Final answer:
The probability that the couple's first 2 children both have cystic fibrosis is 25%, while the probability that their first 5 children all have the disease is 3.125%.
Step-by-step explanation:
Probability of Cystic Fibrosis in Children from Carrier Parents
Cystic fibrosis (CF) is an autosomal recessive disorder where individuals with two copies of the faulty allele (ff) will exhibit the disease. Jill, having cystic fibrosis, possesses two recessive alleles (ff), whereas her husband John, being heterozygous, carries one normal and one faulty allele (Ff).
Probability of Cystic Fibrosis in First 2 Children
Since Jill is homozygous recessive for CF, all of her gametes will carry the recessive allele (f). John, being heterozygous, has a 50% chance of passing on the recessive allele (f) and a 50% chance of passing the dominant allele (F). The probability of their child having CF is when John passes the recessive allele, which is a 50% chance. Hence, for one child, the probability is 50%, or 0.5. For two children to both have CF, we multiply the probabilities: 0.5 × 0.5 = 0.25 or 25%.
Probability of Cystic Fibrosis in First 5 Children
The calculation is similar for five children, each must receive a recessive allele from John, which consecutively occurs at a probability of 0.5. Therefore, the probability that all five children will have CF is 0.5^5, which equals 0.03125 or 3.125%.
These calculations are based on the assumption that each pregnancy is independent of the others, meaning the outcome of one does not affect the outcome of another.