Question

each of two parents has the genotype green/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 green allele, that color will dominate and the childs eye color will be green.
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 green eye color?
a. list the possible outcomes.
a. green/green and blue/blue
b. green/green, green/blue, and blue/blue
c. green/green, green/blue, blue/green, and blue/blue
d. green/blue and blue/green
b. the probability that a child of these parents will have the blue/blue genotype is 0.25 (round to two decimal places as needed.)
c. the probability that the child will have green eye color is (round to two decimal places as needed.)

Explanation:

Step1: Determine total number of outcomes

Each parent has 2 alleles (green and blue), so the total number of possible combinations for the child's genotype is $2\times2 = 4$ (using the multiplication - principle for independent events). The possible genotypes are green/green, green/blue, blue/green, blue/blue.

Step2: Calculate probability of green - eyed child

A child has green eyes if the genotype is green/green, green/blue, or blue/green. There are 3 such favorable outcomes out of 4 total outcomes.
The probability $P$ of an event is given by the formula $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. So, $P=\frac{3}{4}=0.75$.

Answer:

a. C. green/green, green/blue, blue/green, and blue/blue
b. 0.25
c. 0.75