Percentage of Monohybrids in Dihybrid F2

In a dihybrid cross between two heterozygous parents (AaBb x AaBb), the F1 generation will be all heterozygous (AaBb). When the F1 generation is allowed to self-pollinate, the resulting F2 generation will have a phenotypic ratio of 9:3:3:1, which includes different combinations of the two traits (A and B).

Monohybrid Ratio

The monohybrid ratio refers to the phenotypic ratio of a single trait in a cross between two heterozygous parents. For example, if two heterozygous pea plants with yellow seeds (Yy) are crossed, the resulting F2 generation would have a phenotypic ratio of 3:1 for yellow (YY or Yy) to green (yy) seed color.

Percentage of Monohybrids in Dihybrid F2

In a dihybrid F2 generation, there are different possible combinations of alleles that can result in the same phenotype. For example, a plant with a yellow round seed (AaBb) could have inherited the dominant allele for seed color (A) from one parent and the dominant allele for seed shape (B) from the other parent, or vice versa. This means that the phenotypic ratio of the F2 generation (9:3:3:1) does not directly correspond to the monohybrid ratio of either trait.

However, it is possible to calculate the percentage of monohybrids (heterozygous for one trait and homozygous recessive for the other trait) in the dihybrid F2 generation. These individuals would exhibit the dominant phenotype for one trait and the recessive phenotype for the other trait. In a dihybrid cross, the proportion of monohybrids can be calculated using the formula:

P(monohybrid) = 2pq

Where p and q represent the frequencies of the dominant and recessive alleles, respectively, for one of the traits. In this case, we can use the frequency of the recessive allele (q) for the other trait, since the monohybrids are homozygous recessive for that trait.

Assuming that the two traits are independently inherited (i.e., there is no linkage), the frequencies of the dominant and recessive alleles for each trait can be calculated using the Hardy-Weinberg equilibrium equation:

p^2 + 2pq + q^2 = 1

For example, if the frequency of the dominant allele for seed color (A) is 0.75 and the frequency of the recessive allele for seed shape (b) is 0.25, the proportion of monohybrids in the dihybrid F2 generation would be:

P(monohybrid) = 2pq = 2(0.75)(0.25) = 0.375 or 37.5%

Therefore, the correct answer is (a) 12.5%. None of the options provided are correct.