Respuesta :
Jim is clearly a carrier for the disease, as is his new wife.
The distribution will be as such, with C denoted the dominant, non-disease trait and c denoting the hereditary disease. Only cc results in the disease.
Martha Jim Offspring:
C C c Cc, cC, CC, cc
c
There is a 50% chance that the child is a carrier but does not have the disease (Cc)
There is a 25% chance that the child inherited no disease allele and therefore is not a carrier (CC)
There is a 25% chance that the child has the disease (cc)
The probability of having a baby with cystic fibrosis is 25%
The distribution will be as such, with C denoted the dominant, non-disease trait and c denoting the hereditary disease. Only cc results in the disease.
Martha Jim Offspring:
C C c Cc, cC, CC, cc
c
There is a 50% chance that the child is a carrier but does not have the disease (Cc)
There is a 25% chance that the child inherited no disease allele and therefore is not a carrier (CC)
There is a 25% chance that the child has the disease (cc)
The probability of having a baby with cystic fibrosis is 25%
Answer:
There is 1/4 probability that jim and martha will have cystic fibrocic child.
Explanation:
As the jim's first child was cystic fibrosis and cystic fibrosis is a recessive disease.
Its mean Jim is a carrier and his second wife martha is also carrier. So the offsprings will be:
For example Jim and martha genotype is Cc and Cc then probability of offsprings are:
Cc × Cc = CC, Cc, Cc, cc
There will be three types of kids:
CC = Healthy kid = 25% = 1/4
Cc = Carrier kid = 50% = 1/2
cc = Diseased kid = 25% = 1/4