Question

In silkmoths, red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these 2 genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wings. The F1 are crossed with moths that have red eyes and white banded wings in a test cross. The progeny of this test cross are:

wild-type eyes, wild-type wings 418
red eyes, wild-type wings 19
wild-type eyes, white-banded wings 16
red eyes, white-banded wings 426
a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes?b. What is the rate of recombination between the genes for red eyes and those for white-banded wings?

Answer 1

a. The phenotypic proportions that would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes is 1:1:1:1

b. The rate of recombination between the genes for red eyes and those for white-banded wings is 0.04

Step-by-step explanation:

To know that two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.

If the genes for red eyes and white-banded wings were located on different chromosomes, they would assort independently, and hence the expected phenotypic proportions would be 1/4:1/4:1/4:1/4

4/16=1/4 re+re wb+wb

4/16=1/4 rere wb+wb

4/16=1/4 re+re wbwb

4/16=1/4 rere wbwb

In the present example, the genotype, in linked gene format, of the double heterozygote is re+wb+/rewb.

In this way, we might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

In the present example:

Parental) re+ wb+/re wb x re wb/re wb

Gametes) re+wb+ parental type

re wb parental type

re+wb recombinant type

re wb+ recombinante type

Phenotypic class Number of offspring

• wild-type eyes, wild-type wings, re+ wb+, 418

• red eyes, wild-type wings, re wb+, 19

• wild-type eyes, white-banded wings, re+ wb 16

• red eyes, white-banded wings, re wb, 426

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 19 + 16 / 418 + 19 + 16 + 426

P = 35/ 879

P = 0.0398=0.04

The genetic distance between genes is 0.04 x 100= 4 MU.