Two Affected Brothers, an Unaffected Father and a Missing Restriction Site
A couple, both unaffected, attend a genetics clinic. Their second son has had a bleeding disorder since infancy. They also have two unaffected daughters and one unaffected son. The elder daughter has two affected sons and one unaffected daughter; her partner is unrelated and unaffected. The younger daughter has one unaffected son and is 14 weeks pregnant; she has declined carrier testing. The mother's brother is also affected, although her father was not. No female in the family is affected. Population screening in the same region finds the disorder in about 1 in 2500 males. Blood was sent for PCR-RFLP analysis: the disease allele has lost a recognition site for the enzyme, so digestion of the 320 bp product gives 180 bp and 140 bp fragments from the normal allele and an uncut 320 bp fragment from the disease allele. Affected males show only the 320 bp band, unaffected males show 180 and 140 bp, and carrier females show all three bands. The father's digest shows only the 180 and 140 bp bands.
1. Deduce the most likely mode of inheritance and explain why the pedigree alone cannot exclude autosomal recessive. State the single observation that does exclude it.
Read the worked answer
Pattern in the pedigree. Affected individuals are all male; every affected male has unaffected parents; the affected boys in the elder daughter's family are linked to her through the maternal line (her brother and her maternal uncle are affected); no father-to-son transmission occurs; no female is affected. This is the classic signature of an X-linked recessive allele: males are hemizygous so one copy is expressed, while females need two copies, which is rare for a rare allele.
Why autosomal recessive survives the pedigree. If both parents of the affected son were carriers, he could be homozygous. The fatal objection is the elder daughter: for her two sons to be affected under autosomal recessive, she and her unrelated partner would both have to be carriers. With an allele frequency q = 1/2500 = 0.0004, the carrier frequency is about 2q = 1/1250, so the probability that her partner is a carrier is about 0.08%, and each affected son then also needs a 1 in 4 transmission. Two affected sons with an unrelated partner is therefore extremely improbable but not formally impossible, so the pedigree alone leaves a residual doubt.
The clinching observation. The father's digest shows only 180 and 140 bp, with no 320 bp band. Under autosomal recessive, the unaffected father of an affected child must be a carrier and would show the 320 bp band. He does not, so he carries no disease allele; the allele passed only from the mother, which is exactly what X-linked inheritance requires (allowing for the very small possibility of a new mutation or germline mosaicism).
2. Interpret the banding pattern. Explain why an affected male gives one band while a carrier female gives three, and why the single band in a male does not by itself prove X-linkage.
Read the worked answer
What the digest shows. The 320 bp amplified region contains a recognition site in the normal allele. Digestion cuts it into 180 bp and 140 bp, which sum to 320 and so act as an internal check. The disease allele has lost the site, so it stays at 320 bp. Each band therefore represents an allele, not a person.
Male pattern. A male has one X and one Y, so he is hemizygous: there is only one copy of this locus in each cell. An affected male therefore shows only the uncut 320 bp band; an unaffected male shows only 180 and 140 bp. No male can be heterozygous at an X-linked locus, which is why no male shows three bands.
Female pattern. A female has two X chromosomes. A carrier has one normal and one disease allele, so the digest contains both the uncut 320 bp fragment and the two fragments from the normal allele: three bands. The 320 bp band in a carrier should be about half as intense as the corresponding band in an affected male relative to the 180 bp band, though this is only semi-quantitative on a simple gel. X-inactivation in her cells is irrelevant here because PCR amplifies genomic DNA from the whole blood sample, not mRNA.
Why one band is not proof. An affected female homozygous for an autosomal recessive allele would also give a single 320 bp band, and a homozygous normal person gives a single band after digestion in the sense of only the two cut fragments. Band number alone therefore shows genotype at the locus, not chromosome location; the pedigree and the father's result are what establish X-linkage.
3. Calculate the risk that the younger daughter's baby is affected, given that she declined testing. Explain how her test result would change the figure, and comment on whether the 1 in 2500 figure should be used in counselling her.
Read the worked answer
Step 1: the mother's status. Her son is affected and her husband is unaffected, so the son's single X came from her; she must be a carrier.
Step 2: the younger daughter's risk of being a carrier. She is unaffected, and she received one X from a carrier mother, so P(carrier) = 1/2.
Step 3: the pregnancy. If she is a carrier, each male fetus has a 1/2 chance of receiving the disease allele; female fetuses are unaffected (at most carriers). P(affected child) = 1/2 (she is a carrier) × 1/2 (fetus is male) × 1/2 (transmits the allele) = 1/8, or 12.5%. Conditioned on the fetus being male, the risk is 1/4; conditioned on a female fetus, effectively zero.
How testing changes the answer. A carrier test converts her 1/2 into approximately 0 or 1, so the risk becomes about 1/4 for a male fetus or negligible. This is the single most informative test available and should be offered before any discussion of prenatal diagnosis. If she is a carrier and the fetus is male, RFLP analysis of chorionic villus or amniotic cells gives a definitive answer.
The population figure. With q = 1/2500, the female carrier frequency is 2pq ≈ 1/1250. This is an a priori risk for a person with no family history, useful for interpreting the elder daughter's partner under an autosomal hypothesis. It is not the right basis for counselling this family, whose risk is fixed by a known segregation; quoting 1 in 2500 would seriously mislead them.