How to calculate the percentage of heterozygous cells based on a gene map?

Question

I've encountered the following question and am quite stumped by it.A female with genotype AABBCC has been hybridized with a male that has the genotype aabbcc. The first generation (F1) has been hybridized again (backcross) with offsprings that have the genotype aabbcc. Given the following genetic map, what would be the percentage of F2 offsprings that are heterozygous to all three genes:A------- 20 MU ------- B -- 10 MU --- C**How should I approach the following question**, the brute force approach is too difficult, I keep getting things wrong.

Answer 1

It seems you haven't had a chance to get an answer yet, so I'll jump in and let you know how I would teach this question.

In your scenario, AABBCC x aabbcc -> 100% AaBbCc individuals. If they are backcrossed to aabbcc (allowing for you to detect recombination in the F1 heterozygotes), then we get a wide variety of offspring, which would include all of the following (drawn with one chromosome above the line and one below the line to help clarify parental gametes and recombinant gametes):

ABC; abc; Abc; aBC; ABc; abC; aBc; AbC

abc abc abc abc abc abc abc abc

Notice that all of the individuals get "abc" because of the backcross, so we are really only interested in the gametes produced in the F1 (the top line). Moving left to right, they would be the parental games (no recombination) in the first two, the next four are single recombinants, and the last two are double recombinants (assuming B is in the middle, as stated).

Many times students are asked to calculate the map distance based on numbers of offspring, but the question you describe has it going the other way. To figure this, we just have to work in reverse.

The map distances are equivalent to recombination frequencies or %. If the distance between AB is 20 mu, then 20% of the time, there is a recombination event between A and B. We want AB and C to stay together (no recombination between any of them so that the final genotype is ABC/abc), so we are interested in the 80% of the time that we don't have recombination between then. Similarly, between B and C, we are interested in the 90% of time that they don't recombine. So, to calculate the chance that there is no recombination between any of these genes, the formula would be:

(.8)(.9) = 72%

However, we also have to take into account that without recombination, the F1 individual has a 50/50 chance of passing on the ABC gamete or the abc gamete. We're interested in the ABC gametes only (to give a fully heterozygous individual), so we need to divide 72% by 2.

72%/2 = 36% chance of having a fully heterozygous individual in the F2 generation.

Hope that helps!

Heidi