For the purposes of this problem, assume that in humans the gene for brown eyes

Question

For the purposes of this problem, assume that in humans the gene for brown eyes is dominant to that for blue eyes.

A brown-eyed man marries a blue-eyed woman, and they have eight brown-eyed children. What are the genotypes of all the individuals in the family?

If the first child is a brown eyed girl (same parents as stated above), what is the probability that the second child will be a blue-eyed boy?

Now, assuming that the first child of the marriage described above is a brown eyed girl and she marries a heterozygous brown-eyed boy. What is the probabilty that their first three children will be blue-eyed girls and fourth a brown-eyed boy?

Explanation / Answer

1)

Based on the given data, brown eyes is dominant to that for blue eyes. For example, consider the gene symbol:

The genotypes of Brown eye are:

The genotype of Blue eye is:

bb- (Homozygous recessive)

Here, a brown-eyed man (BB) marries a blue-eyed woman (aa) and progeny is:

Thus, all children have Brown eyes (100%). So,

2)

If the genotype of father is (Bb), so the cross would look like

So, the probability that the second child will be a blue-eyed boy is 50%

3)

The genotype of first child is: Bb and the heterozygous brown-eyed boy genotype is: Bb. So, the cross would look like

So, the probability that their first three children will be blue-eyed girls is ¾ or 0.75 and the probability of fourth brown-eyed boy is ¼ or 0.25.

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