4. In Drosophila, three autosomal, recessive mutations are ebony body (e), sepia
ID: 199732 • Letter: 4
Question
4. In Drosophila, three autosomal, recessive mutations are ebony body (e), sepia eye color (s), and hairless bristles (h). A female fly heterozygous at all loci was testcrossed with an ebony, sepia, hairless male; the progeny are shown below. Using the gene symbols stated above, answer the following questions. Phenotype I Observed # of offspring Ebony sepia, hairless ebony, sepia Hairless ebony, hairless Sepia wild type 57 53 344 346 89 91 1000 Total offspring A. Give the genotypes of the flies in the testcross. Write the answer using the gene symbols stated above and using proper notation. B. Deduce the linkage map showing all distances among the three genes. C. Calculate the interference. Show all work.
Explanation / Answer
Question(A).
Answer.
Question (B).
Answer.
Where SCO = Single cross over
DCO = Double cross over.
As by looking at the parental and DCO it is clear that the gene order is "SEH" [ Note the middle alleles moves during crossong over]. Just look at the parental genotype and DCO and try to compare them , you will find that it is the allele "E" which has moved from it's location.
Distance betwwen S and E = No of SCO + No. of DCO / Total X 100%
= 57+ 53+ 9+11/1000 X 100
= 13 cM
Distance between E and H = 89+91+ 9+11/ 1000 X 100
= 20 cM
Therefore linkage map is
S----------------13cM---------------E--------------------------------20cM--------------------------------H
Question (C)
Interference = 1 - coefficient of coincidence
Coefficinet of Coincience(COC) = Observed double cross over / Expected double cross over
= 9+11 / 0.13 X .20 X 1000
= 20/26
= 0.76
Interference = 1- COC
= 1- 0.76
= 0.24.
Phenotype Genotype Number of offsprings Ebony eSH 57 Sepia, hairless Esh 53 ebony , sepia esH 344 Hairless ESh 346 Ebony , hairless eSh 11 sepia EsH 9 Wild type ESH 89 ebony, sepia , Hairless esh 91 Total 1000Related Questions
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