Our genetics professor has posted up working for previous examination answers, but I am not convinced that his answer is correct. My answer is close but may just be due to co-incidence.
Question: There is a dominant "risk" allele with a frequency of 5%; what is the chance that a woman with one copy of this allele, and her husband will have a child with at least one copy of the allele.
Heterozygote frequency: 0.095
Non-risk allele homozygote frequency: 0.9025
Risk allele homozygote frequencyy: 0.0025
0.50 (woman's chance of passing allele) + 0.095/2 (heterozygote frequency in population divided by chance of passing allele) = 0.5475
There are three possible situations, that the husband is homozygous recessive, homozygous dominant and heterozygous. The probability that any child will have the allele in each situation is 50%, 75% and 100% respectively.
Hence my answer = (50% x 0.9025) + (75% x 0.095) + (100% x 0.0025) = 0.525