Question
2. Assuming that a normal (untransformed) E. coli is not resistant to any antibiotic, what will happen if a ligation product that does not have any insert in it enters the bacteria through heat shock transformation?
A. The E. coli will have AmpR
B. The E. coli will have TetR
C. The E. coli will have AmpR and TetR
D. The E. coli will die
Read this for questions 1 through 6: In a gene cloning project, the goal is to produce a recombinant plasmid DNA that harbors the red segment of the genomic DNA specified in the following figure. The genomic DNA is 1600 bp, with multiple BamHI and EcoRI restriction sites. Digesting this DNA with BamHI enzyme results in multiple cut fragments of 300 bp, 750 bp, 75 bp, and other fragments of unknown sizes at both tails. The plasmid DNA, on the other hand, is 2600 bp long and contains only one BamHI restriction site on the Tetracyclin Resistant gene (TetRO, such that insertion of any forign DNA in this site results in "insertional inactivation" of TetR. In a test tube we digest the genomic DNA and plasmid DNA with restriction enzyme BamHI and then use the enzyme ligase with the hope to bring the red segment (750 bp) into the BamHI cut site of the plasmid DNA. Then, frozen E. coli will be heat shocked in the presence of this plasmid such that the bacteria is transformed with plasmid DNA which we hope contains the red-insert segment. We then plate the bacteria solution on to three types of plates A) without any antibiotic, B) with both ampicillin and tetracyclin antibiotic, and C with ampicillin only. 750 bp 300 b 75 bp 420 bp 730 bp 450 bp Bam HI EcoRI 2600 bp
Explanation / Answer
The answer is: C. The E. coli will have AmpR and TetR.
This is because the plasmid will have intact tetR and ampR genes as there is no insert. So it will render the E. coli resistance to both ampicillin and tetracycline after transformation.
