Heterozygote Deficit in an Island Lizard Population
A conservation team manages a closed population of 250 northern fence lizards on a 6-ha island. Albinism is a recessive trait controlled by a single locus, and 9 lizards are albino. To assess genetic health, the team genotypes 80 adults captured at random and finds 12 heterozygotes. The population has been isolated for at least 40 years, and the sex ratio is 1:1. A manager proposes releasing six unrelated lizards from a mainland population to increase genetic diversity. A colleague warns that the island population may be locally adapted and that six animals is ‘too few to matter.’ The team must decide whether to proceed, delay, or gather more data. They have limited funding for one more field season and can either genotype all 250 lizards or run a reciprocal transplant experiment with 40 lizards.
1. Assuming the population is in Hardy–Weinberg equilibrium, calculate the frequency of the albinism allele and the expected number of heterozygotes in an 80-lizard sample. Show your work.
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For a recessive trait, the albino frequency is q². Here q² = 9/250 = 0.036. Then q = √0.036 = 0.19, and p = 1 − 0.19 = 0.81. The expected heterozygote frequency is 2pq = 2(0.81)(0.19) = 0.3078. In a sample of 80 lizards, the expected number of heterozygotes is 0.3078 × 80 = 24.6, or about 25. The observed count is 12, roughly half the expectation. This calculation assumes random mating, no selection, no migration, no mutation, and a large population-assumptions that a small island population may violate.
2. The team observes only 12 heterozygotes in the sample. Identify two mechanisms that could explain the deficiency and explain why the HWE-based allele frequency may be biased.
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Two mechanisms are inbreeding and population subdivision within the island; selection against heterozygotes or genotyping error such as null alleles could also contribute. Inbreeding raises the frequency of homozygotes and lowers heterozygotes at all loci. If inbreeding is present, the albino frequency is q² + Fpq, not q² alone. Solving 9/250 = q² therefore overestimates q: the true q is lower, and the HWE expected heterozygote count of about 25 is too high. The calculated q is a rough upper estimate, and the observed deficit may be partly an artifact of applying HWE to a nonrandom-mating population. Pedigree data, multilocus genotypes, and an estimate of F would separate inbreeding from other causes.
3. A manager proposes releasing six unrelated lizards to increase genetic diversity. Evaluate this proposal using principles of drift and effective population size.
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Six migrants add at most 2.4% new individuals, so the immediate change in allele frequencies is small. However, if the migrants are unrelated and breed, they can raise effective population size and reduce inbreeding in the next generation more than the raw number suggests. Risks include outbreeding depression if the island population is locally adapted, disease introduction, and unequal reproductive success. A defensible plan is a pilot with 10–20 migrants, equal sex ratio, health screening, and genetic monitoring; six may be better than none, but it is not a guaranteed fix. Genotyping all 250 lizards would clarify the heterozygote deficit before release and provide a baseline for measuring any genetic rescue.