2. Iin silkmoths (Bombyx mori), red eyes (re) and white-banded wing (wb) are enc

Question

2. Iin silkmoths (Bombyx mori), red eyes (re) and white-banded wing (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re and we'l: these two genes are on the same chromasome. Amoen homorygous for red eyes and white-banded wings is crossed with a moth homorygous for the wild-type traits. The Pi have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings testcross. The progeny of this testcrossare wild-type eyes wild-type wings red eyes, wild-type wings wild-type eyes, white-banded wings 16 in a 418 19 red eyes, white-banded wings a. What phenotypic on different chromosomes? Be specifie regarding genotypes, phenotypes, 426 e proportions would be expected if the genes for red eyes and for white-banded wings were located and proportions. b. What is the percent recombination between the genes for red eyes and those for white-banded wings? Show all

Explanation / Answer

A) 1?4 wild-type eyes, wild-type wings
1?4 red eyes, wild-type wings
1?4 wild-type eyes, white-banded wings
1?4 red eyes, white-banded wings

B) The F1 heterozygote inherited a chromosome with alleles for red eyes and white-banded wings (re wb) from one parent and a chromosome with alleles for wild-type eyes and wild-type wings (re+ wb+) from the other parent. So, the phenotypes of the nonrecombinant progeny are present in the highest numbers. The recombinants are the 19 with red eyes, wild-type wings and 16 with wild-type eyes, white-banded wings. Now we can calculate the recombination frequency.
RF = recombinants/total progeny × 100% = (19 + 16)/879 × 100% = 4.0%.
So, the distance between the genes is four map units.

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