Labrador retrievers may be black, brown, or golden in color (see the chapter opening photograph on \(\mathrm{p} .53\) ). Although each color may breed true, many different outcomes occur if numerous litters are examined from a variety of matings, where the parents are not necessarily true-breeding. The following results show some of the possibilities. Propose a mode of inheritance that is consistent with these data, and indicate the corresponding genotypes of the parents in each mating. Indicate as well the genotypes of dogs that breed true for each color. (a) black \(\times\) brown \(\longrightarrow\) all black (b) black \(\times\) brown \(\longrightarrow \quad 1 / 2\) black \(1 / 2\) brown (c) black \(\times\) brown \(\longrightarrow \quad 3 / 4\) black \(1 / 4\) golden (d) black \(\quad \times\) golden \(\longrightarrow \quad\) all black (e) black \(\times\) golden \(\longrightarrow \quad 4 / 8\) golden 318 black \(1 / 8\) brown (f) black \(\times\) golden \(\longrightarrow \quad 2 / 4\) golden \(1 / 4\) black \(1 / 4\) brown (8) brown \(\times\) brown \(\longrightarrow \quad 3 / 4\) brown \(1 / 4\) golden (h) black \(\times\) black \(\longrightarrow 9 / 16\) black \(4 / 16\) golden \(3 / 16\) brown

We can start by observing the patterns in the offspring colors. For example, when two black dogs mate, there seem to be three possible outcomes: all black, some golden, and some brown. This suggests that black is a dominant color, as it's the most common outcome. The inheritance pattern also seems to have a predictable ratio in some cases, like when a black dog mates with a brown dog, and the offspring are 3 black and 1 golden.

Now that we have observed the modalities of inheritance, consider using a two-genes model to determine the genetic information of dog coat colors. We will use the following symbols for genotypes: - "B" for the black allele, a dominant allele - "b" for the brown allele, a recessive allele - "G" for the golden allele, a dominant allele With these symbols, we can now represent Labrador retriever colors as: Black: B_G_ Brown: bbG_ Golden: bbgg (Here, the underscore _ denotes that either the dominant or recessive allele can be present.)

(a) B?G? x bbG? ⟶ all black Both parents must have at least one B allele. Thus, the black parent must be B?G? and the brown parent must be bbG?. (b) B?G? x bbG? ⟶ ½ black, ½ brown The parents must be heterozygous for their color. Thus, the black parent must be BbG? and the brown parent must be bbG?. (c) B?G? x bbG? ⟶ ¾ black, ¼ golden Both parents must have at least one B and one G allele. In this case, the black parent must be BbGg and the brown parent must be bbG?. (d) B?G? x bbgg ⟶ all black Both parents must have at least one B allele and be homozygous for the g allele. Thus, the black parent must be B?gg and the golden parent must be bbgg. (e) B?G? x bbgg ⟶ ½ golden, ⅜ black, ⅛ brown Here, black dogs should contain a B allele and one g allele. The golden parent should be homozygous for both. Thus, the black parent must be Bbgg and the golden parent must be bbgg. (f) B?G? x bbgg ⟶ ½ golden, ¼ black, ¼ brown In this cross, the black parent must be heterozygous for each color, while the golden parent should be homozygous for the g allele. Thus the black parent is BbGg and the golden parent is bbgg. (g) bbG? x bbG? ⟶ ¾ brown, ¼ golden Both parents should be bbG?. The offspring inherit gg only when both parents contribute the g allele, creating a ¼ probability. (h) B?G? x B?G? ⟶ 9/16 black, 4/16 golden, 3/16 brown This is a dihybrid cross involving independent assortment of B (or b) and G (or g) alleles. The black parent is BbGg and the other black parent is also BbGg. This produces a classic 9:3:3:1 ratio, with 9 black, 3 golden, and 4 brown offspring. Now we have the genotypes for all parents and offspring in each mating.