About Shweta Mishra Expertise I can answer general college level biology and biochemistry questions for example those related to cells, their structure, signaling etc. I can also answer questions related to Nutrition in various diseases such as Diabetes, Renal disease, Liver disease etc.
Experience 1.5 yrs of cell biology lab experience
Education/Credentials MS Biochemistry; MS Food and Nutrition
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.
