Counselling a family after a recessive condition appears in a healthy couple's child
Both parents in a family are healthy, but their first child, a girl, is born with a condition caused by a fault in a single gene. Neither parent has any symptoms and neither has been tested. The condition is inherited in a recessive way, and in the community about 1 in 25 people carry the faulty allele without knowing it. The couple ask a genetic counsellor two things: what the chance is that their next child will be affected, and whether the fact that their second child is a healthy boy means the risk for a third child has gone down. The counsellor also explains that the healthy boy himself will face a question about carrier status when he grows up and has children of his own.
1. Explain, with a genetic diagram, why the risk to the next child is 1 in 4, and state the genotypes of the parents.
Read the worked answer
Let A be the normal allele and a the faulty allele. The condition is recessive, so an affected child must be aa, receiving one 'a' from each parent. Since neither parent is affected, neither can be aa; both must therefore be Aa - unaffected carriers.
Cross: Aa × Aa gives
| A (father) | a (father) | |
|---|---|---|
| A (mother) | AA | Aa |
| a (mother) | Aa | aa |
Each parent passes 'a' with a probability of ½, and the two events are independent, so the chance of an aa child is ½ × ½ = ¼, or 25 % (1 in 4).
Note what else the diagram shows: 3 in 4 children are expected to be unaffected, but 2 of those 3 are carriers. So an unaffected child of this couple still has a 2/3 chance of carrying the allele - relevant to the boy in later life.
2. The couple's second child is a healthy boy. Does this lower the risk for a third child? Justify your answer.
Read the worked answer
No. Each pregnancy is a separate event. At every conception the father's sperm carries either A or a, and the mother's egg carries either A or a, each with a ½ chance. The combination is decided afresh each time, so the probability of an affected child is 1 in 4 for every child, regardless of what came before.
The common mistake is to think the 1-in-4 risk is a quota that has to be used up - as if after one affected child and one healthy child the family is "due" something. That is the gambler's fallacy. A coin that has landed heads twice is still 50:50 on the next toss.
A related error is the opposite one: thinking that because one child is affected the next is safe, or that because the second child is healthy the faulty allele has been "used up". Neither is true. What does change the picture is new information, not new children - for example, if testing showed one parent's genotype was different from what was assumed. Assuming the parents' genotypes are as deduced, the risk stays at 25 %.
3. The healthy son will one day want to know his own chance of being a carrier, and what that means for his children. Explain where the 2 in 3 comes from, and calculate the risk to a child of his if his partner comes from the community.
Read the worked answer
Where 2/3 comes from: the parents are Aa × Aa, so each child has an equal 1-in-4 chance of being AA, Aa or aa (with Aa counted twice, once from each parent's contribution). The son is healthy, so he is not aa. That leaves three equally likely outcomes - AA, Aa and Aa - and two of them are carriers. His chance of being a carrier is therefore 2/3, not 3/4, because the question is asked after we already know he is unaffected. This is a conditional probability: we are narrowing the sample space, not changing the genetics.
Risk to his child: his partner, taken from the community, has a carrier chance of about 1/25.
Risk = P(he is a carrier) × P(she is a carrier) × P(both pass on 'a') = 2/3 × 1/25 × 1/4 = 2/300 ≈ 1 in 150, about 0.7 %.
If she were instead a known carrier (for example through family history or screening), the middle term becomes 1 and the risk rises to 2/3 × 1 × 1/4 = 1/6, about 17 %. The lesson is that the partner's carrier status matters more than anything else in the calculation.