GENE 412 1st Edition Lecture 3 Outline of Last Lecture I Hardy Weinburg for multiple alleles II Find allele frequency from genotype frequency III Chi Squared for Genetics Problems Outline of Current Lecture I X2 Chi Squared Test II P 05 Concludes Current Lecture I X2 Chi Squared Test a Ho population is in H W equilibrium b Ha Population is not in H W equilibrium c Equation i Summation of O E 2 E where O observed number in population E Expected number in population Accept Ho Reject Ho Ho True Correct risk Ha True Beta error 1 power d Example AA 40 p2 Aa 40 2pq aa 20 q2 Step 1 Find Allele Frequencies Frequency of A 40 20 100 6 Frequency of a 1 p 4 Step 2 These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute Find Chi Squared Value for each genotype and then add together p2 2pq q2 Expected 36 48 16 O E 4 8 4 O E 2 16 64 16 O E 2 E 16 36 64 48 16 16 16 36 64 48 16 16 2 77 Step 3 Calculate Degrees of Freedom DF DF of genotypes of alleles Step 4 If Chi Square value is above set risk or Alpha Error and for the degrees of freedom for experiment then the population is not in Hardy Weinburg H W and therefore one of the assumptions of H W is not met see lecture 1 for H W assumptions II Other Test topics for Chi Squared a Risk factors assume the type of risk you want to take b In genetics usually use p 05 is an acceptable risk to reject Ho c Reasons H W is rejected i Two genes in perfect H W while other is not ii Mate selected based on gene or gene is lethal iii Hard to score H W d Tips for Chi Squared i Never reject H W because it is a law the population just may not be in H W for a particular gene e Vocabulary i Endoreduplicated is 2N amount of DNA ii Inversions Create inversion loops and that is their only way to pair 1 Inversions In Drosophila heterozygotes inhibit the ability to crossover