a chi square tablet for the question. I left a reference down below. Question: 3. In dihybrid cross where both parents were heterozygous for both traits, the following data was obtained: 890 dominant for both traits, 97 recessive for both traits, 315 dominant for trait 1, but recessive for trait 2, and 298 dominant for trait 2, but recessive for trait one. Perform a chi-square analysis to determine if this data agrees with the expected outcomes. State your null hypothesis as well as if the null is rejected or not.

Concepts of Biology
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ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:Samantha Fowler, Rebecca Roush, James Wise
Chapter8: Patterns Of Inheritance
Section: Chapter Questions
Problem 2ACQ: Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are...
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Do a chi square tablet for the question. I left a reference down below. Question: 3. In dihybrid cross where both parents were heterozygous for both traits, the following data was obtained: 890 dominant for both traits, 97 recessive for both traits, 315 dominant for trait 1, but recessive for trait 2, and 298 dominant for trait 2, but recessive for trait one. Perform a chi-square analysis to determine if this data agrees with the expected outcomes. State your null hypothesis as well as if the null is rejected or not.
Part 3: Chi-Square
Directions: Complete the following data table to reject or fail to reject the null hypothesis.
14. Consider a cross between two heterozygous round yellow peas with the following observed results. Calculate the expected results and
chi-square value to determine if the alleles assort independently.
P
Degrees of Freedom
x²= (0-6)²
value
1
2
3
4
5
6
7
8
0.05
3.84
e
5.99
7.81
0.01
6.63
9.21
11.34
9.49 11.07 12.59
13.28 15.09 16.81 18.48 20.09
14.07 15.51
Phenotype
Observed
Expected
(o - e)
(0 - e)²
(0-e)²
e
Round & Yellow
2706
2713.5
-7.5
56.25
0.021
Round & Green
930
904.5
25.5
650.25
0.719
Wrinkled & Yellow
888
904.5
-16.5
272.25
0.301
Wrinkled & Green
300
301.5
-1.5
2.25
0.007
7.81>1.048
Fail to reject the null hypothesis (Accept the null hypothesis)
The experiment does follow the 9:3:3:1 ratio demonstrating independent assortment.
Σ: 1.048
Degrees of Freedom: 3
Transcribed Image Text:Part 3: Chi-Square Directions: Complete the following data table to reject or fail to reject the null hypothesis. 14. Consider a cross between two heterozygous round yellow peas with the following observed results. Calculate the expected results and chi-square value to determine if the alleles assort independently. P Degrees of Freedom x²= (0-6)² value 1 2 3 4 5 6 7 8 0.05 3.84 e 5.99 7.81 0.01 6.63 9.21 11.34 9.49 11.07 12.59 13.28 15.09 16.81 18.48 20.09 14.07 15.51 Phenotype Observed Expected (o - e) (0 - e)² (0-e)² e Round & Yellow 2706 2713.5 -7.5 56.25 0.021 Round & Green 930 904.5 25.5 650.25 0.719 Wrinkled & Yellow 888 904.5 -16.5 272.25 0.301 Wrinkled & Green 300 301.5 -1.5 2.25 0.007 7.81>1.048 Fail to reject the null hypothesis (Accept the null hypothesis) The experiment does follow the 9:3:3:1 ratio demonstrating independent assortment. Σ: 1.048 Degrees of Freedom: 3
P
Chi-Square Table
Degrees of Freedom
Chi-Square
value
1
2
0.05
0.01
3.84
6.64
3
5.99 7.82
9.21 11.34
4
S
9.49 11.07
13.28
15.09
7
8
14.07
12.59
16.81 18.48
15.51
20.09
x²-Σ (0-2)²
3. In dihybrid cross where both parents were heterozygous for both traits, the following data was
obtained: 890 dominant for both traits, 97 recessive for both traits, 315 dominant for trait 1,
but recessive for trait 2, and 298 dominant for trait 2, but recessive for trait one. Perform a chi-
square analysis to determine if this data agrees with the expected outcomes. State your null
hypothesis as well as if the null is rejected or not.
Transcribed Image Text:P Chi-Square Table Degrees of Freedom Chi-Square value 1 2 0.05 0.01 3.84 6.64 3 5.99 7.82 9.21 11.34 4 S 9.49 11.07 13.28 15.09 7 8 14.07 12.59 16.81 18.48 15.51 20.09 x²-Σ (0-2)² 3. In dihybrid cross where both parents were heterozygous for both traits, the following data was obtained: 890 dominant for both traits, 97 recessive for both traits, 315 dominant for trait 1, but recessive for trait 2, and 298 dominant for trait 2, but recessive for trait one. Perform a chi- square analysis to determine if this data agrees with the expected outcomes. State your null hypothesis as well as if the null is rejected or not.
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