Question text In drosophila (fruit flies): The yellow gene is on the X-chromosome, the gene product...

Question text

In drosophila (fruit flies):

The yellow gene is on the X-chromosome, the gene product plays a role in body colour. Wild type flies are a light brown/tan in colour, homozygote yellow mutants have a more yellow shaded body colour. We'll use y⁺ to represent the wild type allele, and y to represent the mutant allele. y⁺ is dominant to y.
The vestigial gene is on chromosome 2. Homozygote vestigial mutants have very short wings, and cannot fly. We'll use vg⁺ to represent the wild type allele, and vg to represent the mutant allele. vg⁺ is dominant to vg.
The scarlet gene is on chromosome 3. Homozygote scarlet mutants have bright red eyes, while wild type flies have darker brownish-red eyes. We'll use st⁺ to represent the wild type allele, and st to represent the mutant allele. st⁺ is dominant to st.
Males are usually XY, and females are usually XX

You cross a female fly who is wild-type in phenotype and heterozygous for all three genes, with a male who has a yellow body, very short wings and bright red eyes. What is the probability their offspring will have very short wings and normal eyes, if body colour doesn't matter (i.e, they can be either yellow or wild-type in colour)? For full marks you must show your work.

Solutions

Expert Solution

Ans: We have to cross between a male and female fly for chromosome 2 (size of wings) and chromosome 3 (colour of eyes). As these are autosomal traits, we take vg⁺ to represent the wild type allele (normal wings), and vg to represent the mutant allele (very short wings). vg⁺ is dominant to vg. Also, we take st⁺ to represent the wild type allele (normal darker brwnish red eyes), and st to represent the mutant allele (bright red eyes). st⁺ is dominant to st. There is no concept of X and Y because these traits are not sexlinked.

Here, it is given that body colour doesn't matter so we exclude the use of body colour in our calculations.

Now, Genotype of female is given heterozygous for both size of wings and colour of eyes, hence, = vg⁺vg, st⁺st

genotype of male is given for size of wings and colour of eyes = vgvg, stst

Gamets of female parent = vg⁺st , vg⁺st⁺ , vgst⁺ , vgst

Gamtes of male parent = vgst

Now lets do the cross,

Female Parent

Male Parent

vg⁺st

vg⁺st⁺

vgst⁺

vgst

vg⁺vg, stst

Normal wings, bright red eyes

vg⁺vg, st⁺st

Normal wings, Normal eyes

vgvg, st⁺st

Very short wings, Normal eyes

vgvg, stst

Very short wings, bright red eyes

Hence, the probability that offspring have very short wings and normal eyes = 1/4 = 0.25.

gladiator answered 1 year ago

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