1. The genes for mohogany eyes and ebony body are approximately 25 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and the resulting F1 phenotypically wild type females were mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes and in what numbers would they be expected?
2. Given that loci A and B in Drosophila are sex-linked and 20 map units apart, what phenotypic frequencies would you expect in male and female offspring resulting from the following crosses? (Assume A and B are dominant to a and b, respectively):
(a) AaBb(cis)female X ab/Y male
(b) AaBb(trans) X ab/Y male
(c) aabb female X AB/Y male
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1. To determine the expected phenotypes and their numbers, we need to consider the inheritance patterns and calculate the probabilities.
In Drosophila, we know that the genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome III. This means that there is a 25% chance of recombination occurring between these two genes during the formation of gametes.
Let's break down the problem step by step:
Step 1: Initial Cross - mahogany-eyed female (MM) x ebony-bodied male (ee)
The female is homozygous for the mahogany eye gene (MM) and the male is homozygous for the ebony body gene (ee).
Gametes produced:
Female (MM): M
Male (ee): e
Possible F1 combinations:
- Me (mahogany eyes, ebony body)
Step 2: F1 Cross - F1 females (Me) x Mahogany (MM) ebony (ee) males
The F1 females from the previous cross are heterozygous for both the mahogany eye gene (M) and the ebony body gene (e). They are crossed with males who are homozygous for the respective genes.
Gametes produced:
Females (Me): M, e
Males (MM): M
Males (ee): e
Possible F2 combinations and their probabilities:
- MMe (mahogany eyes, ebony body) - 25%
- Me (mahogany eyes, wild-type body) - 50%
- Me (wild-type eyes, ebony body) - 50%
- ee (wild-type eyes, wild-type body) - 25%
Calculations for expected numbers:
- Out of 1000 offspring, 25% will have mahogany eyes and ebony body: (25/100) * 1000 = 250
- 50% will have mahogany eyes and wild-type body: (50/100) * 1000 = 500
- 50% will have wild-type eyes and ebony body: (50/100) * 1000 = 500
- 25% will have wild-type eyes and wild-type body: (25/100) * 1000 = 250
Therefore, we would expect 250 offspring with mahogany eyes and ebony body, 500 offspring with mahogany eyes and wild-type body, 500 offspring with wild-type eyes and ebony body, and 250 offspring with wild-type eyes and wild-type body.
2. To determine the phenotypic frequencies of the offspring resulting from the given crosses, we need to consider the inheritance patterns and calculate the probabilities based on the recombination frequency.
Given that loci A and B in Drosophila are sex-linked and 20 map units apart, there is a 20% chance of recombination occurring between these two loci during the formation of gametes.
(a) AaBb(cis) female x ab/Y male
The female is heterozygous for both loci (AaBb), and the male is hemizygous for both loci (ab/Y).
Gametes produced:
Female (AaBb): AB, ab
Male (ab/Y): ab
Possible combinations and their probabilities:
- AB/Y (normal phenotype) - 80%
- Ab/Y (phenotype A, normal phenotype B) - 20%
So, we would expect 80% of the male offspring to have a normal phenotype and 20% to have phenotype A, normal phenotype B.
(b) AaBb(trans) female x ab/Y male
The female is heterozygous for both loci (AaBb), and the male is hemizygous for both loci (ab/Y).
Gametes produced:
Female (AaBb): AB, ab
Male (ab/Y): ab
Possible combinations and their probabilities:
- AB/Y (normal phenotype) - 80%
- ab/Y (normal phenotype A, phenotype B) - 20%
So, we would expect 80% of the male offspring to have a normal phenotype and 20% to have a normal phenotype A, phenotype B.
(c) aabb female x AB/Y male
The female is homozygous recessive for both loci (aabb), and the male is hemizygous for both loci (AB/Y).
Gametes produced:
Female (aabb): ab
Male (AB/Y): AB
Possible combinations and their probabilities:
- AB/Y (normal phenotype A, phenotype B) - 100%
So, we would expect all of the male offspring to have a normal phenotype A, phenotype B.