Final answer:
In Biology, a test cross involving genes A and B with a 12-map unit distance points to a recombination frequency of 12%. The outcome for a test cross between heterozygotes with some gene linkage would alter a typical 9:3:3:1 phenotypic ratio. Forked-line diagrams can help predict the genotypic and phenotypic ratios of the offspring in various genetic crosses.
Step-by-step explanation:
The subject of this question is Biology, specifically genetics, and the study of inheritance patterns through test crosses and dihybrid crosses. When dealing with the map distance between genes, it indicates how often crossing over happens between them; in this case, a 12-map unit (mu) distance between genes A and B suggests a predicted frequency of recombination of 12%.
A test cross often involves crossing an individual with a known genotype (homozygous recessive) with an individual with an unknown genotype expressing the dominant phenotype. The outcome can help determine the genotype of the unknown parent.
For the F₂ generation resulting from a dihybrid cross between AABB (homozygous dominant) and aabb (homozygous recessive) where the F₁ generation is heterozygous AaBb, one would typically expect to see a 9:3:3:1 phenotypic ratio without linkage. However, with a 12mu distance indicating some linkage, this perfect ratio might be altered, though still illustrates a dihybrid cross with some recombinants.
The forked-line method is a handy tool to visualize the expected outcome in a dihybrid or trihybrid cross by considering each gene's segregation independently and then combining the probabilities along each pathway to determine the overall genotypic and phenotypic ratios for the offspring.
The final outcome stated such as the 9:3:4 ratio in epistasis and 27:9:9:9:3:3:3:1 ratio in a trihybrid cross are classic examples in Mendelian genetics, illustrating how multiple genes can interact and affect phenotypic ratios.