Question

The mode of inheritance for the following pedigree is X-linked recessive. - O 1 2 3 4 5 6 N III O 1 2 3 5 6 IV 1 2

Answer 1

It is given that the trait is X-linked recessive. Therefore, a woman who expresses the phenotype, must have the alleles for the trait on both of her X chromosomes. Let the allele for the trait be represented by 'a'

Individuals III-1 and III-2 are unaffected. They are having a son who expresses the phenotype. This means that the mother is a carrier for the trait, that is, she carries the allele on one of her X​​chromosomes, but does not show the phenotype (this is because, in X-linked recessive, the son inherits the trait from his mother). Hence, mother's genotype will be X^{​​​​​a}X. The father is normal, thus, his genotype will be XY.

We can find the probability of their next child to express the trait from the Punnette square given below-

	X	Y
X​​​​​​a	X​​​​​aX (normal daughter)	X​​​​aY (affected son)
X	XX (normal daughter)	XY (normal son)

Thus, out of 4 children, 1 is showing the phenotype. Hence, there is 1/4 or 25% probability, that the next child will express the trait.