CSE182-L18

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CSE182-L18

• Population Genetics

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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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Summary No recombination leads to correlation
between sites
A
B
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.
• The history of a population can be expressed as a
  tree.
• The tree can be constructed efficiently

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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

00000000 11111111 00011111
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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?

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Linkage (Dis)-equilibrium (LD)
A B 0 0 0 1 0 0 0 0 1 1 1 0 1 0 1 0
A B 0 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0

• Extensive Recombination
• PrA,B(0,1)0.125
• Linkage equilibrium

• No recombination
• PrA,B0,1 0.25
• Linkage disequilibrium

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Measuring LD

• Consider two bi-allelic sites A and B, with
  values 0 and 1.
• Let p1 probabilityindividual has allele 1 in
  site A
• q1 probabilityindividual has allele 1 in site
  B
• P11 Prob individual has allele 1 in site A,
  and B
• Linkage Disequilibrium, D P11-p1q1
  P01-p0q1 .
• If D0, sites are uncorrelated, (are in linkage
  equilibrium)
• If D gtgt0, sites are highly correlated (have
  high LD)
• Other measures exist, but they all measure
  similar quantities.

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LD can be used to map disease genes
LD
0 1 1 0 0 1
D N N D D N

• LD decays with distance from the disease allele.
• By plotting LD, one can short list the region
  containing the disease gene.

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Population sub-structure can cause problems in
disease gene mapping
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Population sub-structure can increase LD

• Consider two populations that were isolated and
  evolving independently.
• They might have different allele frequencies in
  some regions.
• Pick two regions that are far apart (LD is very
  low, close to 0)

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Recent ad-mixing of population

• If the populations came together recently (Ex
  African and European population), artificial LD
  might be created.
• D 0.15 (instead of 0.01), increases 10-fold
• This spurious LD might lead one false
  associations
• Other genetic events can cause LD to arise, and
  one needs to be careful

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1
Pop. AB
p10.5 q10.5 P110.1 D0.1-0.250.15
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0
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Determining population sub-structure

• Given a mix of people, can you sub-divide them
  into ethnic populations.
• Turn the problem of spurious LD into a clue.
• Find markers that are too far apart to show LD
• If they do show LD (correlation), that shows the
  existence of multiple populations.
• Sub-divide them into populations so that LD
  disappears.

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Determining Population sub-structure

• Same example as before
• The two markers are too similar to show any LD,
  yet they do show LD.
• However, if you split them so that all 0..1 are
  in one population and all 1..0 are in another, LD
  disappears

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Iterative Algorithm for Population Substructure

• Assume that there are 2 sub-populations
• Randomly partition the individuals into two.
• Select an individual, and compute the
  probabilities Pr(xA), and Pr (xB)
• Assign the individual to A with probability
• Pr(xA)/ (Pr(xA)Pr(xB))
• Continue.

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Iterative algorithm for population sub-structure

• Define
• N number of individuals (each has a single
  chromosome)
• k number of sub-populations.
• Z ? 1..kN is a vector giving the
  sub-population.
• Zik gt individual i is assigned to population
  k
• Xi,j allelic value for individual i in position
  j
• Pk,j,l frequency of allele l at position j in
  population k

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Example

• Ex consider the following assignment
• P1,1,0 0.9
• P2,1,0 0.1

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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Goal

• X is known.
• P, Z are unknown.
• The goal is to estimate Pr(P,ZX)
• Various learning techniques can be employed.
• Here a Bayesian (MCMC) scheme is employed. We
  will only consider a simplified version

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AlgorithmStructure

• Iteratively estimate
• (Z(0),P(0)), (Z(1),P(1)),.., (Z(m),P(m))
• After convergence, Z(m) is the answer.
• Iteration
• Guess Z(0)
• For m 1,2,..
• Sample P(m) from Pr(P X, Z(m-1))
• Sample Z(m) from Pr(P X, P(m-1))
• How is this sampling done?

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Example

• Choose Z at random, so each individual is
  assigned to be in one of 2 populations. See
  example.
• Now, we need to sample P(1) from Pr(P X, Z(0))
• Simply count
• Nk,j,l number of people in pouplation k which
  have allele l in position j
• pk,j,l Nk,j,l / N

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 2 2 1 1 2 1 2 1 2 1 2 2 1 1 2 1 2 2 1
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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Example
0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 2 2 1 1 2 1 2 1 2 1 2 2 1 1 2 1 2 2 1

• Nk,j,l number of people in population k which
  have allele l in position j
• pk,j,l Nk,j,l / N1,1,
• N1,1,0 4
• N1,1,1 6
• p1,1,0 4/10
• p1,2,0 4/10
• Thus, we can sample P(m)

1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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Sampling Z

• PrZ1 1 Pr01 belongs to population 1?
• We know that each position should be in linkage
  equilibrium and independent.
• Pr01 Population 1 p1,1,0 p1,2,1
  (4/10)(6/10)(0.24)
• Pr01 Population 2 p2,1,0 p2,2,1
  (6/10)(4/10)0.24
• Pr Z1 1 0.24/(0.240.24) 0.5

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Sampling

• Suppose, during the iteration, there is a bias.
• Then, in the next step of sampling Z, we will do
  the right thing
• Pr01 pop. 1 p1,1,0 p1,2,1 0.70.7
  0.49
• Pr01 pop. 2 p2,1,0 p2,2,1 0.30.3
  0.09
• PrZ1 1 0.49/(0.490.09) 0.85
• PrZ6 1 0.49/(0.490.09) 0.85
• Eventually all 01 will become 1 population, and
  all 10 will become a second population

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 1 1 2 1 2 1 2 1 1 2 2 2 1 2 2 1 2 2 1
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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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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Other topics
Assembly
Genomic Analysis/ Pop. Genetics
Protein Sequence Analysis
Sequence Analysis
Gene Finding
ncRNA
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ncRNA gene finding

• Gene is transcribed but not translated.
• What are the clues to non-coding genes?
• Look for signals selecting start of transcription
  and translation. Non coding genes are transcribed
  by Pol III
• Non-coding genes have structure. Look for genomic
  sequences that fold into an RNA structure
• Structure Given a sequence, what is the
  structure into which it can fold with minimum
  energy?

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tRNA structure
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RNA structure Basics

• Key RNA is single-stranded. Think of a string
  over 4 letters, AC,G, and U.
• The complementary bases form pairs.
• Base-pairing defines a secondary structure. The
  base-pairing is usually non-crossing.

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RNA structure pseudoknots
Sometimes, unpaired bases in loops form
crossing pairs. These are pseudoknots
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A Static picture of the cell is insufficient

• Each Cell is continuously active,
• Genes are being transcribed into RNA
• RNA is translated into proteins
• Proteins are PT modified and transported
• Proteins perform various cellular functions
• Can we probe the Cell dynamically

Pathways
Gene Regulation
Proteomic profiling
Transcript profiling
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Gene expression

• The expression of transcripts and protein in the
  cell is not static. It changes in response to
  signals.
• The expression can be measured using
  micro-arrays.
• What causes the change in expression?

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Transcriptional machinery

• DNA polymerase (II) scans the genome, initiating
  transcription, and terminating it.
• The same machinery is used for every gene, so
  while Pol II is required, it is not sufficient
  to confer specificity

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TF binding
Transcription factors

• Other transcription factors interact with the
  core machinery and upstream DNA to provide
  specificity.
• TFs bind to TF binding sites which are clustered
  in upstream enhancer and promoter elements.
• The enhancer elements may be located many kb
  upstream of the core-promoter

Upstream elements
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TF binding sites

• TF binding sites are weak signal (about 10 bp
  with 5bp conserved)
• If two genes are co-regulated, they are likely to
  share binding sites
• Discovery of binding site motifs is an important
  research problem.

TCAGGAG
g1
TGAGGAG
g2
g3
TCAGGTG
g4
TGAGGTG
g5
TCAGGTG
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http//www.gene-regulation.com/pub/databases.html
transfac
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Discovering TF binding sites

• Identification of these TF binding sites/switches
  is critical.
• Requires identification of co-regulated genes
  (genes containing the same set of switches).
• How do we find co-regulated genes?

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Idea1 Use orthologous genes from different
species

1. The species are too close (EX humans and
   chimps). Binding non-binding sites are both
   conserved.
2. The species are distant. Binding sites are
   conserved but not other sequence.
3. The species are very distant. Even binding sites
   are not conerved. The genes have alternative
   regulators.

ACGGCAGCTCGCCGCCGCGC
ACGGC-GGGCGCCGCCCCGC
ACGGCAGCTCGCCGCCGC-C
AGTGC-GGGCGCCGCCTCAT
ACGGC-GC-TCGCCGCCGCGC
AT-ACGAAGTAGCGG-ATGGT
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Idea2 Microarray

• Expression level of all genes

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Pathways

• Proteins interact to transduce signal, catalyze
  reactions, etc.
• The interactions can be captured in a database.
• Queries on this database are about looking for
  interesting sub-graphs in a large graph.

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Summary

• Biological databases cannot be understood without
  understanding the data, and the tools for
  querying and accessing these data.
• While database technology (XML, Relational OO
  databases, text formats) is used to store this
  data, its use is (often) transparent for
  Bioinformatics people.
• In this course, we looked at various
  data-streams, and pointed to databases that store
  these data-streams
• Nucleic Acids Research brings out a database
  issue every January

2005 issue