Question

A new gene is being investigated in fruit flies. The recessive allele of this gene (b) causes the wings to develop a blue color, while the dominant allele (b) permits wild-type colorless wings to develop. Preliminary studies indicate that this new gene is located on the X-chromosome. You decided to perform a two-point testcross to determine its position relative to the well-established garnet eyes gene (g). You cross a female heterozygous for both genes with a testcross male fly, and obtain the male offspring results shown in table 1, below. Using this information, answer the following questions: a) is the original female a coupling or repulsion heterozygote? (circle one) b) what is the map distance between genes b and g? CM c) based on these results, gene b must be located between what two genes on the map? (list both possible gene pairs) d) you perform another two-point testcross between gene b and gene v and obtained the results in table 2, below. Now, you can localize gene b to be specifically between which two genes? or Table 1: phenotype #male offspring wild-type 472 garnet eyes 27 blue wings 23 garnet eyes & blue wings 478 Table 2: phenotype wild-type vermillion eyes blue wings vermillion eyes & blue wings #male offspring 79 427 413 81

Answer 1

Ans: The answer is explained in the image enclosed below:

Ans-a- homologic on Nowe female breterozygotes • gence that Coupling refers to the case where dominant alleles are on the same chromosome and both recessive allcts are the other homologue Chromosome. Repulseen Refers to the case where each homologue chromosome has one domenient and one recessive allele from huo is test crossed with recessive male which means wild type male and recessive male by Realtout moule godype an ke 1 and xy out ws.dre gjester offspring clearly heterozygate is coupled because only in this case the phenotype distance (CM) = Number of recombinanty x 100 and here garnat eyes phenotypand Total number of progenici blue wrings Phenotype u rebombeant. Co map distance a which indicate that the can be obtained Map 50 5CM M.D. (CM ) = 217 23 972+4 78 +27 +23 Home the map distance between b and ŭ 5cm gå 11 not given question about the pouble genes x chromosome and their nap distame so can't answer this in the present on the question. The map destance between hou and be in cM is equal to M. CCM) = 79+81 x 100 = 160 x 1 16 cm 927 14 13 +79 781 1970 and the lenkage map So we ab between gene g and o localize can gene will be -16 CM 19 -5 cm b.