Final answer:
When a heterozygous rose comb white rooster is crossed with a homozygous single comb blue Andalusian hen, the offspring will have a 1:1 genotypic and phenotypic ratio, with half having rose combs (Rr) and half with single combs (rr).
Step-by-step explanation:
To answer the question regarding the phenotype and genotype of the expected offspring from a white rooster with a rose comb (heterozygous) crossed with a blue Andalusian hen with a single comb, we need to establish the genotypes for the rose comb (dominant) and the single comb (recessive). Let's denote the rose comb allele as 'R' and the single comb allele as 'r'. Given that the rooster is heterozygous, its genotype is Rr, while the hen, with a recessive single comb, has the genotype rr.
Using a Punnett square, the cross Rr x rr will result in the following combinations: Rr (rose comb) and rr (single comb). This means the expected genotypic ratio of the offspring is 1:1, with half the offspring carrying the heterozygous genotype (Rr) and the other half the homozygous recessive genotype (rr). Consequently, the phenotypic ratio is also 1:1 for rose combed to single combed chickens.
While this question deals exclusively with simple dominance, it's notable that not all traits display this pattern. In cases of incomplete dominance, the heterozygous phenotype is an intermediate blend of the two parental phenotypes, such as the pink flowers from the cross between red (CRCR) and white (CWCW) snapdragons. Here, the heterozygous genotype is CRCW and displays a pink phenotype, showcasing incomplete dominance of the red allele over the white allele, producing a 1:2:1 phenotypic ratio for red:pink:white flowers.