Assume that a female fly that has disrupted wings (dsr) and a speck body (sp) is mated to a male that has cinnabar eyes (cn). Phenotypically wild-type F1 female progeny were mated to males that had speck bodies, disrupted wings and cinnabar eyes, and the following progeny were observed. The double crossover classes will be those that appear least frequently: speck body, sp dsr^+ cn^+ (22) and disrupted body, cinnabar eyes, sp^+ darn (25). Assume that a female fly that has disrupted wings (dsr) and a speck body (sp) is mated to a male that has cinnabar eyes (cn). Phenotypically wild- type F_1 female progeny was mated to males that had speck bodies, disrupted wings and cinnabar eyes, and the following progeny were observed. Next, identify the gene that switches places between the groups. Because the dsr gene switches places between the no crossover and double crossover groups, it is in the middle. With respect to the three genes mentioned in the problem, what are the genotypes of the parents used in making the phenotypically wild-type F_1 heterozygote? What is the map distance between sp and dsr? Solution The gene at the middle is for spec body, and the is linkage between the genes for disrupted wings and speck body showing higher number of offsprings with cinnabar eys (235) and disrupted wings, speck body (241). If dsr gene will be at middle , then there will be double crossing over between sp and cn that can lead to less individuals showing speck body, cinnabar eyes or disrupted wings, which actually is not the case here. therefore the middle gene is sp and there is double cross over between dsr and cn that leads to reduction in the individuals with disrupted body, cinnabar eyes (25) or spec body (22). Thus gene sequence will be dsr-sp-cn or cn-sp-dsr. Part B in order to obtain the phenotypically wild type F1 heterozygote, the genotypes of the parents from the gives genotypes will be 2nd and 3rd. sp dsr cn / sp dsr cn and sp* dsr* cn* / sp* dsr* cn* sp dsr cn* / sp dsr cn* and sp* dsr* cn / sp* dsr* cn Part C Map distance between sp and dsr single cross over between sp and dsr= 46+52 = 48 doble crossover 22+25 = 47 Total recombinants = 98+ 47 = 145 Total offsprings are 850 Percentage of recombination is 145/850 X100 = 17.0588 Therefore the map distance is 17 cM..

1. Assume that a female fly that has disrupted wings (dsr) and a speck body (sp) is mated to a male
that has cinnabar eyes (cn).
Phenotypically wild-type F1 female progeny were mated to males that had speck bodies,
disrupted wings and cinnabar eyes, and the following progeny were observed. The double
crossover classes will be those that appear least frequently: speck body, sp dsr^+ cn^+ (22) and
disrupted body, cinnabar eyes, sp^+ darn (25). Assume that a female fly that has disrupted wings
(dsr) and a speck body (sp) is mated to a male that has cinnabar eyes (cn). Phenotypically wild-
type F_1 female progeny was mated to males that had speck bodies, disrupted wings and
cinnabar eyes, and the following progeny were observed. Next, identify the gene that switches
places between the groups. Because the dsr gene switches places between the no crossover and
double crossover groups, it is in the middle. With respect to the three genes mentioned in the
problem, what are the genotypes of the parents used in making the phenotypically wild-type F_1
heterozygote? What is the map distance between sp and dsr?
Solution
The gene at the middle is for spec body, and the is linkage between the genes for disrupted wings
and speck body showing higher number of offsprings with cinnabar eys (235) and disrupted
wings, speck body (241).
If dsr gene will be at middle , then there will be double crossing over between sp and cn that can
lead to less individuals showing speck body, cinnabar eyes or disrupted wings, which actually is
not the case here.
therefore the middle gene is sp and there is double cross over between dsr and cn that leads to
reduction in the individuals with disrupted body, cinnabar eyes (25) or spec body (22).
Thus gene sequence will be dsr-sp-cn or cn-sp-dsr.
Part B
in order to obtain the phenotypically wild type F1 heterozygote, the genotypes of the parents
from the gives genotypes will be 2nd and 3rd.
sp dsr cn / sp dsr cn and sp* dsr* cn* / sp* dsr* cn*
sp dsr cn* / sp dsr cn* and sp* dsr* cn / sp* dsr* cn
Part C
Map distance between sp and dsr
single cross over between sp and dsr= 46+52 = 48
doble crossover 22+25 = 47

2. Total recombinants = 98+ 47 = 145
Total offsprings are 850
Percentage of recombination is 145/850 X100 = 17.0588
Therefore the map distance is 17 cM.