Final answer:
The phenotypic ratio expected in the progeny of the cross Aabb X aaBb in colorful swamp roses is 1 red : 2 pink : 1 white, corresponding to 25% red, 50% pink, and 25% white colored offspring due to the genes' complete dominance patterns.
Step-by-step explanation:
To determine the phenotypic ratio of the progeny from a cross Aabb X aaBb in colorful swamp roses, where the A allele produces an enzyme to convert white pigment to pink and the B allele produces an enzyme to convert pink pigment to red, we need to perform a dihybrid cross. This cross involves two genes, one for each pigment conversion step. The A gene (white to pink pigment) and the B gene (pink to red pigment).
In this case, the first parent can only pass on the alleles Ab or Ab, as it is homozygous for the b allele. The second parent can pass on aB or ab alleles, as it is heterozygous for both genes. When we set up a Punnett square, the possible combinations are AaBb, Aabb, aabb, and Aabb. However, since we are considering complete dominance, only the presence or absence of the dominant alleles will affect the phenotype.
Owning to the dominance of the A and B alleles over their respective recessive alleles, the cross produces the following phenotypes:
- A_B_ results in red pigment
- A_bb results in pink pigment (no B allele to convert to red)
- aaB_ gives white pigment (no A allele to convert to pink, and thus no pigment to turn red)
- aabb also results in white pigment as there are no dominant alleles present to transform the pigments
The phenotypic ratio for this cross will be:
Thus, we would expect 25% of red, 50% pink, and 25% white colored swamp roses from the given cross.