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Sagot :
To determine the probability that the child will have color-deficient vision given the parent genotypes [tex]\(X^R X^r \times X^r Y\)[/tex], we can use a Punnett square to visually represent the possible genetic combinations of the offspring.
### 1. Set Up Parent Genotypes
- Mother: [tex]\(X^R X^r\)[/tex]
- Father: [tex]\(X^r Y\)[/tex]
### 2. Create Gametes
Each parent produces gametes (sperm or eggs) that carry only one of their two sex chromosomes.
- Mother’s possible gametes: [tex]\(X^R\)[/tex] and [tex]\(X^r\)[/tex]
- Father’s possible gametes: [tex]\(X^r\)[/tex] and [tex]\(Y\)[/tex]
### 3. Construct the Punnett Square
We will pair each of the mother’s gametes with each of the father’s gametes to see all possible outcomes for the offspring.
| Parent \ Partner Gametes | [tex]\(X^r\)[/tex] (Father) | [tex]\(Y\)[/tex] (Father) |
|--------------------------|-------------------|-------------------|
| [tex]\(X^R\)[/tex] (Mother) | [tex]\(X^RX^r\)[/tex] | [tex]\(X^R Y\)[/tex] |
| [tex]\(X^r\)[/tex] (Mother) | [tex]\(X^rX^r\)[/tex] | [tex]\(X^r Y\)[/tex] |
### 4. Determine the Phenotypes
We look at each combination to determine if the child has color-deficient vision.
- [tex]\(X^RX^r\)[/tex]: The child is a carrier female but has normal vision because [tex]\(X^R\)[/tex] is dominant.
- [tex]\(X^R Y\)[/tex]: The child is a normal-vision male because [tex]\(X^R\)[/tex] is dominant.
- [tex]\(X^rX^r\)[/tex]: The child is a color-deficient female because both [tex]\(X^r\)[/tex] alleles carry the recessive trait.
- [tex]\(X^r Y\)[/tex]: The child is a color-deficient male because [tex]\(X^r\)[/tex] carries the recessive trait and there is no corresponding dominant allele on the Y chromosome.
### 5. Calculate the Probability
We summarize the outcomes:
- Total possible outcomes: 4 ([tex]\(X^RX^r\)[/tex], [tex]\(X^R Y\)[/tex], [tex]\(X^rX^r\)[/tex], [tex]\(X^r Y\)[/tex])
- Color-deficient outcomes: 2 ([tex]\(X^rX^r\)[/tex], [tex]\(X^r Y\)[/tex])
Therefore, the probability that a child will have color-deficient vision is the number of color-deficient outcomes divided by the total number of outcomes.
[tex]\[ \text{Probability} = \frac{\text{Number of color-deficient outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5 \][/tex]
### 6. Conclusion
The probability that the child will have color-deficient vision is [tex]\(0.50\)[/tex]. Therefore, the correct answer is:
D. 0.50
### 1. Set Up Parent Genotypes
- Mother: [tex]\(X^R X^r\)[/tex]
- Father: [tex]\(X^r Y\)[/tex]
### 2. Create Gametes
Each parent produces gametes (sperm or eggs) that carry only one of their two sex chromosomes.
- Mother’s possible gametes: [tex]\(X^R\)[/tex] and [tex]\(X^r\)[/tex]
- Father’s possible gametes: [tex]\(X^r\)[/tex] and [tex]\(Y\)[/tex]
### 3. Construct the Punnett Square
We will pair each of the mother’s gametes with each of the father’s gametes to see all possible outcomes for the offspring.
| Parent \ Partner Gametes | [tex]\(X^r\)[/tex] (Father) | [tex]\(Y\)[/tex] (Father) |
|--------------------------|-------------------|-------------------|
| [tex]\(X^R\)[/tex] (Mother) | [tex]\(X^RX^r\)[/tex] | [tex]\(X^R Y\)[/tex] |
| [tex]\(X^r\)[/tex] (Mother) | [tex]\(X^rX^r\)[/tex] | [tex]\(X^r Y\)[/tex] |
### 4. Determine the Phenotypes
We look at each combination to determine if the child has color-deficient vision.
- [tex]\(X^RX^r\)[/tex]: The child is a carrier female but has normal vision because [tex]\(X^R\)[/tex] is dominant.
- [tex]\(X^R Y\)[/tex]: The child is a normal-vision male because [tex]\(X^R\)[/tex] is dominant.
- [tex]\(X^rX^r\)[/tex]: The child is a color-deficient female because both [tex]\(X^r\)[/tex] alleles carry the recessive trait.
- [tex]\(X^r Y\)[/tex]: The child is a color-deficient male because [tex]\(X^r\)[/tex] carries the recessive trait and there is no corresponding dominant allele on the Y chromosome.
### 5. Calculate the Probability
We summarize the outcomes:
- Total possible outcomes: 4 ([tex]\(X^RX^r\)[/tex], [tex]\(X^R Y\)[/tex], [tex]\(X^rX^r\)[/tex], [tex]\(X^r Y\)[/tex])
- Color-deficient outcomes: 2 ([tex]\(X^rX^r\)[/tex], [tex]\(X^r Y\)[/tex])
Therefore, the probability that a child will have color-deficient vision is the number of color-deficient outcomes divided by the total number of outcomes.
[tex]\[ \text{Probability} = \frac{\text{Number of color-deficient outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5 \][/tex]
### 6. Conclusion
The probability that the child will have color-deficient vision is [tex]\(0.50\)[/tex]. Therefore, the correct answer is:
D. 0.50
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