Person #5 in the second generation marries a man who does not have cataracts. two of their four children, a boy and a girl, develop cataracts. what is the chance that their next child will be a girl with cataracts?.

1/4

Explanation:

Person 5 marries a man who does not have cataracts and produced children in which some have cataracts and others do not have. It thus means that person 5 is affected for the disorder and the man she married is a carrier (although phenotypically normal) of the disorder.

Assuming the cataract disorder allele is represented by c, unaffected version of the allele will be C. Person 5 will have the genotype cc while the man she married will have the genotype Cc.

Crossing the two:

Cc   x   cc = Cc, Cc, cc, cc

50% of the offspring are phenotypically free from cataracts while 50% have cataracts.

The probability of having a cataracts infected child is thus 50% or 1/2. This probability remains the same irrespective the number of child they have already produced.

The probability of producing a girl for every marriage is 50% or 1/2.

Hence the chance that their next child will be a girl with cataracts = probability of their next child being a girl and the probability of their next child having a cataracts.

    = 1/2 x 1/2 = 1/4

The chance that their next child will be a girl with cataracts is 1/4.

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