Question

each of two parents has the genotype brown/blue, which consists of the pair of alleles that determine eye color, and each parent contributes one of those alleles to a child. assume that if the child has at least one brown allele, that color will dominate and the childs eye color will be brown.
a. list the different possible outcomes. assume that these outcomes are equally likely.
b. what is the probability that a child of these parents will have the blue/blue genotype?
c. what is the probability that the child will have brown eye color?
b. brown/brown, brown/blue, blue/brown, and blue/blue
c. brown/brown, brown/blue, and blue/blue
d. brown/brown and blue/blue
b. the probability that a child of these parents will have the blue/blue genotype is $\frac{1}{4}$ (round to two decimal places as needed.)

Explanation:

Step1: List possible outcomes

Parents with genotype brown/blue can pass on either brown (B) or blue (b) alleles. Using a Punnett - square, the possible genotypes of the child are: (B, B) for brown/brown, (B, b) for brown/blue, (b, B) for blue/brown, and (b, b) for blue/blue.

Step2: Calculate probability of blue/blue genotype

There are 4 total possible outcomes. The blue/blue genotype is 1 outcome. So the probability $P(\text{blue/blue})=\frac{1}{4}= 0.25$.

Step3: Calculate probability of brown - eyed child

A child has brown eyes if the genotype is brown/brown, brown/blue, or blue/brown. There are 3 such outcomes out of 4 total outcomes. So the probability $P(\text{brown eyes})=\frac{3}{4}=0.75$.

Answer:

a. B. brown/brown, brown/blue, blue/brown, and blue/blue
b. 0.25
c. 0.75