P10-Lecture 9- Hardy-Weinberg- Feb. 22

P10-Lecture 9- Hardy-Weinberg- Feb. 22 - General...

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Unformatted text preview: General Announcements s Turn in Problem Set 1 NOW s Genetics Problem Set 2 x Hardy-Weinberg 3 Posted problems online now. x Due in 1 week- March 1 s Exam 1- Friday March 5 x Exam 1 Review- Wed. March 3 36 7 pm 1130 PLS Hardy-Weinberg as a null model Population Genetics Hardy-Weinberg as a null model - Terms s Alleles: Alternative forms of a gene (e.g. A vs. a) s Genotype: In diploid organisms, the combination of alleles present for a gene (e.g. AA, Aa, aa) s Population: a group of individuals of the same species occupying a given area s Gene pool: All the individual alleles in a population for a given locus Hardy-Weinberg describes non-evolving populations: s The Theorum: Frequencies of alleles and genotypes in a population's gene pool remain constant over generations. xIf: 1. Infinitely large population 2. No mutation 3. Random mating 4. Isolated from other populations 5. All individuals survive & reproduce equally i.e., no natural selection Hardy-Weinberg as a null model - Terms s Allele frequency: The proportion of total alleles composed of a particular allele. E.g. for two alleles: x p = f(A) = (# A alleles)/(# A alleles + # a alleles) x q = f(a) = (# a alleles)/(# A alleles + # a alleles) xp + q = 1 Hardy-Weinberg as a null model - Terms s Genotypic frequency: Proportion of the total number of individuals composed of a particular genotype. E.g. For two alleles: x f(AA) = (# AA)/(total number of individuals) x f(Aa) = (# Aa)/(total number of individuals) x f(aa) = (# aa)/(total number of individuals) An example: Calculating p and q s The free ear lobe allele (F) is dominant to the attached earlobe allele (f) s Suppose we have a population of 1000 individuals x # FF = 700 # Ff = 200 # ff = 100 s What are the allele frequencies? An example: Calculating p and q x # FF = 700 s In a population of 1000 individuals: # Ff = 200 # ff = 100 s What are the allele frequencies? x There are 2000 total alleles (2 per individual) x p = f(F) = (2x(#FF)+(#Ff))/2000 = (1400+200)/2000= 0.8 x q = f(f) = (2x(#ff)+(#Ff))/2000 = (200+200)/2000= 0.2 3 OR 1- p = q (This will always be true!) x (p + q) = (0.8 + 0.2) = 1 An example: Calculating Genotypic frequencies # FF = 700 # Ff = 200 # ff = 100 s What are the genotypic frequencies? x There are 1000 total individuals: x f(FF) = 700/1000 = 0.7 x f(Ff) = 200/1000 = 0.2 x f(ff) = 100/1000 = 0.1 Hardy-Weinberg - What does it do? s If you know the allelle frequencies, it allows you to predict the genotypic frequencies, assuming: x No mutation x No migration x No selection x No drift x Random mating s That is, assume that evolution is not occurring. Hardy-Weinberg - How it works s Assume p = f(F) = 0.8; q = f(f) = 0.2 s If I pull out two alleles at random, what's the chance of getting two F alleles? x The chance of getting one the first time (p=0.8) times the chance of getting one the second time (p=0.8) x Equals p*p = p2 = (0.8)(0.8) = 0.64 Hardy-Weinberg - How it works s Assume p = f(F) = 0.8; q = f(f) = 0.2 s If I pull out two alleles at random, what's the chance of getting two f alleles? x The chance of getting one the first time (q=0.2) times the chance of getting one the second time (q=0.2) x Equals q*q = q2 = (0.2)(0.2) = 0.04 Hardy-Weinberg - How it works s Assume p = f(F) = 0.8; q = f(f) = 0.2 s If I pull out two alleles at random, what's the chance of getting one F and one f? x The chance of getting an F then a f = pq = (0.8)(0.2) = 0.16 + x The chance of getting an f then an F = qp = (0.2) (0.8) = 0.16 x 2pq = 2(0.8)(0.2) = 0.32 Hardy-Weinberg - How it works 25.2 Hardy-Weinberg Equation s The Hardy-Weinberg equation for genotype frequencies: x p2 + 2pq + q2 = 1 This is always true. s Hardy-Weinberg predicts: f(AA) = p2; f(Aa) = 2pq; f (aa) = q2 s Assumes no evolutionary agents are acting x conditions never met in real life... Hardy-Weinberg - Why it's useful s Two distinct ways: s 1) Allows you to determine if evolutionary agents are acting, provided you know p and q! x Assumes no mutation, no migration, no drift, no selection, and random mating is occurring. x If the genotypic frequencies are not what you expect, something interesting must be happening. x I.e., f(FF) p2 3 and/or f(Ff) 2pq 3 If 3 and/or f(ff) q2 Hardy-Weinberg - Why it's useful s 1) Allows you to determine if evolutionary agents are acting, provided you know p and q! x Earlier example: we calculated p=0.8, q=0.2 Genotype Expected Observed f(FF) f(Ff) f(ff) p2 = 0.64 2pq=0.32 q = 0.04 2 Hardy-Weinberg - Why it's useful s 1) Allows you to determine if evolutionary agents are acting, provided you know p and q! x Example, earlier we calculated p=0.8, q=0.2 Genotype Expected Observed f(FF) f(Ff) f(ff) p = 0.64 0.732 2pq=0.32 0.26 q = 0.04 0.008 2 2 Hardy-Weinberg - Why it's useful s Decrease of both Ff & ff plus increase in FF implies: 3 Mutation - usually a small effect 3 Migration - perhaps recessives / heterozygotes migrated out 3 Non-random mating - perhaps those with dominant alleles prefer each other 3 Natural selection perhaps those with the recessive allele don't survive as well 3 Drift - can be important especially in small populations Genetic bottleneck Founder effect Hardy-Weinberg - Why it's useful s 2) Allows you to calculate p and q if you are willing to assume the population is in HardyWeinberg equilibrium (i.e. not evolving) x Example - PKU, due to a recessive allele (a) x Expressed (i.e. homozygous aa) in 1/10,000 people. x Assuming Hardy-Weinberg, q2 = f(aa) = 1/10,000 = 0.0001 x q = SquareRt (.0001) = 0.01 x p = (1- q) = (1 0.01) = 0.99 x Expected genotypic frequencies are: 3 f(AA) = p2 = 0.9801 f(Aa) = 2pq = 0.0198 f(aa) = q2 = .0001 What can cause alleles to shift away from the expected equilibrium? s Changes in allele frequency = Microevolution s Mechanisms of Microevolution: x Mutation x Gene flow x Nonrandom mating x Genetic drift 3 Bottlenecks (assortative) effect x Natural Selection- the only mechanism that results in adaptations to environment. 3 Founder Genetic Drift - Small populations s Drift can be an important factor in changing allele frequencies in small populations. Fixed allele Genetic Drift - Small populations Genetic Drift - Genetic Bottleneck s Survivors of a catastrophe may have allele frequencies that differ from the original population purely due to chance - E.g. Elephant seals; cheetahs Genetic Drift - Founder Effect s A small founding population may have allele frequencies that differ from the parent population purely due to chance x E.g. one form of dwarfism among the Amish in Pennsylvania x 1/60,000 = general pop x 300/60,000 in Amish Genetic Drift - Founder Effect s A small founding population may have allele frequencies that differ from the parent population purely due to chance. x Divergence of Darwin's finches after strays from S. America reached the Galapagos. Gene Flow - Section 25.4 s Movement of alleles from one population to another. x Fitness can increase or decrease x Outcome = allele frequencies equalize 3 Genetic differences are reduced So ur ce p opu la ti on Recipient Population Summary s Important terms: Allele, Genotype, Gene pool, Allele frequency, Genotypic frequency s Hardy-Weinberg assumes no evolutionary agents are acting: No mutation, No migration, No selection, No drift, and Random mating. s Hardy-Weinberg is useful in two distinct ways: x Allows you to determine if evolutionary agents are acting, if you know p and q! Hardy-Weinberg predicts f(AA) = p2, f(Aa) = 2pq, f (aa) = q2 x Allows you to calculate p and q (and genotypic frequencies) without knowing p and q (Assumes HW) Genetics Practice Problems s A boy has hemophilia. Neither his parents nor his grandparents have the disorder. What are the possible genotypes and phenotypes of: x a. the boy? x b. his parents? Mom = Dad = x c. his grandparents? Maternal = Paternal = x d. If this boy grows up and marries his 3rd cousin who is a carrier, what is the probability they will have children with hemophilia? x e. can they have a daughter with hemophilia? Genetics Practice Problems s A boy has hemophilia. Neither his parents nor his grandparents have the disorder. What are the possible genotypes and phenotypes of: x a. the boy? x b. his parents? Mom = Dad = x c. his grandparents? Maternal = Paternal = x d. If this boy grows up and marries his 3rd cousin who is a carrier, what is the probability they will have children with hemophilia? x e. can they have a daughter with hemophilia? What are the chances of a child having AB blood if one of the parents is heterozygous for A blood (IAi) and the other is heterozygous for B (IBi) ? What other genotypes are possible? Recall: Heterozygous = IA i or IB i In humans, hypercholesterolemia is inherited by incomplete dominance. The allele HC calls for abnormally high levels of cholesterol in the blood; the allele HN = normal levels. HC HC persons usually die of a heart attack early in life. HN HN persons are phenotypically normal. When 2 HC HN individuals marry, what is the probability that a given child will have the more serious form of the disease? The milder form? Be completely normal? Could HC HN x HN HN produce a severely diseased baby? ...
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This note was uploaded on 05/11/2010 for the course BIOLOGY 100 taught by Professor Richard during the Spring '10 term at George Mason.

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