Maayan Fishelson some by Nir and most are mine I have slightly edited all slides Genetic linkage analysis Gene Hunting find genes responsible for a given disease ID: 797656
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Slide1
.
Some slides were prepared by Ma’ayan Fishelson, some by Nir, and most are mine. I have slightly edited all slides.
Genetic linkage analysis
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.
Slide22
Human GenomeMost human cells contain 46 chromosomes:
2 sex chromosomes (X,Y): XY – in males. XX – in females.
22 pairs of chromosomes named
autosomes
.
Slide33
Sexual Reproduction
zygote
gametes
sperm
egg
תאי מין
תא זרע
ביצית
ביצית מופרית
Slide44
Chromosome Logical StructureLocus – the location of genes or other 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
Slide55
Genotypes versus PhenotypesAt 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.
Slide66
Recombination PhenomenonA 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).
Slide77
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
.
Genotype
Phenotype
A/A, A/O
A
B/B, B/O
B
A/B
AB
O/O
O
Note that the listed genotypes are unordered
(we don’t know which allele is from the father
and which one is from the mother).
Slide88
Example: ABO near AK1 on Chromosome 9
2
4
5
1
3
A
A
1
/A
1
O
A
2
/A
2
A
A
1
/A
2
O
A
1
/A
2
A
A
2
/A
2
O O
A
1
A
2
A O
A
1
A
2
A O
A
2
| A2O OA2 A2Recombinant
Slide99
Example for Finding Disease Genes
We use a marker with codominant alleles A1/A2.We
speculate a locus
with alleles H (Healthy) / A (affected)
If the expected number of recombinats is low (close to zero), then the speculated locus and the marker are tentatively physically closed.
2
4
5
1
3
H
A
1
/A
1
A
A
2
/A
2
H
A
1
/A
2
A
A
1
/A
2
H
A
2
/A
2
A A
A
1
A
2
H A
A
1
A
2H | AA2 | A2A AA2 A2Recombinant
Slide1010
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.
Slide1111
Comments about the exampleOften:
Pedigrees are larger and more complex.Not every individual is typed.There are more markers and they have
more than two
alleles.
Recombinants cannot always be determined.
Slide1212
Usually recombination can not be simply counted
One can compute the likelihood of data given every location and choose the most likely location.
2
4
5
1
3
A
A
1
/A
1
A
A
2
/A
2
A
A
1
/A
2
A
A
1
/A
2
A
A
2
/A
2
? ?
A
1
A
2
A O
A
1
A
2
A O
A
2 | A2A OA2 A2Recombinant ?Sometimes !
Slide1313
A Bayesian Network ModelL11f
L11m
L
13m
X
11
S
13m
Selector of maternal allele at locus 1 of person 3
Maternal allele at locus 1 of person 3 (offspring)
Selector variables S
ijm
are 0 or 1 depending on whose allele is transmitted to offspring i at maternal locus j.
P(s
13m
) = ½
P(l
13m
| l
11m
, l
11f,
,S
13m
=0) = 1 if l
13m
= l
11m
P(l
13m
| l
11m
, l
11f,
,S
13m
=1) = 1 if l
13m
= l
11f
P(l
13m
| l
11m
, l11f,,s13m) = 0 otherwise
Slide1414
Probabilistic model for two loci
S13m
L
11f
L
11m
L
13m
X
11
S
13f
L
12f
L
12m
L
13f
X
12
X
13
Model for locus 1
S
23m
L
21f
L
21m
L
23m
X
21
S
23f
L
22f
L
22m
L
23f
X
22
X
23
Model for locus 2
Slide1515
Probabilistic model for RecombinationS
23mL
21f
L
21m
L
23m
X
21
S
23f
L
22f
L
22m
L
23f
X
22
X
23
S
13m
L
11f
L
11m
L
13m
X
11
S
13f
L
12f
L
12m
L
13f
X
12
X
13
θ
2
is called the recombination fraction between loci 2 & 1.
Slide1616
Modeling Phenotypes IL11f
L11m
L
13m
X
11
S
13m
Phenotype variables Y
ij
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(y
11
= sick | X
11
= (a,a)) = 1
P(y
11
= sick | X
11
= (A,a)) = 0
P(y
11
= sick | X
11
= (A,A)) = 0
Y
11
Slide1717
Introducing a tentative disease LocusS
23mL
21f
L
21m
L
23m
X
21
S
23f
L
22f
L
22m
L
23f
X
22
X
23
S
13m
L
11f
L
11m
L
13m
X
11
S
13f
L
12f
L
12m
L
13f
X
12
X
13
The recombination fraction
θ
2
is unknown. Finding it can help determine whether a gene causing the disease lies in the vicinity of the marker locus.
Disease locus
: assume
sick means x
ij
=(a,a)
Marker locus
Slide1818
SUPERLINKStage 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
θ
values is calculated. This is done by
by
performing
variable elimination
according to the elimination order determined
in stage 3.