In: Biology
Bar eye (B) is a dominant and sex-linked trait of Drosophila. Curved wings (cu) and black body (bl) are recessive traits, both found on chromosome 2, separated by 24 map units. A homozygous Bar-eyed, black-bodied female is crossed to a true-breeding male expressing curved wings. The heterozygous F1 female is then crossed to a male expressing all three traits. At what frequency among the F2 would you expect to find a completely wild type male?
How often would you find a Bar eyed, curved wing female?
Cross 1: (Black bodied, Bar eyed female crossed to Male with
Curved wings)
bl c+/bl c+; XB/XBx bl+
c/bl+ c; X/Y
F1 --> bl c+/bl+ c ; XB/X (F1 Heterozygous Female)
Cross 2 (F1 Heterozygous female crossed to male expressing all three traits):
bl c+/bl+ c ; XB/X x bl c/bl c ; XB/Y
The female parent of Cross 2 produces the following gametes:
Gamete | Gamete Type | Proportion | Rationale |
---|---|---|---|
bl c+; XB |
Parental | 19% | Since the two loci are 24 map units apart, recombination would occur in 24% of all meioses. Therefore, 76% of all gametes would be parental, with each class of parental gametes represented equally |
bl c+; X |
Parental | 19% | |
bl+ c ; XB |
Parental | 19% | |
bl+ c ; X |
Parental | 19% | |
bl+ c+; XB |
Recombinant | 6% | 24% of all meioses would be recombinant, and all recombinant gametes would be represented equally. |
bl+ c+; X |
Recombinant | 6% | |
bl c ; XB |
Recombinant | 6% | |
bl c ; X | Recombinant | 6% |
Now, the tester male (Male used in Cross 2) would produce half gametes with the X chromosome and half with the Y chromosome.
Therefore, the progeny proportions are:
A completely normal male is 3% of the F2 progeny.
A Bar-eyed, Curved Wing Female would occur in 19% of female progeny.