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 F₃ generation.

The correct answer is 1/4.

Step-by-step Explanation:

1. Determine the Mode of Inheritance:
By analyzing the pedigree chart:

• Only females are represented as carriers (circles with a dot inside). There are no carrier males (squares with a dot).
• In the $F_0$ generation, an affected female (solid black circle) mated with an unaffected male (empty square) produces only carrier females in the $F_1$ generation.
• In the $F_2$ generation, an affected male (solid black square) is born from a carrier female mother ($F_1$) and an unaffected father ($F_1$).

These observations indicate that the trait is inherited in an X-linked recessive pattern.

2. Assign Genotypes:
Let $X^D$ represent the normal allele, and $X^d$ represent the allele containing the disease mutation.

For the mating in the $F_2$ generation that produces the $F_3$ generation:

• Mother: Carrier female with the genotype $X^D{X}^d$.
• Father: Affected male with the genotype $X^{d}Y$.

3. Perform the Genetic Cross:
Crossing the carrier female ($X^D{X}^d$) and the affected male ($X^{d}Y$) yields the following potential offspring genotypes and phenotypes:

$X^D{X}^d$ (Carrier female, unaffected) - Probability = 1/4
$X^d{X}^d$ (Affected female) - Probability = 1/4
$X^{D}Y(Normal\; male,\; unaffected)\; -\; Probability\; =\; 1/4$
$X^{d}Y(Affected\; male)\; -\; Probability\; =\; 1/4$

4. Calculate the Final Probability:
The question asks for the probability of the birth of a child in the $F_3$ generation who:

1. Has no disease.
2. Is a carrier (meaning they have the disease mutation in only one allele).

Only a heterozygous female with the genotype $X^D{X}^d$ meets both criteria (males cannot be carriers because they are hemizygous).

Thus, the probability of having a child who is unaffected and a carrier is:

$\text{Probability}=\frac{1}{4}$