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Sagot :
To solve this problem, let's consider the genetic makeup of both parents and the potential combinations of their offspring's genotypes.
The mother has the genotype [tex]\( X^R X^\top \)[/tex]:
- [tex]\( X^R \)[/tex] denotes the normal vision allele.
- [tex]\( X^\top \)[/tex] denotes the color-deficient vision allele (recessive).
The father has the genotype [tex]\( X^\top Y \)[/tex]:
- [tex]\( X^\top \)[/tex] denotes the color-deficient vision allele (recessive).
- [tex]\( Y \)[/tex] denotes the male chromosome.
To determine the possible genotypes of their children, we need to look at all potential combinations of the parents' alleles.
1. From the mother, the possible gametes are [tex]\( X^R \)[/tex] and [tex]\( X^\top \)[/tex].
2. From the father, the possible gametes are [tex]\( X^\top \)[/tex] and [tex]\( Y \)[/tex].
We need to consider all possible combinations of these alleles from the parents:
1. [tex]\( X^R \)[/tex] from mother and [tex]\( X^\top \)[/tex] from father: [tex]\( X^R X^\top \)[/tex]
2. [tex]\( X^R \)[/tex] from mother and [tex]\( Y \)[/tex] from father: [tex]\( X^R Y \)[/tex]
3. [tex]\( X^\top \)[/tex] from mother and [tex]\( X^\top \)[/tex] from father: [tex]\( X^\top X^\top \)[/tex]
4. [tex]\( X^\top \)[/tex] from mother and [tex]\( Y \)[/tex] from father: [tex]\( X^\top Y \)[/tex]
Now, let's determine which of these combinations result in color-deficient vision.
- [tex]\( X^R X^\top \)[/tex]: Child is female and carries the recessive allele but does not display color-deficient vision (since [tex]\( X^R \)[/tex] is dominant).
- [tex]\( X^R Y \)[/tex]: Child is male and has normal vision (since there's no [tex]\( X^\top \)[/tex] on the Y chromosome to cause the trait).
- [tex]\( X^\top X^\top \)[/tex]: Child is female and will display color-deficient vision (both alleles are recessive).
- [tex]\( X^\top Y \)[/tex]: Child is male and will display color-deficient vision (since the only X chromosome has the recessive allele).
Next, let's count the total number of combinations and how many of those result in color-deficient vision:
- Total combinations: 4 (as listed above).
- Combinations resulting in color-deficient vision: [tex]\( X^\top X^\top \)[/tex] and [tex]\( X^\top Y \)[/tex] give us 1 genotype for color-deficient females and 1 for color-deficient males out of 4 combinations.
Only 1 out of 4 resulted in color-deficient vision.
Hence, the number of favorable outcomes for color-deficient vision is 1.
The probability that the child will have color-deficient vision is therefore:
[tex]\[ \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{4} = 0.25 \][/tex]
Thus, the correct answer is:
C. 0.25
The mother has the genotype [tex]\( X^R X^\top \)[/tex]:
- [tex]\( X^R \)[/tex] denotes the normal vision allele.
- [tex]\( X^\top \)[/tex] denotes the color-deficient vision allele (recessive).
The father has the genotype [tex]\( X^\top Y \)[/tex]:
- [tex]\( X^\top \)[/tex] denotes the color-deficient vision allele (recessive).
- [tex]\( Y \)[/tex] denotes the male chromosome.
To determine the possible genotypes of their children, we need to look at all potential combinations of the parents' alleles.
1. From the mother, the possible gametes are [tex]\( X^R \)[/tex] and [tex]\( X^\top \)[/tex].
2. From the father, the possible gametes are [tex]\( X^\top \)[/tex] and [tex]\( Y \)[/tex].
We need to consider all possible combinations of these alleles from the parents:
1. [tex]\( X^R \)[/tex] from mother and [tex]\( X^\top \)[/tex] from father: [tex]\( X^R X^\top \)[/tex]
2. [tex]\( X^R \)[/tex] from mother and [tex]\( Y \)[/tex] from father: [tex]\( X^R Y \)[/tex]
3. [tex]\( X^\top \)[/tex] from mother and [tex]\( X^\top \)[/tex] from father: [tex]\( X^\top X^\top \)[/tex]
4. [tex]\( X^\top \)[/tex] from mother and [tex]\( Y \)[/tex] from father: [tex]\( X^\top Y \)[/tex]
Now, let's determine which of these combinations result in color-deficient vision.
- [tex]\( X^R X^\top \)[/tex]: Child is female and carries the recessive allele but does not display color-deficient vision (since [tex]\( X^R \)[/tex] is dominant).
- [tex]\( X^R Y \)[/tex]: Child is male and has normal vision (since there's no [tex]\( X^\top \)[/tex] on the Y chromosome to cause the trait).
- [tex]\( X^\top X^\top \)[/tex]: Child is female and will display color-deficient vision (both alleles are recessive).
- [tex]\( X^\top Y \)[/tex]: Child is male and will display color-deficient vision (since the only X chromosome has the recessive allele).
Next, let's count the total number of combinations and how many of those result in color-deficient vision:
- Total combinations: 4 (as listed above).
- Combinations resulting in color-deficient vision: [tex]\( X^\top X^\top \)[/tex] and [tex]\( X^\top Y \)[/tex] give us 1 genotype for color-deficient females and 1 for color-deficient males out of 4 combinations.
Only 1 out of 4 resulted in color-deficient vision.
Hence, the number of favorable outcomes for color-deficient vision is 1.
The probability that the child will have color-deficient vision is therefore:
[tex]\[ \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{4} = 0.25 \][/tex]
Thus, the correct answer is:
C. 0.25
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