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For harebell plants, dominant alleles for two separate genes are needed to produce blue petal color. Suppose gene 1 has alleles P and p and gene 2 has alleles B and b. Only plants with at least one Pallele and at least one B allele will have blue petals; all other genotypes lead to white petals. How many phenotypes will be observed among the offspring of the cross PpBb x PpBb?

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Final answer:

In the cross PpBb x PpBb of harebell plants, there will be two phenotypes observed among the offspring: blue petals and white petals. These result from the necessity of having at least one dominant allele from both gene 1 (P) and gene 2 (B) for blue petal production.

Step-by-step explanation:

The harebell plant presents two phenotypes: one with blue petals and the other with white petals. When considering a Punnett square for the cross PpBb x PpBb, where gene 1 (P and p alleles) and gene 2 (B and b alleles) influence blue petal color, the phenotypic outcome can be determined. A plant requires at least one P and one B allele to exhibit blue petals. Thus, within the resulting genotypes, there can be either of these two phenotypes: plants with blue petals and plants with white petals.

Carrying out a dihybrid cross for the given genotype, we find that the offspring fall into a 9:3:3:1 ratio, which typically represents the distribution of genotypes in a dihybrid cross of heterozygotes. However, since the question focuses on petal color and only the presence of at least one P and B allele is required for blue, the phenotypic ratio simplifies to only two observable phenotypes: blue and white. In conclusion, the plants that have at least one dominant allele P (either PP or Pp) and one dominant allele B (either BB or Bb) will present with blue petals, while the rest will have white. Therefore, among the offspring of a PpBb x PpBb cross, two distinct phenotypes will be observed, pertaining to petal color: blue and white.