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

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.

• if this is homework please tag it as such and we would appreciate to see some previous attempts at tackling the question. We are not here to do your homework :) – AliceD Jul 13 '15 at 12:11
• actually, this is a question from an exam I am preparing for. the exam is tomorrow :) – vondip Jul 13 '15 at 12:13
• What happens when you draw out all the possible offspring genotypes and calculate the percentage chance? – James Jul 13 '15 at 12:23
• I get horribly confused and after several attempts can't seem to come to the correct answer, is there no technique to do it without writing down all possible outcomes? – vondip Jul 13 '15 at 12:30
• +1 thanks for the edits. I'm not a geneticist, so I hope someone else here may help you further! Good luck tomorrow :) – AliceD Jul 13 '15 at 12:39

Units of genetic distance

It took me a little while to understand what MU means. I have never seen MU but rather m.u. for map unit, or cM for centiMorgane.

Solving the problem

This problem got me thinking a little bit. My logic is the following:

Probability of being heterozygous at the first locus

You cross two individuals (we'll call them individual 1 and 2)that are both heterozygote at each locus A, B and C (not to be confused with the alleles A and a, B and b and C and c). We'll start with locus A. The probability of being heterozygous at locus A is the probability that individual 2 gives another allele than individual 1. This occurs with probability 0.5. If the three loci were perfectly independent, then the probability of being heterozygous for a given offspring at all 3 loci would just be $0.5^3 = 0.125$... but that would be too easy so let's keep going with the second locus.

Probability of being heterozygote at the two first loci

So let's assume our offspring of interest, got two different alleles for the locus A. The probability to receive two different alleles at the locus B is the probability that either no individual recombine between A and B or both individuals recombine between A and B. It is therefore: $0.2^2 + 0.8^2$. Put together this probability with the probability that the first locus was heterozygote it gives $0.5 \cdot (0.2^2 + 0.8^2)$.

Probability of being heterozygous for all 3 loci

Finally, we can just keep going with he next locus. Let's assume that the offspring have received two different alleles for both loci A and B. The probability to receive different alleles at locus C is the probability that either both parents recombine (between B and C) or none recombine. This occurs with probability $0.1^2 + 0.9^2$.

Put the whole thing together, the probability of a given offspring to be heterozygote (or the frequency of heterozygotes offsprings if you prefer) is $0.5 \cdot (0.2^2 + 0.8^2) \cdot (0.1^2 + 0.9^2) = 0.2788$

Note that calculating the probability that exactly 2 loci are heterozygous would have been a bit trickier.

Good luck for your exam tomorrow!