Question

Cystic fibrosis is a human recessive autosomal condition. If two heterozygous parents have two children, what is the probability that they will be born in the order: normal boy and affected girl?

a) 1/4
b) 1/8
c) 1/16
d) 1/2

Answer 1

The probability of two heterozygous parents having a normal boy followed by an affected girl with cystic fibrosis is 3/64, based on the independent probabilities of each event. None of the provided answer choices are correct.

Step-by-step explanation:

Cystic fibrosis is an autosomal recessive condition. When calculating probabilities for genetic outcomes, a Punnett square can be used. If both parents are heterozygous carriers (Ff) for cystic fibrosis, the probability of having a child with the disease (ff) is 25%, and the probability of a child being unaffected (FF or Ff) is 75%. However, this question asks for a specific outcome: a normal boy followed by an affected girl. The probability of having a boy or girl is independent of genetic inheritance and is always 1/2 for each gender.

The probability for having a normal boy would be the chance of having a boy (1/2) multiplied by the chance of being normal (3/4), which equals 3/8. For an affected girl, it's the chance of being a girl (1/2) multiplied by the chance of being affected (1/4), which equals to 1/8. Therefore, the probability of having a normal boy followed by an affected girl is 3/8 multiplied by 1/8, which equals 3/64. None of the answer options provided are correct, so there may be a mistake in the options given.