Bob and Sally recently married. Upon deciding to plan a family, both Sally and Bob find out that they are both heterozygous for cystic fibrosis, but neither of them has symptons of the disorder.
Set up and complete a Punnett Square for cystic fibrosis for this couple; turn in the Punnett square. When doing the Punnett Square, assume that C=normal allele and c=allele for cystic fibrosis.
Based on the Punnett square, calculate chances (percentages) for having a healthy child ( not a carrier), a child that is a carrier for the cystic fibrosis trait, and a child with cystic fibrosis. Turn in these percentages.© BrainMass Inc. brainmass.com June 20, 2018, 7:49 am ad1c9bdddf
The parents, Bobby and Sally are heterozygous for cystic fibrosis. So they are carriers of the disease. Hence the genotype for both will be Cc .
This solution is an explanation to the chances of inheritance of cystic fibrosis from parents to offspring based on an example. It is explained with the help of a Punnett Square. It is a single page MS Word document with original contents.