Hi, 1) from the mixture of lambda HindIII fragments, say you wished to colon onl

Question

Hi,

1) from the mixture of lambda HindIII fragments, say you wished to colon only the second largest (9.4kb) into pUC18. if you proceed as in the protocol experiment you may need to screen hundreds of recombinants in order to find the required clone. can you advise an easier solution to this problem? ( clue:  Need to find a way of isolating just the 9.4 kb lambda fragment from the others, then cloning the 9.4 kb insert into a vector.)

2) if lambda EcoRI/BamHI inserts had been used blue colony 132 and white colony 0, what differences would be there be between recombination plasmids from the blue 1,351 and white 9,236 and blue 132 and white 0. (clue: Firstly look at the respective restriction sites. Secondly look at the way(s) that the insert will clone into the vector in the blue 1,351 and white 9,236 and blue 132 and white 0.)

Explanation / Answer

1. I think, you should run those Hind III fragments in agarose gel electrophoresis and pick up that 9.4 kb DNA band selectively. Then, purify this 9.4 kb DNA fragment from the gel using DNA isolation/purification kit and insert into PUC18 using appropriate restriction enzyme through ligation process. Then, you will get selectively 9.4 kb DNA insert containing plasmid clones.

2. Here, EcoRI/BamHI blue colonies -132, but total are 1351. So there is a chance of finding blue colonies when we use EcoRI/XbaI, EcoRI/HincII, EcoRI/SalI, BamH1/XbaI, BamHl/Sall, BamHl/Hincll, XbaI/HincII, XbaI/Sall, HincII/SalI, In which, 1219 blue colonies obtained by using other enzymes that are recombinants. On average they would produce 135 colonies from each pair of restriction enzyme involving recombination or insertion. So EcoRI/BamHI has to produce 135 but it has short of 3 colonies. Coming to white colonies (9236), they are non-recombinant naive lambda plasmids.

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