A Family with a Lysosomal Enzyme Deficiency: Reading Pedigree and Enzyme Data Together
A couple, both clinically unaffected, have a 4-year-old son, Devon, diagnosed at 18 months with a lysosomal storage disorder after episodes of vomiting, an enlarged liver and developmental delay. Enzyme assay results are: Devon 2% of normal activity; mother 48%; father 52%; Devon's 7-year-old sister 100%. The mother's brother died in infancy with similar symptoms; the father's family has no history. Population screening in the parish finds 1 affected infant in 40 000 births. The parents are told each future child has a 1 in 4 chance of being affected. The maternal aunt, who is 10 weeks pregnant, has an enzyme activity of 50% and asks what her baby's risk is; her partner is unrelated and from the same parish. Devon's sister's parents also want to know whether she is a carrier, given her 100% result.
1. Deduce the mode of inheritance, using the enzyme data to discriminate between the possibilities rather than relying on the pedigree alone.
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Autosomal recessive.
Unaffected parents with an affected child establishes recessiveness. The pedigree alone cannot separate autosomal from X-linked, which is what the enzyme data are for.
If the condition were X-linked recessive, the father would be hemizygous for a normal allele, produce normal enzyme and pass the mutant allele to no sons. He has 52% activity, so he carries a mutant allele and transmits it to sons and daughters equally - inconsistent with X-linkage.
Mitochondrial inheritance is excluded for a different reason: mtDNA passes only through the mother, so the father's carrier status would be irrelevant, and an affected child would imply a maternally transmitted variant. His 48/52% pairing with the mother is far better explained by both parents being heterozygous.
The ~50% activities are typical of heterozygotes for a loss-of-function allele (one functioning copy). Devon's 2% residual activity explains the severe, early-onset presentation - the small remainder is probably a misfolded or unstable protein.
The counsellor's 1-in-4 figure is correct, and each pregnancy is independent: a run of healthy children does not reduce the next child's risk.
2. Calculate the risk that the aunt's baby is affected, stating each assumption you make.
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Step 1: population carrier frequency. Using Hardy–Weinberg, q² = 1/40 000 = 2.5 × 10⁻⁵, so q = 0.005 and the carrier frequency 2pq ≈ 2 × 0.995 × 0.005 ≈ 0.00995, about 1 in 100.
Step 2: the aunt's carrier probability. Her 50% activity is the heterozygote signature, so she is almost certainly a carrier - say 0.95 to allow for assay error.
Step 3: the partner. Unrelated, from the same parish, so his carrier risk is 1/100.
Step 4: risk = 0.95 × 0.01 × 0.25 = 0.0024, i.e. about 0.24%, roughly 1 in 420.
Assumptions: Hardy–Weinberg equilibrium in the parish (no consanguinity, no selection, no recent migration), the partner is unrelated to the aunt, the assay reliably identifies heterozygotes, and the condition is fully penetrant in homozygotes.
This is about a hundredfold lower than the parents' 1-in-4 risk. That comparison is the basis for deciding whether partner testing and prenatal diagnosis are warranted.
3. The sister's enzyme result is 100%. Advise the family on what this means for her carrier status, showing the calculation if the assay detects 90% of carriers.
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From the pedigree alone, an unaffected sibling of an affected child has a 2/3 chance of being a carrier, because the expected 1 AA : 2 Aa : 1 aa ratio is conditioned on her not being aa. A 100% activity result is reassuring but does not reduce the risk to zero, because no assay detects every carrier.
Using Bayes' reasoning with 90% sensitivity:
- P(carrier) prior = 2/3; P(normal result | carrier) = 0.10; product = 0.0667
- P(non-carrier) prior = 1/3; P(normal result | non-carrier) = 1.0; product = 0.3333
- P(carrier | normal result) = 0.0667 / (0.0667 + 0.3333) ≈ 17%
With a 99%-sensitive assay the same calculation gives about 2%.
The judgement to communicate is that the residual risk is set by the sensitivity of the test used, and that figure should be quoted by the laboratory rather than assumed. DNA testing of the known family variants would resolve the question more or less completely; enzyme testing alone cannot.