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Case-Control Genetic Association Studies

Case-Control Genetic Association Studies. Genetic Designs of Complex Traits and Diseases. Controlled crosses – backcross, F2, RIL, … - Linkage analysis (recombination fraction) Natural populations - Linkage disequilibrium (LD) (Unrelated) Family design - Joint linkage and LD

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Case-Control Genetic Association Studies

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  1. Case-Control Genetic Association Studies

  2. Genetic Designs ofComplex Traits and Diseases • Controlled crosses – backcross, F2, RIL, … - Linkage analysis (recombination fraction) • Natural populations - Linkage disequilibrium (LD) • (Unrelated) Family design - Joint linkage and LD • (Related) Family design - Joint linkage and LD with identical by descent • Case-control design

  3. Case-Control Association Studies • The basic idea is to compare genetic factors within case (affected) and control populations to identify correlations with a defined phenotype. • The basic approach is to test if markers are more frequent in one population compared to a second population.

  4. Case-Control Association Studies Observed contingency table AA (2) Aa (1) aa (0) Subtotal Cases n2n1n0n Controls m2m1m0m Subtotal h2 h1 h0N Expected contingency table AA (2) Aa (1) aa (0) Freq. Cases p1q2N p1q1N p1q0N p1 Controls p2q2N p2q1N p2q0N p2 Freq. q2 q1 q0 1

  5. Test Case-Control Associations is compared with 2[df=(3-1)(2-1)=2,0.05] In general, let g = number of levels of treatment 1, c = number of levels of treatment 2: df =(g-1)(c-1)

  6. Example • Sample 100 cases and 100 controls from a natural population • Genotype these cases and controls genome-wide or at particular regions • Consider a SNP with two alleles A and a AA Aa aa Cases 45 35 20 Controls 60 30 10 Χ2 = 5.9, compared with Χ20.05(df=2) = 5.99  p-value = 0.053

  7. Integrating quantitative genetic theory into the case-control analysis framework Three genotypes Testing the additive effect AA μ2 = μ + a AA aa Aa μ1 = μ + d Cases 45 20 aa μ0 = μ - a Controls 60 10 a: additive effect Χ2 = ?, compared with Χ20.05(df=1) = 3.84 d: dominant effect Testing the dominant effect 2Aa AA+aa 2a = μ2 – μ0 Cases 70 65 2d = 2μ1 – (μ2 + μ0) Controls 60 70 Χ2 = ?, compared with Χ20.05(df=1) = 3.84

  8. Multiple Testing • Multiple comparisons – Associating multiple SNPs with multiple phenotypes can lead to false positive results • Solutions - Simulation to determine empirical P values - Replication - Bonferroni, FDR, …

  9. Epistasis • Biological definition: The expression of one gene is masked by the second gene AA ≠ Aa ≠ aa, but AABB = AaBB = aaBB • Statistical definition: Deviation from the additive expectation of allelic effects AABB ≠ AA + BB

  10. Why Study Epistasis • Increase genetic diversity and variation as so to better adapt to changing environments (evolution and speciation) • Regulate complex human diseases such as cancer (one of the reasons why cancer is so difficult to study)

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  12. Quantitative Genetic Model of Epistasis BB (a2) Bb (d2) bb (-a2) AA (a1) μ22=μ+a1+a2+iaa μ21=μ+a1+d2+iad μ20=μ+a1-a2-iaa Aa (d1) μ12=μ+d1+a2+ida μ11=μ+d1+d2+idd μ10=μ+d1-a2-ida aa (-a1) μ02=μ-a1+a2-iaa μ01=μ-a1+d2-iad μ00=μ-a1-a2+iaa iaa = additive x additive epistasis iad = additive x dominant epistasis ida = dominant x additive epistasis idd = dominant x dominant epistasis

  13. Genetic Effects

  14. Testing additive x additive

  15. Testing additive x dominant

  16. Testing dominant x additive

  17. Testing dominant x dominant

  18. Calculate x2 test statistics • Compare them to critical value x2(df=1)

  19. Example • 830 unrelated stroke patients • 454 normal unrelated subjects • 27 candidate genes for stroke, located on chromosomes 1, 2, 5, 11, 14, 17, 18, and 21

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