1. Outline Cancer Progression Models
SNPs, Haplotypes, and Population Genetics: Introduction

2. Cancer: Mutation and Selection
Clonal theory of cancer: Nowell (Science 1976)

3. Cancer Genomes
Leukemia
Breast

4. “Comparative Genomics” of Cancer
Mutation, selection
Human genome
Tumor genome
Tumor genome 2
Tumor genome 4
Tumor genome 3
Identify recurrent aberrations
Mitelman Database, >40,000 aberrations
Reconstruct temporal sequence of aberrations
Linear model: Colorectal cancer (Vogelstein, 1988): -5q  12p*  -17p  -18q
Tree model: (Desper et al.1999)
3) Find age of tumor,
time of clonal expansion

5. Observing Cancer Progression
Obtaining longitudinal (time-course) data difficult. t1 t2 t3 t4
Latitudinal data (multiple patients) readily available.
Mutation, selection
Human genome
Tumor genome
Tumor genome 2
Tumor genome 4
Tumor genome 3

6. Multiple Mutations
4 step model for colorectal cancer, Vogelstein, et al. (1988) New Eng. J.Med
-5q  12p*  -17p  -18q
Inferred from latitudinal data in 172 tumor samples.

7. Oncogenetic Tree models (Desper et al. JCB 1999, 2001)
Given: measurements of chromosome gain/loss events in multiple tumor samples (CGH)
Compute: rooted tree that best explains temporal sequence of events.
{+1q}, {-8p}, {+Xq},
{+Xq, -8p}, {-8p, +1q}

8. Oncogenetic Tree models (Desper et al. JCB 1999, 2000)
Given: measurements of chromosome gain/loss events in multiple tumor samples
{+1q}, {-8p}, {+Xq}, {+Xq, -8p}, {-8p, +1q}
L = set of chromosome alterations observed in all samples
Tumor samples give probability distribution on 2L

9. Oncogenetic Tree T = (V, E, r, p, L) rooted tree V = vertices
E = edges
L = set of events (leaves)
r root
p: E  (0,1] probability distribution
T gives probability distribution on 2L e1 e2 e3 e4 e0

10. Results
CGH of 117 cases of kidney cancer

11. Extensions
Oncogenetic trees based on branching (Desper et al., JCB 1999)

12. Extensions

13. Extensions
Oncogenetic trees based on branching (Desper et al., JCB 1999)
Maximum Likelihood Estimation (von Heydebreck et al, 2004)
Mutagenic trees:
mixtures of trees
(Beerenwinkel, et al.
JCB 2005)

14. Heterogeneity within a tumor
Final tumor is clonal expansion of single cell lineage.
Can we date the time of clonal expansion?
Tsao, … Tavare, et al. Genetic reconstruction of individual colorectal
tumor histories, PNAS 2000.

15. Estimating time of clonal expansion
Microsatellite loci (MS), CA dinucleotides.
In tumors with loss of mismatch repair (e.g. colorectal), MS change size.

16. Estimating time of clonal expansion
For each MS locus, measure mean mi and variance si of size.
S2allele = average of s12, …, sL2
S2loci = variance of m1, …, mL

17. Time to clonal expansion?

18. Simulation Estimates of Tumor AgeY2 Y1
Y1 = time to clonal expansion
Tumor age = Y1 + Y2
Branching process simulation. Each cell in population gives birth to 0, 1 or 2 daughter cells with +- 1 change in MS size
(coalescent: forward, backward, forward simulation)
Posterior estimate of Y1, Y2 by running simulations, accepting runs with simulated values of S2allele, S2loci close to observed.

19. Results 15 patients, 25 MS loci
Estimate time since clonal expansion from observed S2allele, S2loci .

20. Cancer: Mutation and Selection
Clonal theory of cancer: Nowell (Science 1976)

21. Population Genetics
C.C. Maley: selective sweeps of mutations in tumor cell populations
Chin and Gray: solid tumors

22. Genetics 101
Humans are diploid: two copies of each chromosome, maternal and paternal
Locus: Region on a chromosome (gene, nucleotide, etc.)
Allele: “Value” at a locus
Genotype: Pair of alleles (maternal and paternal) at loci on a chromosome (homozygous, heterozygous)
Haplotype: Alleles of loci on same chromosome (maternal or paternal)

23. Allele Measurement “Old days” (< 1970?): gene variants
More recently: (1980’s-90’s), various sequence based genetic markers: microsatellites, sequence tagged sites (STS), etc.
Today: single nucelotide polymorphisms (SNPs)

24. Single Nucleotide Polymorphisms
Infinite Sites Assumption:
Each site mutates at most once
By convention, SNPs are biallelic: only two of four possible nucleotides present in population

25. Infinite Sites AssumptionB 3 8 5
The different sites are linked. A 1 in position 8 implies 0 in position 5, and vice versa.
Each sequence has single parent.
The history of a population can be expressed as a tree.
The tree can be constructed efficiently

26. Infinite sites Assumption and Perfect Phylogeny
Each site is mutated at most once in the history.
All descendants must carry the mutated value, and all others must carry the ancestral value i
1 in position i
0 in position i

27. Perfect Phylogeny
Assume an evolutionary model in which only mutation takes place,
The evolutionary history is explained by a tree in which every mutation is on an edge of the tree. All the species in one sub-tree contain a 0, and all species in the other contain a 1. Such a tree is called a perfect phylogeny.
How can one reconstruct such a tree?

28. The 4-gamete condition
A column i partitions the set of species into two sets i0, and i1
A column is homogeneous w.r.t a set of species, if it has the same value for all species. Otherwise, it is heterogenous.
EX: i is heterogenous w.r.t {A,D,E} i
A 0
B 0
C 0
D 1
E 1
F 1 i0 i1

29. 4 Gamete Condition 4 Gamete Condition
There exists a perfect phylogeny if and only if for all pair of columns (i,j), either j is not heterogenous w.r.t i0, or i1.
Equivalent to
There exists a perfect phylogeny if and only if for all pairs of columns (i,j), the following 4 rows do not exist
(0,0), (0,1), (1,0), (1,1)

30. 4-gamete condition: proof
Depending on which edge the mutation j occurs, either i0, or i1 should be homogenous.
(only if) Every perfect phylogeny satisfies the 4-gamete condition
(if) If the 4-gamete condition is satisfied, does a prefect phylogeny exist? i0 i1 i

31. An algorithm for constructing a perfect phylogeny
We will consider the case where 0 is the ancestral state, and 1 is the mutated state. This will be fixed later.
In any tree, each node (except the root) has a single parent.
It is sufficient to construct a parent for every node.
In each step, we add a column and refine some of the nodes containing multiple children.
Stop if all columns have been considered.

32. Inclusion Property For any pair of columns i,j
i < j if and only if i1  j1
Note that if i<j then the edge containing i is an ancestor of the edge containing j i j

33. Example r A B C D E A B C D E
Initially, there is a single clade r, and each node has r as its parent

34. Sort columns
Sort columns according to the inclusion property (note that the columns are already sorted here).
This can be achieved by considering the columns as binary representations of numbers (most significant bit in row 1) and sorting in decreasing order A B C D E

35. Add first column In adding column i 1 2 3 4 5B C D E
In adding column i
Check each edge and decide which side you belong.
Finally add a node if you can resolve a clade r u B D A C E

36. Adding other columns A B C D E
Add other columns on edges using the ordering property r 1 3 E 2 B 5 4 D A C

37. Unrooted case
Switch the values in each column, so that 0 is the majority element.
Apply the algorithm for the rooted case

38. Summary :No recombination leads to correlation between sites 3 8 5
The different sites are linked. A 1 in position 8 implies 0 in position 5, and vice versa.
The history of a population can be expressed as a tree.
The tree can be constructed efficiently

39. Haplotype Phasing Problem
Most sequencing technologies measure genotypes not haplotypes
Pair of haplotypes
Genotype: 2 = heterozygous
Given a set of genotypes, infer the haplotypes.
Use parsimony assumption
Haplotypes satisfy perfect phylogeny (Gusfield)
Find minimum number of haplotypes that explain observed genotypes

40. Recombination

41. Recombination
A tree is not sufficient as a sequence may have 2 parents
Recombination leads to violation of 4 gamete property.
Recombination leads to loss of correlation between columns

42. Studying recombination
A tree is not sufficient as a sequence may have 2 parents
Recombination leads to loss of correlation between columns
How can we measure recombination?

43. Linkage (Dis)-equilibrium (LD)
A B
0 0
0 1
1 1
1 0
A B
0 1
0 0
1 0
No recombination
Pr[A,B=0,1] = 0.25
Linkage disequilibrium
Extensive Recombination
Pr[A,B=(0,1)=0.125
Linkage equilibrium