Counselling a Family with Male-Only Affected Relatives
At a community genetics camp, a couple seeks advice. The woman is unaffected, but her father has a rare condition. The man is unaffected, and his brother has the same condition. The couple already has one unaffected daughter and one affected son. No other relatives are affected, and all affected people in the extended family are male. The condition is known to be caused either by an X-linked recessive allele or by an autosomal recessive allele, depending on the family. The woman is pregnant. The counsellor must explain the risk under both models, decide which model better fits the male-only pattern, and state what single test would most efficiently clarify the situation. The local carrier frequency for the autosomal recessive form is 1/40, but the couple's family history is more informative than the population figure.
1. Under the X-linked recessive model, assign genotypes to the couple and calculate the risk for a son and for a daughter.
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Let XA be normal and Xa the recessive allele. The woman's father is affected, so he is XaY. He passes Xa to all daughters, so the unaffected woman must be XAXa. The unaffected man is XAY.
For a son: he receives Y from the father and either XA or Xa from the mother. Risk of an affected son = 1/2. The existing affected son fits this model.
For a daughter: she receives XA from the father, so she cannot be affected. She receives XA or Xa from the mother, giving 1/2 XAXA and 1/2 XAXa. Risk of an affected daughter = 0; risk of a carrier daughter = 1/2.
The man's affected brother is also consistent: the man's mother could be a carrier, and the unaffected man received XA.
2. Under the autosomal recessive model, assign genotypes and calculate the risk for the next child. Which model better fits the observation that all affected relatives are male?
Read the worked answer
Let A be normal and a the recessive allele. The woman's father is aa, so the unaffected woman must be Aa. The man's brother is aa, so the man's parents were Aa x Aa; an unaffected man has prior probability 2/3 of being Aa. However, the couple already has an affected son aa, so the man must actually be Aa. Both parents are therefore Aa.
Risk for the next child = 1/4 affected (aa), 3/4 unaffected (AA or Aa). This risk is the same for a son or a daughter.
The X-linked model better fits the male-only pattern: affected males are expected, and daughters of a carrier mother and unaffected father are carriers but not affected. Under autosomal recessive inheritance, the three affected males could still occur by chance, but the probability that all three are male is (1/2)^3 = 1/8. That is not negligible, so autosomal recessive inheritance cannot be excluded from the family pattern alone.
3. What single test would most efficiently distinguish the two models, and how would its result change the counselling?
Read the worked answer
The most efficient single test is to test the unaffected man for the known autosomal recessive mutation. Under the autosomal recessive model he must be a carrier because he already has an affected son. Under the X-linked model he is XAY and does not carry the X-linked allele.
If the man is found to carry the autosomal recessive mutation, the autosomal recessive model is confirmed: both parents are carriers and the next-child risk is 1/4 regardless of sex. If he does not carry it, the X-linked model becomes much more likely: the woman is an obligate carrier, sons have a 1/2 risk, and daughters have no risk of being affected but a 1/2 risk of being carriers.
Testing the affected son for the X-linked gene would also help, but testing the man directly answers whether the autosomal recessive model can explain the affected son. If no mutation is found, broader gene testing may still be needed because rare or private mutations can be missed.