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 an 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

chromosome mapping is done to know how the linked genes on a chromosomes alleles combine.
here in the above example the linked genes where the crossing over occured are between the genes disrupted wings [dsr], speck body[sp}, cinnabar eyes {cn].

MAPPING IS DONE TO KNOW THE DISTANCE BETWEEN THE ALLELES sp and dsr;

BY obeserving the progeny obtained when F1 female progeny were matted to the males having the sp body ,dsr and cn the progeny obtained are having different rations ,so the genes are independentof each other.

the total number of offspring obtained are ; 850 progeny were observed

the total number of offspring having the speck body and dsr wings are - 241.   in order to caluculate the map distanc between the two genes we need to consider the progeny having any one of the alleles ie., speck body or dsr.

so we need to caluculate the distance by taking the progeny having the sp and dsr genes by total number of progeny

112+ 52+22+241+25+46+117 = 615 progeny having dsr and sp genes

so ,615/895 = 0.72 morgans .

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