Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where the color can be red, pink (the heterozygote), or white. The second pair leads to personate (dominant) or peloric (recessive) flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed to those that are white, peloric, and dwarf. Determine the \(F_{1}\) genotype(s) and phenotype(s). If the \(\mathrm{F}_{1}\) plants are interbred, what proportion of the offspring will exhibit the same phenotype as the \(\mathrm{F}_{1}\) plants?

The first parent plant is homozygous for red, personate, and tall traits. If we use the symbols \(R\), \(P\), and \(T\) to represent these dominant alleles, and \(W\), \(p\), and \(t\) to represent the recessive alleles (white, peloric, and dwarf), then the genotype of the first parent plant is \(RR PP TT\). The second parent plant is homozygous for white, peloric, and dwarf traits. So its genotype is \(WW pp tt\).

When crossing the two parent plants, each parent will contribute one allele from each gene pair to their offspring. Since both parents are homozygous for their respective traits, the only possible genotype for the F1 offspring is \(RW Pp Tt\).

From the F1 genotype \(RW Pp Tt\), we can determine the phenotype for each trait: - Color: Since the first gene pair exhibits incomplete dominance, \(RW\) corresponds to the pink color (heterozygote) - Flower Shape: The second gene pair is \(Pp\), which leads to the personate flower shape (dominant) - Plant Height: The third gene pair is \(Tt\), which produces the tall trait (dominant) So, the F1 offspring phenotype is pink, personate, and tall.

When interbreeding the F1 plants (\(RW Pp Tt\) x \(RW Pp Tt\)), we will need to calculate the probabilities for each trait separately, and then multiply them together to find the overall probability. 1. Color: - Crossing \(RW \times RW\) results in 1 red (\(RR\)), 2 pink (\(RW\)), and 1 white (\(WW\)), so the probability of the F2 offspring being pink is \(\frac{2}{4} = \frac{1}{2}\). 2. Flower Shape: - Crossing \(Pp \times Pp\) results in 3 personate (\(PP\) or \(Pp\)) and 1 peloric (\(pp\)), so the probability of the F2 offspring having personate shape is \(\frac{3}{4}\). 3. Plant Height: - Crossing \(Tt \times Tt\) results in 3 tall (\(TT\) or \(Tt\)) and 1 dwarf (\(tt\)), so the probability of the F2 offspring being tall is \(\frac{3}{4}\). To find the overall probability for the F2 offspring with the same phenotype: Pink Probability x Personate Probability x Tall Probability = \(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{32}\) So, 9 out of 32 F2 offspring (28.125%) will have the same phenotype as the F1 plants, which is pink, personate, and tall.