The following pedigree chart shows the inheritance of a genetic disorder up to three generations of a family. Observe the chart and answer the questions that follow.

(i) Is the disease sex-linked or autosomal as per the chart ? Give reasons in support of your answer.

(ii) Is it a recessive or a dominant disorder?

(iii) Write the genotypes of the individuals 'C', 'D' and 'H'.

(iv) a) If the female 'D' marries a normal man, what will be the probability of their daughter being a sufferer of this disease ?

or

b) If the mother 'B' is a carrier of the disease, what will be the probability of their daughter being a sufferer of this disease ?

Based on the pedigree chart provided in the image, the correct answers are:
(i) Sex-linked disorder. More males are affected in the family as males have only one X chromosome which, if affected, expresses the disease.
(ii) Recessive disorder.
(iii) Genotypes: C - $X{X}^c$; D - $X{X}^c$; H - $X{X}^c$ (where $X^c$ is the affected allele).
(iv) a) Probability 0% or b) Probability 50%.

Step-by-Step Educational Explanation:

(i) Determining Sex-linked vs. Autosomal Inheritance:
Looking closely at the pedigree chart, we can observe the affected individuals who are represented by shaded square symbols:
• Generation I: Individual 'A' (affected male)
• Generation II: Individual 'E' (affected male)
• Generation III: Individual 'I' (affected male)
All affected individuals in this pedigree are males. Males have only one X chromosome (genotype $XY$). In a sex-linked (specifically X-linked) recessive disorder, a single copy of the mutated allele on the X chromosome is sufficient to cause the disease in males because they lack a second X chromosome to mask it. Females, having two X chromosomes ($XX$), require two copies of the affected allele to show symptoms, making them carriers if they have only one. The high prevalence of affected males supports a sex-linked disorder.

(ii) Determining Dominant vs. Recessive Nature:
The disorder is recessive. We can prove this by looking at individual 'E' (affected father, genotype $X^{c}Y$) and individual 'F' (unaffected mother, genotype $X{X}^c$ or $XX$). Their son, individual 'I', is affected. Since a male inherits his Y chromosome from his father ('E') and his X chromosome from his mother ('F'), individual 'I' must have received the affected X chromosome ($X^c$) from his unaffected mother 'F'. Since mother 'F' carries the disease allele but does not show symptoms, the disease allele must be recessive to the normal allele.

(iii) Determining Genotypes:
Let $X^c$ represent the X chromosome carrying the recessive affected allele and $X$ represent the X chromosome carrying the normal dominant allele.
• Individual 'C': Since her father 'A' is affected ($X^{c}Y$), he must pass his only X chromosome ($X^c$) to all of his daughters. Since 'C' is an unaffected female, she must have received a normal X chromosome from her mother 'B'. Therefore, her genotype is $X{X}^c$.
• Individual 'D': Like her sister 'C', she is an unaffected daughter of the affected father 'A' ($X^{c}Y$). She also inherits $X^c$ from her father and a normal $X$ from her mother 'B', resulting in the genotype $X{X}^c$.
• Individual 'H': She is an unaffected daughter of the affected father 'E' ($X^{c}Y$). She must inherit her father's $X^c$ chromosome and a normal $X$ chromosome from her mother 'F'. Thus, her genotype is $X{X}^c$.

(iv) Probability Analysis:

Case a) If female 'D' marries a normal man:
• Female 'D' genotype: $X{X}^c$ (carrier)
• Normal male genotype: $XY$
Let us perform the genetic cross:
$X{X}^c\times XY$
The possible genotypes of their daughters are:
• $XX$ (normal female)
• $X{X}^c$ (carrier, unaffected female)
Since none of the daughters have the genotype $X^c{X}^c$, the probability of their daughter being a sufferer of this disease is 0%.

Case b) If mother 'B' is a carrier of the disease:
• Mother 'B' genotype: $X{X}^c$ (carrier)
• Father 'A' genotype: $X^{c}Y$ (affected)
Let us perform the genetic cross:
$X{X}^c\times X^{c}Y$
The possible genotypes of their daughters are:
• $X{X}^c$ (carrier, unaffected female) - 50% probability among daughters
• $X^c{X}^c$ (affected female) - 50% probability among daughters
Therefore, the probability of their daughter being a sufferer of this disease is 50%.