Question

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 \(F_{2}\) seneration is produced in a ratio of 1 dark-red: 4 medium-darkred: 6 medium-red: 4 light-red: 1 white. Further crossess reveal that the dark-red and white \(F_{2}\) 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.

Short answer

Additionally, list the possible genotypes for medium-red and light-red phenotypes, and predict the outcome of the F1 and F2 generations in a cross between a true-breeding medium-red plant and a white plant. Answer: Four genes are involved in color production. The number of additive alleles needed for each phenotype is as follows: Dark-red requires 8, Medium-dark-red requires 6, Medium-red requires 4, Light-red requires 2, and White requires 0. Possible genotypes for medium-red and light-red phenotypes are: Medium-red: 1. AABBccdd 2. AABbccDd 3. AABbCcdd 4. AAbbCCdd 5. AAbbccCD 6. AAbbccCcDd Light-red: 1. AAbbccdd 2. AAbbCcDd 3. AAbbccDd 4. AABbccdd The predicted outcome of the F1 and F2 generations in a cross between a true-breeding medium-red plant (AABBccdd) and a white plant (aabbccdd) is as follows: F1 generation - All offspring will have the genotype: AaBbcdCd F2 generation - The phenotypic ratio will be: 1. Dark-red: 1 2. Medium-dark-red: 8 3. Medium-red: 28 4. Light-red: 56 5. White: 70

Step-by-step solution

(a) Identifying the number of genes involved in color production

To determine the number of genes involved, we need to analyze the ratio of phenotypes produced in the \(F_2\) generation which is 1 dark-red:4 medium-dark-red:6 medium-red:4 light-red:1 white. This ratio can be rewritten as 1:4:6:4:1. We can observe that this ratio follows the pattern of a binomial expansion. Now, we need to find the power of the binomial expansion. The binomial expansion \({(a+b)}^n\) where n is the power; in our case, we are searching for the number of genes. Use this formula to find the coefficients: \(\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \binom{n}{3}, \cdots\) Comparing this to our given ratio 1:4:6:4:1, we can say that n = 4. So, there are 4 genes involved in the production of color in this case.

(b) Identifying the number of additive alleles needed to produce each phenotype

Now that we know there are 4 genes, let's identify the number of additive alleles needed for each phenotype. We can use the phenotypic ratio of the \(F_2\) generation to understand this. Since there are 4 genes, each plant contains a combination of 8 alleles (two from each gene). 1. Dark-red: 8 additive alleles (all contributing) 2. Medium-dark-red: 6 additive alleles contributing 3. Medium-red: 4 additive alleles contributing 4. Light-red: 2 additive alleles contributing 5. White: 0 additive alleles contributing (none)

(c) Assign symbols to these alleles and list possible genotypes for medium-red and light-red phenotypes

Let's now assign symbols to these alleles. Since we have 4 genes, we assign symbols A, B, C, and D to them. The alleles that contribute to the color (additive) can be represented as uppercase letters (A, B, C, and D) while those that do not contribute to the color (non-additive) can be represented as lowercase letters (a, b, c, d). Now, let's list the possible genotypes for medium-red and light-red phenotypes. Medium-red (4 additive alleles contributing): 1. AABBccdd 2. AABbccDd 3. AABbCcdd 4. AAbbCCdd 5. AAbbccCD 6. AAbbccCcDd Light-red (2 additive alleles contributing): 1. AAbbccdd 2. AAbbCcDd 3. AAbbccDd 4. AABbccdd

(d) Predict the outcome of the F1 and F2 generations in a cross between a true-breeding medium-red plant and a white plant

Let's predict the outcomes for the cross between true-breeding medium-red and true-breeding white plants. True-breeding medium-red: AABBccdd True-breeding white: aabbccdd In the \(F_1\) generation, all the offspring will inherit one allele from each parent, resulting in the following genotype: AaBbcdCd Now, let's predict the \(F_2\) generation outcome by self-fertilizing the \(F_1\) generation plants. The \(F_2\) generation genotypes and their respective ratios can be predicted by expanding \({(AB+C)}^4\): 1:8:28:56:70:56:28:8:1 Therefore, in the \(F_2\) generation, the phenotypic ratio will be as follows: 1. Dark-red: 1 2. Medium-dark-red: 8 3. Medium-red: 28 4. Light-red: 56 5. White: 70

Key concepts

F2 Generation Phenotypic Ratio

Grasping the pattern of inheritance in wheat color can be fascinating, as it unveils the fundamental principles of genetics. F2 generation phenotypic ratio provides us with a snapshot of how traits are passed on from parents to offspring in the second filial generation. In the given exercise, wheat plants showing a variety of colors were crossbred to analyze the genetic pattern. The resulting ratio of 1 dark-red: 4 medium-dark-red: 6 medium-red: 4 light-red: 1 white in the F2 generation is quintessential for understanding this genetic transmission.

The pattern mirrors a binomial expansion, which suggests a model where multiple genes and alleles contribute to the different shades of red displayed in wheat. This phenotypic ratio helps discern the number of genes at work—meaning the variety of colors in wheat is not controlled by a single gene but rather a combination of them. By understanding these phenotypic ratios, students can unravel how traits are inherited according to Mendelian genetics, and predict outcomes of future crosses.

Additive Alleles

Moving deeper into the complexities of genetics, the concept of additive alleles proves crucial. These alleles, when present, collectively influence the phenotype, leading to variations in the intensity of the wheat color. In this exercise, the wheat plants' coloration intensity is determined by the number of additive alleles they possess.

For a vivid depiction, think of each additive allele as a 'paint drop' contributing to the overall color. The more additive alleles (paint drops), the darker the resulting color. For instance, dark-red wheat plants carry all 8 possible additive alleles; hence they express the darkest shade. Conversely, white wheat lacks additive alleles, presenting no color. The exercise elegantly teaches students that the expression of a trait can be a result of additive effects, rather than just 'on' or 'off' scenarios seen in simple dominant-recessive patterns.

Genotype Designation

Decoding the genetic foundation leading to the colorful display in wheat can be best understood through the lens of genotype designation. In this exercise, specific genotypes are assigned to each phenotype expressible in wheat plants, affirming the role of genetics in determining observable characteristics.

By assigning the alleles A, B, C, and D to represent the additive alleles that affect wheat color, and a, b, c, and d for their non-additive counterparts, students can correlate genotype with phenotype. This assignment shows that for a medium-red phenotype, any genotype with four additive alleles, such as AABBccdd or AABbCcdd, would suffice. Similarly, a light-red phenotype would need just two additive alleles like AAbbccdd or AAbbCcDd. Understanding genotype designations empowers students to visualize the genetic make-up behind a trait and how different combinations of alleles manifest in varied phenotypes.