Answer:
See the answer with the explanation below.
Step-by-step explanation:
Assuming the allele for sickle cell disease is represented by S and the alternate, non-sickle cell version is A; AA would be homozygous normal, AS would be heterozygous normal, and SS would be a sickler.
a. The genotype of the father in the first generation would be AS.
The father is heterozygous, since the mother is affected and the couple produced an affected child. The cross can be illustrated thus:
AS (father) x SS (mother) = AS, AS, SS, SS
b. The genotype of the daughter in the second generation would be SS since she is phenotypically affected for the disease.
c. The genotype of individual 3 in the second generation would be heterozygous, AS.
The cross between the the heterozygous normal father and the sickler mother can only produce one of heterozygous normal individuals or sickle cell diseased individuals. Hence, individual 3 has to be heterozygous since he appeared phenotypically normal in the pedigree.