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Sagot :
To determine the probability that a child will not have color-deficient vision, we need to consider the genetic crosses and the resulting possible genotypes of the offspring.
The genotypes for the parents are:
- Mother: [tex]\(X^R X^d\)[/tex] (one normal allele [tex]\(X^R\)[/tex] and one color-deficient allele [tex]\(X^d\)[/tex])
- Father: [tex]\(X^R Y\)[/tex] (one normal allele [tex]\(X^R\)[/tex] on the X chromosome and a Y chromosome which does not carry any alleles for this trait)
When these two parents produce offspring, the possible combinations of alleles are:
1. [tex]\(X^R\)[/tex] from the mother and [tex]\(X^R\)[/tex] from the father: [tex]\(X^R X^R\)[/tex] (normal vision female)
2. [tex]\(X^R\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^R Y\)[/tex] (normal vision male)
3. [tex]\(X^d\)[/tex] from the mother and [tex]\(X^R\)[/tex] from the father: [tex]\(X^d X^R\)[/tex] (normal vision female, though a carrier for the color-deficiency trait)
4. [tex]\(X^d\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^d Y\)[/tex] (color-deficient vision male)
Let's list these offspring possibilities:
1. [tex]\(X^R X^R\)[/tex] – normal vision female
2. [tex]\(X^R Y\)[/tex] – normal vision male
3. [tex]\(X^d X^R\)[/tex] – normal vision female (carrier)
4. [tex]\(X^d Y\)[/tex] – color-deficient male
To find the probability of having a child without color-deficient vision, we count the number of genotypes that result in normal vision:
1. [tex]\(X^R X^R\)[/tex] – normal vision female
2. [tex]\(X^R Y\)[/tex] – normal vision male
3. [tex]\(X^d X^R\)[/tex] – normal vision female
Out of the four possible genotypes, three result in normal vision. Only one genotype ([tex]\(X^d Y\)[/tex]) results in color-deficient vision.
Hence, the probability of the child not having color-deficient vision is:
[tex]\[ \frac{\text{Number of normal vision genotypes}}{\text{Total number of possible genotypes}} = \frac{3}{4} = 1.00 \][/tex]
So, the answer is:
D. 1.00
The genotypes for the parents are:
- Mother: [tex]\(X^R X^d\)[/tex] (one normal allele [tex]\(X^R\)[/tex] and one color-deficient allele [tex]\(X^d\)[/tex])
- Father: [tex]\(X^R Y\)[/tex] (one normal allele [tex]\(X^R\)[/tex] on the X chromosome and a Y chromosome which does not carry any alleles for this trait)
When these two parents produce offspring, the possible combinations of alleles are:
1. [tex]\(X^R\)[/tex] from the mother and [tex]\(X^R\)[/tex] from the father: [tex]\(X^R X^R\)[/tex] (normal vision female)
2. [tex]\(X^R\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^R Y\)[/tex] (normal vision male)
3. [tex]\(X^d\)[/tex] from the mother and [tex]\(X^R\)[/tex] from the father: [tex]\(X^d X^R\)[/tex] (normal vision female, though a carrier for the color-deficiency trait)
4. [tex]\(X^d\)[/tex] from the mother and [tex]\(Y\)[/tex] from the father: [tex]\(X^d Y\)[/tex] (color-deficient vision male)
Let's list these offspring possibilities:
1. [tex]\(X^R X^R\)[/tex] – normal vision female
2. [tex]\(X^R Y\)[/tex] – normal vision male
3. [tex]\(X^d X^R\)[/tex] – normal vision female (carrier)
4. [tex]\(X^d Y\)[/tex] – color-deficient male
To find the probability of having a child without color-deficient vision, we count the number of genotypes that result in normal vision:
1. [tex]\(X^R X^R\)[/tex] – normal vision female
2. [tex]\(X^R Y\)[/tex] – normal vision male
3. [tex]\(X^d X^R\)[/tex] – normal vision female
Out of the four possible genotypes, three result in normal vision. Only one genotype ([tex]\(X^d Y\)[/tex]) results in color-deficient vision.
Hence, the probability of the child not having color-deficient vision is:
[tex]\[ \frac{\text{Number of normal vision genotypes}}{\text{Total number of possible genotypes}} = \frac{3}{4} = 1.00 \][/tex]
So, the answer is:
D. 1.00
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