Final answer:
The expected genotypes for the gametes of a heterozygous pea plant would be BYR, ByR, bYR, byR, BYr, Byr, bYr, or byr. However, if all three genes are on the same chromosome arm, inheritance patterns likely won't follow Mendelian genetics due to linkage, making it improbable that they will assort independently.
Step-by-step explanation:
A heterozygous pea plant produces violet flowers and yellow, round seeds. Considering Mendelian inheritance, if the plant is heterozygous for each trait, and assuming each gene for these traits is represented by different alleles, we can denote the alleles as follows: 'B' for violet flowers, 'b' for white flowers, 'Y' for yellow seeds, and 'y' for green seeds, 'R' for round seeds, and 'r' for wrinkled seeds. The plant's genotype would be 'BbYyRr'. When forming gametes, each gamete will receive one allele from each gene due to the segregation of genes during meiosis. Therefore, the expected genotypes for the gametes would be BYR, ByR, bYR, byR, BYr, Byr, bYr, or byr.
If all three genes are found on the same arm of one chromosome, it is improbable that inheritance patterns will follow Mendelian genetics strictly due to the phenomenon of linkage. This breaks Mendel's law of independent assortment, as genes located close together on the same chromosome tend to be inherited together. Recombination can occur through crossing over during meiosis, but the likelihood of this can vary depending on the distance between the genes on the chromosome.