Step 1: Cystic fibrosis is autosomal recessive, so a person needs two copies of the mutant allele to be affected. Let the alleles be A (normal) and a (mutant).
Step 2: Both parents are clinically normal but have already produced an affected daughter (genotype aa). The daughter inherited one a from each parent, so each parent must carry one a. Both parents are therefore carriers with genotype Aa.
Step 3: Cross Aa with Aa. The offspring ratio is 1 AA to 2 Aa to 1 aa. The aa genotype is the affected one.
Step 4: So the probability that the next child is affected (aa) is 1 in 4, that is 1/4. Each pregnancy is independent, so the earlier affected child does not change this risk for the next child.
Step 5: The printed key gives option C, which matches this calculation.
Answer: Option C (1/4).