An aabb individual and an aabb individual were crossed, and the resulting aabb f1 individuals were testcrossed to aabb individuals. if the a and b genes are 20 m. u. apart on the same chromosome, what progeny genotype ratios would be expected from the test cross?

0.4 Ab/ab

0.4 aB/ab

0.1 AB/ab

0.1 ab/ab

Explanation:

The genes A/a and B/b are linked, i.e. located in the same chromosome. To solve problems with linked genes it’s important to use a notation that allows us to quickly determine how the alleles are distributed in the chromosomes of an individual.

In our problem, we have two initial individuals: AAbb and aaBB. I will write them down as Ab/Ab and aB/aB respectively (each chromosome contains an alelle of the A/a gene and an alelle of the B/b gene).  

The initial cross is:

The F1 is then test crossed:

There are four possible genotypes in the progeny:

Ab/ab

aB/ab

AB/ab

ab/ab

To determine their ratios, we need to use the information about the distance between the genes A/a and B/b, which is 20 map units.

The formula that relates genetic distance with recombination frequency (RF) is:

Genetic Distance (m.u.)= Recombination Frequency * 100.

Replacing the data in the formula, we have that RF=0.2

That means that out of the total number of individuals in the progeny, 0.2 will have recombinant genotypes for the genes A/b and B/b, and 0.8 will be parental. What does that mean? The parental genotypes are the ones that the parental individuals had. If you take a look at the genotype of the F1, you’ll notice how the alleles are paired in each chromosome: Ab/aB. That means that the parental genotypes in the progeny are Ab/ab and aB/ab, and each of them will appear in the progeny at a 0.4 ratio (in total they will be 0.8).

On the other hand, the AB/ab and ab/ab genotypes in the progeny were not present in the parental individuals of the test cross, and are therefore recombinant. The recombinant frequency is 0.2 in total, so each of those genotypes will have a 0.1 ratio.

layer 3 is younger than layer 6.