Question

Pretend human eye color is determined by a single gene, with brown eyes (B) being dominant over blue (b). A brown-eyed man marries a blue-eyed woman, and they have three children. Two of the children are brown-eyed, and one is blue-eyed. What is the probability the brown-eyed children have the genotype BB?

Answer 1

The probability of the brown-eyed children having the genotype BB is 0% as their father must be heterozygous (Bb) in order to have a blue-eyed child with a blue-eyed woman. The existence of a blue-eyed child confirms that both parents cannot have a homozygous dominant BB genotype.

Step-by-step explanation:

The question revolves around the genetics of eye color, specifically the probability of the brown-eyed children inheriting a certain genotype (BB)

Given that we know the blue-eyed woman must have a genotype of bb (as blue eye color is recessive), and one of their children also has blue eyes, we can conclude that the brown-eyed man must have a genotype of Bb in order to produce a blue-eyed child. This is because two heterozygous Bb parents have a 25% chance of producing a bb (blue-eyed) offspring.

Therefore, with each Bb parent (the brown-eyed man and the blue-eyed woman), we use a Punnett square to predict offspring genotypes. The possible genotypes for their children would be 25% BB, 50% Bb, and 25% bb. However, since one child is blue-eyed (bb), we know the man's genotype cannot be BB. So the possible genotypes for the two brown-eyed children are either BB or Bb. But since BB is not possible, as at least one parent is Bb, the probability of the brown-eyed children having the genotype BB is 0%.