Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. (When two alternative genotypes are possible, list both.) (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c) A normal male and an albino female have six children, three normal and three albino. (d) Construct a pedigree of the families in (b) and (c). Assume that one of the normal children in (b) and one of the albino children in (c) become the parents of eight children. Add these children to the pedigree, predicting their phenotypes (normal or albino).

Two normal parents have five children, four normal and one albino. This means that both parents are carriers for the albino trait (heterozygous Aa) since they have produced an albino offspring (aa). Aa x Aa mating can produce AA (normal), Aa (normal), or aa (albino) offspring. The genotypes of parents: Aa and Aa, and the genotypes of the offspring: AA (normal), Aa (normal), Aa (normal), Aa (normal), and aa (albino).

A normal male and an albino female have six children, all normal. The albino female must have the genotype aa. The normal male must have at least one dominant A allele, and since all the offspring are normal, he must not carry the recessive albino allele. Therefore, the genotypes of the parents are AA (normal male) and aa (albino female), while the offspring are all Aa (normal).

A normal male and an albino female have six children, three normal and three albino. The albino female must have the genotype aa for the recessive trait. The normal male must be heterozygous Aa as three albino offspring are produced (meaning he has passed on his recessive a gene to those children). Therefore, the genotypes of the parents are Aa (normal male) and aa (albino female) while the offspring have genotypes Aa (normal), Aa (normal), Aa (normal), aa (albino), aa (albino), and aa (albino).

Construct a pedigree for families in (b) and (c) by assuming that one of the normal children from (b) and one of the albino children from (c) become parents of eight children. Parents from (b): AA (normal male) x aa (albino female) - All of their children will be Aa (normal). Parents from (c): Aa (normal male) x aa (albino female) - 3 children with genotype Aa (normal) and 3 children with genotype aa (albino). Choose one normal child (Aa) from family (b) and one albino child (aa) from family (c) as the new couple. So, their genotypes will be Aa (normal) x aa (albino). Their possible offspring genotypes are: Aa (normal), Aa (normal), Aa (normal), Aa (normal), aa (albino), aa (albino), aa (albino), and aa (albino). The pedigree will show the following phenotypes: - Family (b) - Parents: Normal male and Albino female. Offspring: 6 Normal children. - Family (c) - Parents: Normal male and Albino female. Offspring: 3 Normal children and 3 Albino children. - New family - Parents: Normal child from family (b) and Albino child from family (c). Offspring: 4 Normal children and 4 Albino children.