The smallest recombinant plasmid in this experiment contains the lambda/HindIII

Question

The smallest recombinant plasmid in this experiment contains the lambda/HindIII 0.13 kb insert. The size of this recombinant is 2.82 kb, which is only slightly larger than the pUC18 vector (2.69 kb). Say you were given purified samples of both plasmids, without being told which was which, and asked to come up with a way to identify each plasmid. How would you accomplish this task? (Note: There are at least three ways of doing this, but only one answer is required).

Hint given: Two ways involving gels, the first a straight gel run of the plasmids but consider the composition of the agarose gel. Second way involves cutting the plasmids before using the same type of gel as for method one.

Explanation / Answer

The recombinant DNA contains 0.13 kb sized lambda/HindIII insert in the pUC18 vector (2.69 kb). As a result the size of the recombinant DNA adds up to 2.82 kb. If purified samples of both plasmids are given, they can be identified by cutting the plasmids with HindIII restriction enzyme. pUC18 vector contains a single site for HindIII, however, the recombinant DNA will have two sites, due to the Lambda DNA insert. Hence, in agarose gel electrophoresis, pUC18 will show single band of linearized DNA, while the recombinant DNA will show 2 bands, a small band of 0.13 kb and bigger band corresponding to linearized pUC18 DNA. Based on the gel pattern, the plasmid DNA in each sample can be then identified.

Agarose gel electrophoresis is based on the migration of DNA molecules under electric current. Small sized DNA migrates faster than larger sized DNA. Further configuration of the DNA molecule affects migration, i.e. circular DNA migrates faster than linear DNA. For detection of the given plasmids, 1.5 to 2.0% agarose would be helpful in detecting DNA molecule as small as 130 bp upto 2690 bp.

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