Question

Mendelian ratios are modified in crosses involving autotet. raploids. 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, pre- dict the \(F_{1}\) and \(F_{2}\) results of such a cross. Assume that synapsis between chromosome pairs is random during meiosis.

Short answer

Question: Predict the F1 and F2 generation results of a cross between two homozygous autotetraploid plants, one with dominant green seeds and the other with recessive white seeds, given that only one dominant allele is needed for the expression of the green seed trait. Answer: The F1 generation will have all green-seeded offspring. The F2 generation will have a predicted phenotypic ratio of approximately 80 green seeds to 1 white seed.

Step-by-step solution

Cross the Parental Generation (P) plants

In this step, we will cross the two homozygous plants to produce the \(F_{1}\) generation. The plant with green seeds is homozygous dominant (WWWW) and the plant with white seeds is homozygous recessive (wwww). Since we are given that synapsis between chromosome pairs is random, when these two plants are crossed, all of the gametes produced by the green seeded plant will carry a W allele, while all the gametes produced by the white seeded plant will carry a w allele. Therefore, when fertilization takes place, the genotype of the resulting \(F_{1}\) offspring will be WWWW x wwww -> WwWwWwWw.

Determine Phenotype of \(F_{1}\) Generation

Now that we have the genotype of the \(F_{1}\) generation (WwWwWwWw), we need to determine their phenotype. Since we are given that only one dominant allele (W) is sufficient to produce green seeds, the presence of any W allele in the genotype will result in green seeds. Thus, all the \(F_{1}\) offspring will have green seeds because they all have at least one W allele present in their genotype.

Cross the \(F_{1}\) Generation to Produce \(F_{2}\) Generation

To produce the \(F_{2}\) generation, we need to cross the \(F_{1}\) generation plants with themselves (WwWwWwWw x WwWwWwWw). In order to predict the phenotypic ratio of the \(F_{2}\) generation, we will use the Binomial theorem. Here's an explanation of how to apply the Binomial theorem in this case: - Since the four homologous chromosomes undergo random synapsis during meiosis, each chromosome in the autotetraploid can pair with any one of the other three homologous chromosomes equally. - Given that W is the dominant allele and w is the recessive allele, we can write the probability of forming a green seed as \(P(W) = (W + w)^4\) which is equivalent to \((a + b)^n\) in the Binomial theorem. - Using the Binomial theorem, we can expand \((W + w)^4\) as follows: \((1 + W)^4 = 1 + {4\choose1}W + {4\choose2}W^2 + {4\choose3}W^3 + W^4\) where \({n\choose k}\) is the binomial coefficient and \(W\) represents the probability of having a dominant allele. - Since the genotype of the \(F_{1}\) generation is (WwWwWwWw), the probability of having a dominant allele in the gamete is 0.5, and therefore we can replace \(W\) with 0.5 in the equation above to obtain the probability of each genotype in the \(F_{2}\) generation.

Calculate the \(F_{2}\) Generation Phenotypes

Once we have expanded and simplified the expression from step 3, we will get the probability distribution of genotypes in the \(F_{2}\) generation: \((1+0.5)^4 = 1 + 4(0.5) + 6(0.5)^2 + 4(0.5)^3 + (0.5)^4 = 1 + 2 + 1.5 + 0.5 + 0.0625\) Now we can determine the \(F_{2}\) generation phenotype ratios based on this distribution: - Green seeds (having at least one W allele): \(1 + 2 + 1.5 + 0.5 = 5\) - White seeds (having all w alleles): \(0.0625 = \frac{1}{16}\) Thus, the predicted phenotypic ratio of the \(F_{2}\) generation is 5 green seeds: \(\frac{1}{16}\) white seeds (approximately 80:1 when rounding the numbers).