In a dihybrid cross, if the F1 generation has genotype RrYy, what proportion of gametes produced by this F1 individual will carry the alleles 'RY'? Justify your answer based on Mendel's laws.

The F1 individual has the genotype RrYy. According to Mendel's Law of Independent Assortment, the alleles for different genes (in this case, R/r for seed shape and Y/y for seed color) assort independently of each other during gamete formation. This means that the segregation of R and r is independent of the segregation of Y and y. For the 'R' gene, the individual Rr will produce gametes with 'R' and 'r' alleles in equal proportions (1/2 R, 1/2 r). For the 'Y' gene, the individual Yy will produce gametes with 'Y' and 'y' alleles in equal proportions (1/2 Y, 1/2 y). To find the proportion of gametes carrying 'RY', we multiply the probabilities of inheriting 'R' and 'Y': P(R) = 1/2 P(Y) = 1/2 P(RY) = P(R) * P(Y) = (1/2) * (1/2) = 1/4 Therefore, 1/4 (or 25%) of the gametes produced by the RrYy individual will carry the alleles 'RY'. The other types of gametes (Ry, rY, ry) will also be produced in equal proportions of 1/4 each.