A blue Andalusian fowl is the product of mating black fowl with white fowl. A cross of blue fowl with another blue fowl results in:

Question

Accepted Answer

The correct option is 1: 1 black: 2 blue: 1 white

The inheritance pattern observed in Andalusian fowl is a classic example of **Incomplete Dominance**, which is a deviation from simple Mendelian genetics. In this pattern, the heterozygous genotype results in a phenotype that is an intermediate blend of the two homozygous parental phenotypes. The initial cross described—mating a black fowl with a white fowl to produce all blue fowl—establishes the parental and $F\_1$ generations. We can represent the alleles as $B^B$ for black and $B^W$ for white. The black fowl's genotype is $B^B B^B$ and the white fowl's genotype is $B^W B^W$. Their offspring, the blue fowl, are all heterozygotes with the genotype $B^B B^W$. The question asks for the result of crossing two of these heterozygous blue fowl ($B^B B^W \times B^B B^W$). According to the rules of Mendelian inheritance, this cross will produce three genotypes: $1 \ B^B B^B : 2 \ B^B B^W : 1 \ B^W B^W$. Due to incomplete dominance, the phenotypic ratio directly mirrors the genotypic ratio: the $B^B B^B$ fowl will be black, the $B^B B^W$ fowl will be blue, and the $B^W B^W$ fowl will be white, resulting in the characteristic phenotypic ratio of 1 black : 2 blue : 1 white.

The incorrect option is 2: 1 black: 2 white: 1 blue

This ratio (1:2:1) is numerically correct but misassigns the phenotypes. Specifically, it suggests that the blue phenotype is the least common and the white phenotype is the most common among the two alternatives, which fundamentally contradicts the known pattern of incomplete dominance where the intermediate phenotype (blue) must be the most frequent, making up 50% of the offspring. In a cross between two heterozygotes, the intermediate color should always appear in double the proportion of either homozygous parental color. Therefore, a 1:2:1 ratio is correct, but the pairing of the ratio with the specified phenotypes in this option is incorrect based on the biological principle of the cross.

The incorrect option is 3: 1 blue: 2 black: 1 white

This ratio (1:2:1) is also numerically correct but incorrectly assigns the blue phenotype to the lowest frequency and the black phenotype to the highest frequency (50% of offspring). If this were the result, it would imply that the black allele ($B^B$) was dominant and the white allele ($B^W$) was recessive, which would be an example of simple Mendelian dominance, not incomplete dominance. Furthermore, under simple dominance, the phenotypic ratio for a heterozygous cross would be 3 (dominant color) : 1 (recessive color), which does not match the phenotypes provided here. In incomplete dominance, the heterozygous blue is the most common phenotype, appearing twice as often as the black or white phenotypes.

The incorrect option is 4: 2 black: 2 blue

This ratio is equivalent to 1 black : 1 blue, representing a 1:1 phenotypic and genotypic ratio. A 1:1 ratio is typically the result of a **test cross**, where a heterozygote ($B^B B^W$) is crossed with a homozygous recessive individual. In the case of Andalusian fowl, this hypothetical cross would be a blue fowl crossed with a black fowl ($B^B B^W \times B^B B^B$) or a blue fowl crossed with a white fowl ($B^B B^W \times B^W B^W$). Neither of these crosses matches the problem description, which explicitly states a cross of "blue fowl with another blue fowl" (two heterozygotes). Therefore, the 1:1 ratio is incorrect because the parents specified in the question will yield a 1:2:1, or 4-part, distribution of offspring, not a 1:1, or 2-part, distribution.