Albinism is an autosomal recessive disorder (aa = affected, AA or Aa = normal). The grandparents, both phenotypically normal, have two children: one affected (aa) and one normal. The normal daughter (who is phenotypically normal) marries an unrelated normal man, and they have three children, two of whom are affected and one is normal. Assuming no new mutations, determine the most likely genotypes of the grandparents, the normal daughter, and her husband.
Answer
The grandparents are heterozygous (Aa); the normal daughter is heterozygous (Aa); her husband is also heterozygous (Aa).
The correct answer is The grandparents are heterozygous (Aa); the normal daughter is heterozygous (Aa); her husband is also heterozygous (Aa). — Therefore, the most consistent genotypes are: grandparents Aa, daughter Aa, husband Aa.
Step-by-step solution
- Because the grandparents are normal but produced an affected child (aa), each must carry one recessive allele: genotype Aa.
- The affected grand‑child (aa) confirms both parents contributed a recessive allele.
- The normal daughter must have received at least one dominant allele from each parent, so her genotype can be AA or Aa.
- Since she later produced two affected children with a normal husband, the husband must also carry a recessive allele; otherwise no aa offspring could appear.
- The only cross that can give two affected children among three is Aa (daughter) × Aa (husband), which yields a 1/4 chance of aa per child.
- Therefore, the most consistent genotypes are: grandparents Aa, daughter Aa, husband Aa.
Common mistake: Assuming a phenotypically normal husband must be AA, ignoring the possibility of being a carrier (Aa).
Concept tested: Autosomal recessive pedigree analysis
Published · How this solution was prepared and checked
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