Having correctly established the \(\mathrm{F}_{2}\) ratio in Problem 16 , predict the \(\mathrm{F}_{2}\) ratio of a "dihybrid" cross involving two independently assorting characteristics (e.g., \(\mathrm{P}_{1}=W W W W A A A A \times\) wwwwaaaa).

To find the F1 generation, we need to find the possible gametes produced by the P1 generation plants. For our example, each plant can only produce one type of gamete since each one is homozygous. The homozygous dominant plant can only produce gametes with WA alleles, and the homozygous recessive plant can only produce gametes with wa alleles.

In a dihybrid cross, the F1 generation is produced by the combination of the gametes from the P1 generation plants. In our example, the cross between the P1 plants (WWWWAAAA x wwwwaaaa) will produce the F1 generation offspring with WwAa.

Now, we have to identify the possible gametes produced by the F1 generation. Since their genotype is WwAa, they can produce four types of gametes: WA, Wa, wA, and wa.

In this step, use the Punnett square to find the combinations of the F2 generation. The Punnett square will be 4x4 because we identified four types of gametes in the F1 generation.

Now, count how many of each genotype appear in the Punnett square. Then, express this as a ratio or proportion of the total offspring. You will find the following genotypes and their respective proportions: 1. WWAA (9/16) 2. WWAa (3/16) 3. WwAA (3/16) 4. WwAa (1/16) 5. WWaa (3/16) 6. Wwaa (1/16) 7. wwAA (3/16) 8. wwAa (1/16) 9. wwaA (1/16) Combine the dominant and recessive allele combinations: 1. 9/16 showing both dominant traits (WWAA, WWAa, WwAA, WwAa) 2. 3/16 showing the dominant trait for the W gene and recessive trait for the A gene (WWaa, Wwaa) 3. 3/16 showing recessive trait for the W gene and dominant trait for the A gene (wwAA, wwAa) 4. 1/16 showing both recessive traits (wwaa) So, the F2 ratio for this dihybrid cross is 9:3:3:1.