Final answer:
In a tri-hybrid cross with potential gene linkage, we expect mostly parental type progeny with the genotype ABC/abc or abc/abc and fewer recombinant progeny. Recombinant gamete frequency depends on gene proximity on the chromosome. Without gene linkage, a dihybrid cross F₁ gamete ratio is 1:1:1:1 and a trihybrid cross F₂ phenotypic ratio is 27:9:9:9:3:3:3:1.
Step-by-step explanation:
The cross described involves an organism that is heterozygous at three loci (ABC/abc) and is crossed with a homozygous recessive organism (abc/abc). In the context of this question, we are considering a tri-hybrid cross with a potential for gene linkage. If genes A and B are linked, the frequency of parental gametes (AB and ab) will be higher than the recombinant gametes (Ab and aB). Therefore, without observing the specific linkage map or recombination frequency, we can't predict the exact genotype ratio, but we do anticipate a majority of offspring displaying the parental types (ABC/abc and abc/abc), with lesser amounts displaying the recombinant types (Ab/abc, aB/abc).
During meiosis, homologous recombination might occur, resulting in nonparental or recombinant gametes (Ab and aB), but the probability of recombination events will depend on the distance between the genes on the chromosome – the closer the genes, the lower the likelihood of recombination.
When considering a dihybrid cross with no gene linkage as in AABB x aabb, and the subsequent crossing of the F₁ heterozygotes (AaBb), we expect F₁ gametes to be produced in a 1:1:1:1 ratio (AB, aB, Ab, ab) due to independent assortment as per Mendelian genetics. In a trihybrid cross with independent assortment, the phenotypic ratio of the F₂ offspring is expected to be 27:9:9:9:3:3:3:1.