Color-deficient vision is a sex-linked recessive trait in humans. Parents with the following genotypes have a child:

[tex]\[ X^R X^{r} \times X^R Y \][/tex]

What is the probability that the child will have color-deficient vision?

A. 0.25
B. 0.50
C. 1.00
D. 0.75



Answer :

To determine the probability that a child will have color-deficient vision based on the given parental genotypes, we need to analyze the possible genotypes of the offspring from the combination of the parents' genes. Let's break down each step systematically.

### Step 1: Identify Parental Genotypes
- Mother's Genotype: [tex]\( X^R X^r \)[/tex], where [tex]\( X^R \)[/tex] represents the normal vision allele and [tex]\( X^r \)[/tex] represents the color-deficient vision allele.
- Father's Genotype: [tex]\( X^R Y \)[/tex], where [tex]\( X^R \)[/tex] represents the normal vision allele.

### Step 2: Determine Possible Offspring Genotypes
We'll set up a Punnett square to determine the possible genotypes of the offspring.

[tex]\[ \begin{array}{c|c|c} & X^R & Y \\ \hline X^R & X^R X^R & X^R Y \\ \hline X^r & X^r X^R & X^r Y \\ \end{array} \][/tex]

From this Punnett square, the possible genotypes of the offspring are:
1. [tex]\( X^R X^R \)[/tex] (normal vision female)
2. [tex]\( X^R Y \)[/tex] (normal vision male)
3. [tex]\( X^r X^R \)[/tex] (carrier female, normal vision)
4. [tex]\( X^r Y \)[/tex] (color-deficient vision male)

### Step 3: Count the Frequency of the Color-Deficient Genotype
We observe that the genotype [tex]\( X^r Y \)[/tex] results in color-deficient vision. We need to identify how often this genotype occurs in the offspring.

From the Punnett square, the possible genotypes are:
- [tex]\( X^R X^R \)[/tex]
- [tex]\( X^R Y \)[/tex]
- [tex]\( X^r X^R \)[/tex]
- [tex]\( X^r Y \)[/tex]

That gives us a total of four possible outcomes. Among these four outcomes, only one ([tex]\( X^r Y \)[/tex]) results in color-deficient vision.

### Step 4: Calculate the Probability
The probability is the number of favorable outcomes (color-deficient vision) divided by the total number of possible outcomes.

[tex]\[ \text{Probability} = \frac{\text{Number of color-deficient genotype outcomes}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25 \][/tex]

### Conclusion
Therefore, the probability that the child will have color-deficient vision is:

A. 0.25