Question

In Exercises 25–32, find the probability and answer the questions.

Genetics: Eye Color 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 eyes will be brown. (The actual determination of eye color is more complicated than that.)

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 eyes?

Short answer

a. The possible outcomes of the genotype of the child are given as brown/brown, brown/blue, blue/brown, and blue/blue.

b. The probability that a child will have blue/blue genotype is equal to 0.25.

c. The probability that the child will have brown eyes is equal to 0.75.

Step-by-step solution

Given information

It is given that each of the two parents’ genotypes is brown/blue.

If one of the alleles is brown, the eye colour will be brown.

Sample space

a.

In probability theory,the list of all possible outcomes of an event is called thesample space.

Here, each of the two parents has the genotype brown/blue and one allele is carried from each of the two parents to the child; therefore, the following possible outcomes for the genotype of the child can be listed:

• If the brown allele is inherited from the mother and the brown allele is inherited from the father, the genotype of the child will be brown/brown.
• If the brown allele is inherited from the mother and the blue allele is inherited from the father, the genotype of the child will be brown/blue.
• If the blue allele is inherited from the mother and the brown allele is inherited from the father, the genotype of the child will be blue/brown.
• If the blue allele is inherited from the mother and the blue allele is inherited from the father, the genotype of the child will be blue/blue.

Therefore, the possible outcomes of the genotype of the child become brown/brown, brown/blue, blue/brown, and blue/blue.

Define probability

Theprobability of an eventis calculated by dividing the number of favourable outcomes of an event by the total number of outcomes.

For an arbitrary event A,

$P\left(A\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}\text{of}\text{A}}{\text{Total}\text{number}\text{of}\text{outcomes}}$

Compute the probability value for the blue/blue genotype

b.

The total number of possible genotypes = 4.

The number of genotypes of the blue/blue combination = 1.

Therefore, the probability that parents have blue/blue genotype is given by the following equation:

$$\begin{gathered} P\left(\text{blue/blue}\text{genotype}\right)=\frac{1}{4} \\ =0.25 \end{gathered}$$

Therefore, the probability that the child will have the blue/blue genotype is equal to 0.25.

Compute of probability that child has brown eyes

c.

It is given that if one of the alleles is brown, the eye colour will be brown.

The following are the type of eye colours for each of the four genotypes:

• brown/brown genotype: brown eye colour
• brown/blue genotype: brown eye colour
• blue/brown genotype: brown eye colour
• blue/blue genotype: blue eye colour

The total number of genotypes that produce brown eye colour = 3.

The total number of genotypes = 4.

The probability that the child will have brown eyes is given by the following equation:

$$\begin{gathered} P\left(browneyes\right)=\frac{\text{Number}\text{of}\text{genotypes}\text{that}\text{give}\text{brown}\text{eyes}}{\text{Total}\text{number}\text{of}\text{genotypes}} \\ =\frac{3}{4} \\ =0.75 \end{gathered}$$

Therefore, the probability that the child will have brown eyes is equal to 0.75.