Question

You add (1.04x10^0) ng of plasmid DNA in 50 µL to 100 µL of competent E. coli cells and incubate the mixture on ice for 20 min. Following a heat-shock treatment, the cells are incubated at 37 C for 30 min. At the end of the incubation you spread (1.2000x10^2) µL of the transformation mix on an agar plate and incubate the plate overnight at 37 C. The next morning you observe (6.8400x10^2) colonies on the plate, each colony having originated from a single, transformed E. coli cell in the plated mixture. What is the transformation efficiency, expressed as transformants per µg of plasmid DNA? Express your answer in scientific notation rounded to exactly three significant figures.

Answer 1

Before starting lets simplify the given figures.

(1.04 x 10^0) = 1.04 ng of plasmid DNA, since 10^0 is 1.

(1.2000x10^2) µL of the transformation mix= 120 ul ,since 10^2 is 100.

(6.8400x10^2) colonies = 684 colonies.

Now we can convert 1.04 ng to ul for calculation. 1.04 ng of plasmid DNA will be equal to 10.4 ul of plDNA ( 1ul = 0.1ng/ul). You are adding 50ul of E.coli cells to 100ul of broth. The total volume will be 10.4 + 50 + 100 = 160.4 ul.

The 160.4 ul contains 1.04ng of plDNA , so the concentration will be 1.04/160.4 = 0.00648 ng/ul.

Now if you are spreading 120ul of 0.00648 ng/ul solution on the plate, you're spreading 120 X 0.0065 = 0.78 ng of plDNA.

converting this to ug will give. 0.78/ 1000 = 0.00078ug.

Now if 120ul gave you 684 colonies, you will get 684 cfu/ 0.00078ug = 876923.076 tranformants/ ug of plDNA.

Hope this is clear and helps.