Determine the probability of birth in a family of brown-eyed children of normal height, if the following is known: the wife was blue-eyed short; the father had brown eyes and short stature, his parents were one – blue-eyed of normal height, and the other – brown-eyed and short. This family already has one child of normal height and blue eyes. Short stature and brown eyes are dominant features.
A – short stature
a – normal growth
B – brown eyes
b – blue eyes
Find: the probability of birth in a family of brown-eyed children of normal height.
The father’s parents had blue eyes and normal height (genotype – aabb) and brown eyes and short stature, which means that the father is a diheterozygote (AaBb).
R: ♀ Aabb (short stature, blue eyes) x ♂ AaBb (short stature, brown eyes)
Since in the task it is given that a blue-eyed child with normal growth was born to these parents during crossing, the mother is heterozygous for growth (since the genotype carrying normal growth is aa (homozygous for recessive).
G: Ab; ab; AB; Ab; aB; ab
F1: AABb (short stature, brown eyes); AAbb (short stature, blue eyes); AaBb (short stature, brown eyes); Aabb (short stature, blue eyes); ABBb (short, brown eyes); Aabb (short stature, blue eyes); aaBb (normal height; brown eyes); aabb (normal height, blue eyes).
Thus, the probability of having children with normal height and brown eyes is: 1/8 = 0.125
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