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Parametric and Non-Parametric
analysis of complex diseases
Lecture #6
Based on: Chapter 25 & 26 in Terwilliger and Ott’s Handbook of
Human Genetic Linkage.
.
Prepared by Dan Geiger.
Complex Diseases
1.
2.
3.
4.
5.
Unknown mode of inheritance (Dominant/recessive)
Several interacting loci (Epistasis)
Unclear affected status (e.g., psychiatric disorders)
Genetic heterogeneity
Non genetic factors
We start by specifying how alternative models look like
using a Bayesian network model.
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Mode of Inheritance
L11m
L12m
L11f
X11
S13m
Specify different
conditional probability
tables between the
phenotype variables Yi
and the genotypes
y1
L12f
X12
y2
L13f
L13m
S13f
X13
y3
L21m
S23m
Recessive, full penetrance:
P(y1 = sick | X11= (a,a)) = 1
P(y1 = sick | X11= (A,a)) = 0
P(y1 = sick | X11= (A,A)) = 0
L22m
L21f
X21
X22
L22f
S23f
L23f
L23m
X23
3
More modes of Inheritance
Dominant, full penetrance:
P(y1 = sick | X11= (a,a)) = 1
P(y1 = sick | X11= (A,a)) = 1
P(y1 = sick | X11= (A,A)) = 0
Dominant, 60% penetrance:
P(y1 = sick | X11= (a,a)) = 0.6
P(y1 = sick | X11= (A,a)) = 0.6
P(y1 = sick | X11= (A,A)) = 0
Dominant, 20% penetrance, 5%
penetrance for phenocopies:
P(y1 = sick | X11= (a,a)) = 0.2
P(y1 = sick | X11= (A,a)) = 0.2
P(y1 = sick | X11= (A,A)) = 0.05
Recessive, 40% penetrance, 1%
penetrance for phenocopies:
P(y1 = sick | X11= (a,a)) = 0.4
P(y1 = sick | X11= (A,a)) = 0.01
P(y1 = sick | X11= (A,A)) = 0.01
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Two or more interacting loci (epistasis)
L11m
X11
S13m
Specify different
conditional probability
tables between the
phenotype variables Yi
and the 2 or more
genotypes of person i.
L12m
L11f
y1
L12f
X12
y2
L13f
L13m
S13f
X13
y3
L21m
S23m
L22m
L21f
X21
Example: Recessive, full penetrance:
P(y11 = sick | X11= (a,a), X21= (a,a)) = 1
P(y11 = sick | X11= (A,a), X21= (a,a)) = 0
P(y11 = sick | X11= (A,A), X21= (a,a)) = 0
6 more zero options to specify.
X22
L22f
S23f
L23f
L23m
X23
5
Unclear affection status
L11m
X11
S13m
Specify a “confusion
matrix” regarding the
process that
determines affected
status.
L12m
L11f
Y1
L12f
X12
S13f
Y2
L13f
L13m
Z1
X13
Z1
Y3
Z1
L21m
S23m
L22m
L21f
X21
X22
S23f
L23f
L23m
P(z1 = measured sick | y1 = sick) = 0.9
P(z1 = measured sick | y1 = not sick) = 0.2
L22f
X23
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Genetic Heterogeneity
1
2
3
Non-Allelic heterogeneity: several independent loci predisposes to the disease .
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Non genetic factors
L11m
L12m
L11f
X11
S13m
L12f
X12
S13f
Liability Class L1
L2
y1
Example: Li = 1 means “old”
Li = 2 means “young”.
y2
L13f
L13m
X13
L3
y3
L21m
S23m
L22m
L21f
X21
L22f
X22
Under liability class 1 (L1=1):
L
L
P(y1 = sick | X11= (a,a), L1 =1) = 1
X
P(y1 = sick | X11= (A,a), L1 =1) = 0.05
P(y1 = sick | X11= (A,A), L1 =1) = 0.05
Under L1 =2 (“young”): the first line changes, say, to 0.3
and the other two lines to, say, 0.
S23f
23f
23m
23
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Parametric versus Non-Parametric
All analyses considered so far are “parametric” meaning that
a mode of inheritance is assumed . In some cases, several
options of modes of inheritance are assumed but still the
analysis uses each option in turn.
For complex diseases it is believed that “non-parametric”
methods might work better. In our context, these are
methods that do not take mode of inheritance into account.
The idea is that computing linkage without assuming mode of
inheritance is more robust to error in model specification.
Clearly, if the model is correct, parametric methods perform
better, but not so if the model is wrong as for complex traits.9
Some Non-Parametric Methods
Definitions: Any two identical copies of an allele l are said to
be identical by state (IBS). If these alleles are inherited
from the same individual then they are also identical by
descent (IBD). Clearly, IBD implies IBS but not vice versa.
Main idea: if affected siblings share more IBD alleles at
some marker locus than randomly expected among siblings,
then that locus might be near a locus of a predisposing gene.
We will consider the following non-parametric methods:
•Affected Sib-Pair Analysis (ASP)
•Extended Affected Sib-Pair Analysis (ESPA)
•Affected Pedigree Member method (APM)
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Identical By Descent (IBD)
1/2
1/3
1/2
1/3
1/2
1/1
1/2
1/3
No allele is IBD.
One allele is IBS.
Exactly one allele IBD.
1/2
1/1
1/1
1/1
At least one allele IBD.
Expected 1.5 alleles IBD.
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Affected Sib-Pair Analysis
The idea is that any two siblings are expected to have one
allele IBD by chance (and at most two IBD alleles, ofcourse).
When a deviation of this pattern is detected, by examining
many sib-pairs, a linkage is established between a disease
gene and the marker location.
This phenomena happens regardless of mode of inheritance,
but its strength is different for each mode.
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Affected Sib-Pair Analysis
1/2
3/4
1/4
1/3
There are 16 combinations of sibling marker genotypes:
SON1 SON2 IBD SON1 SON2 IBD SON1 SON2 IBD SON1 SON2 IBD
1/3
1/3
2
1/4
1/4 2
2/3 2/3
2
2/4
2/4 2
1/3
1/4
1
1/4
1/3 1
2/3 2/4
1
2/4
2/3 1
1/3 2/3
1
1/4
2/4 1
2/3 1/3
1
2/4
1/4
1
1/3 2/4
0
1/4
2/3 0
2/3 1/4
0
2/4
1/3 0
Not surprisingly, the expected number of IBD alleles is (4*2+8*1)/16=1.
But now assume a dominant disease coming from the father and is on the
haplotype with the 1 allele. The only viable options are marked in the table. The
expected IBD is thus (2*2+2*1)/4 = 1.5, which can be detected in analysis.
For a recessive disease linked on the haplotype of 1 and 3, the only viable
pair is 1/3, 1/3 with expected IBD of 2.
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Affected Sib-Pair Analysis
1/2
3/4
1/3
1/4
Standard practice of the ASP method where pedigrees
look like the above (two parents, two children, all
observed), can be done even by hand.
However, one can use general pedigrees, and assume some
family members are not observed, and consider more
distant relatives such as first-cousins, etc.
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Extended Affected Sib-Pair Analysis
(e.g, the ESPA program)
?/?
3/4
1/3
1/4
Compute the probability of alleles of every family
configuration given the other typed persons in the pedigree.
Based on this probabilities compute:
E[IBD] = 1Pr(1 allele IBD) + 2Pr(2 allele IBD)
(The ESPA program currently assumes no loops and at most 5 alleles at a locus.)
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Affected Pedigree Members method (APM)
Computing IBD for distant relatives is considered hard on
large pedigrees so researchers used IBS instead.
Consider one relative to have alleles (A1,A2) and the other to
have (B1,B2). There are four possibilities to have IBS alleles.
Weeks and Lang (1988) used the following statistics zij for
counting IBS status of two individuals:
1 2 2
zij    ( Aa , Bb )
4 a 1 b 1
This measure should be compared to what is expected under no linkage. To use many
pedigrees, a converstion to standard normal variables is used.
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Taking Gene Frequencies into Account
Clearly it is more surprising for affected relatives to share
a rare allele than a common one. So one can use a weighted
average:
1 2 2
zij    ( Aa , Bb ) f ( Aa )
4 a 1 b 1
where f ( Aa )  1
or
f ( Aa )  1 / p Aa
or
f ( Aa )  1 / p Aa
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