Final answer:
A homozygous recessive pea plant with genotype 'yy' will produce gametes with only the 'y' allele. A cross between plants with genotypes RrYY and rrYy requires a 4x4 Punnett square to predict the genotypes and phenotypes of the offspring.
Step-by-step explanation:
If a pea plant is homozygous recessive for yellow seeds (yy), it will produce gametes with only the recessive allele, which is 'y' in this case. Therefore, the correct answer to what type of gametes such a plant will produce is B. y only.
In genetics, and particularly in Mendelian genetics, organisms that are homozygous for a gene have two identical alleles for a trait. When such an organism produces gametes, each gamete receives one allele. Because our subject plant is homozygous recessive (yy), each gamete will carry the recessive allele 'y'.
The cross between a RrYY plant and an rrYy plant would result in offspring with various combinations of the alleles for seed shape and color. To determine the genotypes and phenotypes of the offspring, one would use a Punnett square. A Punnett square is a diagram that is used to predict the genetic makeup of the offspring of a cross. Since this is a dihybrid cross, involving two traits (seed shape and color), we would need a 4x4 Punnett square - totaling 16 squares - to analyze the potential offspring combinations.