Hierarchical Clustering

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1 Hierarchical Clustering Some slides by Serafim Batzoglou 1

From expression profiles to distances From the Raw Data matrix we compute the similarity matrix S. S ij reflects the similarity of the expression patterns of gene i and gene j.

2 From expression profiles to distances From the Raw Data matrix we compute the similarity matrix S. S ij reflects the similarity of the expression patterns of gene i and gene j. genes experiments Expression levels, Raw Data experiments experiments In some situation the input for clustering is only the similarities / distances

3 More generally In K-means and SOM the input was a vector for each item (e.g. a dot in R n ) Here we have a matrix of pairwise distances between items, and we wish to cluster the items. A distance based clustering alg 3

4 An Alternative view of Clustering Form a tree-hierarchy of the input elements satisfying: More similar elements are placed closer along the tree. Or: Tree distances reflect element similarity Note: No explicit partition into clusters. 4

5 Partitioning vs Hierarchical Representations dendrogram 5

6 Hierarchical Representations (2) Ultrametric: rooted tree, all root-leaf distances are equal

7 UPGMA Clustering (unweighted pair group method using arithmetic averages) Approach: Form a tree; closer species according to input distances should be closer in the tree Build the tree bottom up, each time merging two smaller trees All leaves are at same distance from the root 7

8 Hierarchical Clustering: UPGMA Sokal & Michener 58, Lance & Williams 67 UPGMA (unweighted pair group method using arithmetic averages) Given two disjoint clusters C i, C j, 1 d ij = Σ {p Ci, q Cj} d pq C i C j If C k = C i C j, then distance from C k to another cluster C l is: d il C i + d jl C j d kl = C i + C j

9 Algorithm: UPGMA Initialization: Assign each x i into its own cluster C i Define one leaf per sequence, height 0 Iteration: Find two clusters C r, C s s.t. d rs is min Define a new cluster C t = C r C s Define node A rs connecting C r, C s, height d rs /2 Thm: If the input distances match an ultrametric tree UPGMA finds it Delete C r, C s d it =d ti =( C r d ir + C s d is )/( C r + C s ) length(c r, A rs ) = height(a rs ) - height(c r ) length(cs,a rs ) = height(a rs ) - height(c s ) Termination: When all sequences belong to one cluster Time: Naïve: O(n 3 ); Can show O(n 2 logn) (ex.); O(n 2 ) (harder ex.)

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Robert R. Sokal (1926-2012) Ph.D. 1952, University of Chicago. Was at Dept. of Ecology and Evolution, SUNY Stony Brook Member of the National Academy of Sciences & American Academy of Sciences.

11 Robert R. Sokal ( ) Ph.D. 1952, University of Chicago. Was at Dept. of Ecology and Evolution, SUNY Stony Brook Member of the National Academy of Sciences & American Academy of Sciences. Promoted the use of statistics in biology and co-founded the field of numerical taxonomy. Together with P.H.A. Sneath, authored the two defining texts in this field. Along with F. James Rohlf, authored the very popular biostatistics book, Biometry. Editor of the American Naturalist, president of several learned societies. 12

12 Results (2) 10 major groups with similar patterns of cooccurrence, confirming that specific groups of phenotypes co-occur within families. certain malformations co-occur in more than one group, e.g. TGA,AVSD. Some differences from a proposed taxonomy (Houyel 11) (Also: co-occurrence of defects in families is caused by shared susceptibility genes.) A starting point for further biomed research 15

13 Variants on hierarchical clustering Input: Distance matrix D ij; Initially each element is a cluster. Find min element D rs in D; merge clusters r,s Delete elts. r,s, add new elt. t with updated weights Repeat Variants: Average linkage: UPGMA Single linkage: D it = min(d ir, D is ) Max linkage D it = max(d ir, D is ) Sometimes the number of clusters is needed. Methods abound. Sometimes leaf order matters and not only topology. 16

14 Hierarchical clustering of GE data Eisen et al., PNAS 1998 Growth response: Starved human fibroblast cells, added serum Monitored levels of 8600 genes over 13 time-points t ij - level of target gene i in condition j; r ij same for reference D ij = log(t ij /r ij ) D* ij = [D ij E(D i )]/std(d i ) Similarity of genes k,l: S kl =(Σ j D* kj D* lj )/N cond Applied average linkage method Ordered leaves by increasing subtree weight: average expression level, time of maximal induction, other criteria 17

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17 Clustering the same data after randomly permuted within rows (1), columns (2) and both(3) 20

18 Observations Distinct measurements of same genes cluster together Genes of similar function cluster together Many cluster-function specific insights Interpretation is a REAL biological challenge 21

19 Yeast GE data 22

20 Mike Eisen & Pat Brown 23

21 More on hierarchical methods (2) The methods described above agglomerative (bottom up) An alternative approach: Divisive (top down) Advantages: gives a single coherent global picture Intuitive for biologists (from phylogeny) Disadvantages: no single partition; no specific clusters Forces all elements to fit a tree There are other methods that do not assume an ultrametric solution, notably Neighbor Joining. In genomics still UPGMA rules. 24

22 Hierarchical Clustering & Congenital Heart Defects Ellsoe et al. (Soren Brunak lab) European Heart Journal (2017) 25

23 CHD Congenital heart defects (CHD) affect almost 1% of all live born children Number of adults with CHD is increasing Recurrence patterns in families are poorly understood Do cases in the same family tend to have similar types of malformations? 26

24 Study 1163 families, 3080 family members with clinical diagnosis (avg 2.65 CHD cases /family) Each case is identified as having one or more of 41 different types of CHD lesions: AVD, BSD, VSD, 27

25 Concordant & discordant disease pairs Concordant: (ASD,ASD), (ASD,VSD) Discordant: (BAV,BAV) 28

26 Gender ratio, concordance & discordance 29

27 Scoring pairs of defects N(A,B) # families with A, B N(A, B) # families with A, not B N( A,B) # families with B, not A N( A, B) # families with none The odds ratio (OR) between phenotypes A and B: OR(A,B) = N(A,B) N( A, B)/N(A, B)N( A,B)??! Perhaps OR(A,B) = N(A,B)/[N(A, B)+N( A,B)] 30

28 Results 31

Molecular Evolution and Phylogenetic Tree Reconstruction

1 4 Molecular Evolution and Phylogenetic Tree Reconstruction 3 2 5 1 4 2 3 5 Orthology, Paralogy, Inparalogs, Outparalogs Phylogenetic Trees Nodes: species Edges: time of independent evolution Edge length