Albinism in humans is an autosomal recessive disease. for the following families, determine the genotypes of the parents and offspring. when two alternative genotypes are possible, list both.
a) two normally pigmented parents have five children, four normally pigmented, one albino.
b) a normally pigmented male and an albino female have six children, all with normal pigmentation.
c) a normal male and an albino female have six children, three with normal pigmentation, three with albinism.

Normally pigmented: P (dominant)

Albino: a (recessive)

a) Two normally pigmented parents have five children, four normally pigmented, one albino.

If parents are normally pigmented and had an albino child, both need to be heterozygous dominant for normal pigments

Pa x Pa =( PP, Pa, aP, aa) Those are all the probabilities, left to chance.

b) A normally pigmented male and an albino female have six children, all with normal pigmentation.

The albino is a recessive character so, an albino person, it will always aport the albino allele.

If none of the children is albino, you could say the male is heterogygous or not.

PP (male) x aa (female) = (Pa, Pa, aP, aP) phenotypically all will be normally pigmented.

or

Pa (male) x aa (female) = (Pa, Pa, aa, aa) Because of chance, it could be this way but never appear the aa combination of alleles.

c) A normal male and an albino female have six children, three with normal pigmentation, three with albinism.

Now, you have a male with heterozygous alleles.

Pa (male) x aa (female) = (Pa, Pa, aa, aa) Those are all the probabilities, left to chance.

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explanation:

the answer is trees if i'm not mistaken