Without Modification of Normal F2

Gene interaction without modification of normal F2 ratio is explained below.

In the early part of the 20th century, Bateson and Punnett studied this type of gene interaction in case of comb type of poultry.

Comb types and breed used are as follows

Pea comb (brahmas breed)

Rose comb (Wyandotte breed)

Walnut comb

Single comb (leghorn breed)

Cross between F1 F2
Pea X Single Pea 3 pea : 1 single
Rose X Single Rose 3 rose : 2 single
Rose X Pea Walnut 9 walnut : 3 rose : 3 pea : 1 single

From the results of these crosses, it can be observed that, pea type and rose type are monogenic dominant over single.

F1 generation (F2) of walnut type showed segregation, its behavior is based on ratio of dihybrid cross. This suggests that, walnut comb type must be governed by the pair of genes.


Suppose, gene P produces pea comb and gene R produces rose comb. Genes P and R when together give rise to walnut comb and combination of both in recessive form (p and r) produce single comb.

Rose x Pea

We want to test a cross between rose and pea. Now the question is what will be the genotype for both of these? It will be RRpp for pure rose comb type and rrPP for the pure pea comb type.

This is because though R and P are absent in pea and rose type resp. their recessive genes are present in both the cases.

Parents Rose comb (RRpp) Pea comb (rrPP)
F1 Walnut (RrPp)
F2 RRPP (1) 9 walnut
RRPp (2)
RrPP (2)
RrPp (4)
RRpp (1) 3 rose
Rrpp (2)
rrPP (1) 3 pea
rrPp (2)
rrpp (1) 1 single


Conclusion from the above obtained ratio is, gene P results in pea comb and gene R in rose comb, when present alone. Genes P and R in combination produce walnut comb and in recessive combination (p and q) produce single comb.

The phenotype of F1 resulting from the cross is entirely different from parents; because of the interaction between two dominant genes. Both of these genes are inherited independently.

To know other types of gene interactions visit page Gene Interactions