1. CSCI2950-C Lecture 10 Cancer Genomics: Duplications
October 21, 2008

2. Outline Cancer Genomes Cancer Progression Models
Duplications and End Sequence Profiling
Comparative Genomic Hybridization
Cancer Progression Models

3. End Sequence Profiling (ESP) C. Collins and S. Volik (2003)
Pieces of cancer genome: clones ( kb).
Cancer DNA
Sequence ends of clones (500bp).
Map end sequences to human genome.
Because of end sequencing protocol, clones have direction x y
Human DNA
Each clone corresponds to pair of end sequences (ES pair) (x,y).
Retain clones that correspond to a unique ES pair.

4. End Sequence Profiling (ESP) C. Collins and S. Volik (2003)
Pieces of cancer genome: clones ( kb).
Cancer DNA
Sequence ends of clones (500bp). L
Valid ES pairs
Lmin ≤ y – x ≤ Lmax, min (max) size of clone.
Convergent orientation.
Map end sequences to human genome.
Because of end sequencing protocol, clones have direction x y
Human DNA

5. End Sequence Profiling (ESP) C. Collins and S. Volik (2003)
Pieces of cancer genome: clones ( kb).
Cancer DNA
Sequence ends of clones (500bp). L
Map end sequences to human genome.
Because of end sequencing protocol, clones have direction. Some pairs cannot be mapped due to repeats in human genome. x y a b
Human DNA
Invalid ES pairs
Putative rearrangement in cancer
ES directions toward breakpoints (a,b):
Lmin ≤ |x-a| + |y-b| ≤ Lmax a x y b

6. ESP of Normal Cell 2D Representation All ES pairs valid. x y Human DNA
Lmin ≤ y – x ≤ Lmax
2D Representation
Each point (x,y) is ES pair.
Sometimes useful to have a 2D representation. All ES pairs near diagonal.
Genome Coordinate
Genome Coordinate

7. ESP of Tumor Cell Valid ES pairs satisfy length/direction constraints
Lmin ≤ y – x ≤ Lmax
Invalid ES pairs
indicate rearrangements
experimental errors
First show how to deal with noise….

8. What about duplications?
Clusters were biased to a very small region of the genome. These regions are duplicated – can be measured as such independently. Take a duplication centric view
11240 ES pairs
10453 valid (black)
737 invalid
489 isolated (red)
248 form 70 clusters (blue)
33/70 clusters
Total length: 31Mb

9. Structure of Duplications in Tumors?
Duplicated segments may co-localize
(Guan et al. Nat.Gen.1994)
Human genome
Tumor genome
Mechanisms not well understood.

10. Structure of Duplications: Approaches
Model free
Assemble tumor genome
Model based
Use knowledge of duplication mechanisms
Human genome
Could just sequence the genome. We didn’t have enough data – so used a model based approach based on known biology.
Tumor genome

11. Duplication by Amplisome
(Maurer, et al. 1987; Wahl, 1989…)
Other terms:
Episome
Amplicon
Double-minute
Extrachromosomal amplification. Gives single model of duplication.

12. Amplisome Reconstruction Problem
Assume
Tumor genome sequence is known.
Insertions are independent,
i.e. no insertions within insertions
We don’t know the genome sequence. Can we do something similar?
Approach
Identify duplicated sequences A1, …, Am
Amplisome is shortest common superstring of A1, …, Am

13. Analyzing Duplications
Tumor
Human
duplication u A B w C D v E u A B w D C v E u A B w C v D
????

14. Analyzing Duplications
Tumor
Human
duplication u A B w C D v E u A B w D C v E u A B w C v D u A B C D ??
Signature of a duplication. Complicated duplication.
Overlapping duplication
and rearrangement

15. Analyzing Duplications
Tumor
Human
duplication u A B w C D v E u A B w D C v E u A B w C v D A C D B u
Overlapping duplication
and rearrangement
Additional ES pair resolves duplication

16. Duples and Boundary Elements
Tumor
Human
duplication u A B w C D v E u A B w D C v E u A B w C v D A C D B u
Call this configuration a duple with boundary elements v and w.

17. Duplications in ESP graphB C D E u A B w D C v E u v w
boundary elements v,w
are vertices in ESP graph E w
duple D v C B u A

18. Duplications in ESP graphB C D E u A B w D C v E u v w
boundary elements v,w
are vertices in ESP graph E w
duple D v C
Path between
boundary
elements
resolves duple. B u A

19. Duplication ComplicationsB w C v E
???? w
These configurations frequent in MCF7 data. v u

20. Resolving Duplication as PathsB C D E A B u v w u w
Path between
boundary elements
resolves duple. v u

21. Resolving Duplications as PathsB C E A B u v w u w
Multiple paths between duple boundary elements. v u

22. ESP Amplisome Reconstruction Problem
Assume
Insertions are independent,
i.e. no insertions within insertions u A B w C v E
Approach
Identify endpoints of duplications:
(v1, w1), , …, (vm, wm)
Amplisome is shortest common superpath in ESP graph containing subpaths:
v1…w1, v2…w2, …, vm…wm

23. Reconstructed MCF7 amplisome
Chromosomes 1 3 17 20
Would like to tell you that we’ve done this many times – but there just isn’t the data yet. Still waiting on new sequencing technologies. Noticed similarities w/ problem in evolution.
Amplisome model explains
24/33 invalid clusters.
33 clusters
Total length: 31Mb
Raphael and Pevzner (2004) Bioinformatics.

24. DNA Basepairing

25. DNA Microarrays

26. Comparative Genomic Hybridization (CGH)
Measuring Mutations in Cancer
Comparative Genomic Hybridization (CGH)

27. CGH Analysis (1) Divide genome into segments of equal copy number
-0.5
0.5
Log2(R/G)
Genomic position
Copy number of adjacent probes correlated
-0.5
0.5
Deletion
Amplification
Genomic position

28. CGH Analysis (2) Identify aberrations common to multiple samples
Chromosome 3 of 26 lung tumor samples on mid-density cDNA array.
Common deletion located in 3p21 and common amplification – in 3q.
Samples

29. CGH Analysis (1) Divide genome into segments of equal copy number
Copy number profile
Copy number
Genome
coordinate
Numerous methods
(e.g. clustering, Hidden Markov Model, Bayesian, etc.)
Segmentation
Copy number of adjacent probes correlated
Input: yi = log2 Ti / Ri , clone i = 1, …, N
Output: Assignment s(yi)  {S1, …, SK}
Si represent copy number states

30. An Approach to CGH Segmentation
Circular Binary Segmentation (CBS), Olshen et al. 2004
Use hypothesis test to compare means of two intervals using t-test
-0.5
0.5
Deletion
Amplification
Genomic position

31. Interval Score Assume: Xi are independent, normally distributed
Lipson, et al. J. Computational Biology, 2006
Assume:
Xi are independent, normally distributed
µ and  denote the mean and standard deviation of the normal genomic data.
Given an interval I spanning k probes, we define its score as:

32. Significance of Interval Score
Lipson, et al. J. Computational Biology, 2006
Assume:
Xi ~ N(µ, )

33. The MaxInterval Problem
Input: A vector X=(X1…Xn)
Output: An interval I  [1…n], that maximizes S(I )
Other intervals with high scores may be found by recursively calling this function.
Exhaustive algorithm: O(n2)

34. MaxInterval Algorithm I: LookAhead
Assume given:
m : An upper bound for the value of a single element Xi
t : A lower bound on the maximum score
sum
length
score
s = jI Xj k
s+m r
k+r
I = [i,…,i+k-1]
I’ = [i,…,i+k+r-1]
Solve for first r for which S(I ) may exceed t.
Complexity: Expected O(n1.5) (unproved)
Benchmarking
Benchmarking results of the Exhaustive, LookAhead and GFA algorithms on synthetic vectors of varying lengths.
Linear regression suggests that the complexities of the Exhaustive, LookAhead and GFA algorithms are O(n2), O(n1.5), O(n), respectively. 8

35. MaxInterval Algorithm II: Geometric Family Approximation (GFA)
For >0 define the following geometric family of intervals: kj j
(j1)
(j2)
(j3)
Theorem:
Let I* be the optimal scoring interval.
Let J be the leftmost longest interval of  fully contained in I*.
Then S(J) ≥ S(I*)/, where   (1- -2 )-1.
Complexity: O(n)
MaxInterval Algorithm I: LookAhead
Assume you are given:
m – An upper bound for the value of a single element ci
t – A lower bound on the maximum score
If we are currently considering an interval I=[i,…,i+k-1] with a sum of s = jI cj, then the score of I is:
The score of an interval I’ = [i,…,i+k+x-1] is then bounded by:
Complexity:
Expected O(n1.5) (unproved)
Solve for first x for which S(I ) may exceed t.
sum
length
score s k
s+mx
k+x I I’ 6
Benchmarking
Benchmarking results of the Exhaustive, LookAhead and GFA algorithms on synthetic vectors of varying lengths.
Linear regression suggests that the complexities of the Exhaustive, LookAhead and GFA algorithms are O(n2), O(n1.5), O(n), respectively. 8

36. Benchmarking
synthetic vectors of varying lengths
Linear regression suggests that the complexities of the Exhaustive, LookAhead and GFA algorithms are O(n2), O(n1.5), O(n), respectively.

37. Applications: Single Samples-1 1
FRA16B
A2BP1 50
Mbp 25 75
Log2(ratio)
Chromosome 16 of HCT116 colon carcinoma cell line on high-density oligo array (n=5,464). 50 25 75
Mbp 1
ERBB2
Log2(ratio)
Chromosome 17 of several breast carcinoma cell lines on mid-density cDNA array (n=364).

38. Another Approach to CGH Segmentation
Use Hidden Markov Model (HMM) to “parse” sequence of probes into copy number states
-0.5
0.5
Deletion
Amplification
Genomic position

39. Hidden Markov Models 1 2 K … x1 x2 x3 xK

40. Example: The Dishonest Casino
A casino has two dice:
Fair die
P(1) = P(2) = P(3) = P(5) = P(6) = 1/6
Loaded die
P(1) = P(2) = P(3) = P(5) = 1/10
P(6) = 1/2
Casino player switches back-&-forth between fair and loaded die once every 20 turns
Game:
You bet $1
You roll (always with a fair die)
Casino player rolls (maybe with fair die, maybe with loaded die)
Highest number wins $2

41. Question # 1 – Evaluation
GIVEN
A sequence of rolls by the casino player
QUESTION
How likely is this sequence, given our model of how the casino works?
This is the EVALUATION problem in HMMs
Prob = 1.3 x 10-35

42. Question # 2 – Decoding GIVEN A sequence of rolls by the casino player
QUESTION
What portion of the sequence was generated with the fair die, and what portion with the loaded die?
This is the DECODING question in HMMs.
This is what we want to solve for CGH analysis
FAIR
LOADED
FAIR

43. Question # 3 – Learning GIVEN A sequence of rolls by the casino player
QUESTION
How “loaded” is the loaded die? How “fair” is the fair die? How often does the casino player change from fair to loaded, and back?
This is the LEARNING question in HMMs
Prob(6) = 64%

44. Definition of a hidden Markov model
Definition: A hidden Markov model (HMM)
Alphabet  = { b1, b2, …, bM }
Set of states Q = { 1, ..., K }
Transition probabilities between any two states
aij = transition prob from state i to state j
ai1 + … + aiK = 1, for all states i = 1…K
Start probabilities a0i
a01 + … + a0K = 1
Emission probabilities within each state
ei(b) = P( xi = b | i = k)
ei(b1) + … + ei(bM) = 1, for all states i = 1…K 1 2 K …

45. The dishonest casino model
0.05
0.95
0.95
FAIR
LOADED
P(1|F) = 1/6
P(2|F) = 1/6
P(3|F) = 1/6
P(4|F) = 1/6
P(5|F) = 1/6
P(6|F) = 1/6
P(1|L) = 1/10
P(2|L) = 1/10
P(3|L) = 1/10
P(4|L) = 1/10
P(5|L) = 1/10
P(6|L) = 1/2
0.05

46. A HMM is memory-less
At each time step t, the only thing that affects future states is the current state t P(t+1 = k | “whatever happened so far”) = P(t+1 = k | 1, 2, …, t, x1, x2, …, xt) = P(t+1 = k | t) 1 2 K …

47. A parse of a sequence
Given a sequence x = x1……xN, A parse of x is a sequence of states  = 1, ……, N 1 2 K … 1 1 2 K … 1 2 K … 1 2 K … … 2 2 K x1 x2 x3 xK

48. Likelihood of a Parse x1 x2 x3 xK
Simply, multiply all the orange arrows! (transition probs and emission probs) 1 2 K … 1 1 2 K … 1 2 K … 1 2 K … … 2 2 K x1 x2 x3 xK

49. Likelihood of a parse P(x, ) = j=1…n jF(j, x, ) =
Given a sequence x = x1……xN and a parse  = 1, ……, N, To find how likely is the parse: (given our HMM) P(x, ) = P(x1, …, xN, 1, ……, N) = P(xN, N | N-1) P(xN-1, N-1 | N-2)……P(x2, 2 | 1) P(x1, 1) = P(xN | N) P(N | N-1) ……P(x2 | 2) P(2 | 1) P(x1 | 1) P(1) = a01 a12……aN-1N e1(x1)……eN(xN)
A compact way to write
a01 a12……aN-1N e1(x1)……eN(xN)
Number all parameters aij and ei(b); n params
Example:
a0Fair : 1; a0Loaded : 2; … eLoaded(6) = 18
Then, count in x and  the # of times each
parameter j = 1, …, n occurs
F(j, x, ) = # parameter j occurs in (x, )
(call F(.,.,.) the feature counts) Then,
P(x, ) = j=1…n jF(j, x, ) =
= exp[j=1…n log(j)F(j, x, )] 1 2 K … 1 1 2 K … 1 2 K … 1 2 K … … 2 2 K x1 x2 x3 xK

50. Example: the dishonest casino
Let the sequence of rolls be:
x = 1, 2, 1, 5, 6, 2, 1, 5, 2, 4
Then, what is the likelihood of
 = Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair?
(say initial probs a0Fair = ½, aoLoaded = ½)
½  P(1 | Fair) P(Fair | Fair) P(2 | Fair) P(Fair | Fair) … P(4 | Fair) =
½  (1/6)10  (0.95)9 = ~= 0.5  10-9

51. Example: the dishonest casino
So, the likelihood the die is fair in this run
is just  10-9
OK, but what is the likelihood of
 = Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded?
½  P(1 | Loaded) P(Loaded, Loaded) … P(4 | Loaded) =
½  (1/10)9  (1/2)1 (0.95)9 = ~= 0.16  10-9
Therefore, it somewhat more likely that all the rolls are done with the fair die, than that they are all done with the loaded die

52. Example: the dishonest casino
Let the sequence of rolls be:
x = 1, 6, 6, 5, 6, 2, 6, 6, 3, 6
Now, what is the likelihood  = F, F, …, F?
½  (1/6)10  (0.95)9 = 0.5  10-9, same as before
What is the likelihood
 = L, L, …, L?
½  (1/10)4  (1/2)6 (0.95)9 = ~= 0.5  10-7
So, it is 100 times more likely the die is loaded

53. HMM Model for CGH data A+ C+ G+ S1 S2 S3 S4
Fridlyand et al. (2004)

54. A model for CGH data S1 S2 S3 S4 1, 1 K states copy numbers
Homozygous
Deletion
(copy =0)
Heterozygous
Deletion
(copy =1)
Normal
(copy =2)
Duplication
(copy >2)
1, 1
2, 2
3, 3
4, 4
Emissions:
Gaussians
Copy number
Genome
coordinate

55. Sources
BJ Raphael, S Volik, C Collins, PA Pevzner - Reconstructing tumor genome architectures. Bioinformatics, 2003
Raphael and Pevzner. Reconstructing tumor amplisomes. Bioinformatics, 2004
Olshen et al. Circular binary segmentation for the analysis of array-based DNA copy number data. Biostatistics, 2004.
Lipson, et al. Efficient Calculation of Interval Scores for DNA Copy Number Data Analysis. Journal of Computational Biology, 2006.
Fridyland, et al. Hidden Markov models approach to the analysis of array CGH data. Journal of Multivariate Analysis, 2004
(HMM slides)