Question

Mendelian ratios are modified in crosses involving autotetraploids. Assume that one plant expresses the dominant trait green seeds and is homozygous (WWWW). This plant is crossed to one with white seeds that is also homozygous (wwww). If only one dominant allele is sufficient to produce green seeds, predict the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of such a cross. Assume that synapsis between chromosome pairs is random during meiosis.

Short answer

The F1 generation has a phenotypic ratio of 1 green : 0 white seeds since all individuals have the heterozygous genotype (WwWw) and at least one dominant allele is sufficient to produce green seeds. For the F2 generation, after performing a self-cross of the F1 individuals, the phenotypic ratio is 7 greens: 1 white. This result is obtained by counting the different genotypes and phenotypes of the F2 offspring (4 WWWW, 3 WwWw, and 1 wwww).

Step-by-step solution

Determine the gametes produced by each parent

First, we need to find the possible gametes produced by each parent. For the parent with green seeds (WWWW), there is only one type of gamete, which is WW. For the parent with white seeds (wwww), there is only one type of gamete as well, which is ww.

Predict the genotype and phenotype of the F1 generation

Since there is only one type of gamete for each parent, we cross these gametes to get the genotype for the F1 generation: WW x ww = WwWw (which is the heterozygous genotype). Since we are given that only one dominant allele is sufficient to produce green seeds, all individuals of the F1 generation will have green seeds.

Determine the gametes produced by the F1 generation

Now, we need to find the possible gametes produced by the F1 generation. Since chromosome pairs synapse randomly during meiosis, the F1 generation of autotetraploids can produce multiple types of gametes. In this case, the F1 individual can produce gametes: WW, Ww, and ww.

Predict the genotype and phenotype of the F2 generation

To predict the genotype and phenotype of the F2 generation, we perform a self-cross between the gametes from the F1 individuals. We get the following set of outcomes: - WW x WW = WWWW (green seeded plants) - WW x Ww = WWWW, WwWw (green seeded plants) - WW x ww = WwWw (green seeded plants) - Ww x Ww = WWWW, WwWw, wwww (green and white seeded plants) - Ww x ww = WwWw, wwww (green and white seeded plants) - ww x ww = wwww (white seeded plants) Counting the F2 offspring genotypes, we get: - 4 WWWW - 3 WwWw - 1 wwww

Determine the phenotypic ratio in the F2 generation

Now let's count the phenotypes in the F2 generation: - 7 green seeded plants (WWWW and WwWw) - 1 white seeded plant (wwww) So, the phenotypic ratio for the F2 generation is 7 green: 1 white.