Final answer:
Crossing two heterozygous roan cattle (both Rr) will result in a 1:2:1 genotypic ratio and also a 1:2:1 phenotypic ratio, with offspring displaying one red, two roan, and one white coat color, due to incomplete dominance.
Step-by-step explanation:
When crossing a roan male bull (heterozygous for the roan coat color, Rr) with a roan female cow (also heterozygous, Rr), we are dealing with a classic Mendelian monohybrid cross with incomplete dominance. In such a cross between two heterozygous individuals, the expected genotypic ratio in the offspring is 1:2:1, which means there will be one red (RR), two roan (Rr), and one white (rr) offspring for every four calves born. As for the phenotypic ratio, it should also be 1:2:1 due to the incomplete dominance of the roan trait (where the phenotype of the heterozygote is an intermediate of the two homozygotes), resulting in one red, two roan, and one white.
Another situation provided for context is the Aa x Aa cross, which also exhibits a 1:2:1 genotypic ratio and a 3:1 phenotypic ratio if the trait in question shows complete dominance. Furthermore, when we consider more complex traits with epistasis involved, as in the example where the C gene is epistatic to the A gene, we get modified phenotypic ratios like 9:3:4 instead of the simpler Mendelian ratios. Similarly, the independent assortment in dihybrid crosses often leads to a 9:3:3:1 phenotypic ratio, which can be further analyzed in terms of single traits to obtain monohybrid 3:1 ratios.