Female Drosophila heterozygous for three recessive mutations e (ebony body), st

Question

Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained: Phenotype Number wild-type 67 ebony 8 ebony, scarlet 68 ebony, spineless 347 ebony, scarlet, spineless 78 scarlet 368 scarlet, spineless 10 spineless 54 (a) What indicates that the genes are linked? (b) What was the genotype of the original heterozygous females? (c) What is the order of the genes? (d) What is the map distance between e and st? (e) Between e and ss? (f) What is the coefficient of coincidence? (g) Diagram the crosses in this experiment.

Explanation / Answer

(a) Two of the classes (the parental types) vastly outnumber the other six classes (recombinant types);

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(b) st + +/+ ss e;

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(c) st—ss—e;

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(d) [(145 + 122) x 1 + (18) x 2]/1000 = 30.3 cM;

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(e) (122 + 18)/1000 = 14.0 cM;

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(f) (0.018)/ (0.163 x 0.140) = 0.789.

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(g) st + +/+ ss e females –-------x ---- st ss e/st ss e males

2 parental classes and 6 recombinant classes.

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