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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.

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.

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,

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

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.

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.