Problems It's a probability question of genetics
Phenylketonurea in humans is caused by a recessive allele p. If both partners are known to be carrier (Pp), what is the chance in the following combination with five children that
a) All are normal
b) Three are normal and two are affected
c) Two are normal and three are affected
d) Four are normal and one is affected
e) One is normal and four are affected
f) All are affected
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Answer:
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Explanation:
If they are both heterozygous, then one-quarter of their children would have PKU, so the probability that their first child will have PKU is 1/4, and the probability of their being heterozygous and of their first child having PKU is 4/9 × 1/4 = 4/36 = 1/9, which is the answer to the question.
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