In chickens, as we have seen, two pairs of alleles determine comb shape. RR or Rr results in rose comb,whereas

Question

rr produces single comb. PP or Pp produces pea comb, and pp produces single comb. When R and P occur together, they produce a new type of comb: Walnut. What would be the genotype of the (f1) generation resulting from RRpp times rrPP? The Phenotype? If (f1) hybrids were crossbred, what would be the probable distribution of genotypes? Of phenotypes?

Answer 1

Parents .................. RRpp x rrPP
Allele segregation ..(R) (R) (p) (p) x (r) (r) (P) (P)
F1 generation ........ RrPp
Phenotype ............. 100% walnut comb (because R & P occur tog.)

F1 cross ................ RrPp x RrPp
Allele segregation .. (R) (r) (P) (p) x (R) (r) (P) (p)
F2 generation ......... (RRPP), 2(RRPp), (RRpp), 2(RrPP), 4(RrPp), 2(Rrpp), (rrPP), 2(rrPp), (rrpp) ---->[do your punnet square]
F2 phenotype ....... 9 walnut : 3 rose : 3 pea : 1 single

Answer 2

Your question is really really confusing. Let me try to straighten it.

Alright... We know that R stands for rose shape and it's a dominant one. We also know that P stands for Pea shape and it is also another dominant allele. If R and P are found together, they produces a co-dominant comb known as the Walnut shape. Now we also know that if rr or pp is found, that means there is only 1 comb.

Next you want to know the F1 generation by giving the parent alleles. Let's list out everything in simple terms again. RR, Rp = rose, PP, Pp = pea, R + P = walnut, rr or pp = single. Your parent genotypes will be, Rp and rP. After breeding them, you will get RrPp. You will get all of them at 100%. They will all have Walnut shape.

F2 generation is a crossbred of the above F1 and in theory, you will get the 9-3-3-1 result. 9 of them will be at RrPp, 3 of them at Rrpp, 3 of them at rrPp and 1 of them at rrpp. Since R and P are together, you should get 9 walnut shape, 3 of them at single rose shape, 3 at single and 1 just a single comb without any shape stated.

Remember, when you do a dihybrid cross on RrPp kind, it will always follow the mendalian law of 9-3-3-1 as long as there is independent assortment happening. In your first case, it will also always be at 100% heterozygous.