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

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 1EQ: 1. Suppose that, in Example 2.27, 400 units of food A, 600 units of B, and 600 units of C are placed...
icon
Related questions
Question
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,
22
purple
Short,
10
white
Total
160
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
0.216
0.584
2.366
6.251
7.815
9.348
11.345
12.838
0.207
0.484
1.064
3.357
7.779
9.488
11.143
13.377
14.960
15.006
0.412
16.750
0.831
1.610
11.070
4.351
12.832
9.236
0.676
1.237
2.204
5.340
10.645
12.592
14.449
16.812
18.348
0.989
1.690
20.278
2.833
6.346
14.067
12.017
16.013
18.475
15.507
20.090
17.535
21.955
13.362
1.344
3.490
2.180
7.344
16.919
23.589
4.168
21.666
19.023
1.735
14.684
8.343
2.700
25.188
23.309
3.247
4.865
2.156
18.307
15.987
20.483
9.342
19.675
21.920
24.725
26.757
2.603
3.016
17.375
5.578
10.341
26.217 28.300
18.549
23.337
21.026
12
4.404
6.304
11.340
3.074
29.819
27.608
24.736
13
22.362
5.009
3.565
7.042
19.812
12.340
29.141
31.-319
26.119
14
23.685
4.075
5.629
7.790
13.339
21.064
27.488
30.578
32.801
15
6.262
4.601
24.996
8.547
22.307
14.339
P, probability df, degrees of freedom.
"Most scientists assume that, when P<0.05, e significant difference exists between the observed and the expected valees in a chi aquare test
Table 3.7
Genetics A Conceptual Approach, Sixth Edition
2017 WH Freeman and Company
3
4
S
6
7
8
9
10
11
Transcribed Image Text: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, 22 purple Short, 10 white Total 160 160 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 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.377 14.960 15.006 0.412 16.750 0.831 1.610 11.070 4.351 12.832 9.236 0.676 1.237 2.204 5.340 10.645 12.592 14.449 16.812 18.348 0.989 1.690 20.278 2.833 6.346 14.067 12.017 16.013 18.475 15.507 20.090 17.535 21.955 13.362 1.344 3.490 2.180 7.344 16.919 23.589 4.168 21.666 19.023 1.735 14.684 8.343 2.700 25.188 23.309 3.247 4.865 2.156 18.307 15.987 20.483 9.342 19.675 21.920 24.725 26.757 2.603 3.016 17.375 5.578 10.341 26.217 28.300 18.549 23.337 21.026 12 4.404 6.304 11.340 3.074 29.819 27.608 24.736 13 22.362 5.009 3.565 7.042 19.812 12.340 29.141 31.-319 26.119 14 23.685 4.075 5.629 7.790 13.339 21.064 27.488 30.578 32.801 15 6.262 4.601 24.996 8.547 22.307 14.339 P, probability df, degrees of freedom. "Most scientists assume that, when P<0.05, e significant difference exists between the observed and the expected valees in a chi aquare test Table 3.7 Genetics A Conceptual Approach, Sixth Edition 2017 WH Freeman and Company 3 4 S 6 7 8 9 10 11
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage