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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 - PowerPoint PPT Presentation
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Case-Control Genetic Association Studies
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
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.
Case-Control Association StudiesObserved contingency table
AA (2) Aa (1) aa (0) SubtotalCases n2 n1 n0 nControls m2 m1 m0 m
Subtotal h2 h1 h0 N
Expected contingency table
AA (2) Aa (1) aa (0) Freq.Cases p1q2N p1q1N p1q0N p1Controls p2q2N p2q1N p2q0N p2
Freq. q2 q1 q0 1
p2 mN
q2 h2
N
p1 nN
q0 h0
N
q1 h1
N
Test Case-Control Associations
cells all
22
exp.
.)exp(obs.
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)
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 20Controls 60 30 10
Χ2 = 5.9, compared with Χ20.05(df=2) = 5.99 p-value = 0.053
Integrating quantitative genetic theory into the case-control analysis framework
Three genotypes Testing the additive effectAA μ2 = μ + a AA aaAa μ1 = μ + d Cases 45 20aa μ0 = μ - a Controls 60 10a: additive effect Χ2 = ?, compared with Χ2
0.05(df=1) = 3.84 d: dominant effect Testing the dominant effect
2AaAA+aa
2a = μ2 – μ0 Cases 70 65 2d = 2μ1 – (μ2 + μ0) Controls 60 70 Χ2 = ?, compared with Χ2
0.05(df=1) = 3.84
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, …
Epistasis
• Biological definition: The expression of one gene is masked by the second geneAA ≠ Aa ≠ aa, but AABB = AaBB = aaBB
• Statistical definition: Deviation from the additive expectation of allelic effectsAABB ≠ AA + BB
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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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 epistasisiad = additive x dominant epistasisida = dominant x additive epistasisidd = dominant x dominant epistasis
Genetic Effects
Testing additive x additive
Testing additive x dominant
Testing dominant x additive
Testing dominant x dominant
• Calculate x2 test statistics• Compare them to critical value x2
(df=1)
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