Answer:
PpCc; ccpp; ccPp; 1/2
Explanation: The complete correct question is as follows
A polydactylous woman, otherwise normal, marries a healthy normal man. Their four children have the following phenotypes:
Child 1 is normal in all respects; Child 3 has cystic fibrosis, otherwise normal Child 2 is polydactylous, otherwise normal; Child 4 has cystic fibrosis and is polydactylous. A. What is the genotype of the mother? B. What is the genotype of child 3? C. What is the genotype of child 4? D. What is the chance that child 1 is heterozygous for cystic fibrosis?
We represent the traits of the parents with the following
P= Polydactyly
p=normal
C= normal
c= abnormal (cystic fibrosis
A. What is the genotype of the mother
Mother PpCc (polydactylous)
Father: ppCc (healthy normal)
B. What is the genotype of child 3?
Applying the mendel's cross link between the mother's genotype and father's genotype,
PpCc × ppCc
Child 3 will be ccpp (abnormal with cystic fibrosis)
C. What is the genotype of child 4?
Applying the mendel's cross link between the mother's genotype and father's genotype,
PpCc × ppCc
Child 4 will be ccPp
D. What is the chance that child 1 is heterozygous for cystic fibrosis?
Applying the mendel's cross link between the mother's genotype and father's genotype,
PpCc × ppCc
Child will be Cccc
The chance is 1/2
