Question 3: (2 points) The famous scientist Johnny Nerdelbaum Frink decides to investigate a fly with unusual wings. He crosses a male fly with brown eyes and square wings to a female fly that has red eyes and square wings. The offspring consist of the following: 51 red, square 53 brown, square 18 red, oval 16 brown, oval Knowing nothing else about these flies, what could Frink deduce about the genotypes of the parents?

Biology, Science

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EXPERT ANSWER

in this case, the parental phenotypes are- brown eyes, square wings, and red eyes and square wings.

The number of parental types in the given progeny is much higher than that of the recombinants.

So,

Recombination frequency = (Number of recombinants / Total progeny) x 100%

= (18 + 16 / 138) x 100%

= (34 / 138) x 100%

= 24.64%

This means the genetic distance between the two genes controlling these two traits is about 25 cM (centimorgan).

This genetic distance is less than 50 cM. This means that the two genes controlling these traits are present on the same autosome and are LINKED to each other.

They are present on an autosome because, in the progeny, the inheritance of the traits is not affected by the sex of the offspring. This means that both these genes are present on an autosome and not on a sex chromosome.

Also, they are said to be linked because, during the prophase of meiosis, genetic recombination or crossing over between two homologous chromosomes occur. During this process, genetic material is exchanged between the non-sister chromatids of the two homologous chromosomes. This results in the production of recombinants that are different from their parents.

But when two genes are present in close proximity to each other on the same chromosome, they are inherited together as a unit and are thus said to be linked to each other. In this case, the progeny has a higher proportion of the parental combinations as seen in this case.

Suppose eye color is controlled by the gene B and the wing shape is controlled by the gene S,

Then,

The genotype of the male parent = Bb / Ss

The genotype of the female parent = bb / Ss