Answer:
Half would have the genotype Aa.
Explanation:
Given that the genotype of both of the parents was "Aa"; both parents would make two types of gametes. The 50% gametes from each parent would carry allele "A" while the rest 50% gametes from each parent would carry allele "a".
The random fusion of gametes of two parents would give the progeny with genotype ratio 1/4 AA: 1/2 Aa: 1/4 aa.
Answer:
50%
Explanation:
This question involves a gene coding for eye color in humans. The allele for brown eyes (B) are dominant to blue eyes (b). This means that blue-eyed individual will possess genotype: bb.
According to this question, a blue-eyed man (bb) has children with a brown-eyed woman (B_) whose mother has blue eyes (bb). Since the mother of the brown-eyed woman has a blue eye, this means that the woman will be heterozygous for brown eye (Bb).
Hence, the parents in this question will cross as follows: blue eyed man (bb) × brown-eyed woman (Bb). The offspring/children will have the following genotypes (see attached punnet square); Bb, Bb, bb, bb.
Based on the question, 2/4 = 1/2 of the children will be heterozygous for the eye color trait. That is, ½ × 100 = 50%.
