Question

Cystic fibrosis results from the action of a recessive gene. A normal-appearing couple has a child with cystic fibrosis. Construct a Punnett square, determine the parents' genotypes, determine the possible gametes, and indicate the resulting genotypes and phenotypes as ratios from the cross by clicking and dragging the labels to the correct location. (Some labels may not be used.)

A) AA, Aa, aa
B) Aa, Aa
C) Aa, aa
D) aa, aa

Answer 1

Normal-appearing parents with a CF child are both carriers of the autosomal recessive gene (genotype Aa). A Punnett square indicates genotypes AA (non-carrier), Aa (carrier), and aa (CF), yielding a 3:1 phenotype ratio where 1/4 children will have CF.

Step-by-step explanation:

Cystic fibrosis (CF) is an autosomal recessive disorder, meaning that an individual must inherit two copies of the recessive allele to express the disease. A child with CF with normal-appearing parents indicates that both parents are carriers of the recessive gene (genotype Aa), but they do not express the disease phenotype.

Creating a Punnett square with the genotype Aa for both parents yields the following genotypes for offspring: AA, Aa, Aa, and aa. The genotype aa represents the individual with CF. As such, the children can be 1/4 AA (non-carriers), 1/2 Aa (carriers), and 1/4 aa (affected by CF). The phenotype ratio would be 3:1, where three out of four children would be phenotypically normal, but two of those three will be carriers.