Assume that a female fly that has disrupted wings ( dsr ) and a speck body ( sp

Question

Assume that a female fly that has disrupted wings (dsr) and a speck body (sp) is mated to a male that has cinnabar eyes (cn).

Phenotypically wild-type F1 female progeny were mated to males that had speck bodies, disrupted wings and cinnabar eyes, and the following progeny were observed.

Part C

What is the map distance between sp and dsr?

Phenotype Number of offspring wild-type 112 disrupted wings 52 speck body 22 cinnabar eyes 235 disrupted wings, speck body 241 disrupted wings, cinnabar eyes 25 speck body, cinnabar eyes 46 disrupted wings, speck body, withered eyes 117

Explanation / Answer

Map distance is calculated as (# Recombinants)/(Total offspring) X 100.

disrupted wings and cinnabar eye recombinants are 25.

Total no. of recombinants = (241 + 25+ 46 + 117) = 429

map distance = (25 / 429) * 100 = 5.82 LMU

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