CHSE Odisha Elective Biology study materials
CHSE Odisha Elective Biology study resources

Counselling a Family with a Rare Enzyme Deficiency

CHSE Odisha Elective Biology · Put your knowledge to work in context.

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Counselling a Family with a Rare Enzyme Deficiency

A couple from a village near Koraput seeks counselling. Their three-year-old son has progressive neurodegeneration, an enlarged liver and no detectable activity of a lysosomal enzyme on two separate assays. The parents are unrelated but belong to the same endogamous community, in which a screening survey found that 1 in 40 people carry the mutant allele. Enzyme activity in relatives, expressed as a percentage of the normal mean: father 48%, mother 52%, mother's sister 50%, mother's brother 45%, maternal grandmother 50%, maternal grandfather 100%, paternal grandfather 95%, paternal grandmother 100%, paternal uncle 100%. The paediatrician says the parents must both be carriers, but the father objects that no one in his family has ever been affected. Sequencing of the child's gene shows a single-base substitution in exon 4 that changes a glutamine codon to a UAA codon.

1. Deduce the mode of inheritance and identify which relatives are carriers. How do you answer the father's objection?

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The decisive pattern is a child with 0% activity and both parents with about half of normal activity. That is the signature of an autosomal recessive condition: each parent carries one mutant and one normal allele, makes roughly half the normal amount of enzyme, and that half is enough for health; the child inherited the mutant allele from both parents and makes none. The maternal grandmother, the mother's sister and the mother's brother all sit near 50%, so the allele is clearly present on the mother's side. The paternal grandfather's 95% and the paternal grandmother's and uncle's 100% do not exclude carriage in the father, because enzyme assays have measurement error and because carrier status is invisible clinically - a carrier has no symptoms and no family history of affected children unless two carriers happen to marry. His objection rests on the difference between carrying an allele and expressing the disease. The community carrier frequency of 1 in 40 explains why two unrelated carriers can meet. The caveat to state is that 50% values should be confirmed by DNA testing of the family mutation, and that genetic counselling must be non-directive.

2. Give the recurrence risk for the next pregnancy, the chance that a healthy sibling is a carrier, and the population risk, and state what a prenatal test would examine.

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Since both parents are obligatory carriers, each pregnancy has a 1 in 4 chance of an affected child, a 1 in 2 chance of a carrier child and a 1 in 4 chance of a homozygous normal child, independently of the sex of the child. Among children who are healthy, the possible genotypes are homozygous normal and carrier in a 1:2 ratio, so the chance that a healthy sibling is a carrier is 2/3, not 1/2. For comparison, an unrelated couple from the same community would face (1/40) × (1/40) × 1/4 = 1/6400 per pregnancy, so this couple's risk is 1600 times higher; family history, not population frequency, drives the counselling. A prenatal test would either measure the enzyme activity in cultured amniotic fluid cells or chorionic villus cells, or, more reliably, test the DNA for the exact exon 4 substitution found in the child, since the assay may not distinguish a carrier fetus from a normal one with confidence. The reasoning point is that the couple should be told the risk before conception, not only after an affected child is born.

3. Explain the molecular consequence of the exon 4 change and why the same mutation leaves carriers healthy. Also relate the 1 in 40 carrier frequency to how rare the disease is.

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The substitution converts a glutamine codon into UAA, one of the three stop codons: a nonsense mutation. The ribosome terminates translation at that point, so the polypeptide is truncated and lacks the catalytic region of the enzyme; the shortened protein is usually misfolded and degraded, and the mRNA may itself be destroyed by nonsense-mediated decay. The result is the 0% activity measured in the child. In a heterozygote the normal allele is still transcribed and translated, giving roughly half the normal amount of full-length enzyme. That is above the threshold of activity needed for normal lysosomal function, so carriers show about 50% activity but no symptoms - clinical recessiveness here is a matter of enzyme reserve, not of the mutant protein being absent from carriers. The population figures follow from the Hardy–Weinberg relation: a carrier frequency of 1 in 40 means 2pq ≈ 1/40, so q ≈ 1/80 and the disease incidence is q² ≈ 1/6400 births. The disease is rare even though carriers are common, which is why community screening identifies couples at risk before an affected child appears.