Mendelian dihybrid cross involves the study of two different traits that are governed by two different genes present on two different chromosomes. The F2 generation of a dihybrid cross involves the crossing of two heterozygous parents (AaBb x AaBb), which yields a 9:3:3:1 phenotypic ratio.

The possible genotypes of the F2 generation are:

- AA-BB (homozygous dominant for both traits)
- Aa-BB (heterozygous for trait A and homozygous dominant for trait B)
- AA-bb (homozygous dominant for trait A and heterozygous for trait B)
- Aa-bb (heterozygous for both traits)
- aa-BB (homozygous recessive for trait A and homozygous dominant for trait B)
- aa-bb (homozygous recessive for both traits)
- Aa-Bb (heterozygous for both traits)

To determine the number of individuals that are heterozygous for both traits in the F2 generation, we need to consider the Aa-Bb genotype. This genotype can be formed in two ways:

- Aa-Bb can be formed through the cross of Aa-BB x Aa-bb
- Aa-Bb can also be formed through the cross of AA-Bb x aa-BB

Both of these crosses yield the Aa-Bb genotype in the F2 generation. Therefore, the probability of obtaining Aa-Bb in the F2 generation is:

- Aa-BB x Aa-bb: 1/4 x 1/4 = 1/16
- AA-Bb x aa-BB: 1/4 x 1/4 = 1/16

Adding these probabilities together gives us a total probability of 1/16 + 1/16 = 2/16 = 1/8.

Therefore, the number of individuals that are heterozygous for both traits in the F2 generation is 1/8 of the total individuals. Since there are 16 possible genotypes in the F2 generation, the number of individuals that are heterozygous for both traits is:

- 16 x 1/8 = 2 individuals

Therefore, the correct answer is (b) 3/16.