Answer:
given,
rose combed rooster (R) × single combed hen (r)
RR × rr
↓
in F1 generation, all the chicks will be Rr
↓
in F2 generation, mating was done among their own group of Rr
Rr × Rr
↓
25% of RR 50% of Rr and 25% of rr
phenotypes of chicks in f2 generation = 3:1 ( 75% rose combed as R is dominant and 25% single combed)
genotype = 1:2:1 ( 1/4 homozygous rose comb [RR] : 2/4 heterozygous rose comb [Rr] :1/4 homozygous single comb [rr])
