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Phenyllketouria is an injeroyed discase cause by a recessive allele. If a woman and her husband...

Phenyllketouria is an injeroyed discase cause by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following ?
a) If they have 3 children, all will have a normal, unaffected phenotype.

b) If they have 4 children, 3 will be unaffected and 1 will be affected .

c) if they have 3 children, all will be affected
Math   Statistics-And-Probability   
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Answer #1

If both husband and wife are carriers, then the genotypes PP and Pp would be phenotypically normal (3 out of 4) and the genotype pp would express the phenotype for the disease Phenyllketouria (1 out of 4).

Hence probability of a normal child=3/4 and probability of a child who is affected=1/4.

a. P(All will have a normal, unaffected phenotype)=(3/4)3=0.4219 (since whether the child is affected or not does not depend on other child)

b. P(4 children, 3 will be unaffected and 1 will be affected)

c. P(All three will be affected)=(1/4)3=0.0156

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