL). Instead of dissolving 0.46E6 ng of the primer in 4600 μL of water, the entire primer was dissolved in 1 mL water.
Using M₁V₁ = M₂V₂, the volume taken from the solution of 0.46E6 ng in 1 mL of water was calculated to be 217.39 μL. This was transferred to a new tube and filled up with water to a total volume of 1 mL.

V₁ = $\frac{100 ng/ \mu L * 1000 \mu L}{0.46E3 ng/ \mu L}$ = 217.39 μL of the dissolved primer in water

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 * There is limited space in the plastic container (1 mL). Instead of dissolving 0.46E6 ng of the primer in 4600 μL of water, the entire primer was dissolved in 1 mL water.
-* Using M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub>, the volume taken from the solution of 0.46E6 ng in 1 mL of water was calculated to be 217.39 μL.
+* Using M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub>, the volume taken from the solution of 0.46E6 ng in 1 mL of water was calculated to be 217.39 μL. This was transferred to a new tube and filled up with water to a total volume of 1 mL.
-V<sub>1</sub> = <math>\frac{100 ng/ \mu L  *  1000 \mu L}{0.46E3 ng/ \mu L}</math> = 217.39 μL
+V<sub>1</sub> = <math>\frac{100 ng/ \mu L  *  1000 \mu L}{0.46E3 ng/ \mu L}</math> = 217.39 μL of the dissolved primer in water
+*
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