Recombinant Plasmid Case Study

From these results, it cannot be determined if the recombinant plasmid was the desired one or not because no fragments appeared on the gel that allowed for identification of the fragments involved in the recombinant. The only lane that contained a band was lane B which contained the BamHI and HindIII digest of the pAMP plasmid. Due to the location of the single band, its molecular weight falls between 4000 and 5000 base pairs. Since the pAMP plasmid is known to be 4539 bps, it is likely that this band occurred due to some of the pAMP plasmid not getting cut and producing a band that accounts for the full molecular weight of the plasmid. However, only some of the DNA would have remained uncut.
There should have also been two other bands …show more content…
The simplest and most ideal recombinant plasmid would contain only the 3755 bp fragment from pAMP and the 1875 bp fragment from pKAN. These two fragments alone would have been enough to confer resistance to both antibiotics. The presence of other bands would indicate that a different plasmid was constructed instead that consisted of a combination of the other extraneous fragments. A band around 700 would indicate that the 734 bp fragment from pAMP was also incorporated while a band around 2300 would indicate that the 2322 bp fragment from pKAN had been incorporated. The presence of all four bands in the lane would have pointed towards evidence of a super plasmid being formed that contained all of the possible fragments. It is also possible that some of the plasmids, when digested and ligated back together, could have completely reformed the original pAMP and pKAN plasmid instead of recombining. Antibiotic resistance could then be conferred if both a pAMP and a pKAN plasmid made their way into a single cell during the transformation of the E. coli and also would have appeared as four fragments on a