Final answer:
In Mendel's monohybrid crosses, the P generation consisted of homozygous parents, either homozygous dominant (YY) or homozygous recessive (yy), resulting in the correct answer being b) AA and aa. Monohybrid crosses involve a single trait and have simpler Punnett squares compared to dihybrid crosses, which involve two traits and more complex squares. The AaBb self-pollination produces gametes with genotype AB, Ab, aB, ab.
Step-by-step explanation:
Identification of Genotypes in Mendel's Monohybrid Crosses
The genotypes of the P (parental) generation in Mendel's monohybrid crosses were homozygous for the trait being studied. Specifically, these parental plants were either homozygous dominant (YY) or homozygous recessive (yy). When crossing these, all offspring (F1 generation) were heterozygous (Yy), possessing one dominant allele from the yellow parent and one recessive allele from the green parent. Therefore, among the provided options, the correct genotypes for the P generation would be b) AA and aa.
A monohybrid cross involves a single trait and examines the inheritance pattern from parents to offspring. In contrast, a dihybrid cross, like the example of AaBb self-pollination, involves two traits. The alleles from each parent separate independently into the gametes, and the types of gametes produced by an individual with genotype AaBb are c) AB, Ab, aB, ab. These gametes illustrate Mendel's law of independent assortment.
When comparing Punnett squares for monohybrid and dihybrid crosses, monohybrid crosses have four potential combinations in a 2x2 grid, while dihybrid crosses have sixteen in a 4x4 grid due to the involvement of more alleles and traits. The genotypic ratio in a monohybrid cross of Aa x Aa is 1:2:1 and the phenotypic ratio is 3:1, considering the dominant trait masks the recessive one.