Units of genetic distance
It took me a little while to understand what
MU means. I have never seen
MU but rather
map unit, or
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
2)that are both heterozygote at each locus
C (not to be confused with the alleles
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
B or both individuals recombine between
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
B. The probability to receive different alleles at locus
C is the probability that either both parents recombine (between
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!