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

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(PCR mutation)
(PCR mutation)
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* 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.
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V<sub>1</sub> = <math>\frac{100 ng/ \mu L  *  1000 \mu L}{0.46E3 ng/ \mu L}</math>
+
V<sub>1</sub> = <math>\frac{100 ng/ \mu L  *  1000 \mu L}{0.46E3 ng/ \mu L}</math> = 217.39 μL
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Revision as of 01:51, 26 October 2012

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PCR mutation

  • 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.

V1 = \frac{100 ng/ \mu L  *  1000 \mu L}{0.46E3 ng/ \mu L} = 217.39 μL


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