Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Sure, let's solve this step by step.
We are given the genotypes of the parents:
- Parent 1: [tex]\(X^R X^r\)[/tex] (female, one normal vision allele [tex]\(X^R\)[/tex] and one color-deficient allele [tex]\(X^r\)[/tex])
- Parent 2: [tex]\(X^R Y\)[/tex] (male, normal vision allele [tex]\(X^R\)[/tex] and Y chromosome)
Since color-deficient vision is a sex-linked recessive trait, it is carried on the X chromosome. Only males with [tex]\(X^r Y\)[/tex] and females with [tex]\(X^r X^r\)[/tex] will express the trait.
1. Construct the Punnett Square:
Let's consider the alleles that each parent can pass on to their children.
For the mother ([tex]\(X^R X^r\)[/tex]):
- She can pass on [tex]\(X^R\)[/tex] or [tex]\(X^r\)[/tex].
For the father ([tex]\(X^R Y\)[/tex]):
- He can pass on [tex]\(X^R\)[/tex] or [tex]\(Y\)[/tex].
The possible combinations are as follows:
[tex]\[ \begin{array}{c|cc} & X^R & Y \\ \hline X^R & X^R X^R & X^R Y \\ X^r & X^R X^r & X^r Y \\ \end{array} \][/tex]
2. Determine the Genotypic Outcomes:
There are four possible outcomes for the children's genotypes:
- [tex]\(X^R X^R\)[/tex] (female, healthy)
- [tex]\(X^R X^r\)[/tex] (female, carrier)
- [tex]\(X^R Y\)[/tex] (male, healthy)
- [tex]\(X^r Y\)[/tex] (male, color-deficient)
3. Calculate the Probabilities:
Next, we calculate the probability of each genotype by examining the Punnett Square:
- Probability of [tex]\(X^R X^R\)[/tex] (healthy female): 1 out of 4 chance [tex]\((1/4)\)[/tex]
- Probability of [tex]\(X^R X^r\)[/tex] (carrier female): 1 out of 4 chance [tex]\((1/4)\)[/tex]
- Probability of [tex]\(X^R Y\)[/tex] (healthy male): 1 out of 4 chance [tex]\((1/4)\)[/tex]
- Probability of [tex]\(X^r Y\)[/tex] (color-deficient male): 1 out of 4 chance [tex]\((1/4)\)[/tex]
4. Answer the Question:
The question wants to know the probability that the child will have color-deficient vision. Only the genotype [tex]\(X^r Y\)[/tex] results in color-deficient vision.
Therefore, the probability that the child will have color-deficient vision is [tex]\( \frac{1}{4} \)[/tex] or 0.25.
So, the correct answer is:
A. 0.25
We are given the genotypes of the parents:
- Parent 1: [tex]\(X^R X^r\)[/tex] (female, one normal vision allele [tex]\(X^R\)[/tex] and one color-deficient allele [tex]\(X^r\)[/tex])
- Parent 2: [tex]\(X^R Y\)[/tex] (male, normal vision allele [tex]\(X^R\)[/tex] and Y chromosome)
Since color-deficient vision is a sex-linked recessive trait, it is carried on the X chromosome. Only males with [tex]\(X^r Y\)[/tex] and females with [tex]\(X^r X^r\)[/tex] will express the trait.
1. Construct the Punnett Square:
Let's consider the alleles that each parent can pass on to their children.
For the mother ([tex]\(X^R X^r\)[/tex]):
- She can pass on [tex]\(X^R\)[/tex] or [tex]\(X^r\)[/tex].
For the father ([tex]\(X^R Y\)[/tex]):
- He can pass on [tex]\(X^R\)[/tex] or [tex]\(Y\)[/tex].
The possible combinations are as follows:
[tex]\[ \begin{array}{c|cc} & X^R & Y \\ \hline X^R & X^R X^R & X^R Y \\ X^r & X^R X^r & X^r Y \\ \end{array} \][/tex]
2. Determine the Genotypic Outcomes:
There are four possible outcomes for the children's genotypes:
- [tex]\(X^R X^R\)[/tex] (female, healthy)
- [tex]\(X^R X^r\)[/tex] (female, carrier)
- [tex]\(X^R Y\)[/tex] (male, healthy)
- [tex]\(X^r Y\)[/tex] (male, color-deficient)
3. Calculate the Probabilities:
Next, we calculate the probability of each genotype by examining the Punnett Square:
- Probability of [tex]\(X^R X^R\)[/tex] (healthy female): 1 out of 4 chance [tex]\((1/4)\)[/tex]
- Probability of [tex]\(X^R X^r\)[/tex] (carrier female): 1 out of 4 chance [tex]\((1/4)\)[/tex]
- Probability of [tex]\(X^R Y\)[/tex] (healthy male): 1 out of 4 chance [tex]\((1/4)\)[/tex]
- Probability of [tex]\(X^r Y\)[/tex] (color-deficient male): 1 out of 4 chance [tex]\((1/4)\)[/tex]
4. Answer the Question:
The question wants to know the probability that the child will have color-deficient vision. Only the genotype [tex]\(X^r Y\)[/tex] results in color-deficient vision.
Therefore, the probability that the child will have color-deficient vision is [tex]\( \frac{1}{4} \)[/tex] or 0.25.
So, the correct answer is:
A. 0.25
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.