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Computational Problems in
Genetic Linkage Analysis
Dan Geiger
CS, Technion
This talk is based mainly on work with Ma'ayan Fishelson.
See bioinfo.cs.technion.ac.il/superlink/ for more details.
Some slides are due to Gideon Greenspan.
Course homepage:
http://www.cs.technion.ac.il/~fmaayan/cs236524/
.
Requirements
•One homework assignment (10%).
•Mid term progress report.
•Submission in pairs.
•Well documented, tested, useful program.
•Self initiative (by reading and thinking) will be
rewarded.
•One or two excelling projects maybe selected for
continuation next semester under special projects
course, if desired.
2
Outline
Part I:
Part II:
Part III:
Part IV:
Quick look on relevant genetics
Case study: Werner’s syndrome
Relevant Mathematics
Software description / algorithms
Gene Hunting: find genes responsible for a given
disease
Main idea: If a disease is statistically linked with a
marker on a chromosome, then tentatively infer that
a gene causing the disease is located near that
marker.
4
Human Genome
Most human cells contain
46 chromosomes:


2 sex chromosomes
(X,Y):
XY – in males.
XX – in females.
22 pairs of
chromosomes named
autosomes.
5
Chromosome Logical Structure
Marker – Genes, Single Nucleotide
Polymorphisms, Tandem
repeats, etc.
Locus – the location of markers on
the chromosome.
Allele – one variant form (or state)
of a gene/marker at a particular
locus.
Locus1
Possible Alleles: A1,A2
Locus2
Possible Alleles: B1,B2,B3
6
Alleles
genotype
phenotype
b
- dominant allele. Namely, (b,b), (b,w) is Black.
 w - recessive allele. Namely, only (w,w) is White.
This is an example of an X-linked trait.
For males b alone is Black and w alone is white.
7
Genotypes versus Phenotypes
 At
each locus (except for sex chromosomes)
there are 2 genes. These constitute the
individual’s genotype at the locus.
 The expression of a genotype is termed a
phenotype. For example, hair color, weight, or
the presence or absence of a disease.
8
Recombination Phenomenon
Male or female
A recombination between
2 genes occurred if the
haplotype of the individual
contains 2 alleles that
resided in different
haplotypes in the
individual's parent.
(Haplotype – the alleles at
different loci that are
received by an individual
from one parent).
:‫תאי מין‬
‫ או זרע‬,‫ביצית‬
10
An example - the ABO locus.
The ABO locus
determines detectable
antigens on the surface
of red blood cells.
 The 3 major alleles
(A,B,O) interact to
determine the various
ABO blood types.
 O is recessive to A and
B. Alleles A and B are
codominant.

Phenotype
Genotype
A
A/A, A/O
B
B/B, B/O
AB
A/B
O
O/O
Note: genotypes are unordered.
11
Example: ABO, AK1 on Chromosome 9
O
A
A1/A1
2
1
O O
A2 A2
A2/A2
Phase inferred
A O
A1 A2
Recombinant
A
A
A1/A2
4
3
O O
A1 A2
O
A2/A2
A |O
A2 | A2
5
A1/A2
Male recombination fraction 12/100 and female 20/100. These
numbers are measured in centi-morgans. One centi-morgan
means one recombination every 100 meiosis.
Ten centi-morgan corresponds to approx 1M nucleotides (with
large variance) depending on the location and sex.
12
Example for Finding Disease Genes
D
H
A1/A1
2
1
D D
A2 A2
A2/A2
Phase inferred
H D
A1 A2
Recombinant
H
H
A1/A2
4
3
D D
A1 A2
D
A2/A2
H |D
A2 | A2
5
A1/A2
We use a marker with codominant alleles A1/A2.
We speculate a locus with alleles H (Healthy) / D (affected)
If the expected number of recombinants is low (close to
zero), then the speculated locus and the marker are
tentatively physically closed.
13
Recombination cannot be simply counted
H
H
A1/A1
2
1
A2/A2
Phase ???
H D
A1 A2
Possible
Recombinant
H
H
A1/A2
4
3
D D
A1 A2
D
A2/A2
H |D
A2 | A2
5
A1/A2
One can compute the probability that a recombination
occurred and use this number as if this is the real count.
14
Comments about the example
Often:
 Pedigrees
are larger and more complex.
 Not every individual is typed.
 Recombinants cannot always be determined.
 There are more markers and they are
polymorphic (have more than two alleles).
15
Genetic Linkage Analysis


The method just described is called genetic linkage
analysis. It uses the phenomena of recombination in
families of affected individuals to locate the vicinity of
a disease gene.
Recombination fraction is measured in centi morgans
and can change between males and females.
(Linkage ) 0    P(Recombinat ion )  0.5 ( No Linkage)

Next step: Once a suspected area is found, further
studies check the 20-50 candidate genes in that area.
16
Part II: Case study
Werner’s Syndrome
A successful application of
genetic linkage analysis
17
The Disease
 First
references in 1960s
 Causes premature ageing
 Autosomal recessive
 Linkage studies from 1992
 WRN gene cloned in 1996
 Subsequent discovery of mechanisms involved in
wild-type and mutant proteins
18
One Pedigree’s Data (out of 14)
Pedigree
number
1
1
1
1
1
1
1
1
1
115
126
111
122
125
121
135
131
141
0
0
0
111
0
111
126
121
131
Individual
ID
Father’s
ID
0
0
0
115
0
115
122
125
135
2
1
1
2
2
1
2
1
2
1
1
1
1
1
1
1
1
2
Sex: 1=male
2=female
0
0
0
0
0
2
0
0
1
Mother’s
ID
1
0
1
0
0
0
0
0
2
0
1
0
0
0
1
0
0
1
1
0
1
0
0
2
0
0
1
2
1
2
0
0
1
0
0
1
1
2
0
0
0
2
0
0
1
2
1
2
0
0
3
0
0
1
Unknown marker alleles
3
2
0
0
0
3
0
0
1
3
3
3
0
0
1
0
0
1
1
3
1
0
0
2
0
0
1
2
1
2
0
0
1
0
0
1
Status: 1=healthy
2=diseased
1
2
1
0
0
1
0
0
1
1
1
1
0
0
1
0
0
1
1
1
1
0
0
3
0
0
1
3
1
3
0
0
2
0
0
1
2
3
1
0
0
2
0
0
1
2
2
2
0
0
1
0
0
1
1
2
1
0
0
0
0
0
1
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
1
1
0
1
0
0
0
0
0
1
0
1
0
0
0
1
0
0
1
1
0
1
0
0
0
0
0
1
0
1
0
0
0
1
0
0
1
1
1
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
1
Known marker alleles
19
Marker File Input
1 disease
locus + 13
markers
Recessive
disease
requires 2
mutant
genes
14 0 0 5
0 0.0 0.0 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2
First
0.995 0.005 First marker
has 6 alleles
1
marker’s
0 0 1
name
3 6
# D8S133
0.0200 0.3700 0.4050 0.0050 0.0500 0.0750
...[other 12 markers skipped]...
First
marker
founder
allele
frequencies
Recombination
distances between
markers
0 0
10 7.6 7.4 0.9 6.7 1.6 2.5 2.8 2.1 2.8 11.4 1 43.8
1 0.1 0.45
20
Genehunter Output
Putative distance
of disease gene
from first marker
in recombination
units
Most ‘likely’
position
position LOD_score
0.00 -1.254417
1.52
2.836135
information
0.224384
0.226379
...[other data skipped]...
18.58
19.92
21.26
22.60
22.92
23.24
23.56
13.688599
14.238474
14.718037
15.159389
15.056713
14.928614
14.754848
0.384088
0.401992
0.426818
0.462284
0.462510
0.463208
0.464387
...[other data skipped]...
81.84
1.939215
90.60 -11.930449
Log likelihood of
placing disease
gene at distance,
relative to it being
unlinked.
Maximum log
likelihood score
0.059748
0.087869
21
Final Location
Marker
D8S259
location of marker
D8S339
Marker
D8S131
WRN Gene
final location
Error in location by genetic linkage of about 1.25M base pairs.
22
Part III: Relevant Mathematics
23
The Maximum Likelihood Approach
The probability of pedigree data Pr(data |  ) is a function
of the known and unknown recombination fractions denoted
collectively by .
How can we construct this likelihood function ?
The maximum likelihood approach is to seek the value of 
which maximizes the likelihood function Pr(data | ) . This
value is called the ML estimate.
The main computational difficulty is to compute Pr(data|)
for a specific value of .
24
Constructing the Likelihood function
First, we need to determine the variables
that describe the problem. There are several
possible choices. Some variables we can
observe and some we cannot.
Lijm (Lijf) = Maternal (paternal) allele at
locus i of person j. The values of this
variables are the possible alleles li at
locus i.
Xij = Unordered allele pair at locus i of
person j. The values are pairs of ith-locus
alleles (li,l’i).
25
Constructing the Likelihood function
L11f
L11m
Selector of
maternal allele
at locus 1 of
person 3
X11
S13m
P(s13m) = ½
L13m
Maternal allele at locus 1 of person 3 (offspring)
Selector variables Sijm are 0 or 1 depending on whose
allele is transmitted to offspring i at maternal locus j.
P(l13m | l11m, l11f,,S13m=0) = 1 if l13m = l11m
P(l13m | l11m, l11f,,S13m=1) = 1 if l13m = l11f
P(l13m | l11m, l11f,,s13m) = 0 otherwise
26
L11m
L12m
L11f
L12f
Probabilistic model for two loci
X11
S13m
X12
L13f
L13m
X13
Model for locus 1
L21m
S23m
L22m
L21f
X21
X22
L22f
S23f
L23f
L23m
Model for locus 2
S13f
X23
27
Probabilistic model for Recombination
L11m
L12m
L11f
X11
S13m
X12
X13
L21m
S23m
S13f
L13f
L13m
females
L12f
males
L22m
L21f
X21
X22
L22f
S23f
L23f
L23m
X
2 
1   2
P( s23t | s13t , 2 )  
where t  {m,f}

1  2 
 2
2 is called the recombination fraction between loci 2 & 1.
23
28
The Data
The data consists of an assignment to a
subset of the variables {Xij}. In other words
some (or all) persons are genotyped at some
(or all) loci.
29
Constructing the likelihood function
P(l11m,l11f,x11,l12m,l12f,x12,l13m,l13f,x13, l21m,l21f,x21,l22m,l22f,x22,l23m,l23f,x23,
s13m,s13f,s23m,s23f | 2) =
= P(l11m) P(l11f)
Product over all local probability tables
P(x11 | l11m, l11f,) … P(s13m) P(s13f) P(s23m | s13m, 2) P(s23m | s13m, 2)
Prob(data| 2) = P(x11 ,x12 ,x13 ,x21 ,x22 ,x23) =
Probability of data (sum
over all states of all
hidden variables)
Prob(data| 2) = P(x11 ,x12 ,x13 ,x21 ,x22 ,x23) =  l11m, l11f … s23f [P(l11m)
P(l11f) P(x11 | l11m, l11f,) … P(s13m) P(s13f) P(s23m | s13m, 2) P(s23m | s13m, 2)
]
The result is a function of the recombination fraction.
The ML estimate is the 2 value that maximizes this
function.
31
Modeling Phenotypes
L11f
L11m
X11
S13m
Y11
L13m
Phenotype variables Yij are 0 or 1 depending on whether a
phenotypic trait associated with locus i of person j is
observed. E.g., sick versus healthy. For example model of
perfect recessive disease yields the penetrance probabilities:
P(y11 = sick | X11= (a,a)) = 1
P(y11 = sick | X11= (A,a)) = 0
P(y11 = sick | X11= (A,A)) = 0
32
Standard usage of linkage
There are usually 5-15 markers. 20-30% of the persons in
large pedigrees are genotyped (namely, their xij is
measured). For each genotyped person about 90% of the
loci are measured correctly. Recombination fraction
between every two loci is known from previous studies
(available genetic maps).
The user adds a locus called the “disease locus” and places
it between two markers i and i+1. The recombination
fraction ’ between the disease locus and marker i and ”
between the disease locus and marker i+1 are the unknown
parameters being estimated using the likelihood function.
This computation is done for every gap between the given
markers on the map. The MLE hints on the whereabouts of
a single gene causing the disease (if a single one exists).
33
Part IV: Software and Algorithms
•
•
•
•
•
Fastlink v4.1 (Each person’s genotype is one variable)
Vitesse v1,v2 (Only loopless Bayesian networks allowed)
GeneHunter/Alegro (exponential in number of persons)
Many more specific packages (e.g., affected siblings)
Superlink: Why is it better ?
For a list, See http://linkage.rockefeller.edu/soft/list.html
34
SUPERLINK

Stage 1: each pedigree is translated into a Bayesian
network.

Stage 2: value elimination is performed on each
pedigree (i.e., some of the impossible values of the
variables of the network are eliminated).

Stage 3: an elimination order for the variables is
determined, according to some heuristic.

Stage 4: the likelihood of the pedigrees given the 
values is calculated using variable elimination
according to the elimination order determined in stage 3.
35
Experiment A
• Same topology (57 people, no loops)
• Increasing number of loci (each one with 4-5 alleles)
• Run time is in seconds.
Files No. of Run Time Run Time Run Time Run Time
Loci Superlink Fastlink Vitesse Genehunter
A0
2
0.03
0.12
0.27
A1
5
0.1
3.77
0.31
A2
6
0.14
79.32
0.39
A3
7
0.42
0.69
A4
8
0.36
2.81
A5
10
1.19
84.66
A6
12
4.65
A7
14
3.01
A8
18
20.98
A9
37
8510.15
A10
38
10446.27
A11
40
over 100 hours
Out-of-memory
Pedigree size
Too big for
Genehunter.
36
Experiment C
• Same topology (5 people, no loops)
• Increasing number of loci (each one with 3-6 alleles)
• Run time is in seconds.
Bus error
Out-of-memory
37
The computational task at hand
n
P(data |  )    P ( xi | pai )
xk
x3
i 1
x1
Cij   Aik Bkj
Example: Matrix multiplication:
k
A50x50 B50x1 C 1x50 versus A50x50 B50x1 C 1x50 
Multidimensional multiplication/summation:
Yij  
n
  A
ikl
m
l
BkjmClmn
k
38
Some options for improving efficiency
n
P(data |  )    P ( xi | pai )
xk
x3
x1
i 1
1.
Performing approximate calculations
of the likelihood.
2.
Multiplying special probability
matrices efficiently.
3.
Grouping alleles together and
removing inconsistent alleles.
4.
Optimizing the elimination order of
variables in a Bayesian network.
39
Projects
Project
No.
Project Subject
1
Performing approximate likelihood computations by using the
method Iterative Join-Graph Propagation. (ps)
2
Performing haplotyping on the input data, i.e.,
inferring the most likely haplotypes for the
individuals in the input pedigrees. (ps)
3
Performing approximate likelihood computations by
using a heuristic which ignores extreme markers in
the likelihood computation. (ps)
40