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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Question
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
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