GENE 412 1st Edition Lecture 2Outline of Last Lecture 1. Hardy-Weinburg EquationA. Hardy-Weinburg AssumptionsB. Population Assumptions for Hardy-Weinburg2. Different Ways to be an Allele3. Allele Frequencies for Future GenerationsOutline of Current Lecture I. Hardy-Weinburg for multiple allelesII. Find allele frequency from genotype frequencyIII. Chi-Squared for Genetics ProblemsCurrent LectureI. Hardy-Weinburg for multiple allelesa. Many population have genes that have multiple alleles or more than just two allelesi. Some examples is Blood type in Humans and alcohol metabolism in fruit fliesb. Equation for multiple allelesi. (p+q+r)2 = p2+2pq+q2+2qr+2pr+r2ii. Number of genotypes= .5n(n+1) where n = number of alleles II. Find allele frequency from genotype frequencyTEST QUESTION FOR SUREThree Alleles F, M, & S Number of genotypes= .5(3)(3+1) = 6 genotypesFF FM MM FS MS SS40 20 10 10 10 10Find 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=.55III. Chi-Squared for Genetics Problemsa. Equation; (Observed-Expected)/(Expected)2i. Other ways to calculate; (observed-mean)/(standard deviation)2b. Ho is the expected genotypic frequencies of Hardy-Weinburgc. Ha is the hypothesis being tested to see if population is NOT in Hardy-Weinburg Equilibriumd. 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 success1. Example: Testing an anti-cancer drug on mice in a labii. Risk should be high if looking to be precise 1. Example: Testing humans in an ani-cancer drug
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