Answer :
Let's analyze the genotypes of the parents and the possible outcomes for their offspring.
1. Mother's Genotype: \(X^R X^r\)
- The mother is a carrier for the color-deficient vision gene.
- She carries one normal vision allele (\(X^R\)) and one color-deficient vision allele (\(X^r\)).
2. Father's Genotype: \(X^R Y\)
- The father has normal vision and carries one normal vision allele (\(X^R\)) and one Y chromosome.
To determine the possible genotypes of the offspring, we can set up a Punnett square. The potential combinations of the alleles from each parent are as follows:
| Mother \(X^R\) | Mother \(X^r\) |
|:--------------:|:-------------:|
| Father \(X^R\) | Father \(X^R\) |
Here are the potential genotypes of the offspring:
- From \(X^R\) (mother) and \(X^R\) (father): \(X^R X^R\) - Normal Vision (female)
- From \(X^r\) (mother) and \(X^R\) (father): \(X^R X^r\) - Carrier (female, but with normal vision)
- From \(X^R\) (mother) and \(Y\) (father): \(X^R Y\) - Normal Vision (male)
- From \(X^r\) (mother) and \(Y\) (father): \(X^r Y\) - Color-deficient vision (male)
Now, let's list the possible combinations and count them:
1. \(X^R X^R\) - female with normal vision
2. \(X^R X^r\) - female carrier with normal vision
3. \(X^R Y\) - male with normal vision
4. \(X^r Y\) - male with color-deficient vision
Out of these 4 combinations, only one results in a child with color-deficient vision (\(X^r Y\)).
The probability that the child will have color-deficient vision is calculated by dividing the favorable outcomes by the total possible outcomes:
[tex]\[ \frac{\text{Number of outcomes for color-deficient vision}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25 \][/tex]
Thus, the probability that the child will have color-deficient vision is:
[tex]\[ \boxed{0.25} \][/tex]
Therefore, the answer is:
C. 0.25
1. Mother's Genotype: \(X^R X^r\)
- The mother is a carrier for the color-deficient vision gene.
- She carries one normal vision allele (\(X^R\)) and one color-deficient vision allele (\(X^r\)).
2. Father's Genotype: \(X^R Y\)
- The father has normal vision and carries one normal vision allele (\(X^R\)) and one Y chromosome.
To determine the possible genotypes of the offspring, we can set up a Punnett square. The potential combinations of the alleles from each parent are as follows:
| Mother \(X^R\) | Mother \(X^r\) |
|:--------------:|:-------------:|
| Father \(X^R\) | Father \(X^R\) |
Here are the potential genotypes of the offspring:
- From \(X^R\) (mother) and \(X^R\) (father): \(X^R X^R\) - Normal Vision (female)
- From \(X^r\) (mother) and \(X^R\) (father): \(X^R X^r\) - Carrier (female, but with normal vision)
- From \(X^R\) (mother) and \(Y\) (father): \(X^R Y\) - Normal Vision (male)
- From \(X^r\) (mother) and \(Y\) (father): \(X^r Y\) - Color-deficient vision (male)
Now, let's list the possible combinations and count them:
1. \(X^R X^R\) - female with normal vision
2. \(X^R X^r\) - female carrier with normal vision
3. \(X^R Y\) - male with normal vision
4. \(X^r Y\) - male with color-deficient vision
Out of these 4 combinations, only one results in a child with color-deficient vision (\(X^r Y\)).
The probability that the child will have color-deficient vision is calculated by dividing the favorable outcomes by the total possible outcomes:
[tex]\[ \frac{\text{Number of outcomes for color-deficient vision}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25 \][/tex]
Thus, the probability that the child will have color-deficient vision is:
[tex]\[ \boxed{0.25} \][/tex]
Therefore, the answer is:
C. 0.25
