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?

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?

4/16
1/16
7/16
10/16
8/16
3/16
9/16
black
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yellow
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brown

Answer 1

The frequency of the three phenotypes can be predicted using a Punnett square to determine the possible genotypes that can result from the mating of two heterozygous dogs (BbEe).

The possible genotypes are:
- Black (BBEE, BBEe, BbEE, BbEe) - 9/16
- Brown (BBee, Bbee, Bbee, Bbee) - 3/16
- Yellow (bbEE, bbEe, bbee, bbee) - 4/16

Therefore, the frequency of the three phenotypes would be:
- Black: 9/16
- Brown: 3/16
- Yellow: 4/16

Overall, the correct answer is:
9/16 black
3/16 brown
4/16 yellow