Question

in labrador dogs, coat color is controlled by the genotypes of two genes. in one gene, the dominant allele, b, produces black fur, and the recessive allele, b, produces brown fur. however, if a second gene possesses two recessive alleles, ee, the dog produces yellow fur, regardless of the genotype of the first gene. if two dogs that are heterozygous for both genes, bbee mated, what would be the frequency of the three phenotypes, black, brown, and yellow?

Explanation:

Step1: Determine phenotype - genotype relationships

Black: B - E - (at least one B and one E); Brown: bbE - (two b's and at least one E); Yellow: - - ee (two e's).

Step2: Count genotypes from Punnett - square

There are 16 genotypes in the Punnett - square.
For black (B - E -): BBEE (1), BBEe (2), BbEE (2), BbEe (4), total = 9.
For brown (bbE -): bbEE (1), bbEe (2), total = 3.
For yellow (- - ee): BBee (1), Bbee (2), bbee (1), total = 4.

Step3: Calculate frequencies

Frequency of black = $\frac{9}{16}$.
Frequency of brown = $\frac{3}{16}$.
Frequency of yellow = $\frac{4}{16}=\frac{1}{4}$.

Answer:

Frequency of black: $\frac{9}{16}$, Frequency of brown: $\frac{3}{16}$, Frequency of yellow: $\frac{1}{4}$