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1 GENETIC EPIDEMIOLOGY Institute of Clinical Medicine Po-Hsiu Kuo – 04/16/2008 Outline Overview of gene mapping Allele frequency Quantitative genetics Genetic component and Heritability Hardy-Weinberg Equilibrium Linkage Disequilibrium Gene mapping LOCALIZE and then IDENTIFY a locus that regulates a trait Linkage analysis Association analysis Allele frequency and genotype frequency 4 Allele (gene) Sample Allele 30 60 10 100 A1A1 A1A2 A2A2 Total A1 60 60 0 120 A2 0 60 20 80 Genotype frequency? 0.3, 0.6, 0.1 Allele frequency f(A1) ? 120/200 = 0.6 Allele frequency f(A2) ? 80/200 = 0.4 Factors influence on allele frequency 5 Population size Reproduce rate and Survival rate Immigration and mutation Pattern of assortative mating Quantitative genetics Study of continuous traits (such as height or weight) and its underlying mechanisms Combined effect of the many underlying genes results in a continuous distribution of phenotypic values Quantitative genetics is not limited to continuous traits, but to all traits that are determined by many genes Continuous traits are quantitative traits with a continuous phenotypic range. They are usually polygenic, and may also have a significant environmental influence Traits with ordinal numbers, such as number of bristles on a fruit fly. These traits can be either treated as approximately continuous traits or as threshold traits Some qualitative traits can be treated as if they have an underlying quantitative basis, expressed as a threshold trait (or multiple thresholds) Population level Genotype frequencies (Random mating) Allele 1 Allele 2 A (p) a (q) A (p) AA (p2) Aa (pq) a (q) aA (qp) aa (q2) Hardy-Weinberg Equilibrium frequencies P (AA) = p2 P (Aa) = 2pq P (aa) = q2 p2 + 2pq + q2 = 1 Phenotype level Quantitative traits g==-1 g==0 .128205 .072 Fraction AA g==-1 .128205 .128205 g==-1 g==0 g==1 g==0 -3.90647 Fraction .128205 0 Fraction Fraction Aa 0 g==1 .128205 0 0 g==1 -3.90647 -3.90647 .128205 -3.90647 2.7156 2.7156 qt Histograms by g aa 0 -3.90647 2.7156 qt e.g. cholesterol levels 0 -3.90647 0 -3.90647 2.7156 qt Histograms by g 2.7156 qt Histograms by g Components of Phenotypic Values Phenotype (P) = Genotype (G) + Environment (E) Considering variances (Var), this becomes: Var(P) = Var(G) + Var(E) + 2 Cov(G,E) G = A (additive genotypic value) + D (dominance deviation) A: breeding value (i.e. the sum of the average effects of the two alleles) D: intra-allelic interaction In planned experiments, we can often take Cov(G,E) = 0 Heritability Heritability Proportion of phenotypic variation in a population that is attributable to genetic variation among individuals Variation among individuals may be due to genetic and/or environmental factors Heritability analyses estimate the relative contributions of differences in genetic and non-genetic factors to the total phenotypic variance in a population. Resemblance between relatives VA VD VEC Parents & offspring (PO) 1/2 0 VEC(PO) Full - siblings (FS) 1/2 1/4 VEC(FS) Half - siblings (HS) 1/4 0 VEC(HS) Fraternal twins (DE) ½ 1/4 VEC(DE) Identical twins (ME) 1 1 VEC(ME) Phenotypic covariance between rMZ = rDZ = 1 rMZ = 1, rDZ = 0.5 E E e rMZ = 1, rDZ = ^ C c C A a A Q Q q q Twin 1 mole count a Twin 2 mole count c e Twin study One of behavioral genetics study design Highlighting the role of environmental and genetic causes on behavior Shared environmental influences common to members of family —class, parenting styles, education etc Shared genes, inherited from parents Compares the similarity between twins (MZ vs. DZ) Modern history of the twin study derives from Sir Francis Galton's pioneering use of twins to study the role of genes and environment on human development and behavior The Law of Hardy’s 14 Assumption Radom mating, no selection, no mutation, no immigrant Question: What’s the allele frequency between generations? Two alleles (A, a) F(A) = p, f(a) = q, p+q=1 Frequency of genotype distribution AA, Aa, aa? Allele frequency in the next generation 15 f(A) = p2 + ½ (2pq) = p(p+q) =p f(a) = q2 + ½ (2pq) = q(p+q) =q Therefore, under previous assumptions, allele frequency is unchanged over generations – i.e. reached equilibrium Note: a German physician Wilhelm Weinberg made the same conclusion as Hardy so called Hardy-Weinberg Equilibrium (HWE) Test for Hardy-Weinberg Equilibrium 16 Test for genotype frequency Example: examining MN blood typing in 6129 Americans MM MN NN Total 1787 3707 1305 6129 Under HWE, test for χ2 = Σ(O-E)2/ E First: f(M) = p = 0.539 f(N) = q = 0.461 Test for HWE (conti.) P=0.539, q=0.461 17 f(MM) = p2 = 0.291, f(MN) = 2pq = 0.496, f(NN) = q2 = 0.212 MM MN NN Total Observed 1787 3707 1305 6129 Expected 0.291*(6129) 0.496*(6129) 0.212*(6129) 6129 (O-E)2/E χ2 = 0.00489 P(χ2,df=1 ) = 0.9 Conclusion: MN blood patterns in this sample are in HWE Linkage phase 18 For two loci, features of being in equilibrium means p(AB) = 2 p(A)p(B) Linked two loci, say A1A2/B1B2 Coupling phase Repulsion phase A B 1 2 A1 B1 A2 B2 A2 B1 Linkage Disequilibrium 19 Gametic Association equilibrium, f(AB gamete) = PAB = PA× PB Consider 2 alleles are at each of 2 loci with random mating, then there are ten possible zygotes If r is the probability that a gamete is recombinant In each generation, the amount of gametic disequilibrium, measured by PAB - PA PB , is reduced by a factor of r PAB‘ = PAB – r D (where D = p11 p22 - p12 p21) At D: a quantitative measure of the amount of linkage disequilibrium Linkage Disequilibrium (conti.) 20 Dt=(1-r)t D0 Decay of linkage disequilibrium parameters between two loci depends on recombination fraction and time (generation) Dmax = min (p1q2, p2q1), the largest positive value When p11= p22 = 0.5 & p12= p21= 0, max D = 0.25 D´= D / Dmax, defined by Lewontin (1988) 21 Falconer and Mackay (1996) Population Admixture Admixture Population 1 Population 2 m 1-m A1 p1 P1 A2 q1 Q1 B1 p2 P2 B2 q2 Q2 A1B1 m p 1 p2 (1-m) P1 P2 A1B2 m p 1 q2 (1-m) P1 Q2 A2B1 m q 1 p2 (1-m) P2 Q1 A2B2 m q 1 q2 (1-m) Q1 Q2 Mixture proportion Allele frequency Gametic frequency D = PA1B2 PA2B1-PA1B1 PA2B2 = [m p1 q2+(1-m) P1 Q2]‧[m q1 p2+(1-m) P2 Q1] -[m p1 p2+(1-m) P1 P2]‧[m q1 q 2+(1-m) Q1 Q2] = m (1-m) (p1-P1) (p2-P2) Only when p1 = P1 or p2 = P2 or (p1 = P1 and p2 = P2), D=0