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Parametric and Non-Parametric analysis of complex diseases

Lecture #8

Based on: Chapter 25 & 26 in Terwilliger and Ott’s Handbook of Human Genetic Linkage.

Prepared by Dan Geiger.

2

Complex Diseases

1. Unknown mode of inheritance (Dominant/recessive)2. Several interacting loci (Epistasis)3. Unclear affected status (e.g., psychiatric disorders)4. Genetic heterogeneity (allelic and non-allelic)5. Non genetic factors

The result is that computing the likelihood of data given a model and a location of disease loci (specified by ) may be completely off because the model we use is wrong !

We start by specifying how alternative models look like using the Bayes net model we developed.

3

Mode of Inheritance

S23m

L21fL21m

L23m

X21 S23f

L22fL22m

L23f

X22

X23

Specify different conditional probability tables between the phenotype variables Yi and the genotypes

S13m

L11fL11m

L13m

X11 S13f

L12fL12m

L13f

X12

X13

y3

y2y1

Recessive, full penetrance:P(y1 = sick | X11= (a,a)) = 1P(y1 = sick | X11= (A,a)) = 0P(y1 = sick | X11= (A,A)) = 0

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More modes of Inheritance

Dominant, 60% penetrance:P(y1 = sick | X11= (a,a)) = 0.6P(y1 = sick | X11= (A,a)) = 0.6P(y1 = sick | X11= (A,A)) = 0

Dominant, full penetrance:P(y1 = sick | X11= (a,a)) = 1P(y1 = sick | X11= (A,a)) = 1P(y1 = sick | X11= (A,A)) = 0

Recessive, 40% penetrance, 1% penetrance for phenocopies:P(y1 = sick | X11= (a,a)) = 0.4P(y1 = sick | X11= (A,a)) = 0.01P(y1 = sick | X11= (A,A)) = 0.01

Dominant, 20% penetrance, 5% penetrance for phenocopies:P(y1 = sick | X11= (a,a)) = 0.2P(y1 = sick | X11= (A,a)) = 0.2P(y1 = sick | X11= (A,A)) = 0.05

5

Two or more interacting loci (epistasis)

Specify different conditional probability tables between the phenotype variables Yi and the 2 or more genotypes of person i.

Example: Recessive, full penetrance:P(y11 = sick | X11= (a,a), X21= (a,a)) = 1P(y11 = sick | X11= (A,a), X21= (a,a)) = 0P(y11 = sick | X11= (A,A), X21= (a,a)) = 06 more zero options to specify.

S23m

L21fL21m

L23m

X21 S23f

L22fL22m

L23f

X22

X23

S13m

L11fL11m

L13m

X11 S13f

L12fL12m

L13f

X12

X13

y3

y2y1

There are 512 possible patterns of zeros and ones for two loci !Not enough data to try them all.

6

Unclear affection status

S23m

L21fL21m

L23m

X21S23f

L22fL22m

L23f

X22

X23

S13m

L11fL11m

L13m

X11 S13f

L12fL12m

L13f

X12

X13

Specify a “confusion matrix” regarding the process that determines affected status.

Y3

Y2Y1

P(z1 = measured sick | y1 = sick) = 0.9P(z1 = measured sick | y1 = not sick) = 0.2

Z1

Z1

Z1

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Genetic Heterogeneity

Allelic heterogeneity: more than one allele at a locus predisposes to a disease.Non-Allelic heterogeneity: several independent loci predisposes to the disease .

Si3

m

Li1

fL

i1m

Li3

m

Xi1

Si3

f

Li2

fL

i2m

Li3

f

Xi2

Xi3

1 2

3

8

Non genetic factors

S23m

L21fL21m

L23m

X21 S23f

L22fL22m

L23f

X22

X23

S13m

L11fL11m

L13m

X11 S13f

L12fL12m

L13f

X12

X13

y3

y2y1

Under liability class 1 (L1=1):P(y1 = sick | X11= (a,a), L1 =1) = 1P(y1 = sick | X11= (A,a), L1 =1) = 0.05P(y1 = sick | X11= (A,A), L1 =1) = 0.05

Liability Class L1

Example: Li = 1 means “old”Li = 2 means “young”.

L3

L2

Under L1 =2 (“young”): the first line changes, say, to 0.3 and the other two lines to, say, 0.

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Trying all options can be bad

If we try repeatedly many models with various parameters the likelihood of data for one of these models can occasionally be high, just by pure chance.

The usual requirement that the LOD score be higher than 3 should no longer be used; it becomes too permissive.

LOD() = log10[ Pr(data| ) / Pr(data| =½) ] > 3

The results of analyzing some Schizophrenia pedigrees in the genetics literature yielded LOD score of 6.49 implying that true gene locations were found with very high confidence. Further scrutiny could not replicate these findings.

10

But we did use repeated values !?

Even for Mendelian (i.e., monogenetic) diseases, and assuming we have the correct model (say, dominant), we still try many values to find the best location. Why is this practice OK ?

Answer: because whenever we get a negative answer, we eliminate part of the genome. Assuming the model is correct, eliminating 50% of the genome, doubles the prior probability of finding linkage in the remaining part.

However: For complex diseases this answer no longer applies; the model is knowingly wrong.

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Correction factor by Kidd & Ott (1984)

For a fixed model with various values, use the requirement

LOD() > 3 + log10(m) where m is the number of values tried.

The length of the human genome is about 5000cM and a distance of 50cM or so is enough to declare two sites to be independent. So with 100 independent markers, one can cover the genome. So the threshold for LOD score is between 3 and 5.

When we try many models, this correction idea translates to:

LOD() > 5 + log10(n) where n is the number of models tried.

In the Schizophrenia study 18 models were reported so the LOD

score threshold should be 5+log10(18)= 6.25. The results of the

study 6.49 are not that surprising any more. ONE MUST COUNT ALL MODELS ONE TRIED, even the unpublished attempts.

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Affected only analysisProblem: If we analyze a complex disease using a single locus model, attempting to find one of the influential predisposing gene locations, we get into the following problem: Individual reported to be unaffected could actually posses some predisposing alleles. So affection status is more indicative when the individual is affected (say 9/10 cases) but less so when the individual is reported unaffected (say 1/10 cases).Idea: consider unaffected pedigree members as unobserved (regarding affection status).

The two values for Yi are: affected, not affected. Removing Y2 from the Bayesian network means making unaffected be coded as unknown (unobserved) because y2 p(y2|x12)=1.

S13m

L11fL11m

L13m

X11 S13f

L12fL12m

L13f

X12

X13

Y3

Y2Y1

Unaffected; variable removed

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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.

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Some Non-Parametric Methods

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.

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.

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)

Exactly one allele IBD.

1/2

1/2 1/1

1/3 1/2

1/2 1/3

1/3

No allele is IBD. One allele is IBS.

1/1

1/1 1/1

1/2

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

1/3 1/4

3/4

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

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.

Not surprisingly, the expected number of IBD alleles is (4*2+8*1)/16=1.

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

1/3 1/4

3/4

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)

?/?

1/3 1/4

3/4

Compute the probability of alleles of every family configuration given the other typed persons in the pedigree. Based on this probabilities compute:

E[IBD] = 1Pr(1 allele IBD) + 2Pr(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 or impossible 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:),(

4

1 2

1

2

1b

a baij BAz

This measure should be compared to what is expected under no linkage. To use many pedigrees, a conversation 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:

)(),(4

1 2

1

2

1ab

a baij AfBAz

/1)( or /1)(or 1)(aa AaAaa pAfpAfAf where

A popular method that needs to be mentioned. Terwilliger & Ott state

however that: “The reliability and robustness of this method is unclear,

because it is much more dependent on gene frequencies estimates than

standard linkage analysis”.

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Using SUPERLINK’s BayesNet Model

What query do we need to answer to decide how likely two affected relatives A and B share a common allele ?

Answer: Find all the paths of the form Afounder B that connects them and compute the likelihood of the assignment to the selector variables that activate these paths.

It is not needed to resort to IBS calculations.

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Example

?/?

1/3 1/4

?/?

The expected IBD is given using the posterior

distribution of the four selector variables SAf,SAm,SBf,SBm

given the data in the pedigree. I.e.,

Pr(SAf=0,SAm =0,SBf =0,SBm =0 | data)*2+

Pr(SAf=0,SAm =0,SBf =0,SBm =1 | data)*1 +

Etc, 16 terms all together.Note that we only treat one Marker at a time. Standard for APM method.

BA

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Pedigree’s Data (fn.ped)

1 115 0 0 2 1 1 1 0 1 2 1 2 3 3 1 2 1 1 1 3 2 2 1 0 0 1 0 1 0 1 0 1 126 0 0 1 1 1 0 1 0 1 2 1 2 3 3 1 2 1 1 1 3 2 2 1 0 0 1 0 1 1 0 1 111 0 0 1 1 1 1 0 1 2 0 2 0 3 1 2 1 1 1 3 1 2 1 0 0 1 0 1 0 0 0 1 122 111 115 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 125 0 0 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 121 111 115 1 1 1 0 1 2 1 2 3 3 1 2 1 1 1 3 2 2 1 0 0 1 0 1 0 1 0 1 1 135 126 122 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 131 121 125 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 141 131 135 2 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Pedigree number

Individual ID

Mother’s ID

Father’s ID

Sex: 1=male 2=female

Status: 1=healthy 2=diseased

Unknown marker alleles

Known marker alleles

Liability class (in this file 1 class only)

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Marker File (fn.dat)

14 0 0 30 0.0 0.0 01 2 3 4 5 6 7 8 9 10 11 12 13 141 20.995 0.00510 0 13 6 0.0200 0.3700 0.4050 0.0050 0.0500 0.0750

...[other 12 markers skipped]...

0 010 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

1 disease locus + 13 markers

locus type = affection (coded 1)

Number of alleles = 2

Normal and mutated allele frequencies

Which program is used

locus type = “allele numbers” (coded 3)

Number of alleles = 6

Last markerfounder allele frequencies

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Marker File (fn.dat)

14 0 0 50 0.0 0.0 01 2 3 4 5 6 7 8 9 10 11 12 13 141 20.995 0.00510 0 13 6 0.0200 0.3700 0.4050 0.0050 0.0500 0.0750

...[other 12 markers skipped]...

0 010 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

Recessive disease, full penetrance

Recombination distances between

markers

Number of liability classes