Expert: Shweta Mishra - 6/21/2009

Question
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 symptoms 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, C = normal allele; and c = allele for cystic fibrosis.

Questions:

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?  

Answer
The question tells that both Sally and Bob are heterozygous which means they both are carriers of CF which is an autosomal recessive disease. Therefore the genotype of
Bob and Sally is Cc. So the combinations that can be obtained by crossing Cc vs Cc are

Cc    x    Cc = CC,      Cc,     Cc,    and cc

(I was not able to draw a sqaure or any table in this application  so I just did in words....but its really easy to put this in a Punnet square, I am sure you will be able to do it once you get the concept)

CC = Normal offspring (HEALTHY) = There is a 1/4 or 25% probability of getting such an offspring.

Cc = Two CARRIER offsprings without CF trait =  = There is a (1/4 +1/4 = 1/2) or 50% probability of getting such an offspring.

cc = Offspring who has CF trait  = There is a 1/4 or 25% probability of getting such an offspring.

Hope its clear, let me know
Shweta