The answer to the question is option 'A', 6/16.

Explanation:

In a Mendelian dihybrid cross, two individuals are crossed that are heterozygous for two different traits. In this case, we are considering a cross between two individuals that are heterozygous for both traits.

Let's consider two traits, A and B, in this dihybrid cross. The dominant allele for trait A is represented by 'A', and the recessive allele is represented by 'a'. Similarly, the dominant allele for trait B is represented by 'B', and the recessive allele is represented by 'b'.

When the two individuals are crossed, the possible combinations of alleles in the F1 generation are as follows:

AB x AB

Possible gametes from individual 1: AB, Ab
Possible gametes from individual 2: AB, Ab

Possible combinations in the F1 generation: AABB, AABb, AaBB, AaBb

These individuals are then crossed to produce the F2 generation. In the F2 generation, there are multiple possible combinations of alleles.

The possible combinations of alleles in the F2 generation can be represented using a Punnett square. The Punnett square for this dihybrid cross would be a 4x4 grid, with the genotypes of the offspring listed in each square.

The possible genotypes in the F2 generation are:

AABB, AABb, AaBB, AaBb, AAbb, aaBB, Aabb, aabb

Now, let's consider the phenotypes. The dominant phenotype for both traits is determined by having at least one dominant allele. The recessive phenotype is determined by having two recessive alleles.

The dominant phenotypes in the F2 generation are:

AABB, AABb, AaBB, AaBb, AAbb, aaBB, Aabb

The recombinant phenotypes in the F2 generation are:

AAbb, aabb

Thus, out of the 16 possible genotypes in the F2 generation, 6 of them are recombinant phenotypes. Therefore, the proportion of recombinant phenotypes in the F2 offspring of this Mendelian dihybrid cross is 6/16, which simplifies to 3/8 or approximately 0.375.