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.
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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
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.
[(cn +) + (+ bw ]/ 46256 = 41%
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.