101k views
0 votes
In humans, cystic fibrosis is caused by a recessive allele on chromosome 7. A person must inherit two of these alleles to have the disease. What are the chances of a child having the disease if both of the parents are heterozygous?

a) 0% chance
b) 25% chance
c) 50% chance
d) 100% chance

User Nepete
by
7.4k points

1 Answer

4 votes

Final answer:

The likelihood of a child having cystic fibrosis when both parents are heterozygous carriers (Ff) is a 25% chance, as determined by the inheritance patterns of autosomal recessive disorders.

Step-by-step explanation:

The chances of a child having cystic fibrosis if both parents are heterozygous for the disease is 25% chance (option b).

In humans, cystic fibrosis is an autosomal recessive disorder caused by a recessive allele on chromosome 7. A child must inherit two copies of this recessive allele to exhibit the disease. When both parents are heterozygous, meaning they each have one normal allele and one mutant allele (Ff), there are four possible combinations for their offspring: FF (normal), Ff (carrier), fF (carrier), and ff (cystic fibrosis). Using a Punnett square, it can be shown that there is a 25% chance for the child to inherit the ff combination, resulting in cystic fibrosis. The remaining combinations result in either a healthy child or a carrier.

User Tristan Nemoz
by
8.5k points