The next two questions go together. You cross true-breeding lines of tall, purple-flowered pea plants with short, white-flowered pea plants and find that all the F₁ offspring are tall and purple-flowered. When an F₁ is self-pollinated, you get 109 tall, purple: 19 tall, white: 22 short, purple: 10 short, white plants. Your null hypothesis is that the offspring fit a model of two independently-assorting loci, with tall dominant to short and purple dominant to white, and that deviations from the 9:3:3:1 ratio are due to chance. Fill out the table to calculate the x2 value for the data. For the last column, where decimals are involved, write your answer to one decimal point (e.g. 5.2). Phenotype Observed Expected o-e (o-e)² Tall, purple (o-e)²/e 109 Tall, white 19 Short, purple Short, white Total 160 ↓ How many degrees of freedom are there? [Select] What is the appropriate critical value of the x2 statistic at p=0.05? [Select] What is the p-value range (for example 0.5-0.1 or 0.01-0.005) your x² value falls into? [Select] Do you reject or fail to reject the hypothesis that this type of epistasis is acting? [Select] V TABLE 3.7. Critical values of the X distribution P df 0.995 0.975 0.9 0.5 0.1 0.05 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.584 6.251 2.366 0.216 9.348 7.815 11.345 12.838 0.207 0.484 1.064 3.357 7.779 9.408 11.143 13.377 14.960 0.412 12.933 4.351 0.831 9.236 1.610 11.070 16.750 15.006 0.676 1.237 2.204 5.340 10.645 12.592 14.449 16.812 18.548 14.007 12.017 1.690 20.278 6.346 2.833 10.013 0.989 18.475 15.507 17.535 7.344 20.090 1.344 13.362 21.955 3.490 2.180 14.684 8.343 19.023 16.919 21.666 23.589 4.168 1.735 2.700 18.307 25.100 20.483 23.209 15.987 9.342 3.247 2.156 4.865 19.675 21.920 26.757 24.725 17.375 2.603 5.578 3.016 10.341 18.549 23.337 21.026 28.300 26.217 11.340 12 6.304 3.074 4.404 29.319 24.736 22.362 27.608 19.012 12.340 13 3.565 5.009 7.042 26.119 23.685 29.141 31.319 21.064 4.075 14 13.339 5.629 7.790 27.488 24.996 22.307 30.578 32.901 15 4.601 14.339 8.547 6.262 P.probability: df, degrees of freedom. "Most scientists assume that, when P<0.05, a significant difference exists between the observed and the expected values in a chi aquare test Table 3.7 Genetics A Conceptual Approach, Sixth Edition 2017 WH Freeman and Company 3 4 5 6 7 8 9 10 11 22 10 160