A dihybrid cross involves parents that differ in two traits, controlled by two different genes that assort independently (Mendel's Law of Independent Assortment).
Let the two genes be A/a and B/b, with A and B being dominant alleles.
If the parental cross is, for example, AABB (dominant for both traits) \(\times\) aabb (recessive for both traits):
- P generation: AABB \(\times\) aabb
- Gametes from P: AB from first parent, ab from second parent.
- F\(_1\) generation: All offspring are AaBb (dihybrid, heterozygous for both genes). They will all express the dominant phenotype for both traits.
Now, consider the F\(_2\) generation obtained by selfing or intercrossing the F\(_1\) generation: AaBb \(\times\) AaBb.
The gametes produced by an AaBb individual are AB, Ab, aB, ab (each with probability 1/4, due to independent assortment).
Using a Punnett square (4x4 grid) for the F\(_2\) generation, we get 16 possible genotypic combinations. These combinations result in four distinct phenotypes in the following ratio, assuming complete dominance for both genes:
- 9/16 will have the dominant phenotype for both traits (e.g., A\_B\_: AABB, AABb, AaBB, AaBb)
- 3/16 will have the dominant phenotype for the first trait and recessive for the second (e.g., A\_bb: AAbb, Aabb)
- 3/16 will have the recessive phenotype for the first trait and dominant for the second (e.g., aaB\_: aaBB, aaBb)
- 1/16 will have the recessive phenotype for both traits (e.g., aabb)
Thus, the classic F\(_2\) phenotypic ratio for a dihybrid cross with independent assortment and complete dominance is
9:3:3:1.
Options (b) and (d) have typographical errors in the ratio order. Option (c) correctly states 9:3:3:1.
\[ \boxed{\text{9:3:3:1}} \]