Difference between revisions of "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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* 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.  
 
* 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. This was transferred to a new tube and filled up with water to a total volume of 1 mL.
 
* 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 of the dissolved primer in water
 
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

Revision as of 22:55, 25 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


[math]\frac{0.46E6 ng}{x \mu L}[/math] = [math]\frac{100 ng}{1 \mu L}[/math] 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 = [math]\frac{100 ng/ \mu L * 1000 \mu L}{0.46E3 ng/ \mu L}[/math] = 217.39 μL of the dissolved primer in water