Pedigree Plus Population Data: Deciding Mode of Inheritance
A family shows an inherited condition in 4 of 22 members across three generations. Affected individuals include both sexes; affected fathers have unaffected daughters, and unaffected parents occasionally have an affected child. No affected individual has an affected parent in two of the four cases. A student notes that the allele is rare in the general population (about 1 in 5,000) and argues the pattern "must be recessive because it skips generations." A second student argues it "must be X-linked recessive because more males are affected in the pedigree" (3 males, 1 female).
1. Assess both students' reasoning and identify what additional evidence would discriminate among hypotheses.
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"Skips generations" is consistent with autosomal recessive but does not prove it; reduced penetrance or a new mutation can mimic skipping under dominant inheritance, and rare recessive conditions can appear in unaffected carriers. The sex-ratio argument is weak: with only 4 affected individuals, 3:1 is well within chance, and X-linked recessive would predict no affected daughters from an affected father (unless the mother is a carrier). Key discriminators: (1) whether affected fathers ever transmit to daughters; (2) whether all sons of affected mothers are affected (X-linked recessive prediction); (3) consanguinity raising recessive risk; (4) carrier testing or molecular data; (5) whether unaffected parents of affected children are related. The correct answer is to test predictions of each model against the pedigree rather than infer from a small sample.
2. Using the population frequency, estimate carrier frequency and explain how it bears on the interpretation.
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If the condition is autosomal recessive with q² ≈ 1/5000, then q ≈ 1/√5000 ≈ 0.014, and carrier frequency 2pq ≈ 2(0.986)(0.014) ≈ 0.028, or about 1 in 36. That is high enough that unrelated carrier parents are plausible, weakening the need to invoke consanguinity. If instead the condition were dominant, the allele frequency would be near the disease frequency (about 0.0002), making an affected parent far more likely than two carrier parents. Comparing these expected frequencies against the observed pedigree (e.g., whether affected individuals typically have an affected parent) helps choose between models. This requires applying Hardy-Weinberg reasoning to a specific case, not reciting the equation.