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To determine the probability that a child will have color-deficient vision given the parents' genotypes, we should analyze the possible genotypes of the children. The genotypes of the parents are:
Mother: [tex]\(X^R X^r\)[/tex]
Father: [tex]\(X^r Y\)[/tex]
Since color-deficient vision is a sex-linked recessive trait, it is important to remember that the gene is located on the X chromosome, and the trait is expressed when there is no dominant allele present. Hence, for females, the color-deficient vision is present if they have the genotype [tex]\(X^r X^r\)[/tex]. For males, since they have only one X chromosome, the presence of the [tex]\(X^r\)[/tex] allele will result in color-deficient vision, making the critical male genotype [tex]\(X^r Y\)[/tex].
We'll start by listing all possible combinations of the children's genotypes when considering the segregation of the X and Y chromosomes during reproduction.
The mother can pass on either [tex]\( X^R \)[/tex] or [tex]\( X^r \)[/tex], and the father can pass on either [tex]\( X^r \)[/tex] or [tex]\( Y \)[/tex].
Here are the possible combinations:
1. [tex]\(X^R\)[/tex] from the mother and [tex]\(X^r\)[/tex] from the father: [tex]\(X^R X^r\)[/tex]
2. [tex]\(X^R\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^R Y\)[/tex]
3. [tex]\(X^r\)[/tex] from the mother and [tex]\(X^r\)[/tex] from the father: [tex]\(X^r X^r\)[/tex]
4. [tex]\(X^r\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^r Y\)[/tex]
Next, we'll determine which of these combinations result in color-deficient vision:
1. [tex]\(X^R X^r\)[/tex] - This is a female with one dominant allele [tex]\(X^R\)[/tex] and one recessive allele [tex]\(X^r\)[/tex]. The presence of the dominant [tex]\(X^R\)[/tex] allele means that this individual will have normal vision.
2. [tex]\(X^R Y\)[/tex] - This is a male with only one X chromosome carrying the [tex]\(X^R\)[/tex] allele. Since [tex]\(X^R\)[/tex] is dominant, this male will have normal vision.
3. [tex]\(X^r X^r\)[/tex] - This is a female with two recessive alleles [tex]\(X^r\)[/tex]. She will have color-deficient vision because she lacks a dominant [tex]\(X^R\)[/tex] allele.
4. [tex]\(X^r Y\)[/tex] - This is a male with one recessive allele [tex]\(X^r\)[/tex]. Since there is no other X chromosome to counter the recessive allele, this male will have color-deficient vision.
Counting the combinations, we have a total of 4:
- Two of the combinations ( [tex]\(X^r X^r\)[/tex] and [tex]\(X^r Y\)[/tex] ) lead to color-deficient vision.
- The other two combinations ( [tex]\(X^R X^r\)[/tex] and [tex]\(X^R Y\)[/tex] ) lead to normal vision.
Therefore, the probability that the child will have color-deficient vision is calculated by dividing the number of color-deficient cases by the total number of possible combinations:
[tex]\[ \text{Probability} = \frac{\text{Number of color-deficient cases}}{\text{Total number of combinations}} = \frac{2}{4} = 0.5 \][/tex]
So, the probability that the child will have color-deficient vision is [tex]\(0.50\)[/tex].
Therefore, the answer is:
D. 0.50
Mother: [tex]\(X^R X^r\)[/tex]
Father: [tex]\(X^r Y\)[/tex]
Since color-deficient vision is a sex-linked recessive trait, it is important to remember that the gene is located on the X chromosome, and the trait is expressed when there is no dominant allele present. Hence, for females, the color-deficient vision is present if they have the genotype [tex]\(X^r X^r\)[/tex]. For males, since they have only one X chromosome, the presence of the [tex]\(X^r\)[/tex] allele will result in color-deficient vision, making the critical male genotype [tex]\(X^r Y\)[/tex].
We'll start by listing all possible combinations of the children's genotypes when considering the segregation of the X and Y chromosomes during reproduction.
The mother can pass on either [tex]\( X^R \)[/tex] or [tex]\( X^r \)[/tex], and the father can pass on either [tex]\( X^r \)[/tex] or [tex]\( Y \)[/tex].
Here are the possible combinations:
1. [tex]\(X^R\)[/tex] from the mother and [tex]\(X^r\)[/tex] from the father: [tex]\(X^R X^r\)[/tex]
2. [tex]\(X^R\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^R Y\)[/tex]
3. [tex]\(X^r\)[/tex] from the mother and [tex]\(X^r\)[/tex] from the father: [tex]\(X^r X^r\)[/tex]
4. [tex]\(X^r\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^r Y\)[/tex]
Next, we'll determine which of these combinations result in color-deficient vision:
1. [tex]\(X^R X^r\)[/tex] - This is a female with one dominant allele [tex]\(X^R\)[/tex] and one recessive allele [tex]\(X^r\)[/tex]. The presence of the dominant [tex]\(X^R\)[/tex] allele means that this individual will have normal vision.
2. [tex]\(X^R Y\)[/tex] - This is a male with only one X chromosome carrying the [tex]\(X^R\)[/tex] allele. Since [tex]\(X^R\)[/tex] is dominant, this male will have normal vision.
3. [tex]\(X^r X^r\)[/tex] - This is a female with two recessive alleles [tex]\(X^r\)[/tex]. She will have color-deficient vision because she lacks a dominant [tex]\(X^R\)[/tex] allele.
4. [tex]\(X^r Y\)[/tex] - This is a male with one recessive allele [tex]\(X^r\)[/tex]. Since there is no other X chromosome to counter the recessive allele, this male will have color-deficient vision.
Counting the combinations, we have a total of 4:
- Two of the combinations ( [tex]\(X^r X^r\)[/tex] and [tex]\(X^r Y\)[/tex] ) lead to color-deficient vision.
- The other two combinations ( [tex]\(X^R X^r\)[/tex] and [tex]\(X^R Y\)[/tex] ) lead to normal vision.
Therefore, the probability that the child will have color-deficient vision is calculated by dividing the number of color-deficient cases by the total number of possible combinations:
[tex]\[ \text{Probability} = \frac{\text{Number of color-deficient cases}}{\text{Total number of combinations}} = \frac{2}{4} = 0.5 \][/tex]
So, the probability that the child will have color-deficient vision is [tex]\(0.50\)[/tex].
Therefore, the answer is:
D. 0.50
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