Final Answer:
1. Individuals I-1, II-3, II-4, III-2, III-4, III-5, III-6, IV-1, IV-2, IV-3, IV-4, IV-5
2. Huntington’s disease is dominant (because affected individuals have at least one affected parent; no skipping of generations).
3. 4 girls: II-3, II-4, III-2, III-4
4. III-1 and III-2 have Huntington’s; III-3 and III-4 do not → 2 with disease
5. III-2 and II-4 are mother and daughter → related as parent–child
6. The pedigree shows autosomal dominant inheritance (affected individuals in every generation, both sexes affected, no unaffected parents having affected children)
7. Because III-1 and III-2 are affected and have an unaffected child (III-3), the trait cannot be recessive — recessive traits skip generations and require two unaffected carriers to produce an affected child. Here, affected parents have unaffected kids, so it must be dominant.
8. III-1 and III-2 are both affected and have unaffected children (III-3 and III-4), so they must both be heterozygous (Hh × Hh).
9. III-3 and III-4 (unaffected) → genotype hh
10. IV-1 and IV-2 (affected) → genotype Hh (since their parent III-3 is hh, they must have inherited h from her and H from father III-2)
11. No — because Huntington’s is autosomal dominant, a carrier (heterozygous) *has* the disease. There is no asymptomatic carrier state; if you have the allele, you get the disease (with near-complete penetrance by adulthood).
Parent Tip: Review the logic above to help your child master the concept of sickle cell anemia worksheet answers.
