Question

12 Examine the following X-linked recessive pedigree: X-linked recessive a How many females in the pedigree are heterozygous for the trait being followed? b. What is the probability that a male child of III-5 and III-6 will exhibited the recessive phenotype?

Answer 1

12. a) As the disease is X-linked, it will never pass from father to son & all daughters of affected father will be carrier. Further, Father transfer their Y chromosome to sons & X chromosome to their daughters.

Let, X = normal allele, X^d = Disease allele

So, genotype of affected male will be X^dY & genotype of affected female will be X^dX^d. All normal male will have genotype XY & all normal female can have genotype either XX^d or XX.

So, II-3, III-5 & IV-1 will have X^dY genotype. As III-1 is normal it will have XY genotype. So, III-2 must be heterozygous. In this way we completed the pedigree. This is shown in below picture.

X-linked recessive XY XY XY xxd XX or XX

From the picture we can tell that number of heterozygous female will be 5 (I-2, II-1, II-5, III-2 & III-4). But there is possibility that all of the 8 females will be heterozygous as we can't surely tell the genotype of II-4, III-6 & IV-2.

b) The only possibility that male child of III-5 & III-6 will get the disease if III-6 has XX^d genotype. The cross between X^dY & XX^d is shown in below Punnett square.

Gametes	X	Xd
Xd	XXd (Normal female)	XdXd (Female with recessive phenotype)
Y	XY (Normal male)	XdY (Male with recessive phenotype)

So, probability of male child with recessive phenotype = 1/4 = 0.25

So, overall probability of the male child of III-5 & III-6 with recessive phenotype = Probability of III-6 has XX^d genotype x Probability of male child with recessive phenotype = 0.5 x 0.25 = 0.125 = 12.5%