phenotypes is as follows: Phenotype AB Number 138 a b а в A b 132 69 61 (i) Are the genes assorting independently? Test the hypothesis with a chi-square test. State your hypothesis, estimate the p value using the x' table below and clearly state your conclusion. Table 1: Chi-square table Chi Square Values and Probability Degrees of Freedom P= 0.99 0.95 0.80 0.50 0.20 0.05 0.01 1 0.000157 0.00393 0.0642 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3 0.115 0.352 1.005 2.366 4.642 7.815 11.345 4 0.297 0.711 1.649 3.357 5.989 9.488 13.277 (ii) If the genes are linked, calculate the map distance between the genes in cM. '.

Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Question
In pea plants, a dihybrid for the recessive a and b is testerossed. The distribution of the
phenotypes is as follows:
Phenotype
AB
Number
138
a b
132
a B
69
A b
61
(i)
Are the genes assorting independently? Test the hypothesis with a chi-square test.
State your hypothesis, estimate the p value using the x table below and clearly state
your conclusion.
Table 1: Chi-square table
Chi Square Values and Probability
Degrees of
Freedom
P = 0.99
0.95
0.80
0.50
0.20
0.05
0.01
0.000157
0.00393
0.0642
0.455
1.642
3.841
6.635
2
0.020
0.103
0.446
1.386
3.219
5.991
9.210
3.
0.115
0.352
1.005
2.366
4.642
7.815
11.345
0.297
0.711
1.649
3.357
5.989
9.488
13.277
(ii)
If the genes are linked, calculate the map distance between the genes in cM. '.
Transcribed Image Text:In pea plants, a dihybrid for the recessive a and b is testerossed. The distribution of the phenotypes is as follows: Phenotype AB Number 138 a b 132 a B 69 A b 61 (i) Are the genes assorting independently? Test the hypothesis with a chi-square test. State your hypothesis, estimate the p value using the x table below and clearly state your conclusion. Table 1: Chi-square table Chi Square Values and Probability Degrees of Freedom P = 0.99 0.95 0.80 0.50 0.20 0.05 0.01 0.000157 0.00393 0.0642 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3. 0.115 0.352 1.005 2.366 4.642 7.815 11.345 0.297 0.711 1.649 3.357 5.989 9.488 13.277 (ii) If the genes are linked, calculate the map distance between the genes in cM. '.
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