Answer :
To determine the probability that the child will have color-deficient vision, we need to understand sex-linked inheritance and the given genotypes of the parents.
The mother's genotype is [tex]\(X^R X^r\)[/tex], indicating she has one dominant allele (non-color deficient) and one recessive allele (color deficient). The father's genotype is [tex]\(X^r Y\)[/tex], indicating he has the recessive allele for color deficiency on his X chromosome and does not carry another X chromosome for this trait since he has a Y chromosome.
Let's determine the possible genotypes of their children:
1. Mother's possible alleles: [tex]\(X^R\)[/tex], [tex]\(X^r\)[/tex]
2. Father's possible alleles: [tex]\(X^r\)[/tex], [tex]\(Y\)[/tex]
By combining these, we can form the potential genotypes of the offspring. We have the following combinations and their respective phenotypes:
1. [tex]\(X^R X^r\)[/tex]: This child receives [tex]\(X^R\)[/tex] from the mother and [tex]\(X^r\)[/tex] from the father, resulting in a non-color deficient female.
2. [tex]\(X^R Y\)[/tex]: This child receives [tex]\(X^R\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father, resulting in a non-color deficient male.
3. [tex]\(X^r X^r\)[/tex]: This child receives [tex]\(X^r\)[/tex] from both parents, resulting in a color deficient female.
4. [tex]\(X^r Y\)[/tex]: This child receives [tex]\(X^r\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father, resulting in a color deficient male.
We have four equally likely genotypes for the children. To find the probability of a child being color-deficient, we look at the combinations that result in color deficiency:
- [tex]\(X^r X^r\)[/tex] (color deficient female)
- [tex]\(X^r Y\)[/tex] (color deficient male)
There are 2 combinations that result in color deficiency out of the 4 possible genotypes. Thus, the probability that a child will have color-deficient vision is:
[tex]\[ \frac{\text{Number of color-deficient genotypes}}{\text{Total number of possible genotypes}} = \frac{2}{4} = 0.50 \][/tex]
Therefore, the correct answer is:
A. 0.50
The mother's genotype is [tex]\(X^R X^r\)[/tex], indicating she has one dominant allele (non-color deficient) and one recessive allele (color deficient). The father's genotype is [tex]\(X^r Y\)[/tex], indicating he has the recessive allele for color deficiency on his X chromosome and does not carry another X chromosome for this trait since he has a Y chromosome.
Let's determine the possible genotypes of their children:
1. Mother's possible alleles: [tex]\(X^R\)[/tex], [tex]\(X^r\)[/tex]
2. Father's possible alleles: [tex]\(X^r\)[/tex], [tex]\(Y\)[/tex]
By combining these, we can form the potential genotypes of the offspring. We have the following combinations and their respective phenotypes:
1. [tex]\(X^R X^r\)[/tex]: This child receives [tex]\(X^R\)[/tex] from the mother and [tex]\(X^r\)[/tex] from the father, resulting in a non-color deficient female.
2. [tex]\(X^R Y\)[/tex]: This child receives [tex]\(X^R\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father, resulting in a non-color deficient male.
3. [tex]\(X^r X^r\)[/tex]: This child receives [tex]\(X^r\)[/tex] from both parents, resulting in a color deficient female.
4. [tex]\(X^r Y\)[/tex]: This child receives [tex]\(X^r\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father, resulting in a color deficient male.
We have four equally likely genotypes for the children. To find the probability of a child being color-deficient, we look at the combinations that result in color deficiency:
- [tex]\(X^r X^r\)[/tex] (color deficient female)
- [tex]\(X^r Y\)[/tex] (color deficient male)
There are 2 combinations that result in color deficiency out of the 4 possible genotypes. Thus, the probability that a child will have color-deficient vision is:
[tex]\[ \frac{\text{Number of color-deficient genotypes}}{\text{Total number of possible genotypes}} = \frac{2}{4} = 0.50 \][/tex]
Therefore, the correct answer is:
A. 0.50
