User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/16: Difference between revisions
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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. | ||
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 22:51, 25 October 2012
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PCR mutation
0.46 mg = 0.46E6 ng
V1 = [math]\displaystyle{ \frac{100 ng/ \mu L * 1000 \mu L}{0.46E3 ng/ \mu L} }[/math] = 217.39 μL |