Linkage DisequilibriumLinkage Disequilibrium

Granovsky Ilana and Berliner Yaniv

Computational Genetics

19.06.03

What is Linkage Disequilibrium?What is Linkage Disequilibrium?• When the occurrence of pairs of specific

alleles at different loci on the same haplotype is not independent, the deviation form independence is termed linkage disequilibrium

• In general, linkage disequilibrium is usually seen as an association between one specific allele at one locus and another specific allele at a second locus

LinkageLinkage Disequilibrium Coefficient Disequilibrium Coefficient DefinitionsDefinitions

Marker 2

Marker1

Allele1

(probability = p2)

Allele2

(probability = 1-p2)

Allele1

(probability = p1)

X1

p1*p2+D11

X2

p1*(1-p2)-D11

Allele2

(probability = 1-p1)

X3

(1-p1)*p2-D11

X4

(1-p1)*(1-p2)+D11

•Xi-number of observations in cell i (X1+X2+X3+X4)=n

•D11-coefficient of gametic linkage disequilibrium

between allele 1 at locus 1 and allele 1 at locus 2

D11=E[X1X4-X2X3|n=1]

Population-based sampling and the Population-based sampling and the EH programEH program

• We wish to test the absence of disequilibrium between allele A at locus 1 and allele B at locus 2 (DAB=0)

• The sample of individuals we have consist of genotyping data with no possibility to fully distinguish all of the haplotypes in each individual

Table of all possible two-locus Table of all possible two-locus genotypesgenotypes

Locus2

Locus 2

AA Aa aa

BB k1 k2 k3

Bb k4 k5 k6

bb k7 k8 k9

In cell 5 there can be either of two phases, AB/ab or Ab/aB

Analysis of likelihoodAnalysis of likelihood

• We maximize the log likelihood of the data observed:

• For cell 1: p1=[P(A B)] • For cell 4: p4=2P(A B)P(A b)• For cell 5: p5=P(A B/a b)+P(A b/a B) =

=2P(A B)P(a b)+2P(A b)P(a B)

1 2

1

ln[ ( )] ln( )a a

i ii

L data pk

2

2

Table of probabilities in each cellTable of probabilities in each cell

Locus 1

Locus 2

AA Aa aa

BB p(A B) 2p(A B)p(a B) P(a B)

Bb 2p(A B)p(A b)

2P(A B)P(a b)+

+2P(A b)P(a B)

2p(a B)p(a b)

bb P(A b) 2p(A b)p(a b) P(a b)

2

2

2

2

Analysis of likelihoodAnalysis of likelihood

• We maximize the likelihood above over the possible haplotype frequencies (p(A), p(B) and DAB.

• This likelihood is then compared with the maximum likelihood when DAB is set equal to 0 (absence of linkage disequilibrium)

ExampleExample Locus 1

Locus 2

AA Aa aa

BB K1=10 K2 = 10 K3=3

Bb K4=15 K5=50 K6=13

bb K7=5 K8=13 K9=10

A a

B 45 29

b 38 46

A a

B 0.28 0.18

b 0.24 0.29

*When censoring k5 all the haplotypes can be uniquely determined

Example cont.Example cont.

• P(A) = 0.28+0.24 = 0.525

• P(B) = 0.28+0.18 = 0.468• DAB = p(A B) –p(A)p(B) = 0.28 – 0.525*0.468

= 0.0387

* Biased example due to the elimination of the 50 observations in k5.

EH program input file formatEH program input file format

• EH = estimated haplotype.– Input file EH.dat

Line 1: Number of alleles at each of the two loci

Line 2: k1 k4 k7

Line 3: k2 k5 k8

Line 4: k3 k6 k9

EH program output fileEH program output file• Output – Estimates of Gene Frequencies

(including k5)

AlleleLocus

1 2

1 0.515 0.484

2 0.480 0.519

# of typed Individuals: 129

EH program output fileEH program output file

Allele at locus 1

Allele at locus 2

Haplotype frequencyIndependent w/association

1 1 0.248 0.328

1 2 0.268 0.188

2 1 0.232 0.153

2 2 0.252 0.332

Chi square testChi square test

df Ln(L) Chi-square

H0: No association 2 -252.68 0.00

H1: Allelic association allowed

3 -248.23 8.89

•The difference between the 2 chi-square is 8.89

• The P-value associated with chi-square (with 1 df) is 0.002873

• It is clear the k5 contributes siginificant information

Haplotype frequencies

Without k5 With k5Haplotype Indepe

ndentassociate Indepe

ndentassociate

A B 0.246 0.284 0.247 0.327

A b 0.279 0.24 0.267 0.187

a B 0.222 0.183 0.232 0.152

a B 0.252 0.291 0.251 0.331

p(A) 0.525 0.515

p(B) 0.468 0.48

Dab 0.038 0.079

SummarySummary

Multiallelic genotype information in EH Multiallelic genotype information in EH programprogram

Locus 2Locus 1 1/1 1/2 2/2 1/3 2/3 3/3

1/1 a1 b1 c1 d1 e1 f1

1/2 a2 b2 c2 d2 e2 f2

2/2 a3 b3 c3 d3 e3 f3

1/3 a4 b4 c4 d4 e4 f4

2/3 a5 b5 c5 d5 e5 f5

3/3 a6 b6 c6 d6 e6 f6

Line 1: Number of alleles at each locus

Subsequent lines:

Multilocus genotype dataMultilocus genotype data

Locus 3

Locus 1 Locus 2 1/1 1/2 2/2

1/1 1/1 a1 b1 c1

1/2 a2 b2 c2

2/2 a3 b3 c3

1/2 1/1 a4 b4 c4

1/2 a5 b5 c5

2/2 a6 b6 c6

2/2 1/1 a7 b7 c7

1/2 a8 b8 c8

2/2 a9 b9 c9

Ex. 23Ex. 23• Full data Solution file: • Censored data solution file.

Censored data

1/1 haplotype data

Locus 2Locus 1

1/1 1/2 1/3 1/4 2/2 2/3 2/4 3/3 3/4 4/4

1/1 10 5 6 4 1 2 3 1 2 0

1/2 6 3 3 3 1 2 1 1 2 1

2/2 12 9 8 11 3 2 5 1 0 3

1/3 1 2 2 1 1 1 1 0 4 2

2/3 0 2 2 8 2 2 9 3 6 8

3/3 8 6 4 10 3 3 8 5 9 13

Haplotypes from censored genotype dataHaplotypes from censored genotype data

Allele at locus 2

Allele at locus 1 1 2 3 4

1 42 14 13 12

2 58 25 16 31

3 37 26 29 63

Allele at locus 2

Allele at locus 1

1 2 3 4

1 0.11 0.038 0.035 0.032

2 0.158 0.068 0.044 0.085

3 0.10 0.07 0.079 0.172

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