Question

A mating is set up between two pure breeding strains of plants. One parent has long leaves and long shoots. The other parent has short leaves and stubby shoots. \(\mathrm{F}_{1}\) plants are collected, and all have long leaves and long shoots. \(\mathrm{F}_{1}\) plants are self-crossed, and \(1,000 \mathrm{~F}_{2}\) plants are phenotyped. The data is as follows: $$ \begin{aligned} &\text { Phenotype }\\\ &\\# \text { of } \end{aligned} $$ $$ \begin{array}{|l|r|} & \mathbf{F}_{2} \\ \hline \text { Long leaves, long shoots } & 382 \\ \hline \begin{array}{l} \text { Long leaves, stubby } \\ \text { shoots } \end{array} & 109 \\ \hline \text { Short leaves, long shoots } & 112 \\ \hline \begin{array}{l} \text { Short leaves, stubby } \\ \text { shoots } \end{array} & 397 \\ \hline \text { Total } & 1,000 \\ \hline \end{array} $$ Are the genes for leaf and shoot length segregating independently? (A) Yes; the degrees of freedom are \(3,\) and the calculated \(\chi^{2}\) value is small. (B) No; the degrees of freedom are 3, and the calculated \(\chi^{2}\) value is large. (C) Yes; the degree of freedom is \(1,\) and the calculated \(\chi^{2}\) value is small. (D) No; the degree of freedom is \(1,\) and the calculated \(\chi^{2}\) value is large.

Short answer

(D) No; the degree of freedom is 1, and the calculated χ2 value is large.

Step-by-step solution

Phenotype ratio for independent assortment

If the genes are segregating independently, we expect to see a 9:3:3:1 phenotypic ratio in the F2 generation when considering two independently assorting traits. In our case, these phenotypes are: 1. Long leaves, long shoots (9/16) 2. Long leaves, stubby shoots (3/16) 3. Short leaves, long shoots (3/16) 4. Short leaves, stubby shoots (1/16)

Calculate the expected number of each phenotype

Using the expected phenotypic ratios and the total number of plants (1,000), we can calculate the expected number of plants for each phenotype: 1. Long leaves, long shoots: (9/16) * 1000 = 562.5 2. Long leaves, stubby shoots: (3/16) * 1000 = 187.5 3. Short leaves, long shoots: (3/16) * 1000 = 187.5 4. Short leaves, stubby shoots: (1/16) * 1000 = 62.5

Calculate the Chi-squared value

We will now calculate the Chi-squared value using the observed and expected values for each phenotype. The Chi-squared formula is: \(\chi^2 = \sum\frac{(observed - expected)^2}{expected}\) Applying the formula: \(\chi^2 = \frac{(382-562.5)^2}{562.5} + \frac{(109-187.5)^2}{187.5} + \frac{(112-187.5)^2}{187.5} + \frac{(397-62.5)^2}{62.5} = 77.07 \)

Calculate the degrees of freedom

The degrees of freedom for the test are calculated using the following formula: degrees of freedom = (number of categories - 1) * (number of traits - 1) In our case, we have: degrees of freedom = (2-1) * (2-1) = 1

Choose the correct answer

Since the Chi-squared value is large (77.07) and the degrees of freedom is 1, the correct answer is: (D) No; the degree of freedom is 1, and the calculated χ2 value is large.

Key concepts

Chi-squared Test

The Chi-squared test is a statistical tool used to determine if there's a significant difference between observed and expected frequencies in categorical data. It's particularly helpful in genetics for testing hypotheses about inheritance patterns.

This test calculates a Chi-squared (\( \chi^2 \)) value using the formula: \[ \chi^2 = \sum \frac{(observed - expected)^2}{expected} \]Here's how it works:

• "Observed" refers to the data collected from experiments or observations.
• "Expected" is what you predict to see based on a hypothesis or theory, like Mendel's laws.

A high Chi-squared value suggests that there's a significant deviation between the observed and expected data. This might indicate that the independent assortment hypothesis doesn't fit the results, as in the problem from the exercise.

Independent Assortment

Independent assortment is one of the fundamental principles of genetics first outlined by Gregor Mendel. It refers to how alleles of different genes segregate independently of one another when forming gametes.

Here's what it means:

• For example, if a plant has genes determining leaf length and shoot length, independent assortment suggests these genes are distributed randomly.
• This randomness can lead to various combinations in offspring, which is why Mendel could predict a 9:3:3:1 ratio in dihybrid crosses (crosses analyzing two traits).

However, in the exercise above, the observed phenotypic ratios varied significantly from expectations, indicating that these traits might not assort independently.

Phenotypic Ratios

Phenotypic ratios describe the relative number of offspring manifesting a particular trait or combination of traits. In Mendelian genetics, certain predictable ratios arise from specific types of genetic crosses.

For instance:

• The expected 9:3:3:1 ratio results from a dihybrid cross, where two traits are each influenced by two alleles.
• This ratio essentially reflects the independent assortment and interaction of alleles from each parent during reproduction.

In the exercise, the observed phenotypic ratio of the \( \mathrm{F}_2 \) generation did not align with the expected 9:3:3:1 ratio. This suggests a deviation, which is essential for determining if independent assortment occurs.

Degrees of Freedom

Degrees of freedom refer to the number of independent values or quantities that can be assigned to a test statistic in a calculation. In a Chi-squared test, degrees of freedom are vital for determining the critical value for statistical significance.

To calculate them:

• Count the number of categories minus one, and then multiply by the number of traits analyzed minus one. In this exercise, that would be \((2-1) \times (2-1) = 1\).

The degree of freedom impacts the interpretation of the Chi-squared value. With only 1 degree of freedom, the test indicated a significant divergence between observed and expected phenotypic ratios in the \( \mathrm{F}_2 \) generation. This further supports the conclusion that independent assortment may not be occurring as expected.