Question

17 dihybrid cross in corn
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19 table 10.2
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21 observed purple, smooth 158 purple, wrinkled 69 yellow, smooth 76 yellow, wrinkled 12 total 315
22 expected 177.1875 59.0625 59.0625 19.6875 315
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24 chi - squared
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26 p value
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30

Explanation:

Step1: Calculate chi-squared components

For each phenotype, compute $\frac{(O-E)^2}{E}$:

• Purple, smooth: $\frac{(158-177.1875)^2}{177.1875} = \frac{(-19.1875)^2}{177.1875} = \frac{368.16015625}{177.1875} \approx 2.078$
• Purple, wrinkled: $\frac{(69-59.0625)^2}{59.0625} = \frac{(9.9375)^2}{59.0625} = \frac{98.75390625}{59.0625} \approx 1.672$
• Yellow, smooth: $\frac{(76-59.0625)^2}{59.0625} = \frac{(16.9375)^2}{59.0625} = \frac{286.87890625}{59.0625} \approx 4.857$
• Yellow, wrinkled: $\frac{(12-19.6875)^2}{19.6875} = \frac{(-7.6875)^2}{19.6875} = \frac{59.09765625}{19.6875} \approx 2.991$

Step2: Sum components for chi-squared

$\chi^2 = 2.078 + 1.672 + 4.857 + 2.991$
$\chi^2 \approx 11.598$

Step3: Find degrees of freedom

Degrees of freedom (df) = number of phenotypes - 1 = $4-1=3$

Step4: Determine p-value

Using a chi-squared distribution table or calculator for $\chi^2 \approx 11.6$, df=3:
The p-value falls between 0.01 and 0.005, approximately 0.009.

Answer:

chi-squared: $\approx 11.60$
p value: $\approx 0.009$