Class04-ClassSlides-Spring 2011-03242011

Class04-ClassSlides-Spring 2011-03242011 - Genetic Analysis...

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1 Genetic Analysis II (385) – Population Genetics Lecture 04 Date: 03/24/11 Instructor: Professor Derek Gordon Office: Life Sciences Building, Room 128 Phone: 445-3386 Phone: 445 3386 E-mail: gordon@biology.rutgers.edu Office Hours: Tues, Thurs: 12:00pm – 1:15pm, and by appt. Review 1.Population genetics is concerned with mathematical properties of alleles and mathematical properties of alleles and genotypes. 2.Allele frequencies may always be determined from genotype frequencies. 3.Genotype frequencies may only be 2 determined from allele frequencies under restrictive assumptions known as panmixia . PopGen 4 - 2011 3/24/2011
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2 In Class Assessment AB Ab aB ab .03 .67 .07 .23 1. What are the allele arrays for the A and B loci? 2. What is the expected gametic array at HWE? 3 3. Are the A and B loci jointly in HWE? If not, why ? PopGen 4 - 2011 3/24/2011 A. A (.7 + .3), B(.1 + .9). B (.1 .9). B. (p + q) * (r + s) = (.7+.3) *(.1+.9) = .07 + .63 + .03 + .27. C. D = f(AB)f(ab) - f(Ab)f(aB) D = ( .03*.23)-(.67*.07) = -.04; NO. 4 PopGen 4 - 2011 3/24/2011
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3 Hardy-Weinberg Equilibrium and Different Numbers of Males and Females. An underlying assumption of HWE to date has been that the allele frequencies in the males and females are the allele frequencies in the males and females are the same and that there are the same number of males and females. This does not have to be the case. 5 How can one calculate the allele frequencies in the population, if one knows the allele frequencies in each sex? PopGen 4 - 2011 3/24/2011 Hardy-Weinberg Equilibrium and Different Numbers of Males and Females. This is fairly straight forward . 1. Calculate the allele frequencies in males and females separately. 6 PopGen 4 - 2011 3/24/2011
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4 Hardy-Weinberg Equilibrium and Different Numbers of Males and Females. 2. Then sum the separate allele frequencies weighted by the proportion of males and weighted by the proportion of males and females in the population . p = [f(males) x p m ] + [f(females) x p f ] 7 if f(males) = f(females) = .5 then p = [p m + p f ]/2. Important Concept PopGen 4 - 2011 3/24/2011 In-Class Problem In a population of 4000 Elementary Penguins (3000 females and 1000 males) one observes the following females and 1000 males) one observes the following genotypic arrays. Female Genotype (Counts): AA(1200) + Aa(1200) + aa(600), Male Genotype (Counts): AA (600) + Aa (200) + aa(200). 8 What is the population allele frequency of A and a? PopGen 4 - 2011 3/24/2011
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5 In-Class Problem - Step Through To solve this problem you need to find four things: f (males): f (females): p m and q m : 9 m p f and q f : PopGen 4 - 2011 3/24/2011 In-Class Problem - Step Through To solve this problem you need to find four things: f (males): There are 4000 total Elementary Penguins. 1000 of them are male. Therefore 1000/4000 = .25 of the Elementary Penguins are male. Thus f(males) = .25.
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Class04-ClassSlides-Spring 2011-03242011 - Genetic Analysis...

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