2. F2 phenotype ration for comb shape in chicken was observed to be walnut, rose, peanut and single as 8:3:4:1. What is probable genotype of parents?
(1) PPrr X PPrr          (2) PpRr X PpRr
(3) PpRR X PpRr       (4) ppRr X pprr

The F2 phenotypic ratio of comb shapes in chicken observed as walnut:rose:pea:single = 8:3:4:1 is explained by the interaction of two genes, R and P, which show epistasis. The parental genotypes producing such a ratio are most likely heterozygous for both genes, i.e., PpRr x PpRr. This is option (2).

Explanation of Genotype and Phenotype Ratios

Chicken comb shape is controlled by two interacting genes:

• R (Rose comb gene)

• P (Pea comb gene)

The phenotypes result from the interaction of these genes as follows:

• Walnut comb corresponds to the genotype where both dominant alleles are present (R-P-).

• Rose comb results when only R is dominant and P is recessive (R-pp).

• Pea comb occurs when P is dominant and R is recessive (rrP-).

• Single comb is recessive for both genes (rrpp).

The genotype-phenotype relationship is:

• R-P- → Walnut comb

• R-pp → Rose comb

• rrP- → Pea comb

• rrpp → Single comb

Why PpRr x PpRr?

Crossing two heterozygotes for both genes (PpRr x PpRr) produces gametes in equal proportions (PR, Pr, pR, pr) and the classic dihybrid cross yields a phenotypic ratio approximating 8 walnut : 3 rose : 4 pea : 1 single comb. This is due to epistatic interactions rather than the typical Mendelian 9:3:3:1 ratio.

Explanation of Other Options

• (1) PPrr x PPrr: Both parents are homozygous for P and recessive for R, so all offspring would show the phenotype related to P dominant only, missing variation for R gene, not matching the observed ratio.

• (3) PpRR x PpRr: Both parents have homozygous RR (rose comb dominant), so walnut (requiring R and P) and single comb (recessive for both) would not appear in expected ratios.

• (4) ppRr x pprr: Both parents lack dominant P allele, so walnut and pea combs should not appear; this cannot produce the observed phenotypic ratio.

Introduction

The variety of comb shapes in chickens, including walnut, rose, pea, and single, is controlled by two interacting genes, R and P. These genes exhibit epistatic interaction, producing unique phenotypic ratios in offspring. Understanding the F2 phenotypic ratio, especially the 8:3:4:1 distribution, helps elucidate the likely parental genotypes involved in this classic genetic cross.

Genetics of Chicken Comb Shapes

Chicken comb shape inheritance involves two genes: the rose comb gene (R) and the pea comb gene (P). The walnut comb results when both dominant alleles are present together (R-P-). The rose comb appears when only the rose gene is dominant (R-pp), while the pea comb results from only the pea gene being dominant (rrP-). The single comb is the recessive phenotype (rrpp). These genetic interactions explain the diverse phenotype combinations.

Determining the Parental Genotype

The F2 phenotype ratio of walnut:rose:pea:single = 8:3:4:1 notably diverges from the classical 9:3:3:1 dihybrid Mendelian ratio, suggesting gene interaction or epistasis. The most probable parental genotype is heterozygous for both genes (PpRr x PpRr). Such crosses produce gametes that combine to yield the observed phenotypic ratio due to the combined effects of dominance and epistatic interactions.

Analysis of Other Given Genotypes

• PPrr x PPrr cannot yield the full phenotypic variety since R gene is absent in dominant form.

• PpRR x PpRr would skew phenotypes towards rose comb with less variety.

• ppRr x pprr lacks dominant P alleles, making walnut and pea combs improbable.

Thus, heterozygous PpRr parents best explain the observed ratio.

This explanation clarifies the genetics behind comb shape variation in chickens and highlights the importance of gene interactions in inheritance patterns.

For further detailed genetics questions and exam preparation, focusing on epistasis and dihybrid crosses with interacting genes, this example offers a practical case study.