Question

In the fruit fly Drosophila, there is a dominant gene for normal wings and its recessive allele for vestigial wings. At another gene locus on the same chromosome, there is a dominant gene for red eyes and its recessive allele for purple eyes. A male that was heterozygous at both gene loci was mated with a female that was homozygous for both recessive alleles and the following results were observed among the offspring: Normal wings and red eyes - 420 Vestigial wings and red eyes - 80 Normal wings and purple eyes - 70 Vestigial wings and purple eyes - 430 According to these data, what is the distance, in centimorgans, between these 2 gene loci? (Enter the number only without the units. For example, 100 cM would be entered as 100)

Answer 1

Answer:

15

Explanation:

The genes for vestigial wings (vg) and purple eyes (pr) are linked on the same chromosome.

The following cross was made:

Heterozygous male for both genes   X female vg pr / vg pr  

The male could be either vg+pr+/vg pr or vg+ pr/vg pr+.

The female only produces vg pr gametes.

The following offspring was observed:

420 Normal wings and red eyes (vg+ pr+/vg pr)

430 Vestigial wings and purple eyes (vg pr / vg pr)

80 Vestigial wings and red eyes (vg pr+ / vg pr)

70 Normal wings and purple eyes (vg+ pr / vg pr)

Total: 1000

The individuals observed in the highest frequency have the parental genotypes. The recombinant individuals are vg pr+ / vg pr and vg+ pr / vg pr.

Distance = Number of recombinants × 100 / Total

Distance = (80 + 70)×100 /  1000

Distance = 15 cM.

Just a small note, male drosophila do not experience crossing over in meiosis, so there would be no recombinants observed from this cross. The problem is probably mixed up, the female should be heterozygous and the male the homozygous recessive.