In cattle, coats may be solid white, solid black, or black-and-white spotted. When true-breeding solid whites are mated with truebreeding solid blacks, the \(\mathrm{F}_{1}\) generation consists of all solid white individuals. After many \(\mathrm{F}_{1} \times \mathrm{F}_{1}\) matings, the following ratio was observed in the \(\mathrm{F}_{2}\) generation: \(12 / 16\) solid white \(3 / 16\) black-and-white spotted \(1 / 16\) solid black Explain the mode of inheritance governing coat color by determining how many gene pairs are involved and which genotypes yield which phenotypes. Is it possible to isolate a true-breeding strain of black-and-white spotted cattle? If so, what genotype would they have? If not, explain why not.

From the given information, we have the result of a cross between true-breeding solid white and true-breeding solid black cattle: \(\text{True-breeding solid white} \times \text{True-breeding solid black} \rightarrow \mathrm{F}_1: \text{All solid white individuals}\) Now, after many \(\mathrm{F}_{1} \times \mathrm{F}_{1}\) matings, the following ratio was observed in the \(\mathrm{F}_{2}\) generation: - \(12 / 16\) solid white - \(3 / 16\) black-and-white spotted - \(1 / 16\) solid black The phenotypic ratio in the F2 generation suggests that there is a single gene pair involved, with one dominant and one recessive allele. Let's denote the dominant allele with "W" (white) and the recessive allele with "B" (black). Now let's analyze the genotypes associated with each phenotype in the F2 generation.

We know from the given information that, in the F1 generation, all individuals were solid white, which means they all had the "W" allele (the dominant one). Hence, their genotype must be heterozygous "WB". Now, when they mate, look at the possible combinations: \( WB \times WB \) The Punnett square for this cross would be like this: | | W | B | |---|----|----| | W | WW | WB | | B | WB | BB | From this, we can obtain the genotypes associated with each phenotype: - Solid white -> WW or WB (9 + 3 in the F2 ratio) - Black and white spotted -> BB with one dominant W copy from the mother and one dominant W copy from the father, making it WWB (3 in the F2 ratio) - Solid black -> BBB (1 in the F2 ratio)

To have a true-breeding strain, all offspring must display the same phenotype when self-crossed. In this case, black-and-white spotted cattle have the WWB genotype. When they mate, let's see the possible combinations: \( WWB \times WWB \) The Punnett square for this cross would be like this: | | WW | WB | B | |---|-----|-----|----| | WW| WW^2| WWB | WB | | WB| WWB | W^2B | BB | | B | WB | BB | B^2| The resulting offspring will be either solid white, black-and-white spotted, or solid black. This means that isolating a true-breeding strain of black-and-white spotted cattle isn't possible, as their offspring display different coat color patterns.