Homework Answers
ebony and rough eyes are closely linked to each other so that recombination does not occur between genes.
let the alleles be e+ - wildtype body e- ebony body
r+ - wildtype eyes, r- rough eyes
given that these genes are located on autosomes, so
genotype of the homozygous ebony female: eer+r+
genotype of homozygous rough eyed male: e+e+rr
eer+r+ * e+e+rr
er+/e+r
since these genes are linked closely to each other such that recombination does not occur between these genes, so er+/e+r can produce only two types of gametes er+ and e+r. gametes have the same combination of alleles as the chromosomes, one chromosome has e and r+ alleles and other chromosome has e+ and r alleles so it produces er+ and e+r gametes.
F1 females are crossed to ebony, rough males, the genotype of ebony rough males is eerr, it can produce only one type of gamete er.
er+/e+r * eerr
| er+ | e+r | |
| er | er+/er ( ebony body) | e+r/er (rough eyes) |
so half of the progenies have ebony body and the other half have rough eyes. that is 50% of progeny have ebony body and 50 of progeny have rough eyes.
so the answer is d) ebony 50%, rough 50%, wildtype 0%, rough, ebony 0%.
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