In dogs, the black color of the coat dominates over the coffee color, and the short coat over the long one.

In dogs, the black color of the coat dominates over the coffee color, and the short coat over the long one. What percentage of black short-haired puppies can be expected from crossing two individuals that are heterozygous for both traits? How to determine if a black short-haired dog carries coffee-colored and long-haired genes? What phenotypic and genotypic partner should be taken for crossing for this test?

Let the black color of the dog’s coat be determined by the C gene, and the coffee color by the c gene.

Let the short coat of dogs be determined by the dominant gene S, and the long coat by the gene s.

Dogs that are heterozygous for both traits will be CcSs. They will produce sex cells of four types – cs, CS, cS, Cs.

The offspring obtained by crossing such dogs will look like this:

short-haired black puppies (CCSS, 2CcSS, 2CCSs, 4CcSs);

short-haired coffee-colored puppies (2Ccss, CCss);

puppies with long black hair (ccSS, 2ccSs);

puppies with coffee-colored long hair (ccss).

Black Shorthair puppies in the litter will be 9:16, or 56.25%.

To check the purebredity of a black dog with short hair (homozygosity for both traits – CCSS, sex cells – CS), one should resort to crossing, called analyzing: cross this dog with a coffee long-haired individual (ccss, sex cells – cs).

If the analyzed dog is purebred in the offspring, all puppies will be black and short-haired – CcSs.

In the presence of recessive genes in the dog’s genotype, puppies with phenotypically realized recessive traits will be obtained in the offspring. For example, a dog is СсWW, its sex cells are СS, cS. The offspring in the analyzing cross are black short-haired puppies (СсSs) – 50% and coffee short-haired (ccSs) – 50%.