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 child's 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, green/blue, blue/green, and blue/blue B. green/blue and blue/green C. green/green and blue/blue D. green/green, green/blue, and blue/blue b. The probability that a child of these parents will have the blue/blue genotype is (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.)
Solution
Step 1 ：The possible combinations of alleles that the child could inherit from the parents are green/green, green/blue, blue/green, and blue/blue. These combinations are equally likely.
Step 2 ：To find the probability of each outcome, we need to count the number of ways each outcome can occur and divide by the total number of outcomes. The total number of outcomes is 4.
Step 3 ：The probabilities of the child inheriting each possible combination of alleles are all equal at 0.25. This means that the probability of the child having the blue/blue genotype is \(\boxed{0.25}\).
Step 4 ：However, the child will have green eyes if they inherit at least one green allele. This can occur in three out of the four possible outcomes: green/green, green/blue, and blue/green. Therefore, the probability of the child having green eyes is \(\boxed{0.75}\).
