In a lab strain of Drosophila, cinnabar(cn) and brown(bw) are recessive eye colo
ID: 41773 • Letter: I
Question
In a lab strain of Drosophila, cinnabar(cn) and brown(bw) are recessive eye color mutations known to be 41 map units apart on chromosome 2. When similar mutant alleles were induced in a strain from nature, the same linkage of cn and bw was observed. However, when a wild-type strain from nature was crossed with a cn bw/cn bw lab strain to create the genotype + +/cn bw, and females of this type were test crossed to cn bw/cn bw males from the lab strain, the following phenotypic proportions were obtained in the progeny
+ + 25,200
cn bw 21,009
cn + 11
+ bw 36
a. What is unexpected about these results?
b. Assuming there is no more than one crossing-over within each tetrad, what is the most likely explanation for the results? How does your explanation account for the origin of the cn + and the +bw progenies.
Please help! If not with b then a because I don't know where to start. will give best answer as soon as possible
Explanation / Answer
According to the given data, the following are the genotypes of the F1 parent and the male test crossed with:
F1 female - ++ / cn bw
Male – cn bw / cn bw
The female parent produces the following gametes (since the genes are said to be 41 mu apart, parental genotypes should be 59% and the recombinants should be 41%):
Parental genotypes:
+ + / cn bw (should be 24.5%)
cn bw / cn bw (should be 24.5%)
Recombinants:
cn + / cn bw (should be 20.5%)
+ bw / cn bw (should be 20.5%)
The total number of progeny is 46,256. The following should are the expected ans observed values:
Genotype
Expected
Observed
+ + / cn bw
11333
25,200
cn bw / cn bw
11333
21,009
cn + / cn bw
9483
11
+ bw / cn bw
9483
36
However, the progeny are not obtained in the expected ratio. The recombinants are far less than the expected. The genes might be closely linked than expected.
Genotype
Expected
Observed
+ + / cn bw
11333
25,200
cn bw / cn bw
11333
21,009
cn + / cn bw
9483
11
+ bw / cn bw
9483
36
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