Heping Zhang, Xueqin Wang and Yuanqing Ye Department of Epidemiology and Public Health

A Transmission/disequilibrium Test for Ordinal Traits in Nuclear Families and a Unified Approach for Association Studies

A Transmission/disequilibrium Test for Ordinal Traits in Nuclear Families and a Unified Approach for Association Studies

Heping Zhang, Xueqin Wang and Yuanqing Ye

Department of Epidemiology and Public Health

Yale University

Presented at Workshop on Genomics, NUS

November 14, 2005

November 14, 2005 2

OutlineOutline

• Data structure

• TDT, Q-TDT, S-TDT, etc.

• O-TDT for ordinal traits

• Simulations

• Data analysis

• Discussion and conclusion

November 14, 2005 3

Data StructureData Structure

… …

n families

siblings insiblings 1n siblings nn

November 14, 2005 4

Linkage Analysis – Null HypothesisLinkage Analysis – Null Hypothesis

To test for linkage, the null hypothesis is that the marker locus is not linked to any trait locus.

Marker

Trait locus

Linked Unlinked

November 14, 2005 5

Linkage Analysis - Recombination FractionLinkage Analysis - Recombination Fraction

Marker

Trait locus

2

1 2

1

fractionion recombinat

thebe Let 2

2

November 14, 2005 6

Coefficient of Linkage DisequilibriumCoefficient of Linkage Disequilibrium

Marker

Trait locus

d

d1

m

m1

Allele Frequency Haplotype Frequency?

)1( md

md )1(

)1)(1( md

dm

Freq( , ) - Freq( )Freq( )=

November 14, 2005 7

Null Hypothesis – Linkage DisequilibriumNull Hypothesis – Linkage Disequilibrium

0)2-(1

TDT is to test for linkage in presence of association

or test for association in presence of linkage

(Spielman et al. 1993; Ewens and Spielman 1995).

The null hypothesis of haplotype relative risk (Falk

and Rubinstein, 1987) being 1 is:

November 14, 2005 8

Transmission/Disequilibrium Test (TDT)Transmission/Disequilibrium Test (TDT)

• Eliminate the confounding effects caused by

population stratification/admixture, and other

factors

• A McNemar’s test

November 14, 2005 9

TDT-McNemar TestTDT-McNemar TestSuppose two heterozygous parents and an affected child are genotyped.

A Aa a

Father Mother

marker

Trans Nontrans Father Mother

Aax

AaxAax

Aax

November 14, 2005 10


A Aa a

Father Mother

marker

Trans Nontrans FatherAax Nontrans

A aTrans A a

0 10 0



A Aa a

Father Mother

marker

Trans Nontrans Mother

AaxNontransA aTrans

A a

0 01 0



A Aa a

Father Mother

marker

Trans Nontrans Trans Nontrans

NontransA aTrans

A a

0 11 0

AaxAax


TDT-McNemar TestTDT-McNemar Test

Nontransmitted

Transmitted A a Total

A

a

Total

11n 12n

21n 22n

2111 nn 2212 nn

1211 nn

2221 nn

)/()( 21122

21122 nnnnc

n2

Combinations of Transmitted and Nontransmitted Marker Alleles A and a among 2n Parents of n Affected Children


Further DevelopmentsFurther Developments

• Q-TDT proposed by Allison (1997)

• Q-TDT further investigated by Rabinowitz

(1997)

• S-TDT (Spielman and Ewens 1998)

• FBAT (Lunetta et al. 2000; Rabinowitz and

Laird 2000)

• Many other extensions


General Test StatisticGeneral Test Statistic

Assume that there are n nuclear families. In the

family, there are siblings, i=1,…, n. For the

child in the family, the trait value is and the

genotype is . is the number of allele A in the

genotype . The linkage/association test statistic

can be constructed as follows:

where is a weight of the phenotype .

thi

in

ijythi

thj

ijgijX

ijg

n

i

n

jijij

n

ii

i

XWTT1 11

,

ijW ijy


ExampleExample

• For a sample of affected child-parent triads, let

then is the TDT

introduced by Spielman et al. (1993).

• For a sample of nuclear families with quantitative

trait values , let , where is the average

of trait values, then is the

Q-TDT introduced by Rabinowitz (1997)

• For ordinal trait?

,1ijW

ijy yyW ijij y

)|(/))|(( 2 YTVarYTET

)|(/))|(( 2 YTVarYTET


TDT for Ordinal TraitsTDT for Ordinal Traits

Let be the count of children whose trait values greater

or less than y and , the test statistic for

ordinal traits is

Under the null hypothesis, follows

.

yR

ijij yyij RRW

n

i

n

jijyy

i

ijijXRRT

1 1

)(

)(/))(( 2 TVarTET 21


Model and MethodModel and Method

,1,...,1 ),())|((logit :Model KkgIgkYP k

. genotypein allele of copies ofnumber theis )(

effect. genetic is and ,parameters level are ' where

gDgI

sk

• Di-allelic maker with possible alleles A and a.

• Assume that there is a trait increasing allele , and

we use to denote the wild type allele(s)

•Consider a trait taking values in ordinal responses

1,…, K.


Two Common AssumptionsTwo Common Assumptions

• The trait and marker loci are closely linked

such that, given the family’s genotypes at a trait

locus , the family’s phenotypes and marker

genotypes are independent;

• Given disease genotypes, the traits of the

family members are conditionally independent.


Conditional LikelihoodConditional Likelihood

}.|{}{

}|{}{}|{

:familyth in the phenotypes observed given the

genotypes ssibling' for the likelihood lconditiona The

ig i

iiii MgP

yP

gyPMPyMP

i

}){2()]1()(1[}){1}({

|})|{log( 0

APXyyAPAP

yMP

ijj

ijij

ii

The score function


Score StatisticScore Statistic

.)(1 1

n

i

n

jijyy

i

ijijXRRT

After plugging in the estimates for the nuisance parameters, the score function under the null hypothesis is , where)|( YTET


Expectation and VarianceExpectation and Variance

Following the idea of Rabinowitz and laird (2000),

we can compute or estimate the conditional

expectation and the conditional variance given

the observed trait values under null hypothesis in

the following three cases:

(a) both parental marker information is available;

(b) only one of parental marker information is

available; and

(c) none of parental marker information is available.


Expectation and VarianceExpectation and Variance

)},()]()()][()([2

}{)]()([{}|{

}{)]()([}|{

2

,

ikijjk

ikikijij

i jijijij

ijji

ijij

XXCovyRyRyRyR

XVaryRyRYTVar

XEyRyRYTE


Both Parents GenotypedBoth Parents GenotypedWhen both parents’ genotypes are observed, the children’s genotypes are conditionally independent.

Parental Genotypes Expectation Variance

(AA, AA) 2 0

(AA, Aa) 3/2 1/4

(AA, aa) 1 0

(Aa, Aa) 1 1/2

(Aa, aa) 1/2 1/4

(aa, aa) 0 0


One Parent GenotypedOne Parent GenotypedParental

Genotype

Children’s Possible

Genotypes

Cond. Probability Joint Conditional Genotype Distribution of

Two SibsAA Aa aa

AA {AA} 1 P{AA, AA}=1

{Aa} 1 P{Aa, Aa}=1

{AA, Aa} 1/2 1/2 P{AA, Aa}=

P{AA, AA}=P(Aa, Aa}=

aa {Aa} 1 P{Aa, Aa}=1

{aa} 1 P{aa, aa}=1

{Aa, aa} P{AA, Aa}=

P{Aa, Aa}=P{aa, aa}=

Aa {AA} 1 P{AA, AA}=1

{Aa} 1 P{Aa, Aa}=1

{aa} 1 P{aa, aa}=1

{AA, Aa} P{AA, AA}=

P(Aa, Aa}=

P{AA, Aa}=

)22/(2 2 nn

)22/()12( 2 nn

)22/(2 2 nn

)22/()12( 2 nn

nnAA / nnAa / )1(/)1( nnnn AAAA

)1(/)1( nnnn AaAa)1(/ nnnn AaAA


One Parent Genotyped (continued)One Parent Genotyped (continued)Parental

Genotype

Children’s

Possible

Genotypes

Cond. Probability Joint Conditional Genotype

Distribution of Two SibsAA Aa aa

Aa {Aa, aa} P{Aa, Aa}=

P(aa, aa}=

P{Aa, aa}=

{AA,aa} P{AA, AA}=P{aa, aa}=

P{AA, Aa}/2= P{Aa, aa}/2

=

P{AA, aa}=

P{Aa, Aa}=

{AA, Aa,

aa}

nnaa /nnAa /

nnn

nn

2324

34 11

nnn

nnn

2324

23442 11

nnn

nn

2324

34 11

)2324/()34( 22 nnnnn

nnn

n

2324

4 2

nnn

nnn

2324

2384 21

ly.respective aa, and Aa, AA,

genotypeith children w ofnumber theis and ,, aaAaAA nnn

)1(/ nnnn aaAa

)1(/)1( nnnn aaaa

)1(/)1( nnnn AaAa


No Parental GenotypeNo Parental GenotypeChildren’s

Possible

Genotypes

Cond. Probability Joint Conditional Genotype Distribution

of Two SibsAA Aa aa

{AA} 1 P{AA, AA}=1

{Aa} 1 P{Aa, Aa}=1

{aa} 1 P{aa, aa}=1

{AA, Aa} P{AA, AA}=

P(Aa, Aa}=

P{AA, Aa}=

{Aa, aa} P{Aa, Aa}=

P(aa, aa}=

P{Aa, aa}=

{AA,aa} P{AA,AA}=P{aa,aa}=P{AA,Aa}/2=

P{Aa, aa}/2=

P{AA, aa}=

P{Aa, Aa}=

{AA, Aa, aa}

nnAA / nnAa /

)1(/ nnnn AaAA

nnAa / nnaa /)1(/ nnnn aaAa

nnn

nn

2324

34 11

nnn

nnn

2324

23442 11

)2324/()34( 11 nnnnn )2324/()34( 22 nnnnn

nnn

n

2324

4 2

)2324/(2384( )21 nnnnnn

)1(/)1( nnnn AAAA

)1(/)1( nnnn AaAa

)1(/)1( nnnn AaAa

)1(/)1( nnnn aaaa


Simulation StudiesSimulation Studies

• Assess the type I error of our score test with respect

to specific nominal levels (0.05, 0.01, and 0.0001) to

validate the asymptotic behavior of the test statistic.

• Compare the power of our test with other test

statistics.

• Choose the ordinal level K=3, 4, or 5.


Simulation DesignSimulation Design

• Generate the parent’s genotypes for given the haplotype frequencies

11.0

Haplotype Frequency

AD 0.2

Ad 0.1

aD 0.1

ad 0.6


Simulation DesignSimulation Design

• Given the parental genotypes, generate the offspring genotypes assuming unlinked (null) or linked (1cM, alternative) trait and marker loci

• Conditional on the trait genotype, use the proportional odds model to generate the ordinal trait.

• 200 or 400 families are generated


Three models to generated trait valuesThree models to generated trait values

(a) A proportional odds model is used to

generate an ordinal trait;

(b) A non-proportional odds model is also used

to generate an ordinal trait to assess the

robustness of our score test with respect to

the proportionality assumption;

(c) A Gaussian model is used to generate a

quantitative trait to evaluate the performance

of O-TDT for the quantitative trait.


1,...,1for

,1 and 2, and 1, 0, are of valuesPossible

alleles.

increasing trait of copies ofnumber theis where

,1,...,1,)]|([logit

model odds

alproportion by the generated are traitsordinal The

Kk

kt

t

KkttkYP

k

k

Ordinal Traits Generated from a Proportional Odds Model (a)Ordinal Traits Generated from a Proportional Odds Model (a)


05.0 01.0

Type I Errors Based on 10,000 Replications (a)Type I Errors Based on 10,000 Replications (a)

#of

families

K

Q-TDT O-TDT TDT Q-TDT O-TDT TDT Q-TDT O-TDT TDT

200 3 0.0509

0.0506 0.0508 0.0100 0.0099 0.0095 6e-005 8e-005 9e-005

4 0.0506 0.0501 0.0508 0.0099 0.0101 0.0091 7e-005 6e-005 9e-005

5 0.0509 0.0513 0.0512 0.0102 0.0099 0.0098 6e-005 0.0001 7e-005

400 3 0.0519 0.0518 0.0507 0.0106 0.0107 0.0098 0.0001 0.0001 8e-005

4 0.0508 0.0500 0.0500 0.0105 0.0103 0.0098 0.0001 9e-005 0.0001

5 0.0520 0.0507 0.0502 0.0107 0.0105 0.0098 9e-005 0.0001 7e-005

0001.0


Figure: Power comparison (a)Figure: Power comparison (a)


K=3 P(Y1|dd)=.7

P(Y2|dd)=.9

P(Y1|dD)=.3

P(Y2|dD)=.6

P(Y1|DD)=.1

P(Y2|DD)=.5

P(Y=1)=0.478

P(Y=2)=0.260

P(Y=3)=0.262

K=4 P(Y1|dd)=.7

P(Y2|dd)=.8

P(Y3|dd)=.9

P(Y1|dD)=.3

P(Y2 |dD)=.5

P(Y3 |dD)=.7

P(Y1|DD)=.1

P(Y2 |DD)=.35

P(Y3 |DD)=.6

P(Y=1)=0.478

P(Y=2)=0.155

P(Y=3)=0.156

P(Y=4)=0.211

K=5 P(Y1|dd)=.7

P(Y2|dd)=.77

P(Y3|dd)=.85

P(Y4|dd)=.92

P(Y1 |dD)=.2

P(Y2 |dD)=.45

P(Y3 |dD)=.65

P(Y4 |dD)=.8

P(Y1 |DD)=.05

P(Y2 |DD)=.35

P(Y3 |DD)=.55

P(Y4 |DD)=.75

P(Y=1)=0.431

P(Y=2)=0.166

P(Y=3)=0.141

P(Y=4)=0.115

P(Y=5)=0.146

Non-Proportional Odds Model (b)Non-Proportional Odds Model (b)

Conditional and marginal distribution for ordinal trait


05.0 01.0

Type I Errors Based on 10,000 Replications (b)Type I Errors Based on 10,000 Replications (b)

#of

families

K

Q-TDT O-TDT TDT Q-TDT O-TDT TDT Q-TDT O-TDT TDT

200 3 0.0492 0.0493 0.0495 0.0098 0.0097 0.0095 7e-005 7e-005 7e-005

4 0.0497

0.0499 0.0495 0.0096 0.0097 0.0095 4e-005 7e-005 7e-005

5 0.0497 0.0504 0.0485 0.0105 0.0106 0.0093 5e-005 7e-005 5e-005

400 3 0.0503 0.0501 0.0481 0.0097 0.0097 0.0093 0.0001 0.0001 0.0001

4 0.0509 0.0504 0.0481 0.0096 0.0099 0.0093 8e-005 9e-005 0.0001

5 0.0514 0.0508 0.0475 0.0096 0.0094 0.0090 0.0001 9e-005 0.0001

0001.0


Figure: Power comparison (b)Figure: Power comparison (b)


Performance for Quantitative Traits (c)Performance for Quantitative Traits (c)

Our test can serve as a unified test for any trait. For

quantitative trait, the weights in our test are the

functions of quantiles. Simulations show that our

test is competitive with, but slightly less powerful

than Q-TDT.


Type I Errors for Quantitative Traits Based on 100,000 Replications (c)Type I Errors for Quantitative Traits Based on 100,000 Replications (c)

# of

FamilyQ-TDT O-TDT Q-TDT O-TDT Q-TDT O-TDT

200 0.05407 0.05268 0.01041 0.01015 0.00013 8e-005

400 0.0538 0.05206 0.01099 0.01040 0.00013 0.00012

05.0 01.0 01.0


Power: Quantitative Trait Power: Quantitative Trait

2.= and 1,= 0,= DDDddd

Data are simulated similarly to the experiments for assessing type I error, except the following. Given the genotype at the trait locus, the quantitative trait follows the normal distribution with mean proportional to the number of the trait increasing allele and unit variance. Namely,


Figure: Power comparison (c)Figure: Power comparison (c)


Data (Dr. Ming Li) Data (Dr. Ming Li)

• Identify candidate SNPs through association

analysis

• Nicotine dependence was measured in 313

families with 1,396 subjects. 12 SNPs were

genotyped for GPR51 gene (suggested from

Framingham Heart Study samples).

• One ordinal trait with 8 levels was assessed by

Fagerstrom test for nicotine dependence (FTND)

• FBAT was also used for comparison


FTNDFTND

1. How many cigarettes a day do you usually smoke? (0-3 points)

2. How soon after you wake up do you smoke your first cigarette? (0-3 points)

3. Do you smoke more during the first two hours of the day than during the rest of the day? (0,1)

4. Which cigarette would you most hate to give up? (0,1)

5. Do you find it difficult to refrain from smoking in places where it is forbidden, such as public buildings, on airplanes or at work? (0,1)

6. Do you still smoke even when you are so ill that you are in bed most of the day? (0,1)

TOTAL POINTS =


GPR51 GeneGPR51 Gene

• G protein-coupled receptor 51 (on 5q24 on rat genome and 9p22.33 on human genome)• Combines with GABA-B1 to form functional GABA-B receptors• Inhibits high voltage activated calcium ion channels


ResultsResults

SNP ID PBAT-GEE O-TDT Q-TDT

Pooled AA EA Pooled AA EA Pooled AA EA

rs2304389 .03 .81 .14 .85 .84 .96 .89 .97 .86

rs1435252 .13 .09 .004 .50 .67 .037 .68 .53 .059

rs3780422 .21 .49 .04 .26 .29 .67 .29 .29 .72

rs1537959 .07 .02 .38 .48 .24 .65 .72 .43 .60

rs2491397 .07 .11 .12 .23 .82 .044 .36 .98 .062

rs2779562 .38 .44 .008 .76 .21 .012 .49 .32 .0048

rs3750344 .35 .74 .006 .13 .49 .026 .17 .69 .013


Discussion and ConclusionDiscussion and Conclusion

• We propose a score test statistic for Linkage analysis.• Although it is derived from a proportional odds model for ordinal traits, power comparisons reveal that it can serve as a unified approach for dichotomous, quantitative, and ordinal traits.• The score based Q-TDT test yields lower power than O-TDT for ordinal traits, but the difference ranges from a few to tens of percents, depending on the distribution of the ordinal traits.

Documents

Heping Zhang, Xueqin Wang and Yuanqing Ye Department of Epidemiology and Public Health