Carrier Frequency and Hardy-Weinberg in a Founder Population
A genetic counselor serves an isolated island population of 1,000 people with a high rate of a recessive metabolic disorder caused by a single founder variant. Over 10 years, newborn screening identifies 8 affected infants among 1,000 births. The counselor genotypes 500 adults: 55 are heterozygous carriers, and no homozygous affected adults are found. The variant is absent from mainland populations. Consanguineous marriages are common on the island; about 20% of couples are first cousins. A woman from the island is a known carrier (she had an affected child) and plans to have a child with a mainland man with no family history. Assume the mainland carrier frequency for this disorder is 1/100. The counselor must interpret the observed carrier frequency, assess the couple's risk, and explain how consanguinity changes genotype frequencies.
1. Calculate the carrier frequency expected under Hardy-Weinberg from the newborn incidence. Compare it with the observed 55 carriers among 500 adults, and propose two plausible explanations for the discrepancy.
Read the worked answer
q^2 = 8/1000 = 0.008; q = sqrt(0.008) = 0.089; p = 0.911; expected carrier frequency = 2pq = 2 × 0.911 × 0.089 = 0.162, or 16.2%. Observed = 55/500 = 0.11, or 11%. Two explanations: (1) Consanguinity raises the frequency of affected homozygotes above q^2, so using the newborn incidence overestimates q and the carrier frequency. (2) The adult sample may not represent the population, or the affected cases may include other genetic causes; genotyping only one variant can miss carriers.
2. Explain how consanguinity affects allele and genotype frequencies. Then estimate the risk of an affected child for the known carrier woman if she has a child with (a) a mainland man and (b) her first cousin.
Read the worked answer
Consanguinity increases homozygosity (affected births) but does not change allele frequencies. For a mainland man: risk = P(woman transmits variant) × P(man is carrier) × P(man transmits variant) = 1/2 × 1/100 × 1/2 = 1/400 = 0.25%. For her first cousin: if she is a known carrier, one parent is a carrier; that parent's sibling has 1/2 chance of carrying the variant; the cousin has 1/2 of that = 1/4 chance. Risk = 1/2 × 1/4 × 1/2 = 1/16 = 6.25%.
3. Could negative selection against affected homozygotes explain a lower-than-expected adult carrier frequency? Use mutation-selection balance in your answer.
Read the worked answer
Negative selection can reduce allele frequency, but only slowly for a recessive allele because carriers are unaffected. At mutation-selection balance, q ≈ sqrt(μ/s) for a recessive lethal allele, so selection removes alleles only in homozygotes. A lower adult carrier frequency than the HWE estimate is more likely explained by consanguinity inflating the newborn incidence, sampling, or genetic heterogeneity than by strong selection in one generation. Selection alone would not produce such a large discrepancy quickly.