Question

Two genes, which we will call gene A and B, are known to be 10 map units apart. Let’s suppose cross was made between an AAbb and aaBB individuals to produce AaBb F1 offspring. The F1 offspring were then crossed to aabb individuals to yield an F2 generation. What would be the genotype(s) of F2 offspring that carry recombinant chromosomes? (Reminder: recombinant chromosomes are the product of crossing over). What percentage of F2 offspring would be Aabb?

Answer 1

Genotype of F2 offspring that carry recombinant chromosomes would be:

Aabb and aaBb

5% of F2 offspring would be Aabb

Step-by-step explanation:

Given that the two genes A and B are 10 map units apart. According to the concept of linkage, 1 map unit = 1 % recombination. Hence, there is 10% recombination in the F2 progeny. F2 progeny will get one type of gamete ab from the aabb parent and will get four types of gametes from AaBb parent: AB, ab, aB and Ab. So the four types of offspring that will be prduced are:

AaBb = parental

aabb = parental

Aabb = recombinant

aaBb = recombinant

Two types of recombinant i.e. Aabb and aaBb make 10% of total recombinant progeny so one of them i.e. Aabb will make 5% of the total progeny.