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Association tests for correlating genotypes against phenotypes. Basics of association testing. Consider the evolutionary history of individuals proximal to the disease carrying mutation. Association testing.

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## Association tests for correlating genotypes against phenotypes

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**Association tests for correlating genotypes against**phenotypes**Basics of association testing**• Consider the evolutionary history of individuals proximal to the disease carrying mutation.**Association testing**• The goal of association testing is to identify SNPs that ‘associate’ (are correlated) with the phenotype. • Recall that spatially close SNPs are correlated because of LD. • As we go further, recombination changes evolutionary history, and the SNPs are no longer correlated.**Statistical hypothesis testing**• Example (from wiki) • An individual claims to be clairvoyant. To test this, pick 25 cards from a deck (with replacement) and ask him to guess the color each time. • He guesses correctly c times • Is he clairvoyant • If c=25? • If c= 6? • If c= 10?**Statistical hypothesis testing**• Goal is to take observations and reach a conclusion. The conclusion is often a decision between two hypotheses. • H0: (Null) the individual is not clairvoyant • H1: (Alternative) The individual is clairvoyant**Decision**• Probability of error (of the first kind) • Probability (reject H0| H0 is valid) • In this case**Tests for association: Pearson**Cases Controls O1 MM Mm mm • Case-control phenotype: • Build a 3X2 contingency table • Pearson test (2df)= O2 O3 O4 O5 O6**The χ2 test**Cases Controls O1 O2 MM O3 O4 Mm O5 O6 mm • The statistic behaves like a χ2 distribution. • A p-value can be computed directly**Χ2 distribution properties**A related distribution is the F-distribution**Likelihood ratio**• Another way to check the extremeness of the distribution is by computing a (log) likelihood ratio. • We have two competing hypothesis. Let N be the total number of observations**LLR**• An LLR value close to 0, implies that the null hypothesis is true. Asymptotically, the LLR statistic also follows the chi-square distribution.**Exact test**• The chi-square test does not work so well when the numbers are small. • How can we compute an exact probability of seeing a specific distribution of values in the cells? • Remember: we know the marginals (# cases, # controls,**Fischer exact test**Cases Controls a b MM c d Mm e f mm • Num: #ways of getting configuration (a,b,c,d,e,f) • Den: #ways of ensuring that the row sums and column sums are fixed**Fischer exact test**• Remember that the probability of seeing any specific values in the cells is going to be small. • To get a p-value, we must sum over all similarly extreme values. How?**Test for association: Fisher exact test**Cases Controls a b MM c d Mm e f mm • Here P is the probability of seeing the exact count. • The actual significance is computed by summing over all such tables that are at least this extreme.**Continuous outcomes**• Instead of discrete (Case/control) data, we have real-valued phenotypes • Ex: Diastolic Blood Pressure • In this case, how do we test for association**Continuous outcome ANOVA**• Often, the phenotypes are not offered as case-controls but like a continuous variable • Ex: blood-pressure measurements • Question: Are the mean values of the two groups significantly different? MM mm**Two-sided t-test**• For two categories, ANOVA is also known as the t-test • Assume that the variables from the two sets are drawn from Normal distributions • Different means, equal variances • Null hypothesis is that they are both from the same distribution**Two-sample t-test**• As the variance is not known, we use an estimate S, defined by • The T-statistic is given by • Significant deviations from 0 are used to reject the Null hypothesis**Two-sample t-test (unequal variances)**• If the variances cannot be assumed to be equal, we use • The t-statistic is given by • Significant deviations from 0 are used to reject the Null hypothesis**Continuous outcome ANOVA**• How do we extend the t-test when we have multiple groups? MM mm**F-statistic for 2 groups**explained variance (with m+n-1 – (m+n-2) = 1 df) • Under the alternative hypothesis, the variance is reduced Unexplained variance (with m+n-2 df)**A generic ANOVA strategy**• Consider a null model (p1 parameters), and an alternative model (p2> p1 parameters) • The alternative model can be parameter free (ex: groupings of the phenotype values according to genotypes), or based on a model (ex: additive) • If based on a model, compute the optimum parameters • Compute the reduction in variance. • Use an F-test for association**Haplotype testing**• Why test with multiple SNPs? • Pros: haplotypes might be better correlated with disease outcome • The tests are similar, except that instead of 3 rows, we have a certain number (k) of haplotypes.**Haplotype testing**• Any of the tests described before can be used for haplotype based contingency tables. • What are the Pros and cons of using haplotypes?**Linear regression**• Sometimes, we have additional information on phenotype values • Ex: the phenotype value might be additive in the number of alleles**Linear regression**• The parameters can be estimated using linear regression analysis • Let Xijbe the phenotypic value of the j-th individual in class i (genotype i) • Xij=+i+ij • i=0 • Generally, • X=C+ • Goal is to estimate so that |||| is minimized • Why is this useful? • How do we optimize the choice of ?**Why: Linear regression testing**• Recall that we want to test if the genotype is useful in predicting phenotype (X) • If not, then the null model Xij=+ij should have the same amount of variance in the residual ij**Linear regression**• Linear regression methods can be used to estimate the parameters of • X = C+ • To test for association, estimate the parameters for two models • Ex: Xij=+i+ijvsXij=+’ij • Note that both , ’ are assumed to be random variables with mean 0, and that Var()<=Var(’) • We can test for association by asking if the reduction in variance Var(’)-Var() is significant • This can be done parametrically (Ex: F-test) • Or, non-parametrically, using a permutationtest**How: Solving for least squares**• Min||Cβ-x||2 • It is solved by**Association test summary (Single locus)**• Discrete outcomes (case-control) • Pearson’s/Fischer exact test • Continuous variables • T-test (2 categories) • ANOVA (multiple categories) • Linear regression (multiple categories with linearity assumption) • Single locus can be extended to haplotypes • Multiple correlated SNPs • Only change is that the number of categories expands.**Epistatic and gene environment interactions**• The typical Mendelian disorder assumes that there is a single causal variation. • Having the variation pre-disposes you to a certain phenotype • For complex disease, this may not be a correct model • Different variants may combinatorially interact**Two-way ANOVA**• Suppose that there are two ways of classifying individuals. • Ex: genotypes at two loci • Ex: genotype versus sex • Ex: genotype versus environment • Assume that there are sufficient individuals in each cell. • Estimate the means/variances in each cell • An ANOVA test may be used to determine if the values can are significantly different M F aa Aa AA**2-way ANOVA model**• Xijk: phenotype value for the k-th individual in cell (i,j) • Assume that Xijk=+i+j+ij+ijk • i j are fixed parameters contributing to class i,j • ij is a parameter corresponding to interaction between class i,j • i nii =0, njj =0,nij ij =0**ANOVA model**• We have two questions: • Are the loci associated with the disease? • To answer this, test this model against the null model Xijk=+ijk • Is epistatic interaction important • Test this model against Xijk=+i+j+ijk • (Set ij = 0 in the null hypothesis)**Algorithmic issues in multi-locus genome-wide association**mapping**Detecting multiple loci**• The most naïve strategy, is to look at all pairs of loci (or all k-tuples) that influence a complex disease. • This is computationally intensive, and also has a problem with multiple testing. • Other strategies: • Consider a subset S of SNPs that show an association individually. • Limit association testing to pairs: • At least one of the SNPs comes from S • Both SNPs come from S**Two locus testing results**• The power represents the fraction of times the test succeeded in detecting the right pair. • The pair-wise models often do much better than the other models. Model 1 Model 2 Model 3**Margin based filtering**Controls Cases Controls 0 1 Genotypes at X Cases 0 1 Genotypes at X Control Cases 0 1 Genotypes at Y Genotypes at Y • Consider only those locus pairs that show a marginal effect. Ex: Marchini et al.**Decomposition of 2X2X2**Controls Cases Controls 0 1 Cases 0 1 0 1 Control Cases 0 1**Pairwise interactions**Chi-square(x,y,d) is high Chi-square (x,d) is high OR Chi-square (y,d) is high OR Chi-square (x,y) when limited to cases is high OR Chi-square (x,y) when limited to controls is high. When restricted to cases, X and Y show high correlation. But, testing requires nm2 time Cases T A -n/8 n/8 -n/8 n/8 A G**Paired Interactions (3X3X2 contingency)**Controls Cases 0 1 2 Genotypes at X So, where is the problem? 0 1 2 Genotypes at Y

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