Answer:
See the answer with the explanation below.
Explanation:
Assuming the allele for sickle cell disease is represented by S and the alternate, non-sickle cell version is A; AA would be <em>homozygous normal</em>, AS would be <em>heterozygous normal</em>, and SS would be a <em>sickler</em>.
a. <em>The genotype of the father in the first generation would be </em><em>AS</em>.
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. <em>The genotype of the daughter in the second generation would be </em><em>SS</em><em> </em>since she is phenotypically affected for the disease.
c. <em>The genotype of individual 3 in the second generation would be heterozygous, </em><em>AS</em>.
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.