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Sagot :
To determine the probability that a child will have color-deficient vision, we need to understand the genetics behind the traits and how they are inherited.
Key Concepts:
1. Sex Chromosomes: In humans, females have two X chromosomes (XX), and males have one X and one Y chromosome (XY).
2. Inheritance of Color-Deficient Vision: This trait is carried on the X chromosome and is recessive. This means:
- [tex]\(X^R\)[/tex] represents the allele for normal vision.
- [tex]\(X^r\)[/tex] represents the allele for color-deficiency.
- For females to be color-deficient, they must have two [tex]\(X^r\)[/tex] alleles ([tex]\(X^rX^r\)[/tex]).
- Males only need one [tex]\(X^r\)[/tex] allele ([tex]\(X^rY\)[/tex]) since they have only one X chromosome.
Given the genotypes of the parents:
- Mother: [tex]\(X^R X^r\)[/tex] (carrier of color-deficient vision)
- Father: [tex]\(X^r Y\)[/tex] (color-deficient vision)
We need to find the possible genotype combinations for their children and identify which of them results in color-deficient vision.
Step-by-Step Solution:
1. Determine Possible Combinations:
Create a Punnett square to cross the genotypes:
| | [tex]\(X^R\)[/tex] (mother) | [tex]\(X^r\)[/tex] (mother) |
|--------|------------------|-------------------|
| [tex]\(X^r\)[/tex] (father) | [tex]\(X^R X^r\)[/tex] | [tex]\(X^r X^r\)[/tex] |
| [tex]\(Y\)[/tex] (father) | [tex]\(X^R Y\)[/tex] | [tex]\(X^r Y\)[/tex] |
2. Identify Offspring Genotypes:
- [tex]\(X^R X^r\)[/tex]: Female carrier (normal vision)
- [tex]\(X^r X^r\)[/tex]: Female with color-deficient vision
- [tex]\(X^R Y\)[/tex]: Male with normal vision
- [tex]\(X^r Y\)[/tex]: Male with color-deficient vision
3. Determine Probabilities:
There are four possible outcomes, each equally likely. Let's analyze them:
- Offspring with [tex]\(X^R X^r\)[/tex] (Normal female, carrier): 1 out of 4
- Offspring with [tex]\(X^R Y\)[/tex] (Normal male): 1 out of 4
- Offspring with [tex]\(X^r X^r\)[/tex] (Color-deficient female): 1 out of 4
- Offspring with [tex]\(X^r Y\)[/tex] (Color-deficient male): 1 out of 4
4. Calculate Probability of Color-Deficiency:
- The children with color-deficient vision are [tex]\(X^r X^r\)[/tex] and [tex]\(X^r Y\)[/tex], which account for 2 out of the 4 possible genotypes.
Probability = [tex]\(\frac{Number\, of\, color-deficient\, outcomes}{Total\, number\, of\, outcomes}\)[/tex]
= [tex]\(\frac{2}{4}\)[/tex] = 0.50
Thus, the probability that a child will have color-deficient vision is 0.50.
Answer:
D. 0.50
Key Concepts:
1. Sex Chromosomes: In humans, females have two X chromosomes (XX), and males have one X and one Y chromosome (XY).
2. Inheritance of Color-Deficient Vision: This trait is carried on the X chromosome and is recessive. This means:
- [tex]\(X^R\)[/tex] represents the allele for normal vision.
- [tex]\(X^r\)[/tex] represents the allele for color-deficiency.
- For females to be color-deficient, they must have two [tex]\(X^r\)[/tex] alleles ([tex]\(X^rX^r\)[/tex]).
- Males only need one [tex]\(X^r\)[/tex] allele ([tex]\(X^rY\)[/tex]) since they have only one X chromosome.
Given the genotypes of the parents:
- Mother: [tex]\(X^R X^r\)[/tex] (carrier of color-deficient vision)
- Father: [tex]\(X^r Y\)[/tex] (color-deficient vision)
We need to find the possible genotype combinations for their children and identify which of them results in color-deficient vision.
Step-by-Step Solution:
1. Determine Possible Combinations:
Create a Punnett square to cross the genotypes:
| | [tex]\(X^R\)[/tex] (mother) | [tex]\(X^r\)[/tex] (mother) |
|--------|------------------|-------------------|
| [tex]\(X^r\)[/tex] (father) | [tex]\(X^R X^r\)[/tex] | [tex]\(X^r X^r\)[/tex] |
| [tex]\(Y\)[/tex] (father) | [tex]\(X^R Y\)[/tex] | [tex]\(X^r Y\)[/tex] |
2. Identify Offspring Genotypes:
- [tex]\(X^R X^r\)[/tex]: Female carrier (normal vision)
- [tex]\(X^r X^r\)[/tex]: Female with color-deficient vision
- [tex]\(X^R Y\)[/tex]: Male with normal vision
- [tex]\(X^r Y\)[/tex]: Male with color-deficient vision
3. Determine Probabilities:
There are four possible outcomes, each equally likely. Let's analyze them:
- Offspring with [tex]\(X^R X^r\)[/tex] (Normal female, carrier): 1 out of 4
- Offspring with [tex]\(X^R Y\)[/tex] (Normal male): 1 out of 4
- Offspring with [tex]\(X^r X^r\)[/tex] (Color-deficient female): 1 out of 4
- Offspring with [tex]\(X^r Y\)[/tex] (Color-deficient male): 1 out of 4
4. Calculate Probability of Color-Deficiency:
- The children with color-deficient vision are [tex]\(X^r X^r\)[/tex] and [tex]\(X^r Y\)[/tex], which account for 2 out of the 4 possible genotypes.
Probability = [tex]\(\frac{Number\, of\, color-deficient\, outcomes}{Total\, number\, of\, outcomes}\)[/tex]
= [tex]\(\frac{2}{4}\)[/tex] = 0.50
Thus, the probability that a child will have color-deficient vision is 0.50.
Answer:
D. 0.50
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