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s Turn in Problem Set 1 NOW s Genetics Problem Set 2
x HardyWeinberg
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 HardyWeinberg as a null model Population Genetics HardyWeinberg 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 HardyWeinberg describes nonevolving 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 HardyWeinberg 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 HardyWeinberg 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 HardyWeinberg  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. HardyWeinberg  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 HardyWeinberg  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 HardyWeinberg  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 HardyWeinberg  How it works 25.2 HardyWeinberg Equation
s The HardyWeinberg equation for genotype frequencies: x p2 + 2pq + q2 = 1 This is always true. s HardyWeinberg predicts: f(AA) = p2; f(Aa) = 2pq; f (aa) = q2
s Assumes no evolutionary agents are acting
x conditions never met in real life... HardyWeinberg  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 HardyWeinberg  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 HardyWeinberg  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 HardyWeinberg  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 Nonrandom 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 HardyWeinberg  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 HardyWeinberg, 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 HardyWeinberg assumes no evolutionary agents are acting: No mutation, No migration, No selection, No drift, and Random mating. s HardyWeinberg is useful in two distinct ways: x Allows you to determine if evolutionary agents are acting, if you know p and q! HardyWeinberg 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.
 Spring '10
 richard
 Genetics

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