To determine the probability that the child will not have color-deficient vision, we need to analyze the genetic information provided.
Firstly, let's understand the genotypes of the parents:
- The mother’s genotype is [tex]\( X^R X \)[/tex], meaning she has one normal vision allele ([tex]\( X^R \)[/tex]) and one allele without specifying if it's normal or mutant (implicitly normal for simplicity).
- The father's genotype is [tex]\( X^R Y \)[/tex], meaning he carries a normal vision allele ([tex]\( X^R \)[/tex]) and a Y chromosome (which does not carry the trait for color vision).
Next, we need to construct a Punnett square to visualize the possible genotype combinations of their offspring.
The possible gametes from each parent are:
- Mother: [tex]\( X^R \)[/tex] or [tex]\( X \)[/tex]
- Father: [tex]\( X^R \)[/tex] or [tex]\( Y \)[/tex]
Forming the Punnett square, we have:
[tex]\[
\begin{array}{c|cc}
& X^R & X \\
\hline
X^R & X^R X^R & X^R X \\
Y & X^R Y & X Y \\
\end{array}
\][/tex]
The possible genotypes of the children are:
1. [tex]\( X^R X^R \)[/tex] - Normal vision (female)
2. [tex]\( X^R X \)[/tex] - Normal vision (female, carrier for color deficiency)
3. [tex]\( X^R Y \)[/tex] - Normal vision (male)
4. [tex]\( X Y \)[/tex] - Color-deficient vision (male)
Now, let's count the total outcomes and determine which ones lead to a child without color-deficient vision:
- There are 4 possible combinations in the Punnett square.
- Three out of these four combinations ( [tex]\( X^R X^R \)[/tex], [tex]\( X^R X \)[/tex], and [tex]\( X^R Y \)[/tex] ) result in a child without color-deficient vision.
Therefore, the probability that the child will not have color-deficient vision is:
[tex]\[
\frac{\text{Number of non color-deficient outcomes}}{\text{Total number of outcomes}} = \frac{3}{4} = 0.75
\][/tex]
Thus, the probability that the child will not have color-deficient vision is [tex]\( 0.75 \)[/tex].
The correct answer is:
B. 0.75
