To determine the probability of the child having color-deficient vision, let's analyze the parental genotypes:
- Parent 1 (Mother): [tex]\( X'X \)[/tex]
- Parent 2 (Father): [tex]\( X'Y \)[/tex]
Color-deficient vision is a sex-linked recessive trait carried on the X chromosome. In women, one recessive allele ( [tex]\( X' \)[/tex] ) masked by a dominant normal vision allele ( [tex]\( X \)[/tex] ) makes them carriers but they do not express the trait. Men with one recessive allele ([tex]\( X' \)[/tex]) on their only X chromosome will express color-deficient vision, as they have no second X chromosome to offset the trait.
Let's calculate the possible genotypes for their children by determining all possible allele combinations:
1. Mother ( [tex]\( X'X \)[/tex] ):
- Possible alleles: [tex]\( X' \)[/tex] and [tex]\( X \)[/tex]
2. Father ( [tex]\( X'Y \)[/tex] ):
- Possible alleles: [tex]\( X' \)[/tex] and [tex]\( Y \)[/tex]
Now we create the Punnett square to visualize the possible combinations for their offspring:
[tex]\[
\begin{array}{c|cc}
& X' & Y \\ \hline
X' & X'X' & X'Y \\
X & XX' & XY \\
\end{array}
\][/tex]
From the Punnett square, we identify the possible genotypes of their children:
1. [tex]\( X'X' \)[/tex] - Female with color-deficient vision
2. [tex]\( X'Y \)[/tex] - Male with color-deficient vision
3. [tex]\( XX' \)[/tex] - Female carrier, normal vision
4. [tex]\( XY \)[/tex] - Male, normal vision
Now, let's count how many of these combinations result in color-deficient vision:
- [tex]\( X'X' \)[/tex]: 1 combination
- [tex]\( X'Y \)[/tex]: 1 combination
Out of a total of 4 possible combinations, there are 2 combinations that result in color-deficient vision.
Therefore, the probability that the child will have color-deficient vision is:
[tex]\[
\frac{\text{Number of color-deficient combinations}}{\text{Total number of combinations}} = \frac{0}{4} = 0.0 = 0\%
\][/tex]
The answer to the question is:
B. 0.00
