Let us assume that hairy toes hh and brittle ear wax ww are

Let us assume that hairy toes (hh) and brittle ear wax (ww) are both recessive traits in humans. In families of three children where both parents are heterozygote\'s at both loci (they have smooth toes and sticky ear wax) what is the probability that the family will consist of one hairy-toed, brittle ear-waxed child and two smooth-toed, sticky ear-waxed children?

Solution

Male is--HhWw(XY) x Female-HhWw(XX), where h-recessive hairy gene, H-dominant smooth gene, w-recessive brittle wax ear, W-dominant sticky wax ear.

The probability of male child-1/2, probability of female child-1/2, the number of families in hardy-weinberg law=3. so the binomial distribution formula is-(p+q)³, p=1/2, q=1/2, the probabilty of male=1/2 ,q=1-p=1-1/2=1/2.

(p+q)³=p³+q³+3p²q+3pq ^2.

the probability of getting one male and two female=1/2 (50%)x 1/4(25%) x 1/4(25%)=1/32

probability of one hairy toed and brittle ear wax male=p=1/2 x 50%(chance of occurence of this child)(1/2)=1/2 x 1/2=1/4,q=1-1/4=3/4.

so the probability of having one hairy toed and brittle wax ear male=P=1/2x 1/4 x 1/4=1/32

the probability of two smooth toed and sticky ear female is Q=1/2 x 3/4 x 3/4=9/32.

ANSWER--The over all probability =probability of occurence of one male and two female x one hairy toed and brittle wax ear x two smooth toed and sticky ear child=3 x probability of male child x probabilty of two female child=3 X P X Q²=3 X 1/32 x 9/32 X 9/32=243/32768=0.0074.