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Title: Gene mapping & recombination frequencies Post by: aoctaviani on May 5, 2015 A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,785 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.
(https://biology-forums.com/gallery/47/144193_05_05_15_7_37_26.png) (https://biology-forums.com/index.php?action=gallery;sa=view&id=21664) a) Determine the gene order on the homologous X chromosomes in the trihybrid females b)Calculate the recombination frequencies between each of the gene pairs. c)Compare the recombination frequencies and speculate about the source of any discrepancies in the recombination data. d)Use chi-square analysis to demonstrate that the data in this experiment are not the result of independent assortment. Title: Re: Gene mapping & recombination frequencies Post by: fionamk on May 12, 2015 Hi,
Theses are my answers for a and b.......... I'm no expert so hopefully someone else can confirm :-) Also I'm assuming there is a typo in the question and the progeny = 20765 (not 20785) . The parental types are most frequent : Scute crossveinless (8808) and Ech (8576) DCO least frequent - scute cross and Ech (4) and wildtype (1) total DCO = 5 a) Order would then be Scute Ech Cross (compare one DCO with one parental to get order) b) recomb freq = (single crossovers in region + DCO ) / total progeny so region 1 scute-ech recomb freq = (scute ech + cross + DCO )/20765 = (681 + 716 + 5) / 20765 = 0.0675 and region 2 ech-cross = 0.0955 HTH |