Answered: Would it have made a difference if…

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter1: A Perspective On Human Genetics

Section: Chapter Questions

Problem 9QP: If your father were diagnosed with an inherited disease that develops around the age of 50, would...

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1. Would it have made a difference if natural selection had acted on the dominant allele instead of the recessive one? Why or why not Only answer question 1 below is an information about the question. Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical value (5.99), Thus, we reject the null hypothesis. c. The population is not in Hardy-Weinberg Equilibrium, and evolution is indeed taking place or does not conform to null hypothesis. Data result of Testing Hardy-Weinberg Equilibrium without natural selection Chi-square of results from bean model F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 0.21 The resulting chi-squared value is 0.21. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 0.21 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis Chi-square of results from bean model F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 2.52 The resulting chi-squared value is 2.52. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 2.52 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis

TABLE 2 Comparison of expected and observed genotypic and allelic frequencies for the F1
generation produced by the bean model:
Observed and Expected Genes in F1
Component
Observed
Expected
Frequency
Number
Frequency
Number
Alleles
A
0.43
43
0.40
40
0.57
57
0.60
60
Genotypes
AA
0.24
12
0.16
8
Aa
0.38
19
0.48
24
aa
0.38
19
0.36
18
without natural
selection
Observed and Expected Genes in F2
Component
Observed
Expected
Frequency
Number
Frequency
Number
Alleles
A
0.52
52
0.40
40
a
0.48
48
0.60
60
Genotypes
AA
0.28
14
0.16
8
Aa
0.48
24
0.48
24
aa
0.24
12
0.36
18

TABLE 6
Observed and Expected Genes in F1
Component
Observed
Expected
Frequency
Number
Frequency
Number
Alleles
A
0.69
69
0.70
70
0.31
31
0.30
30
Genotypes
AA
0.48
24
0.49
24.5
Aa
0.42
21
0.42
21
aa
0.1
0.09
4.5
WITH ΝATURAL SELΕCT ION
TABLE 7
Observed and Expected Genes in F2
Component
Observed
Expected
Frequency
Number
Frequency
Number
Alleles
A
0.62
62
0.69
69
a
0.38
38
0.31
31
Genotypes
AA
0.4
20
0.48
24
Aa
0.44
22
0.42
21
aa
0.16
8
0.1

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