CSE182-L17

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CSE182-L17

• Clustering
• Population Genetics Basics

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Unsupervised Clustering

• Given a set of points (in n-dimensions), and k,
  compute the k best clusters.
• In k-means, clustering is done by choosing k
  centers (means).
• Each point is assigned to the closest center.
• The notion of best is defined by distances to
  the center.
• Question How can we compute the k best centers?

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Distance

• Given a data point v and a set of points X,
• define the distance from v to X
• d(v, X)
• as the (Euclidean) distance from v to
  the closest point from X.
• Given a set of n data points Vv1vn and a set
  of k points X,
• define the Squared Error Distortion
• d(V,X) ?d(vi, X)2 / n 1 lt i lt n

v
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K-Means Clustering Problem Formulation

• Input A set, V, consisting of n points and a
  parameter k
• Output A set X consisting of k points (cluster
  centers) that minimizes the squared error
  distortion d(V,X) over all possible choices of X
• This problem is NP-complete in general.

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1-Means Clustering Problem an Easy Case

• Input A set, V, consisting of n points.
• Output A single point X that minimizes d(V,X)
  over all possible choices of X.
• This problem is easy.
• However, it becomes very difficult for more
  than one center.
• An efficient heuristic method for k-Means
  clustering is the Lloyd algorithm

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K-means Lloyds algorithm

• Choose k centers at random
• X x1,x2,x3,xk
• Repeat
• XX
• Assign each v ? V to the closest cluster j
• d(v,xj) d(v,X) ? Cj Cj ? v
• Recompute X
• xj ? (? v ? Cj v) /Cj
• until (X X)

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Conservative K-Means Algorithm

• Lloyd algorithm is fast but in each iteration it
  moves many data points, not necessarily causing
  better convergence.
• A more conservative method would be to move one
  point at a time only if it improves the overall
  clustering cost
• The smaller the clustering cost of a partition of
  data points is the better that clustering is
• Different methods can be used to measure this
  clustering cost (for example in the last
  algorithm the squared error distortion was used)

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Microarray summary

• Microarrays (like MS) are a technology for
  probing the dynamic state of the cell.
• We answered questions like the following
• Which genes are coordinately regulated (They have
  similar expression patterns in different
  conditions)?
• How can we reduce the dimensionality of the
  system?
• Using gene expression values from a sample, can
  you predict if the sample is normal (state A) or
  diseased (state B)
• The techniques employed for classification/cluster
  ing etc. are general and can be employed in a
  number of contexts.

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Microarray non-summary

• We did not cover
• How are the gene expression values measured (the
  technology)? (CSE183)
• How do you control variability across different
  experiments (normalization)? (CSE183)
• What controls the expression of a gene (gene
  regulation), or a set of genes? (CSE 181)

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Population Genetics

• The sequence of an individual does not say
  anything about the diversity of a population.
• Small individual genetic differences can have a
  profound impact on phenotypes
• Response to drugs
• Susceptibility to diseases
• Soon, we will have sequences of many individuals
  from the same species. Studying the differences
  will be a major challenge.

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Population Structure

• 377 locations (loci) were sampled in 1000 people
  from 52 populations.
• 6 genetic clusters were obtained, which
  corresponded to 5 geographic regions (Rosenberg
  et al. Science 2003)

Oceania
Eurasia
East Asia
America
Africa
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Population Genetics

• What is it about our genetic makeup that makes us
  measurably different?
• These genetic differences are correlated with
  phenotypic differences
• With cost reduction in sequencing and genotyping
  technologies, we will know the sequence for
  entire populations of individuals.
• Here, we will study the basics of this
  polymorphism data, and tools that are being
  developed to analyze it.

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What causes variation in a population?

• Mutations (may lead to SNPs)
• Recombinations
• Other genetic events (Ex microsatellite
  repeats)
• Deletions, inversions

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Single Nucleotide Polymorphisms
Infinite Sites Assumption Each site mutates at
most once
00000101011 10001101001 01000101010 01000000011 00
011110000 00101100110
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Short Tandem Repeats
GCTAGATCATCATCATCATTGCTAG GCTAGATCATCATCATTGCTAGTT
A GCTAGATCATCATCATCATCATTGC GCTAGATCATCATCATTGCTAG
TTA GCTAGATCATCATCATTGCTAGTTA GCTAGATCATCATCATCATC
ATTGC
4 3 5 3 3 5
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STR can be used as a DNA fingerprint

• Consider a collection of regions with variable
  length repeats.
• Variable length repeats will lead to variable
  length DNA
• Vector of lengths is a finger-print

4 2 3 3 5 1 3 2 3 1 5 3
individuals
positions
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Recombination
00000000 11111111 00011111
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What if there were no recombinations?

• Life would be simpler
• Each sequence would have a single parent
• The relationship is expressed as a tree.

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The Infinite Sites Assumption
0 0 0 0 0 0 0 0
3
0 0 1 0 0 0 0 0
5
8
0 0 1 0 1 0 0 0
0 0 1 0 0 0 0 1

• The different sites are linked. A 1 in position
  8 implies 0 in position 5, and vice versa.
• Some phenotypes could be linked to the
  polymorphisms
• Some of the linkage is destroyed by
  recombination

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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
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Perfect Phylogeny

• Assume an evolutionary model in which no
  recombination takes place, only mutation.
• 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?

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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
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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)

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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?

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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.

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Inclusion Property

• For any pair of columns i,j
• i lt j if and only if i1 ? j1
• Note that if iltj then the edge containing i is an
  ancestor of the edge containing i

i
j
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Example
1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1 0 D
0 0 1 0 1 E 1 0 0 0 0
Initially, there is a single clade r, and each
node has r as its parent
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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

1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1
0 D 0 0 1 0 1 E 1 0 0 0 0
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Add first column
1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1
0 D 0 0 1 0 1 E 1 0 0 0 0

• 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
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Adding other columns
1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 1 0 D
0 0 1 0 1 E 1 0 0 0 0

• Add other columns on edges using the ordering
  property

r
1
3
E
B
2
5
4
D
A
C
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Unrooted case

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

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Handling recombination

• A tree is not sufficient as a sequence may have 2
  parents
• Recombination leads to loss of correlation
  between columns

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Linkage (Dis)-equilibrium (LD)

• Consider sites A B
• Case 1 No recombination
• PrA,B0,1 0.25
• Linkage disequilibrium
• Case 2Extensive recombination
• PrA,B(0,1)0.125
• Linkage equilibrium

A B 0 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0
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