Final answer:
In a monohybrid cross of two heterozygous Yy pea plants, 50% of the offspring will be heterozygous (Yy), as determined by a 1:2:1 genotypic ratio observed in Mendel's experiments and predicted using a Punnett square.
Step-by-step explanation:
When you conduct a monohybrid cross of two heterozygous Yy pea plants, you can predict the outcome using a Punnett square. Each heterozygous plant has two alleles for the trait: one dominant (Y) and one recessive (y). The offspring can have one of four possible allele combinations: YY, Yy, yY, or yy. There are two different ways to obtain the Yy genotype, either a Y from the egg and a y from the sperm or vice versa, hence, two out of the four possible combinations are heterozygous.
Since the combinations in a monohybrid cross are equally likely, the genotypic ratio of YY:Yy:yy offspring will be 1:2:1. This means that out of four offspring, one will be YY, two will be Yy, and one will be yy. Therefore, the percentage of the offspring that will be heterozygous (Yy) is 50%. Mendel's experiments confirmed this genotypic ratio in the F2 generation for individual traits.