
Answer:
given,
rose combed rooster (R)  ×  single combed hen (r) Â
                   RR × rr
                     ↓
in F1 generation, all the chicks will be Rr
                      ↓
in F2 generation, mating was done among their own group of Rr Â
     Rr × Rr
        ↓
25% of RR  50% of Rr  and 25% of rr
phenotypes of chicks in f2 generation = Â 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])