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Practice 1. Cross a female fly with the eyeless mutation for eye shape but normal bristle to a male fly with wild type eye sh3. Draw a map that shows the map distance in centiMorgans between the locus for the shaven bristle allele and the locus for t

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Answer #1

1. Eye less and normal bristles, Wild eyes and shaven bristles are the parental phenotypes.

Explanation : because in the test cross the parental phenotypes are generally more expressed. in this case these phenotypes are more in number.That is 473 and 447.

2. Wild eyes and normal bristles.Eyeless shaven bristles are the recombinant phenotypes.

Because the recombinant phenotypes are underexpressed and less in number.

3. Map distance can be calculated by the following formula :( where 1% recombination frequency = 1 map unit)

recombinants/total no.of phenotypes x 100

the distance between shaven bristles (s and e) eyeless  = 5+7/473+447+5+7 x 100 =1.28 m.u.

s----------------e

1.28m.u.

map distance is mainly used to know the order of genes.

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