3. You have co-infected E. coli with two strains of T4 phage. One is.pdf

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3. You have co-infected E. coli with two strains of T4 phage. One is minute (m), rapid lysis (r) and turbid (tu). The other is wildtype for these genes. The products of this infection are plated and classified for each phenotype, with the following results: m r tu 3467 + + + 3729 m r + 853 m + tu 162 m + + 520 + r tu 474 + r + 172 + + tu 965 a. Determine recombination frequencies for each pair of genes. b. What gene order best explains these results? c. What are the distances between the genes in terms of recombination frequencies? Solution (1) First, try to know the parental genotype Parent 1: + + + Parent 2: m r tu Now to know the distance between genes, we need to know the recombinant progeny Distance between m and r Recombinant progeny Number m + tu 162 m + + 520 + r tu 474 + r + 172 Total 1328 The map distance between m and r is Recombination frequency= (Number of recombinants/Total progeny)x100=1328/10,342x100 = 12.8 m.u. Recombinant progeny Number m r + 853 m + tu 162 + r + 172 + + tu 965 Total 2152 Using the same approach, the r–tu distance is Recombination frequency= (Number of recombinants/Total progeny)x100 =(2152)/10,342x100 = 20.8 m.u Recombinant progeny Number m r + 853 m + + 520 + r tu 474 + + tu 965 total 2812 Using the same approach, the m -tu distance is Recombination frequency= (Number of recombinants/Total progeny)x100 =(2812)/10,342x100 = 27.2 m.u. b. As the genes m and tu are the farthest apart, you can easily predict the gene order m r tu c. m--------12.8 m.u---------r---------------------20.8 m.u--------------------------tu Recombinant progeny Number m + tu 162 m + + 520 + r tu 474 + r + 172 Total 1328.

Education

172
+ + tu
965
Total
2152
Using the same approach, the r–tu distance is Recombination frequency= (Number of
recombinants/Total progeny)x100 =(2152)/10,342x100 = 20.8 m.u
Recombinant progeny
Number
m r +
853
m + +
520
+ r tu
474
+ + tu
965
total
2812
Using the same approach, the m -tu distance is Recombination frequency= (Number of
recombinants/Total progeny)x100 =(2812)/10,342x100 = 27.2 m.u.
b. As the genes m and tu are the farthest apart, you can easily predict the gene order m r tu
c. m--------12.8 m.u---------r---------------------20.8 m.u--------------------------tu
Recombinant progeny
Number
m + tu
162
m + +
520
+ r tu
474
+ r +
172
Total
1328

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