Transformation of lux and pUC18 into E.coli Danielle Hewitt 4013851 Lab Partners: Jessica Palacio Patrick Foucauld April 8th, 2015 Abstract: In the experiment, two different plasmids were used, plasmid lux, and control plasmid, pUC18. The goal for this experiment was to introduce plasmid lux and pUC18 into the bacterium, Escherichia coli, by transformation. (Albert, Pitzer, and Calero et al., 2012). In order for the plasmids to be transformed into E.coli, the E.coli cell has to become a competent cell. By becoming a competent cell, E.coli has the ability to uptake the plasmid DNA, resulting in transformation. The E.coli was then added to both plasmids in two different test tubes, along with two other test tubes with …show more content…
Fraction of DNA spread=(Volume(μL)spread on LB/〖Amp〗^C plate)/(Total sample volume (μL)in control DNA tube) =130μL/408μL =0.319μL Total amount of DNA present on plate: Total amount(μg) of DNA=μg of DNA× Fraction of DNA spread =0.009μg ×0.319 =0.002871 μg Transformation Efficiency: Transformation Efficiency=(Total number of colonies on the LB/〖Amp〗^C plate)/(Total amount of DNA spread on LB/〖Amp〗^C plate) =215/0.002871 =74886.799 Calculating the Transformation Efficiency: LB/Amp lux Total amount of the plasmid DNA: μg DNA=concentration (μg/μL)of DNA × volume of DNA (μL) =0.003 μg/μL× 3μL =0.009 μg Total volume of cell suspension: Total volume(μL)= amount(μL)of plasmid+amount(μL)of LB+amount(μL)of cell suspension =3μL+275μL+130μL =408μL Fraction of DNA: Fraction of DNA spread=(Volume(μL)spread on LB/〖Amp〗^lux plate)/(Total sample volume (μL)in control DNA tube) =130μL/408μL =0.319μL Total amount of DNA present on plate: Total amount(μg) of DNA=μg of DNA× Fraction of DNA spread =0.009μg ×0.319 =0.002871 μg Transformation