Microb Test ಠ_ಠ

Microbiology Mid-sem test sample:
You have cloned into E. coli, a gene that may encode resistance to the antibiotic chloramphenicol. This conclusion is based on the fact that the DNA sequence of this gene has similarity to that of other choramphenicol resistance genes. Explain how you could use gene disruption analysis to determine whether the gene does indeed confer resistance to chloramphenicol in the strain of bacteria from which you cloned it.

sample answer:

(a) Transposon method: Assume the original host is resistant to chloramphenicol. A suicide plasmid carrying a transposon would be transferred cells of the host bacterium by transformation or conjugation from an E. coli culture. Let us assume that the transposon also carried a tetracycline resistance gene. The transformed celles (or the tranconjugants) would then be plated on growth medium containing tetracycline. For the tetracycline gene to survive, the transposon must hop out of the suicide plasmid into othe host DNA. Since this random event will occur in many millions of cells, some cells are likely to contain transposon insertions in the putative Cm gene. We can select for cells containing insertions in the Cm gene by plating all tetracycline resistant cultures on duplicate plates containing both tetracycline plus chloramphenicol. Cultures that do not grow in the presence of Chloramphenicol must contain insertions in DNA involved with Cm resistance. To confirm that the transposon had inserted into the Cm gene, we could use PCR technology to sequence DNA flanking the transposon to determine whether it is identical to that obtained previously (see question).

(b) Single crossover (or cointegrate method). Assume the original host is resistant to chloramphenicol. A suicide plasmid containing part of the cloned gene could be introduced into the original host by transformation or conjugation. Assume the plasmid also has a tetracycline resistance gene for selection purposes. When the transconjugants or transformants are grown on medium containing tetracycline, the only way the plasmid DNA can survive is by integrating into the host DNA. This will be facilitated by recombination between the part copy of the Cm gene and the wild type copy in the host. This event will result in integration of the entire plasmid into the host DNA, thereby creating an insertional mutation. Tetracycline resistant cultures would then be plated on medium containing tetracycline + chloramphenicol. Lack of growth would confirm that the Cm gene had been insertionally inactivated. This could be confirmed by PCR amplification of DNA encompassing the point of insertion of the plasmid DNA followed by sequencing of the host DNA flanking the insertion. If the sequence data was identical to that obtained previously (see question) then you could say that the gene encoded Chloramphenicol resistance.

(c) Allelic replacement. Assume the original host is resistant to chloramphenicol. The cloned copy of the gene would first be insertionally inactivated in E. coli by cloning a copy of another gene eg tetracycline resistance gene into the Cm gene. The construct would then be cloned into a suicide plasmid (carrying an ampicillin resistance gene for selection of the plasmid) and transferred to the host cells by conjugation or transformation. Transformants or transconjugants would be selected by plating on tetracycline medium. Tetracycline resistant colonies would then be plated separately on growth media containing tetracycline + chloramphenicol and Tetracycline + ampicillin. If a double crossover recombination event had resulted in exchange of the mutated copy of the Cm gene into the host, then no growth should be observed. This would be evidence that the Cm gene encoded chloramphenicol resistance. This could be confirmed by sequencing across the junction between the tetracycline gene and flanking DNA. Any cultures that were able to grow on Tet + ampicillin would represent cointegrates formed by single crossover recombination events. These should also be chloramphencicol sensitive.

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I was liek....omgbbq this is what I'm supposed to answer in the test??? I can hardly remember reading those things in the lecture notes, probably because the lecturer is so boring, or I actually miss the lectures in week 5 when I was infected by the virus...
revision, revision...uh 20% of final marks for Microbiology, seems like I need to cram like Konata :3