Question

wmlwm White eye Miniature wing w m/YO Red eye Full wing W W m wmlwm Red eye Full wing wm/Yd White eye Miniature wing Phenotypes and genotypes of the F Females Males Number observed w m w*m/wm Red eye Full wing w m /Y Red eye Full wing W m wm GP 750 W m wmlwm White eye Miniature wing wm/Y White eye Miniature wing W m w mlwm Red eye Miniature wing w m/Y Red eye Miniature wing W m * W m * 455 W m wmlwm White eye Full wing wm./V White eye Full wing 2441

Answer 1

Wild type dominant traits are red eye (w+) and full wings (m+)

recessive traits are white eye (w) and miniature wings (m)

2. Males have XY chromosomes and females have XX chromosomes. Male have single X chromosome, so the phenotype will be expression due to single chromosome, where recessive traits will also be expressed and individuals are homozygous for the traits. Hence, Yes, we can infer genotype on the basis of phenotype in males

3. In F2 progeny, after crossing over, female can have will have combination of traits. Due to presence of XX chromosomes, we can expect heterozygosity in traits, of which, dominant charactes will be expressed. We can identfy genotype of homozygous recessive females for each trait on the basis of phenotype, but can not distinguish between homozygous dominant and heterozygous genotypes. So, in F2 progeny, we can not infer the genotype for the phenotype in females.

3. As both loci m+ and w+ are located on the long arm of X chromosome. If recombination occur during meiosis at position between these two loci on long arm of X chromosome, during formation of gametes in female, then linkage will change. Note: females can have two types of allele for gene due to presence of XX

4. During gamete formation in male (XY), chromosomes will be replicated, to form gametes X and Y, these are non-homologous chromosomes. Since males have only one type of allele, linkage beteen w+ and m+ will not get modified.