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