Two affected males in one family: counselling an X-linked recessive disease
Meera and Raj, both healthy, bring their 6-year-old son Arjun to a genetic counselling clinic. Arjun has progressive muscle weakness, a waddling gait and enlarged calves; a muscle biopsy and a creatine kinase level confirm Duchenne muscular dystrophy (DMD), an X-linked recessive condition. The counsellor takes a three-generation history. Meera's brother died of the same illness at 19; her father is healthy at 68; her mother is healthy at 64. Meera has one sister, Kavya, who is 29, healthy, and 12 weeks pregnant. Kavya's husband is healthy and has no family history of muscle disease, and no female in the family has ever shown symptoms. Kavya asks three things: whether she can be a carrier, what the chance is that her baby is affected, and whether the answer changes if the baby is a girl. She also mentions that a neighbour told her that since only boys get DMD, the pregnancy "is not a worry if it is a girl".
1. Assign genotypes to the family members. Is Meera a carrier, and what is the probability that Kavya is a carrier? Justify both answers.
Read the worked answer
Write X^D for the normal allele and X^d for the DMD allele.
- Arjun is X^d Y. His X^d must have come from Meera, so Meera is an obligate carrier, X^D X^d - her healthy phenotype is expected, since her second X is normal.
- Meera's brother is also X^d Y. His X^d came from his mother, so Meera's mother is a carrier. Meera's healthy father contributed his X^D to Meera (and his Y to her brother), which is fully consistent. The fact that the mother is healthy proves nothing, because carrier females are usually asymptomatic.
- Kavya is the daughter of a carrier mother and a healthy father. At conception she had a 1/2 chance of inheriting X^d.
Being healthy does not reduce that 1/2, because carrier females are phenotypically normal in most cases. Kavya's prior carrier probability is therefore 1/2, and it can only be resolved by DNA testing for the family's specific deletion.
2. State the risk to Kavya's fetus overall and separately for a male and a female fetus, and say what testing could be offered.
Read the worked answer
Risk = P(Kavya is a carrier) × P(she transmits the allele) × P(expression in the child).
- Male fetus: 1/2 × 1/2 = 1/4 (25%) of being affected - any male receiving X^d is affected.
- Female fetus: her father contributes X^D, so she cannot be affected; but she has 1/2 × 1/2 = 1/4 (25%) chance of being a carrier like her mother.
- Sex unknown: half of fetuses are male, so the overall risk is 1/4 × 1/2 = 1/8 (12.5%).
Offer, in order: (i) test Kavya herself first, so counselling rests on her actual genotype rather than a 1/2 prior; (ii) determine fetal sex - ultrasound after about 14 weeks, or cell-free fetal DNA earlier; (iii) if she is a carrier, offer chorionic villus sampling or amniocentesis with deletion analysis on the sample. Note that the risk figure for a male fetus does not change with gestational age, so the decision is about information, not about probability.
3. Evaluate the neighbour's statement that a female fetus is "not a worry". Also explain whether the high rate of new mutations in DMD changes the reasoning here.
Read the worked answer
The neighbour has merged two different claims.
Correct part: a daughter cannot be affected in this family, because DMD requires two X^d alleles and her father supplies a normal X^D.
Wrong part: "not a worry" assumes phenotype risk is the only risk. A daughter still has a 25% chance of being a carrier, and a carrier daughter will face the same 25% risk for each of her sons, plus the reproductive decisions that follow. Sex therefore changes the phenotype risk, not the genetic risk, and the family should still be counselled.
New mutations: roughly one third of DMD cases arise from new mutations, but that is a population average. Here, two affected males in one sibship make an inherited deletion far more likely, so the 1/2 carrier prior for Kavya is appropriate. A negative carrier test must also be interpreted against that prior, because a test that detects only the family's known deletion will not exclude a rare second mutation.