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Title: The genes for miniature wings Post by: sam01 on Feb 12, 2015 The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in Drosophila.
Phenotypically wild-type females (m + g / mg +) were mated to miniature-winged males with garnet eyes. Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event? A.wild type B.miniature wings, garnet eyes C.miniature wings D.garnet eyes If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes? __wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes Enter your answer as the number of flies of each phenotype separated by a colon (example: 100:300:100:300). Title: Re: The genes for miniature wings Post by: raneensayegh on Feb 17, 2015 part 1:
Part A Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event? Select all that apply. these two: miniature wings, garnet eyes wild type part 2: Part B If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes? __wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes Enter your answer as the number of flies of each phenotype separated by a colon (example: 100:300:100:300). 32:368:368:32 Correct If the genes for miniature wings (m) and garnet eyes (g) are linked with 8 map units between them, then the two recombinant classes must add up to 8% of the total (4% each) and the two parental classes must add up to 92% of the total (46% each). As a result, the following distribution would be expected: wild type: 4% x 800 = 32 miniature wings: 46% x 800 = 368 garnet eyes: 46% x 800 = 368 miniature wings, garnet eyes: 4% x 800 = 368 |