21. In phage T4, 11 recessive mutations were mapped by pairwise recom
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HBO Small Business en Ondernemerschap
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HBO Small Business En Ondernemerschap
21. In phage T4, 11 recessive mutations were mapped by pairwise recombination tests to 0 .2 4
.6 .8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 a single small region of the phage chromosome, which contains 4
genes, as shown in the map of that region. All required an amber mutant host to grow, failing to
grow in...
21. In phage T4, 11 recessive mutations were mapped by pairwise recombination tests to 0 .2 4
.6 .8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 a single small region of the phage chromosome, which contains 4
genes, as shown in the map of that region. All required an amber mutant host to grow, failing to
grow in a WT host. Double infections of a wild type host by pairs of these mutants were carried
out to test for complementation. Predict for each 51 6 34 7 118 10 map units e to lyse the WT
host (-) 6 8 9 10 11 new 8 10 new A new amber-dependent mutant was found that mapped very
far away, outside the region in our map (new). In the last column, predict the complementation
pattern for new when crossed with the eleven known mutants individually.
Solution
Table above showing possable outcomes of coinfection of each pair of mutant phage. +
indiccates complimentation which gives wild phenotype that able to lysis the host. _ indicates the
phenotype not able to lysis the host.
Here the coinfection of pair of mutatants able to show wild phenotype(+) when they are present
in the diffrerent gene, and the complementation between their wild phenotypes gives compleate
chromosome with out deletion. so resulted wild type phase able to lysis the WT host.
The coinfection of pair of mutatants not able to show host lyisis(_) when they are present in the
same gene, so one mutant will not complement the other as both mutants have loss of the same
region in the chromosome.
1234567891011new_+++__++++++1_++++++_+++2__++++++++3_++++++++4__++++++5_
++++++6__+__+7_+__+8_+++9__+10_+11+new
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