Question

A yeast transformation was performed using AH109 cells resuspended in lithium acetate solution. To prepare the yeast cells, 5 ml of AH109 overnight culture was initially pelleted. The pelleted cells were resuspended in 800 microliters of lithium acetate solution. He then used 100 microliters of these yeast cells in his transformation reaction. If the overnight culture had 6.0x10^3 CFU/mL, how many yeast cells are in that transformation tube?

After incubating the cells in PEG and 0.2 micrograms of plasmid DNA, he used a centrifuge to college the yeast and resuspend them in 100 microliters sterile water. All 100 microliters of the yeast were plated on selection media. After incubating the plate for 48 hours at 30 degrees celcius, he counted 280 colonies. What is the transformation efficiency of his competent cells, in cells/microgram?

please help!

Answer 1

A - Given the number of CFU/ml in overnight culture 6.0x10^3 CFU/mL. Total volume = 5 ml.

So, total number of cell in 5ml = 5 * 6.0x10^3 = 30*10^3 cells.

These cells were resuspended in 800ul of lithium acetate solution.

So CFU/ml in 800ul lithium acetate solution = 30*10^3 /

The cell in 100ul transformation reaction = 30*10^3 *100/800 = 3.75 *10^3 cell.

So total cells in the transformation tube are - 3.75 *10^3 cell.

B - Transformation efficiency = Number of colonies/amount of DNA in ug * Total volume of transformation mix/volume plated on the plate.

The number of colonies = 280.

Amount of DNA = 0.2ug

The total volume of transformation mix = 100ul

volume plated on the plate = 100ul

Transformation efficiency = 280/0.2 = 1400 colonies / ug of DNA or 1400 cells/ug of DNA

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