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Phage


dttom

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This is actually I encounter when I am doing an exercise. Now there are 2 phage mutant but we don't know if they occur at the same locus. The usual practice is to conduct a complementation test (to cross them in a restrictive host and see if lysis occurs). But, because it is provided that this phage is T even and replicate with headful packaging (genomes concate and every capsid accomodates 1 genome plus some more bits), my question is why can't we use a permissive host to do the job. Let say if the two mutations occurs at the same locus, in a permissive host no matter how they perform recombination they cannot produce wildtype phage, so that when the progeny is spread on a restrictive lawn, there is no plaque expected. The result should be different if they occur at different loci that, recombination or headful packaging might result in a wildtype genome which then gives rise to plaque.

For different loci mutation, wildtype could be generated, in my own idea, by:

recombination:

A--B' x A'--B --> A--B (WT) + A'--B'

headful packaging:

A--B' + A'--B --> A'--B--A--B' (concatemer)--> A'-|-B--A--|B' (1 unit + bits, of WT)

 

But my instructor says my method cannot tell between 'two loci mutation' and 'single locus mutation', the exact phrasing is:

"No matter if the mutations in the tested mutant phage strains are in the same gene or in different genes, the mutant phage strains can co-infect the (permissive) host cells, and their genomes can recombine to generate the wild-type phage progeny in such an assay."

which I don't understand, can anyone help? Any would be helpful, thanks.

 

[solved, thanks]

Edited by dttom
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