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.

Step by step solution

01

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)

02

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

03

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: \(\chi^2 = \frac{(382-562.5)^2}{562.5} + \frac{(109-187.5)^2}{187.5} + \frac{(112-187.5)^2}{187.5} + \frac{(397-62.5)^2}{62.5} = 77.07 \)

04

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

05

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.

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