# Chance of child inheriting at least one copy of an allele - heterozygote mother, unknown father

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

• What is a risk genotype in a autosomal dominant disease - just the probability of being diseased? – cel Oct 7 '15 at 6:10
• To me, it is unclear whether the "disease" allele is dominant or recessive. Your statements "Homozygous recessive frequency: 0.9025" and "risk genotype (autosomal dominant)" could be impying that the disease is dominant, while the professor's answer is indicating that it's recessive. The answer also depends on whether you know the disease/carrier status of the father, and you and your professor are treating this information differently. [cnd...] – fileunderwater Oct 7 '15 at 8:42
• [cnd:] You could argue that the disease status of the father will be known (irrespectively of being recessive or dominant). Your professor's answer is based on that information while you are treating the fathers status as completly unknown. – fileunderwater Oct 7 '15 at 8:42
• I think the professor may have forgotten to also add the probability that the father is homozygous for the risk allele – rg255 Aug 21 '16 at 7:27

It seems that what you are trying to do is calculate the probability of the child having at least one copy of the "risk allele" which I will denote as "A" (and it's counterpart "a").

First of all you know that the mother is Aa, and the child will get one allele at random. The probability that the mother passes on the A allele is then 0.5.

The tricky bit is the seemingly unknown genotype of the father. However, you do know genotype frequencies, which are the probability for the fathers three possible genotypes.

• If the father is AA then it is certain the child will get the risk allele, and the probability of the father being AA is 0.0025.
• If the father is heterozygous then there is a probability of 0.5 he will pass on the A allele, and the probability of the father being Aa is 0.095.
• If the father is aa there is a probability if 0 that he will pass on the A allele.

Combined these give the probability that the father carries at least one A and will pass it to his child.

$$(1 \times 0.0025) + (0.5 \times 0.095) + (0 \times 0.9025) = 0.05$$

The probability of the child getting at least one A is then the probability of getting it from the mother plus the probability of getting it from the father.

$$0.5 + 0.05 = 0.55$$

I wrote out every possibility, and I agree with yours.

Mom passes good allele 50.00% Dad is homo wt 90.25% Odds of 0 bad alleles 45.13%

Mom passes good allele 50.00% Dat is het 9.50% Dad passes good allele 50.00% Odds of 0 bad allele 2.375000%

Mom passes good allele 50.00% Dad is diseased 0.25% Odds of 1 bad allele 0.12500%