1. A mutant fruit fly strain has short wings. Males ofthis strainare mated to females that have long (wild type) wings. All of theF1 offspring have long wings. When F1 males and female are mated,they producelong andshort wing offspring in a 3:1ratio. Is the short wing mutation dominant or recessive withrespectto thewild type wing length?Explain.
Assign allele designations and write out the genotypes forthe parentaland F1 crosses. What are the genotypes of the F2offpsring?
2. A male fruit fly with bar-shaped eyes is mated to afemale with wild type round eyes. All the F1female offspringhave bar-shaped eyes and all of the F1 male offspring have roundeyes. When the F1 males and females are mated with each other, theF2 offspringare ofthe following phenotypes and ratios:
Is the short wing mutation dominant or recessive withrespectto thewild type wing length?Explain.
Assign allele designations and write out the genotypes forthe parentaland F1 crosses. What are the genotypes of the F2offpsring?
2. A male fruit fly with bar-shaped eyes is mated to afemale with wild type round eyes. All the F1female offspringhave bar-shaped eyes and all of the F1 male offspring have roundeyes. When the F1 males and females are mated with each other, theF2 offspringare ofthe following phenotypes and ratios:
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In Drosophila, the gene for bar-shaped (B) eyes and the gene for scalloped wings (sc) are linked on the X-chromosome. (The normal or wild type eye shape is oval, and the wild type wing is rounded.) Heterozygous, bar-eyed females were mated to homozygous, scalloped-wing males. The offspring are as follows:
| phenotype | female | male |
| scallop | 183 | 186 |
| bar | 189 | 188 |
| wild type | 12 | 10 |
| scallop & bar | 9 | 12 |
-What is the map distance between the two genes?
-Give the exact phenotypes of the F1 flies.
