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 darkred: 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 symbols to these alleles and list possible genotypes that give rise to the medium-red and light-red phenotypes. (d) Predict the outcome of the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) generations in a cross between a true-breeding medium-red plant and a white plant.

To determine the number of genes involved, observe the ratio of the phenotypes in the F2 generation: 1 dark-red : 4 medium-dark-red : 6 medium-red : 4 light-red : 1 white. This looks like a modified Mendelian ratio that occurs when traits are not determined by a single gene but are decided by multiple genes. In this case, 2 genes are involved (A and B). Considering the F2 generation ratio, we can assume that each gene has two additive alleles, i.e., one dark allele and one white allele.

The phenotypes and their ratios suggest that the alleles contribute to the color in an additive fashion. We can now assign the number of additive alleles in each phenotype as follow: - Dark-red: 4 additive alleles - Medium-dark-red: 3 additive alleles - Medium-red: 2 additive alleles - Light-red: 1 additive allele - White: 0 additive alleles

Let's denote the dark-red alleles as A and B and the white alleles as a and b. The possible genotypes for medium-red and light-red phenotypes will be: - Medium-red (2 additive alleles): AABB, AaBB, aaBB, AABb, AAbb - Light-red (1 additive allele): AaBb, aaBb, Aabb

Let's first define the genotypes of the true-breeding medium-red and white plants: - Medium-red: AABB - White: aabb Now, let's predict the outcome of the F1 and F2 generations: F1 generation cross: AABB x aabb All F1 offspring: AaBb (medium-red phenotype) F2 generation cross: Assuming self-fertilization of the F1 plants, we can make a Punnett Square for this cross (AaBb x AaBb): | | AB | Ab | aB | ab | |----|----|----|----|----| | AB | AABB | AABb | AaBB | AaBb | | Ab | AABb | AAbb | AaBb | Aabb | | aB | AaBB | AaBb | aaBB | aaBb | | ab | AaBb | Aabb | aaBb | aabb | F2 Phenotypes: - 1/16 Dark-red (AABB) - 4/16 Medium-dark-red (AABb, AaBB, AAbb, AaBb) - 6/16 Medium-red (AaBb, AaBb, AaBb, AAbb, AAbb, aaBB) - 4/16 Light-red (AAab, Aaab, aABB, aaBb) - 1/16 White (aabb) The predicted outcome of the F1 generation in a cross between a true-breeding medium-red plant and a white plant would be all medium-red plants. The predicted outcome of the F2 generation would have the following phenotype ratio: 1 dark-red : 4 medium-dark-red : 6 medium-red : 4 light-red : 1 white.