1. DNA Sequencing

2. Whole Genome Shotgun Sequencing
cut many times at random
plasmids (2 – 10 Kbp)
forward-reverse paired reads
known dist
cosmids (40 Kbp)
~500 bp
~500 bp

3. Steps to Assemble a Genome
Some Terminology
read a long word that comes
out of sequencer
mate pair a pair of reads from two ends
of the same insert fragment
contig a contiguous sequence formed
by several overlapping reads
with no gaps
supercontig an ordered and oriented set
(scaffold) of contigs, usually by mate
pairs
consensus sequence derived from the
sequene multiple alignment of reads
in a contig
1. Find overlapping reads
2. Merge some “good” pairs of reads into longer contigs
3. Link contigs to form supercontigs
4. Derive consensus sequence
..ACGATTACAATAGGTT..

4. 2. Merge Reads into Contigs
Overlap graph:
Nodes: reads r1…..rn
Edges: overlaps (ri, rj, shift, orientation, score)
Reads that come
from two regions of
the genome (blue
and red) that contain
the same repeat
Note:
of course, we don’t
know the “color” of
these nodes

5. 2. Merge Reads into Contigs
repeat region
Unique Contig
Overcollapsed Contig
We want to merge reads up to potential repeat boundaries

6. 2. Merge Reads into Contigs
repeat region
Ignore non-maximal reads
Merge only maximal reads into contigs

7. 2. Merge Reads into Contigs
Remove transitively inferable overlaps
If read r overlaps to the right reads r1, r2, and r1 overlaps r2, then (r, r2) can be inferred by (r, r1) and (r1, r2)

8. 2. Merge Reads into Contigs

9. Overlap graph after forming contigs
Unitigs:
Gene Myers, 95

10. Repeats, errors, and contig lengths
Repeats shorter than read length are easily resolved
Read that spans across a repeat disambiguates order of flanking regions
Repeats with more base pair diffs than sequencing error rate are OK
We throw overlaps between two reads in different copies of the repeat
To make the genome appear less repetitive, try to:
Increase read length
Decrease sequencing error rate
Role of error correction:
Discards up to 98% of single-letter sequencing errors
decreases error rate
 decreases effective repeat content
 increases contig length

11. 3. Link Contigs into Supercontigs
Normal density
Too dense
 Overcollapsed
Inconsistent links
 Overcollapsed?

12. 3. Link Contigs into Supercontigs
Find all links between unique contigs
Connect contigs incrementally, if  2 links
supercontig
(aka scaffold)

13. 3. Link Contigs into Supercontigs
Fill gaps in supercontigs with paths of repeat contigs

14. 4. Derive Consensus Sequence
TAGATTACACAGATTACTGA TTGATGGCGTAA CTA
TAGATTACACAGATTACTGACTTGATGGCGTAAACTA
TAG TTACACAGATTATTGACTTCATGGCGTAA CTA
TAGATTACACAGATTACTGACTTGATGGCGTAA CTA
TAGATTACACAGATTACTGACTTGATGGGGTAA CTA
TAGATTACACAGATTACTGACTTGATGGCGTAA CTA
Derive multiple alignment from pairwise read alignments
Derive each consensus base by weighted voting
(Alternative: take maximum-quality letter)

15. Some Assemblers PHRAP Celera Arachne Phusion Euler
Early assembler, widely used, good model of read errors
Overlap O(n2)  layout (no mate pairs)  consensus
Celera
First assembler to handle large genomes (fly, human, mouse)
Overlap  layout  consensus
Arachne
Public assembler (mouse, several fungi)
Phusion
Overlap  clustering  PHRAP  assemblage  consensus
Euler
Indexing  Euler graph  layout by picking paths  consensus

16. Quality of assemblies—mouse
Terminology: N50 contig length
If we sort contigs from largest to smallest, and start
Covering the genome in that order, N50 is the length
Of the contig that just covers the 50th percentile.
7.7X
sequence
coverage

17. Quality of assemblies—dog
7.5X
sequence
coverage

18. History of WGA 1982: -virus, 48,502 bp 1995: h-influenzae, 1 Mbp
1997
1982: -virus, 48,502 bp
1995: h-influenzae, 1 Mbp
2000: fly, 100 Mbp
2001 – present
human (3Gbp), mouse (2.5Gbp), rat*, chicken, dog, chimpanzee, several fungal genomes
Let’s sequence the human genome with the shotgun strategy
That is impossible, and a bad idea anyway
Phil Green
Gene Myers

19. Genomes Sequenced

20. Phylogeny Tree Reconstruction1 4 3 2 5
Phylogeny Tree Reconstruction

21. Phylogenetic Trees Nodes: species Edges: time of independent evolution
Edge length represents evolution time
AKA genetic distance
Not necessarily chronological time

22. Inferring Phylogenetic Trees
Trees can be inferred by several criteria:
Morphology of the organisms
Can lead to mistakes!
Sequence comparison
Example:
Orc: ACAGTGACGCCCCAAACGT
Elf: ACAGTGACGCTACAAACGT
Dwarf: CCTGTGACGTAACAAACGA
Hobbit: CCTGTGACGTAGCAAACGA
Human: CCTGTGACGTAGCAAACGA

23. Phylogeny and sequence comparison
Basic principles:
Degree of sequence difference is proportional to length of independent sequence evolution
Only use positions where alignment is certain – avoid areas with (too many) gaps

24. Distance between two sequences
Given sequences xi, xj,
Define
dij = distance between the two sequences
One possible definition:
dij = fraction f of sites u where xi[u]  xj[u]
Better scores are derived by modeling evolution as a continuous change process – not covered here
Jukes Kantor, Kimura, etc.

25. A simple clustering method for building tree
UPGMA (unweighted pair group method using arithmetic averages)
Or the Average Linkage Method
Given two disjoint clusters Ci, Cj of sequences, 1
dij = ––––––––– {p Ci, q Cj}dpq
|Ci|  |Cj|
Claim that if Ck = Ci  Cj, then distance to another cluster Cl is:
dil |Ci| + djl |Cj|
dkl = ––––––––––––––
|Ci| + |Cj|
Proof
Ci,Cl dpq + Cj,Cl dpq
dkl = ––––––––––––––––
(|Ci| + |Cj|) |Cl|
|Ci|/(|Ci||Cl|) Ci,Cl dpq + |Cj|/(|Cj||Cl|) Cj,Cl dpq
= ––––––––––––––––––––––––––––––––––––
(|Ci| + |Cj|)
|Ci| dil + |Cj| djl
= –––––––––––––

26. Algorithm: Average Linkage
Initialization:
Assign each xi into its own cluster Ci
Define one leaf per sequence, height 0
Iteration:
Find two clusters Ci, Cj s.t. dij is min
Let Ck = Ci  Cj
Define node connecting Ci, Cj,
& place it at height dij/2
Delete Ci, Cj
Termination:
When two clusters i, j remain,
place root at height dij/2 1 4 3 2 5 1 4 2 3 5

27. Example v w x y z v w xyz vw xyz v w x yz 6 8 4 2 6 8 8 4 6 8 4 3 2 16 8 4 2 v w
xyz 6 8 vw
xyz 8 4 v w x yz 6 8 4 3 2 1 v w x y z

28. Ultrametric Distances and Molecular Clock
Definition:
A distance function d(.,.) is ultrametric if for any three distances dij  dik  dij, it is true that
dij  dik = dij
The Molecular Clock:
The evolutionary distance between species x and y is 2 the Earth time to reach the nearest common ancestor
That is, the molecular clock has constant rate in all species
The molecular clock results in ultrametric distances
years 1 4 2 3 5

29. Ultrametric Distances & Average Linkage1 4 2 3 5
Average Linkage is guaranteed to reconstruct correctly a binary tree with ultrametric distances
Proof: Exercise

30. Weakness of Average Linkage
Molecular clock: all species evolve at the same rate (Earth time)
However, certain species (e.g., mouse, rat) evolve much faster
Example where UPGMA messes up:
Correct tree
AL tree 3 2 4 1 4 2 3 1

31. Additive Distances 1
d1,4 12 4 8 3 13 7 9 5 11 10 6 2
Given a tree, a distance measure is additive if the distance between any pair of leaves is the sum of lengths of edges connecting them
Given a tree T & additive distances dij, can uniquely reconstruct edge lengths:
Find two neighboring leaves i, j, with common parent k
Place parent node k at distance dkm = ½ (dim + djm – dij) from any node m  i, j

32. d(x, y) + d(z, w) < d(x, z) + d(y, w) = d(x, w) + d(y, z)
Additive Distances z x w y
For any four leaves x, y, z, w, consider the three sums
d(x, y) + d(z, w)
d(x, z) + d(y, w)
d(x, w) + d(y, z)
One of them is smaller than the other two, which are equal
d(x, y) + d(z, w) < d(x, z) + d(y, w) = d(x, w) + d(y, z)

33. Reconstructing Additive Distances Given Tx T D y 5 4 v w x y z 10 17 16 15 14 9 3 z 3 4 7 w 6 v
If we know T and D, but do not know the length of each leaf, we can reconstruct those lengths

34. Reconstructing Additive Distances Given Tx T D y v w x y z 10 17 16 15 14 9 z w v

35. Reconstructing Additive Distances Given Tx v w x y z 10 17 16 15 14 9 T y z a w D1 v
dax = ½ (dvx + dwx – dvw) a x y z 11 10 9 15 14
day = ½ (dvy + dwy – dvw)
daz = ½ (dvz + dwz – dvw)

36. Reconstructing Additive Distances Given Tx a x y z 11 10 9 15 14 T y 5 4 b 3 z 3 a c 4 7 w D2 6
d(a, c) = 3
d(b, c) = d(a, b) – d(a, c) = 3
d(c, z) = d(a, z) – d(a, c) = 7
d(b, x) = d(a, x) – d(a, b) = 5
d(b, y) = d(a, y) – d(a, b) = 4
d(a, w) = d(z, w) – d(a, z) = 4
d(a, v) = d(z, v) – d(a, z) = 6
Correct!!! v a b z 6 10 D3 a c 3

37. Neighbor-Joining
Guaranteed to produce the correct tree if distance is additive
May produce a good tree even when distance is not additive
Step 1: Finding neighboring leaves
Define
Dij = (N – 2) dij – ki dik – kj djk
Claim: The above “magic trick” ensures that Dij is minimal iff i, j are neighbors 1 3
0.1
0.1
0.1
0.4
0.4 2 4

38. Algorithm: Neighbor-joining
Initialization:
Define T to be the set of leaf nodes, one per sequence
Let L = T
Iteration:
Pick i, j s.t. Dij is minimal
Define a new node k, and set dkm = ½ (dim + djm – dij) for all m  L
Add k to T, with edges of lengths dik = ½ (dij + ri – rj), djk = dij – dik
where ri = (N – 2)-1 ki dik
Remove i, j from L;
Add k to L
Termination:
When L consists of two nodes, i, j, and the edge between them of length dij

39. Parsimony – direct method not using distances
One of the most popular methods:
GIVEN multiple alignment
FIND tree & history of substitutions explaining alignment
Idea:
Find the tree that explains the observed sequences with a minimal number of substitutions
Two computational subproblems:
Find the parsimony cost of a given tree (easy)
Search through all tree topologies (hard)

40. Example: Parsimony cost of one column
Final cost C = 1
{A}
{A, B}
Cost
C+=1 A B A B A A
{A}
{B}
{A}
{A}

41. Parsimony Scoring Given a tree, and an alignment column u
Label internal nodes to minimize the number of required substitutions
Initialization:
Set cost C = 0; node k = 2N – 1 (last leaf)
Iteration:
If k is a leaf, set Rk = { xk[u] } // Rk is simply the character of kth species
If k is not a leaf,
Let i, j be the daughter nodes;
Set Rk = Ri  Rj if intersection is nonempty
Set Rk = Ri  Rj, and C += 1, if intersection is empty
Termination:
Minimal cost of tree for column u, = C

42. Example {B} {A,B} {A} {B} {A} {A,B} {A} A A A A B B A B {A} {A} {A}

43. Traceback to find ancestral nucleotides
Choose an arbitrary nucleotide from R2N – 1 for the root
Having chosen nucleotide r for parent k,
If r  Ri choose r for daughter i
Else, choose arbitrary nucleotide from Ri
Easy to see that this traceback produces some assignment of cost C

44. inadmissible with Traceback
Example
Admissible with Traceback x B
Still optimal, but
inadmissible with Traceback A
{A, B} A B x
{A} B
{A, B} A B A B x B x A B A B A
{A}
{B}
{A}
{B} A B A B A x A x A B A B

45. Probabilistic Methods
xroot t1 t2 x1 x2
A more refined measure of evolution along a tree than parsimony
P(x1, x2, xroot | t1, t2) = P(xroot) P(x1 | t1, xroot) P(x2 | t2, xroot)
If we use Jukes-Cantor, for example, and x1 = xroot = A, x2 = C, t1 = t2 = 1,
= pA¼(1 + 3e-4α) ¼(1 – e-4α) = (¼)3(1 + 3e-4α)(1 – e-4α)

46. Probabilistic Methods
xroot xu x2 xN x1
If we know all internal labels xu,
P(x1, x2, …, xN, xN+1, …, x2N-1 | T, t) = P(xroot)jrootP(xj | xparent(j), tj, parent(j))
Usually we don’t know the internal labels, therefore
P(x1, x2, …, xN | T, t) = xN+1 xN+2 … x2N-1 P(x1, x2, …, x2N-1 | T, t)

47. Felsenstein’s Likelihood Algorithm
To calculate P(x1, x2, …, xN | T, t)
Initialization:
Set k = 2N – 1
Iteration: Compute P(Lk | a) for all a  
If k is a leaf node:
Set P(Lk | a) = 1(a = xk)
If k is not a leaf node:
1. Compute P(Li | b), P(Lj | b) for all b, for daughter nodes i, j
2. Set P(Lk | a) = b,c P(b | a, ti) P(Li | b) P(c | a, tj) P(Lj | c)
Termination:
Likelihood at this column = P(x1, x2, …, xN | T, t) = aP(L2N-1 | a)P(a)
Let P(Lk | a) denote the prob.
of all the leaves below node k,
given that the residue at k is a

48. Probabilistic Methods
Given M (ungapped) alignment columns of N sequences,
Define likelihood of a tree:
L(T, t) = P(Data | T, t) = m=1…M P(x1m, …, xnm, T, t)
Maximum Likelihood Reconstruction:
Given data X = (xij), find a topology T and length vector t that maximize likelihood L(T, t)

49. Current popular methods
HUNDREDS of programs available!
Some recommended programs:
Discrete—Parsimony-based
Rec-1-DCM3
Tandy Warnow and colleagues
Probabilistic
SEMPHY
Nir Friedman and colleagues