Question

The color of a person’s eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed, or if one is blueeyed and the other brown-eyed, then the person will have brown eyes. The brown-eyed gene is therefore said to be dominant over the blue-eyed gene. A newborn child independently receives one eye gene from each of its parents, chosen at random from the two eye genes belonging to each parent. Suppose that Smith and both of his parents have brown eyes, but Smith’s sister has blue eyes.

(a) What is the probability that Smith possesses a blue-eyed gene? Now suppose that Smith’s wife has blue eyes.
(b) What is the probability that their first child will have blue eyes?
(c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

Answer 1

A) 50%

B) 25%

C) Same 25%

Step-by-step explanation:

a) Smith parents should be heterozygous to the eye color gene, this means, they each should carry a copy of the “brown eye” gene (that we will call B), and one of the “blue eyed” one (that we will call b). Since the blue eye is a recessive trait, you can assume that they both inherited b genes to the daughter. If we draw a Punnet square (Bb vs Bb), we will see the possibilities of Smith having a copy of the blue eye gene (Bb), and it is 50%.

b) Again, with a Punnet square (Bb vs bb) we see that the possibility of having a blue eyed baby is 50% IF Smith carries the copy of the b gene. But since that possibility is established to be 50%, therefore the possibility of the baby to have blue eyes is reduced by half to 25%

c) The probability is the same, it does not have an additive effect, so there is no effect on the number of children