Question

Show the full cross and provide justification in each case, draw punnet square of F1 and F2 generation in each case:

a) An F1 x F1 self-fertilization gives a 9:7 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1?

Answer 1

The F1 generation exhibits a 9:7 phenotypic ratio due to two genes interacting in a complementary fashion, with genotypes represented by AaBb.

Step-by-step explanation:

The student's question involves a 9:7 phenotypic ratio in the F2 generation from F1 self-fertilization, a pattern indicative of two genes interacting in a complementary fashion. To create a Punnett square for the F1 generation's self-cross and for the F2 generation, you would start by determining the F1 genotype that could produce such a ratio. Let's use AaBb to represent the heterozygous genotype of the F1 plants, assuming A and B are the alleles required for the phenotype.

The F1 x F1 cross Punnett square would give us a ratio where 9/16 of the offspring have both dominant alleles (A- B-), 3/16 have one dominant allele and one recessive allele for the second gene (A- bb), 3/16 with one recessive allele for the first gene and one dominant for the second gene (aa B-), and 1/16 are double recessive (aa bb). This results in a 9:7 phenotypic ratio because only the A- B- genotypes show the dominant phenotype.

When performing a test-cross of the F1, where the F1 is crossed with a homozygous recessive individual (aabb), all possible genotypes are AaBb, Aabb, aaBb, and aabb, with 1:1:1:1 genotypic and phenotypic ratios, as every offspring has an equally likely chance of inheriting either an A or a B allele from the F1 parent.

Punnett squares for both of these crosses can be visually drawn to show the result of these genetic crosses.