Respuesta :
Answer:
1: 4
Explanation:
Since both the parents in this case are the carrier of the cysytic fibrosis. Thus if a cross is carried out between the two parents , the punnet square will be as follows –
Genes of Parents X X
X XX XX
Y XY XY
Here the X is the carrier’s gene in both the parents
From the cross results it is clear that only one child (XX) is having both the mutated gene. Thus it will have the disease.
So the probability of an affected child is 1:4