A dark-red strain and a white strain of wheat are crossed and produce an intermediate, medium-red \(\mathrm{F}_{1}\). When the \(\mathrm{F}_{1}\) plants are interbred, an \(\mathrm{F}_{2}\) generation is produced in a ratio of 1 dark-red: 4 medium-dark-red: 6 medium-red: 4 light-red: 1 white, Further crosses reveal that the dark-red and white \(\mathrm{F}_{2}\) plants are true breeding. (a) Based on the ratios in the \(\mathrm{F}_{2}\) population, how many genes are involved in the production of color? (b) How many additive alleles are needed to produce each possible phenotype? (c) Assign symhols to these alleles and list pnssible genotypes that give rise to the medium-red and light-red phenotypes. (d) Predict the outcome of the \(F_{1}\) and \(F_{2}\) generations in a cross between a true-breeding medium-red plant and a white plant.

In the \(\mathrm{F}_{2}\) generation, we are given the ratio 1:4:6:4:1 (dark-red: medium-dark-red: medium-red: light-red: white). This ratio suggests that there are two genes involved, as this ratio is derived from the binomial expansion \((a+b)^{n}\), where \(n\) is the number of genes involved. In this case, we have \((a+b)^{4}\) which gives us the ratio mentioned.

To determine the number of additive alleles needed for each phenotype, we can count the number of \(a\) and \(b\) components in each term of the \((a+b)^{4}\) expansion: $\text{Dark-red}: a^4 \\ \text{Medium-dark-red}: 4a^3b \\ \text{Medium-red}: 6a^2b^2 \\ \text{Light-red}: 4ab^3 \\ \text{White}: b^4$ Now we can determine the number of additive alleles, considering \(a\) as a dark-red allele and \(b\) as a white allele: Dark-red: 4 dark-red alleles (\(a\)) Medium-dark-red: 3 dark-red alleles and 1 white allele Medium-red: 2 dark-red alleles and 2 white alleles Light-red: 1 dark-red allele and 3 white alleles White: 4 white alleles (\(b\))

Let's assign the symbols R and r for dark-red and white alleles respectively. Now we can list the possible genotypes for each phenotype: Medium-red: \(R_1R_2r_1r_2\) and \(r_1r_2R_1R_2\) (2 dark-red alleles and 2 white alleles) Light-red: \(R_1r_1r_2r_2\), \(r_1R_1r_2r_2\), \(r_1r_1R_2r_2\), and \(r_1r_1r_2R_2\) (1 dark-red allele and 3 white alleles)

A true-breeding medium-red plant will have the genotype \(R_1R_2r_1r_2\) and a white plant will have the genotype \(r_1r_1r_2r_2\). For the \(F_{1}\) generation, the offspring from the cross between these two plants will have the genotype \(R_1r_1R_2r_2\). All offspring in the \(F_{1}\) generation will have a light-red phenotype as per the genotype created. For the \(F_{2}\) generation, we will cross two \(F_{1}\) generation plants with genotype \(R_1r_1R_2r_2\). To calculate the possible genotypes of the offspring in the \(F_{2}\) generation, we can use the forked-line method or Punnett square. The resulting \(F_{2}\) generation will have the following genotypes in a ratio of 1:2:1 (\(R_1R_2r_1r_2\): \(R_1r_1R_2r_2\): \(r_1r_1R_2r_2\)): $\text{Medium-red}: R_1R_2r_1r_2 (1) \\ \text{Light-red}: R_1r_1R_2r_2 (2) \\ \text{White}: r_1r_1R_2r_2 (1)$ The predicted outcome for the \(F_1\) generation in a cross between a true-breeding medium-red plant and a white plant is a light-red phenotype, and for the \(F_{2}\) generation, the phenotype ratio will be 1 medium-red: 2 light-red: 1 white.