GENE 412 1st Edition Lecture 2 Outline of Last Lecture 1 Hardy Weinburg Equation A Hardy Weinburg Assumptions B Population Assumptions for Hardy Weinburg 2 Different Ways to be an Allele 3 Allele Frequencies for Future Generations Outline of Current Lecture I Hardy Weinburg for multiple alleles II Find allele frequency from genotype frequency III Chi Squared for Genetics Problems Current Lecture I Hardy Weinburg for multiple alleles a Many population have genes that have multiple alleles or more than just two alleles i Some examples is Blood type in Humans and alcohol metabolism in fruit flies b Equation for multiple alleles i p q r 2 p2 2pq q2 2qr 2pr r2 ii Number of genotypes 5n n 1 where n number of alleles II Find allele frequency from genotype frequencyTEST QUESTION FOR SURE Three Alleles F M S Number of genotypes 5 3 3 1 6 genotypes FF 40 FM 20 MM 10 FS 10 MS 10 SS 10 Find Frequency for F allele 40x2 20x1 20x1 Totalx2 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 110 200 55 III Chi Squared for Genetics Problems a Equation Observed Expected Expected 2 i Other ways to calculate observed mean standard deviation 2 b Ho is the expected genotypic frequencies of Hardy Weinburg c Ha is the hypothesis being tested to see if population is NOT in Hardy Weinburg Equilibrium d The level at which you reject H0 is rejected when H0 is still true is called risk alpha error i Risk should be low if looking for any level of success 1 Example Testing an anti cancer drug on mice in a lab ii Risk should be high if looking to be precise 1 Example Testing humans in an ani cancer drug facility