1 / 94

Association tests for correlating genotypes against phenotypes

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.

Télécharger la présentation

Association tests for correlating genotypes against phenotypes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.


Presentation Transcript

  1. Association tests for correlating genotypes against phenotypes

  2. Basics of association testing • Consider the evolutionary history of individuals proximal to the disease carrying mutation.

  3. 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.

  4. 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?

  5. 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

  6. Decision • Probability of error (of the first kind) • Probability (reject H0| H0 is valid) • In this case

  7. 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

  8. 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

  9. Χ2 distribution properties A related distribution is the F-distribution

  10. 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

  11. LLR • An LLR value close to 0, implies that the null hypothesis is true. Asymptotically, the LLR statistic also follows the chi-square distribution.

  12. 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,

  13. 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

  14. 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?

  15. 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.

  16. 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

  17. 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

  18. 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

  19. t-test continued

  20. 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

  21. 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

  22. Continuous outcome ANOVA • How do we extend the t-test when we have multiple groups? MM mm

  23. 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)

  24. F-statistic for 2 groups

  25. T-test again

  26. F-statistic for 2 groups

  27. F-statistic for g groups

  28. 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

  29. 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.

  30. 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?

  31. Linear regression • Sometimes, we have additional information on phenotype values • Ex: the phenotype value might be additive in the number of alleles

  32. 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 ?

  33. 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

  34. 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

  35. How: Solving for least squares • Min||Cβ-x||2 • It is solved by

  36. Using partial derivatives

  37. 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.

  38. 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

  39. 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

  40. 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 nii =0, njj =0,nij ij =0

  41. 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)

  42. Algorithmic issues in multi-locus genome-wide association mapping

  43. 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

  44. 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

  45. 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.

  46. Margin Filtering is not sufficient

  47. Decomposition of 2X2X2 Controls Cases Controls 0 1 Cases 0 1 0 1 Control Cases 0 1

  48. 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

  49. Efficient detection of interactions

  50. Paired Interactions (3X3X2 contingency) Controls Cases 0 1 2 Genotypes at X So, where is the problem? 0 1 2 Genotypes at Y

More Related