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?
a. list the possible outcomes.
a. brown/blue and blue/brown
b. brown/brown, brown/blue, blue/brown, and blue/blue
c. brown/brown, brown/blue, and blue/blue
d. brown/brown and blue/blue

Explanation:

Step1: Analyze allele - combination possibilities

Each parent has alleles brown (B) and blue (b). Using a Punnett - square or combinatorial analysis, the possible combinations when parents contribute one allele each are: (B from one parent and B from the other), (B from one parent and b from the other), (b from one parent and B from the other), (b from one parent and b from the other). So the genotypes are brown/brown, brown/blue, blue/brown, blue/blue.

Step2: Calculate probability of blue/blue genotype

There are 4 possible genotypes (brown/brown, brown/blue, blue/brown, blue/blue). The probability of a specific genotype is the number of favorable outcomes divided by the total number of outcomes. The number of favorable outcomes for blue/blue is 1, and the total number of outcomes is 4. So the probability $P(\text{blue/blue})=\frac{1}{4}$.

Step3: Calculate probability of brown - eye color

Brown - eye color occurs when the genotype is brown/brown, brown/blue, or blue/brown. The number of favorable outcomes for brown - eye color is 3, and the total number of outcomes is 4. So the probability $P(\text{brown eye color})=\frac{3}{4}$.

Answer:

a. B. brown/brown, brown/blue, blue/brown, and blue/blue
b. $\frac{1}{4}$
c. $\frac{3}{4}$