punnett square? how do i figure this out, cystic Fibrosis? ?

[?K4yl33?]

A person who inherits only one copy of allele will be unaffected, however if two such people have a child the child could inherit two copies of allele.....
Cc
Cc
Key: C normal allele/c disease allele

1. A healthy man and a healthy woman have a child who has cystic fibrosis. Identify the genotypes for this trait for both parents and for their child.
2. This man and woman want another child. Calculate the probability that the second child would have cystic fibrosis.
3. Calculate the probability that a third child would have cystic fibrosis.
4. More serious genetic diseases are caused by recessive alleles then by dominant alleles. Explain why this is the case.


I know how to make a punnett square, i just don't know how to calculate the probabilities, im not sure what a genotype is

 

[Lauren]

1. In order for the child to have cystic fibrosis, the parents must both be carriers of the disease allele. This means that the child has a genotype of c/c and the parents are both heterozygous with C/c genotypes.
2. There is a 1/16 chance that the parents will have a second child with cystic fibrosis. Because they have a child that has cystic fibrosis, and they are healthy, they now know that they are carriers for the disease. A simple punnet square of two C/c genotypes will show that 1/4 of the children will be c/c which gives the disease. Therefore, because the questions is asking what is the chance that they will have two children with CF, then you take 1/4 times 1/4 getting 1/16.
3. Same principle. You take the chance of getting a child with CF and multiply it three times. So it's 1/4 times 1/4 times to get 1/64.
4. If the serious diseases were dominant than no one in the population would live to reproduction age. This is why that lethal diseases are recessive. However, you can have serious diseases that are dominant, but then the population would be mostly composed of recessive individuals.
Hope this helps.