A blue-eyed dark-haired child was born in the family, similar in these features to his father. Mother – brown-eyed dark-haired; maternal grandmother – blue-eyed dark-haired; grandfather – brown-eyed blonde; paternal grandparents – brown-eyed dark-haired. What are the genotypes of all the persons mentioned?
Let’s denote the genes from the problem statement:
A – dark hair gene;
a – light hair gene;
B – gene for brown eyes;
b – gene for blue eyes.
First, it is necessary to determine the genotypes of grandparents. On the mother’s side: the genotypes are different, hence the grandmother is AAbb, the grandfather is aaBB. In this case, a mother with the AaBb genotype will be born. Now it is necessary to determine the genotype of the father. Because paternal grandparents have dark hair and brown eyes, so that a child with blue eyes is born, they must both be heterozygous for this trait. Hence, grandmother – AABb – dark hair and brown eyes, grandfather – AABb – dark hair and brown eyes. Then a child with the AAbb genotype will be born – dark hair and blue eyes.
AAbb * AaBb, then the following genotypes of children are obtained: AABb – dark hair and brown eyes, Aabb – dark hair and blue eyes. the probability of having a baby with dark hair and blue eyes is 50%.
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