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Problem Set 1

# The f 1 generation is all gg the f 2 generation is gg

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The F 1 generation is all Gg. The F 2 generation is Gg x Gg, resulting in a 3:1 phenotype (3 G_ : 1 gg). If any of this is unclear, do the Punnett square. Detailed answer for part a) There are a total of 8023 plants. Of these 8023, there’s an expected 3:1 ratio of yellow:green plants. Therefore, we expect: 8023 x 0.75 = 6017.25 yellow plants 8023 x 0.25 = 2005.75 green plants We observe 6000 yellow plants, and thus also observe 2023 green plants. So our “e” is 6017.25 yellow plants and 2005.75 green plants. Our “o” in part a) is 6000 yellow plants and 2023 green plants. Do the chi-squared: Σ (o-e) 2 /e (6000-6017.25) 2 /6017.25 + (2023-2005.75) 2 /2005.75 = 0.049 + 0.148 = 0.197 = chi-squared statistic Use the chi-squared table to find the p-value where the calculated chi-squared statistic falls. Remember, degrees of freedom (df) = # categories – 1; since there are two categories (phenotypes), the df = 1. You’ll find that the p-value for calculated chi-squared statistic for part a) is somewhere between 0.75 and 0.5. (Note: you do NOT need to calculate the exact p-value for the corresponding chi-squared statistic for this class.) Since the p-value is between 0.75 and 0.5, the null hypothesis (H o ) is accepted.

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Part a) Chi-squared value = 0.197; p-value between 0.75 & 0.5; H o accepted Part b) Chi-squared value = 0.922; p-value between 0.5 & 0.25; H o accepted Part c) Chi-squared value = 1.484; p-value between 0.25 & 0.1; H o accepted Part d) Chi-squared value = 5.060; p-value between 0.025 & 0.01; H o rejected Part e) Chi-squared value = 6.287; p-value between 0.025 & 0.01; H o rejected
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The F 1 generation is all Gg The F 2 generation is Gg x Gg...

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