In the Mendelian dihybrid cross, how many progenies in the F2 generation possess genotype rryy?

Explanation

Yes, in a Mendelian dihybrid cross, 1/16 of the progeny in the F2 generation will possess the genotype rryy. In a dihybrid cross, two genes are considered. In this case, the genes are for seed shape (R = round, r = wrinkled) and seed color (Y = yellow, y = green). The parents in the cross are heterozygous for both traits, meaning they have one dominant allele for each trait and one recessive allele for each trait. When the parents are crossed, the offspring will inherit one allele from each parent for each trait. The possible outcomes are RRYY: This offspring will be homozygous dominant for both traits. RRyy: This offspring will be heterozygous for seed shape and homozygous dominant for seed color. Rryy: This offspring will be heterozygous for both traits. rryY: This offspring will be heterozygous for seed color and homozygous recessive for seed shape. rryy: This offspring will be homozygous recessive for both traits. The phenotypic ratio of the offspring will be 9:3:3:1, with 9 offspring showing the dominant phenotype for both traits, 3 offspring showing the dominant phenotype for seed shape and the recessive phenotype for seed color, 3 offspring showing the dominant phenotype for seed color and the recessive phenotype for seed shape, and 1 offspring showing the recessive phenotype for both traits. The genotype rryy is the homozygous recessive genotype for both seed shape and seed color. This means that the offspring with this genotype will have wrinkled seeds and green seeds. The frequency of the rryy genotype in the F2 generation is 1/16. This is because the recessive alleles for both seed shape and seed color must be present on the same allele for the offspring to have the rryy genotype. The probability of this happening is 1/2 * 1/2 = 1/4.