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%.
Macronutrients are, in the most basic terms, the components of food that ... gram, and that's not including the additional calories from sugary additives, etc. ... absorbed by the body, making it possible for you to still be deficient.
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