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Chapter 3: Problem 4

Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and offspring for the following families. When two alternative genotypes are possible, list both. (a) Two parents without albinism have five children, four without albinism and one with albinism. (b) A male without albinism and a female with albinism have six children, all without albinism.

Short Answer

Expert verified
Answer: In the first scenario, both parents have a genotype of Aa and their children can have genotypes of AA, Aa, or aa. In the second scenario, the male parent has a genotype of AA and the female parent has a genotype of aa. All their children have a genotype of Aa.

Step by step solution

01

Identify the genotype of the child with albinism

The child with albinism must have the genotype "aa", as it requires two copies of the recessive allele to express the trait.
02

Determine the parents' genotypes from their children

Since the parents have a child with albinism, both parents must carry the recessive allele "a". This means that the parents are both heterozygous (Aa) for the trait, as they both do not express albinism.
03

Determine the genotypes of the other children

These children could have either of the two genotypes without albinism: "AA" or "Aa". To determine the probabilities for each genotype, we can use a Punnett square or calculate probabilities directly. Since both parents are heterozygous (Aa), each child has a 25% chance of being "AA", a 50% chance of being "Aa", and a 25% chance of being "aa" (expressing albinism). In this case, four children without albinism could have either the "AA" or "Aa" genotype. (b) A male without albinism and a female with albinism have six children, all without albinism.
04

Identify the genotype of the female with albinism

Given that the female expresses albinism, her genotype must be "aa".
05

Determine the genotype of the male without albinism from their children

Since all six children are without albinism and have received one allele from each parent, the male without albinism must have the genotype "AA", so his children inherit the "A" allele and do not express albinism.
06

Determine the genotypes of the children

Since the father has the genotype "AA" and the mother has the genotype "aa", each child must inherit one "A" allele from the father and one "a" allele from the mother. Therefore, all six children have the genotype "Aa" without expressing albinism.

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Most popular questions from this chapter

Distinguish between homozygosity and heterozygosity.

Consider three independently assorting gene pairs, \(A / a, B / b,\) and $C / c,\( where each demonstrates typical dominance \)(A-, B-, C-)$ and recessiveness \((a a, b b, c c) .\) What is the probability of obtain ing an offspring that is \(A A B b C c\) from parents that are \(A a B b C C\) and \(A A B b C c ?\)

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