Question

Suppose a married man and woman both carry a gene for cystic fibrosis but don’t have the disease themselves. According to the laws of genetics, the probability that any child they have will develop cystic fibrosis is 0.25. Explain what this probability means. *

Answer 1

The 25% probability of a child developing cystic fibrosis from carrier parents refers to the likelihood that the child will inherit two recessive alleles for the disorder. This is consistent with the typical Mendelian 3:1 ratio expected for autosomal recessive inheritance in offspring from two carriers.

Step-by-step explanation:

When both parents carry a gene for cystic fibrosis, a recessive genetic disorder, and do not exhibit any symptoms themselves, there is a 25% probability that their child will develop the disease. This probability is derived from a Punnett square, which shows that there is a one in four chance of a child inheriting two recessive alleles (ff), required to express the condition. Autosomal recessive disorders like cystic fibrosis manifest only when an individual has two copies of the mutant gene. With one affected allele and one normal one, the parents are carriers, meaning they possess the gene without showing symptoms. Ff (carrier) crosses between two carriers typically result in a 25% chance for ff (affected by cystic fibrosis), a 50% chance for Ff (carrier), and a 25% chance for FF (unaffected and not a carrier).