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# Drosophilla crosses

In a lab strain of Drosophila, cinnabar (cn) and brown (bw) are recessive eye colour mutations known to be 41 m.u. 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 observed in the progeny:
+ + 25,200
cn bw 21,009
cn + 11
+ bw 36
(a) What is unexpected about these results?
(b) What is the most likely explanation? (Draw a diagram)
(c) On the diagram show the precise origin of the cn + and the + bw classes.

Ideas are expressed.

#### Solution Preview

a) Calculate the expected recombination frequencies for 2 loci 41 map units apart. Total number of progeny is 25,200 + 21,009 + 11 + 36 = 46,256
<br>Without any recombination, you would expect the Mendellian crosses to give 50% ++ and 50% cn bw. The recombinant groups of progeny are cn + and +bw.
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<br>[(cn +) + (+ bw ]/ 46256 = 41%
<br>It is reasonable to assume that there is an equal likelyhood of either recombinant occurring, so let's make the bracketed portion equal to ...

#### Solution Summary

An approach to Drosophilla crosses is featured.

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