A kanamycin-resistant, ampicillin-resistance plasmid (3 kb) is cleaved with the

Question

A kanamycin-resistant, ampicillin-resistance plasmid (3 kb) is cleaved with the PstI enzyme, which cuts within the ampicllin resistance gene. The digested plasmid is ligated to a 5-kb PstI DNA fragment from Drosophila (fruit fly). The DNA mixture was then transformed into E.coli host cells.

(i). Which antibiotic would you put into the medium (nutrient agar plate) to select for cells that contain the plasmid?

(ii). Explain step by step, how you would select for cells containing the recombinant Drosophila DNA.

(iii) If different colonies from the plate in part (ii) are cultured in nutrient broth and DNA was isolated from each of the bacterial cultures, what restriction patterns would you expect to see following PstI digestion and agarose gel electrophoresis?

Explanation / Answer

Please find the answers below:

Part 1: According to the information, the ampicillin resistance gene has been cleaved by the PstI enzyme and the same digest has been ligated to the 5 kb PstI DNA fragment. Thus, the DNA mixture so produced will carry recombinant plasmid which LACKS an ampicillin resistant gene. Thus, the bacteria transformed from this plasmid will be sensitive to ampicillin and the nutrient agar plate must be incorporated with ampicillin to screen out for recombinant colonies.

Part 2: The selection steps can be found as below:

Part 3: If different colonies from the culture plates are isolated and their DNA was subjected to restriction digestion, two different patterns of nucleic acid digest would appear on the electrophoresis gel. The non-transformed bacterial DNA will be sensitive to ampicillin and thus restriction digestion. On the other hand, the transformed colonies will contain PstI DNA inserted between the ampicillin gene in the host genome. Thus, two different band patterns of 3 kb and 5 kb will appear as ampicillin gene plasmid and the other without it, respectively.

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