Cloning Blue and White Screening

Theoretical Background / Context

After the preparation of the E. coli JM101 to be competent cells, they are transformed with:

• The recombinant molecule which contains the DNA fragment of interest. (‘expt/comp’)

• Uncut pUC19, prepared as a control. (‘uncut pUC19/comp’)

• Sterile water, also prepared as a control. (‘no DNA/comp’)

• Then, the spread-plate technique is used to plate the culture of the transformed bacteria onto a medium containing Ampicillin, X-Gal, and IPTG (isopropyl-β-D-thiogalactopyranoside).

• E. coli JM101 plated cannot survive and grow in the presence of the antibiotic ampicillin.

• The pUC19 plasmid carries the ‘Ampr’ gene that codes for resistance to the antibiotic ampicillin.

• Therefore, any cell transformed with the pUC19 plasmid will be converted to the ampicillin-resistant phenotype (Ampr) and will be able to grow in the presence of ampicillin. Thus, Ampicillin is used for the selection of successful transformants.

• The control plate ‘only LB agar’ plated with ‘no DNA/comp’ shows white colonies ‘lawn’ since the JM101 can grow in the absence of ampicillin.

• The control plate ‘LB/Amp/X-Gal/IPTG’ plated with ‘no DNA/comp’ should show no growth of bacteria.

• Colonies appear in two colors, blue or white. Blue–white screening explained by the presence or absence of a functional lacZ gene encoding β-galactosidase.

• Blue colonies are the result of hydrolysis of X-Gal present in the medium by β-galactosidase.

• β-galactosidase is produced only in the presence of a functional lacZ gene. E. coli JM101 lacks a functional lacZ gene.

• The control plate ‘only LB agar’ plated with ‘no DNA/comp’ shows white colonies due to the absence of the lacZ gene.

• When pUC19 plasmid (not ligated to yeast DNA fragment) is inserted into the JM101, a functional lacZ gene is formed by α-complementation.

This cell produces blue colonies, this occurs in:

• The plate ‘LB/Amp/X-Gal/IPTG’ is plated with ‘uncut pUC19/comp’ (control plate).

• The plate ‘LB/Amp/X-Gal/IPTG’ is plated with ‘expt/comp’ sample but there was a failure of ligation of pUC19 to the recombinant molecule (experimental plate).

• When pUC19 containing the recombinant molecule is inserted into the JM101, a functional lacZ gene is not created. β-galactosidase will not be produced, and colonies will appear white.

• X-Gal/IPTG are thus used to select successful ligation of yeast DNA fragment of interest to the pUC19 plasmid through blue white selection in cloning. This occurs in the plate ‘LB/Amp/X-Gal/IPTG’ plated with ‘expt/comp’ sample.

• To sum up, the colonies formed by non-recombinant cells appear blue while the recombinant ones appear white. The calculation of the efficiency of transformation can be done using blue/white screening.

• Successful cloning shows transformation efficiencies for E. coli at least in the range of 106 transformants per µg of plasmid DNA.

Blue White Screening Principle of Work

•  In this lab following the blue white screening protocol, you will observe the plates to calculate the transformation efficiency of your transformation reaction.

• You will comment on each plate—both control and experimental—and fill in the table provided.

Then you will calculate the transformation efficiency from the plate ‘uncut pUC19/comp’ using the following equation:

• Transformation efficiency = number of colonies on plate / concentration of DNA plated (ng) * 10000 = -------- ng/µg.