Gene A is 10 mU from Gene B on chromosome 2, and the alleles are in coupling. An individual with the genotype AaBb is testcrossed. What fraction of the offspring would be expected to have genotype aabb?

The answer is supposed to be 45% but I'm unclear on how to get there. Thanks!

Parental chromosomes A----b and a-----B when fertilized with ab will produce AabB and aaBb - these will be progeny with parental chromosome --- so no recombination.

You are told that map distance is 20cM which is also telling you that recombination frequency is 20% , so % progeny showing recombination = 20%, so % progeny not showing recombination is 100= 20 = 80%. This will be composed of the 2 parental categories, as above so % aaBb = 40%

Explanation:

Gene A is 10 mU from Gene B on chromosome 2, and the alleles are in coupling. An individual with the genotype AaBb is testcrossed. What fraction of the offspring would be expected to have genotype aabb? The answer is supposed to be 45% but I'm unclear on how to get there. Thanks!

Respuesta :

Otras preguntas

Gene A is 10 mU from Gene B on chromosome 2, and the alleles are in coupling. An individual with the genotype AaBb is testcrossed. What fraction of the offspring would be expected to have genotype aabb?

The answer is supposed to be 45% but I'm unclear on how to get there. Thanks!