Question: 152. With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in \mathrm{F}_{3} generation.
(1) 1 / 4
(2) 1 / 2
(3) 1 / 8
(4) Zero
Answer: Option (1)
Explanation:
The pedigree shows an affected male in \mathrm{F}_{2} (genotype X^{r}Y,
where X^{r} is the mutant X-linked recessive allele) mating with a carrier female in \mathrm{F}_{2} (genotype X^{r}X^{R},
where X^{R} is the normal allele). This pattern (affected males and carrier females) is characteristic of an X-linked recessive trait.
Set up the cross: affected male X^{r}Y × carrier female X^{r}X^{R}. The possible gametes are:
Father: X^{r} or Y (but daughters receive X^{r} from father)
Mother: X^{r} or X^{R} (each with probability 1/2)
Possible children genotypes and their probabilities:
Daughters (father gives X^{r}, mother gives X^{r} or X^{R}):
– X^{r}X^{r} (affected female): probability = 1/2 \times 1/2 = 1/4
– X^{r}X^{R} (carrier female, unaffected): probability = 1/2 \times 1/2 = 1/4
Sons (father gives Y, mother gives X^{r} or X^{R}):
– X^{r}Y (affected male): probability = 1/2 \times 1/2 = 1/4
– X^{R}Y (normal male): probability = 1/2 \times 1/2 = 1/4
We are asked for the probability that a child in \mathrm{F}_{3} is unaffected and a carrier.
That corresponds to the carrier female genotype X^{r}X^{R}, whose probability from this cross is 1/4.
Therefore the required probability = 1/4, Option (1).