CSCI 1820 Notes. Scribes: tl40. February 26 - March 02, Estimating size of graphs used to build the assembly.

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1 CSCI 1820 Notes Scribes: tl40 February 26 - March 02, 2018 Chapter 2. Genome Assembly Algorithms 2.1. Statistical Theory 2.2. Algorithmic Theory Idury-Waterman Algorithm Estimating size of graphs used to build the assembly. de Bruijn graphs combinatorics. DNA Assembly Problem We have a G bp-long DNA target molecule, which we do not know so want to reconstruct. We know sequences of a number of fragments of the target DNA, f 1, f 2,, f N. Assume that they are all of length L. Note that the same result holds when L is the average length of the fragments. We can record their consecutive order. We use the de Bruijn graphs for algorithms used to assembly the target DNA sequence. Remark 1. The graph consists of nodes as (k 1)-mers and directed edges as k-mers, so depends on the choice of k. 2. The graph is Eulerian, which means that the in-degree and out-degree are same for every node, so there exists at least one Eulerian cycle - actually, we have exponentially many different such ones. 1

2 Estimate the size of the de Bruijn graph based on k-mers Setting: k: size of the mers. G: size of the target DNA. N: number of fragments. L: length of the fragments. a = NL G : coverage. T = f 1 + f f N N(L k + 2): number of (k 1)-mers in the fragments. r: DNA single base error rate (sequencing error rate), about 2% in substitutions. Theorem Let L = L k + 2 and R = 1 (1 r) k 1. Also, let us classify the (k 1)-mers in the de Bruijn graph (sequence graph) as TRUE if they are derived from a correctly used fragment region, and FALSE otherwise. Assume the following: 1. The error rate is small. 2. No two fragments generate the same FALSE (k 1)-mer, that is, FALSE (k 1)- mers appear at most once among fragments. Then, in the associated de Bruijn graph (sequence graph), the expected number of vertices is E[ V ] = RT + [1 e a(1 R) ]L. Proof Let X α be the number of fragments that cover the (k 1)-long region α, α + 1,, α + k 2 of the target DNA sequence. Then, each X α is a Poisson random variable with mean a, and E[ α X α] = T. Since the uniform error rate on the probability of a base being used correctly by the sequencing machine is 1 r, the probability that an entire (k 1)-mer is used correctly is (1 r) k 1, which implies that the probability that an entire (k 1)-mer being used incorrectly is R = 1 (1 r) k 1 : we can say that R is the (k 1)-mer error rate, and, by the second assumption, E[ F ALSE ] = RT 2

3 The expected number of TRUE (k 1)-mers is equal to the number of positions α such that at least one of the fragments contains no sequencing errors in the (k 1)-long region α, α + 1,, α + k 2, so, for a Poisson random variable X, E[ T RUE ] = L (1 R i )P (X = i) ( e = L (P (X = i) P (X = i)r i a ) = L a i ( e a = L a i ) ( e a (ar) i = L e a a i e a = L (1 e a e ar ) = L (1 e a(1 R) ) ) e a a i R i ) (ar) i Therefore, E[ V ] = E[ T RUE ] + E[ F ALSE ] = RT + (1 e a(1 R) )L. Theory of Combinatorics on Words Definition Let A be the alphabet, A be the set of sequences over A including an empty word ɛ, A + be the set of sequences without ɛ, and u m = uu u for a sequence u. w A + is called primitive if w = u m for u A + then m = 1. Two words x and y are conjugate if there exist words u, v such that x = uv and y = vu (cyclic shifts of one another). Definition A Lyndon word is a primitive word that is strictly smaller than any of its proper suffixes (or rotations) in lexicographic order. Lemma Let L be the set of all Lyndon words. If l < m for l, m L, then lm L. Theorem (Factorization) Any word can be factorized uniquely as a non-increasing product of Lyndon words. Example abracadabra (abracad)(abr)(a) Theorem (Fredrickson & Maiorana) Let l 1 < l 2 < < l m be the increasing sequence of Lyndon words of length dividing n. Then l 1 l 2 l m is the lex-first de Bruijn cycle of order n (all n-mers are there). 3

4 De Bruijn Graphs Combinatorics & Eulerian Cycles (Building Genome Assemblies) Example A de Bruijn graph of order n = 3 over A = {a, b}: Definition For a given graph G = (V, E) and v V, d (v)= the in-degree = number of edges coming in. d + (v)= the out-degree = number of edges going out. Definition A spanning tree directed at vertex v has the follwing properties: 1. It is connected. 2. For every vertex u v, there is a directed path in the spanning tree from u to v. Remark (How many Eulerian orders are in the de Bruijn graph?) There is a connection between Eulerian cycles and spanning trees directed towards a fixed node v, which can be used for giving a relationship between the number of Eulerian cycles and that of spanning trees. Example Two directed spanning trees directed towards the vertex bb: 4

5 Definition A directed graph G = (V, E) is strongly connected if u, v V, a directed path from u to v. Theorem A strongly connected graph G = (V, E) is Eulerian if and only if d (v) = d + (v) for any v V. Definition An Eulerian cycle is a cycle on a graph that visits every edge exactly once. Theorem (BEST) For a graph G = (V, E), define the following: Π(G) = v V (d+ (v) 1)!. For a vertex v V, t(g) is the number of spanning trees oriented towards v. Then, the number of Eulerian cycles is equal to t(g) Π(G). Theorem The number of de Bruijn cycles of order n (all n-mers in the de Bruijn cycle) over k-letter alphabets is N(n, k) = (k!)kn 1 k n. Table 1: Some Examples n N(n, 2) N(n, 3) N(n, 4) N(n, 5) 24 Theorem (The Matrix-Tree Theorem) Let G be a graph on a set of vertices V, M = (M v w) be an adjacency matrix of G, where M vw is the number of edges between v and w, D be a diagonal matrix such that D vv = w V M vw, L = D M be the Laplacian matrix of G, and K v (G) be the determinant of the matrix C v, obtained from L by removing the row and column indexed v. Then, for any v V, the number of spanning trees of G oriented towards v is K v (G). Example Let G be a de Bruijn graph of order n = 3 over A = {a, b}, so V = {aa, ab, ba, bb}. Then, the adjacency matrix M, D, and the Laplacian matrix L are given by following: M = , D = , L = Therefore, the number of spanning trees directed towards bb is 2 by following: 5

6 1 1 0 C bb = det C bb =