1. Linkage and Association in Mx
Sarah Medland

2. Linkage

3. In biometrical modeling A is correlated at 1 for MZ twins and
In biometrical modeling A is correlated at 1 for MZ twins and .5 for DZ twins
.5 is the average genome-wide sharing of genes between full siblings (DZ twin relationship)

4. In linkage analysis we will be estimating an additional variance component Q
For each locus under analysis the coefficient of sharing for this parameter will vary for each pair of siblings
The coefficient will be the probability that the pair of siblings have both inherited the same alleles from a common ancestor

5. Genotypic similarity between relatives
IBD Alleles shared Identical By Descent are a copy of the same ancestor allele
Pairs of siblings may share 0, 1 or 2 alleles IBD
The probability of a pair of relatives
being IBD is known as pi-hat M1 M2 M3 M3 Q1 Q2 Q3 Q4 M1 M2 M3 M3 Q1 Q2 Q3 Q4
IBS
IBD M1 M3 M1 M3 2 1 Q1 Q3 Q1 Q4

6. PTwin1 PTwin2 MZ=1.0 DZ=0.5 MZ & DZ = 1.0 Q A C E E C A Q q a c e e c

7. Distribution of pi-hat
Adult Dutch DZ pairs: distribution of pi-hat at 65 cM on chromosome 19
Model resemblance (e.g. correlations, covariances) between sib pairs, or DZ twins, as a function of DNA marker sharing at a particular chromosomal location

8. DZ by IBD status
Variance = Q + F + E
Covariance = πQ + F + E

9. Covariance Statements
G2: DZ IBD2 twins
Matrix K 1
Covariance
F+Q+E | _
| F+Q+E;
G3: DZ IBD1 twins
Matrix K .5
G4: DZ IBD0 twins
F+Q+E | F_
F | F+Q+E;

10. DZ by IBD status + MZ

11. Covariance Statements +MZ
G2: DZ IBD2 twins
Matrix K 1
Covariance
A+C+Q+E | _
| A+C+Q+E;
G3: DZ IBD1 twins
Matrix K .5
G4: DZ IBD0 twins
A+C+Q+E |
| A+C+Q+E;
G5: MZ twins
A+C+Q+E | A+C+Q _
A+C+Q | A+C+Q+E;

12. Using the full distribution
More power if we use all the available information
So instead of dividing the sample we will use as a continuous coefficient that will vary between sib-pair across loci

13. DZ twins using full distribution of pihat
Covariance
F+E+Q |
| F+E+Q ;
DZ=1 1 1 1 1 1 1 Q F E E F Q q f e e f q
PTwin1
PTwin2

14. The example data
Lipid data: apoB

15. Pihat.mx
!script for univariate linkage - pihat approach
!DZ/SIB
#loop $i 1 4 1
#define nvar 1
#NGroups 1
DZ / sib TWINS genotyped
Data NInput=324
Missing =
Rectangular File=lipidall.dat
Labels sample fam ldl1 apob1 ldl2 apob2 …
Select apob1 apob2
ibd0m$i
ibd1m$i
ibd2m$i ;
Definition_variables
This use of the loop command allows you to run the same script over and over moving along the chromosome
The format of the command is:
#loop variable start end increment
So…#loop $i 1 4 1
Starts at marker 1 goes to marker 4 and runs each locus in turn
Each occurrence of $i within the script will be replaced by the current number ie on the second run $i will become 2
With the loop command the last end statement becomes an exit statement and the script ends with #end loop

16. Pihat.mx
!script for univariate linkage - pihat approach
!DZ/SIB
#loop $i 1 4 1
#define nvar 1
#NGroups 1
DZ / sib TWINS genotyped
Data NInput=324
Missing =
Rectangular File=lipidall.dat
Labels sample fam ldl1 apob1 ldl2 apob2 …
Select apob1 apob2
ibd0m$i
ibd1m$i
ibd2m$i ;
Definition_variables
This use of the ‘definition variables’ command allows you to specify which of the selected variables will be used as covariates
The value of the covariate displayed in the mxo will be the values for the last case read

17. Pihat.mx Begin Matrices; X Lower nvar nvar free ! residual familial F
Z Lower nvar nvar free ! unshared environment E
L Full nvar 1 free ! qtl effect Q
G Full 1 nvar free ! grand means
H Full ! scalar, .5
K Full ! IBD probabilities (from Merlin)
J Full ! coefficients 0.5,1 for pihat
End Matrices;
Specify K
ibd0m$i
ibd1m$i
ibd2m$i
Matrix H .5
Matrix J
Start .1 X 1 1 1
Start .1 L 1 1 1
Start .1 Z 1 1 1
Start .5 G 1 1 1
!script for univariate linkage - pihat approach
!DZ/SIB
#loop $i 1 2 1
#define nvar 1
#NGroups 1
DZ / sib TWINS genotyped
Data NInput=324
Missing =
Rectangular File=lipidall.dat
Labels sample fam ldl1 apob1 ldl2 apob2 …
Select apob1 apob2
ibd0m$i
ibd1m$i
ibd2m$i ;
Definition_variables

18. You need to fix this before you run the script
Pihat.mx
Means
G| G ;
Covariance
F+E+Q |
| F+E+Q ;
Option NDecimals=4
Option RSiduals
Option Multiple Issat
!End
!test significance of QTL effect
! Drop L 1 1 1
Exit
#end loop
Begin Algebra;
F= X*X';
! residual familial variance
E= Z*Z';
! unique environmental variance
Q= L*L';
! variance due to QTL
V= F+Q+E;
! total variance
T= F|Q|E;
! parameters in one matrix
S= F%V| Q%V| E%V;
! standardized variance component estimates
P= ???? ;
! estimate of pihat
End Algebra;
Labels Row S standest
Labels Col S f^2 q^2 e^2
Labels Row T unstandest
Labels Col T f^2 q^2 e^2
You need to fix this before you run the script

19. Run the script across your region and add the -2LL to the xls file

20. Converting chi-squares to LOD scores
For univariate linkage analysis
(where you have 1 QTL estimate)
Χ2/4.6 = LOD

21. Converting chi-squares to p values
Complicated
Distribution of genotypes and phenotypes
Boundary problems
For univariate linkage analysis
(where you have 1 QTL estimate)
p(linkage)=

23. What does it mean?

24. Alternative approach is a summary statistic Use all the information
Convenient
Loss of information
.5 can mean p.ibd0=0 p.ibd1=1 p.ibd2=0 or
p.ibd0=.2 p.ibd1=.6 p.ibd2=.2
Use all the information

25. Alternative approach Model each of the possible outcomes
IBD0 IBD1 IBD2
Weight each of the models by the probability that it is the correct model
The pairwise likelihood is equal to the sum of likelihood for each model multiplied by the probability it is the correct model
The combined likelihood is equal to the sum of all the pairwise likelihoods

26. DZ pairs
* pIBD2 +
* pIBD1
* pIBD0

27. Tells Mx we will be using 3 different means and variance models
How to do this in mx?
Script mixture.mx
Tells Mx we will be using 3 different means and variance models
DZ / sib TWINS genotyped
Data NInput=324 NModel=3
Missing =
Rectangular File=lipidall.dat
Labels ….
Select apob1 apob2
ibd0m$i
ibd1m$i
ibd2m$i ;
Definition_variables

28. How to do this in mx? This script runs an FQE model
L is the QTL VC path coefficent
Begin Matrices;
X Lower nvar nvar free ! residual familial F
Z Lower nvar nvar free ! unshared environment E
L Full nvar 1 free ! qtl effect Q
G Full 1 nvar free ! grand means
H Full ! scalar, .5
K Full ! IBD probabilities (from Merlin)
U Unit 3 1 …
Matrix H 0.5
Specify K
ibd0m$i
ibd1m$i
ibd2m$i
We are placing the prob. Of being IBD 0 1 & 2 in the K matrix

29. How to do this in mx? Begin Algebra;
F= X*X'; ! residual familial variance
E= Z*Z'; ! unique environmental variance
Q= L*L'; ! variance due to QTL
V= F+Q+E; ! total variance

30. How to do this in mx? Means U@G| U@G; Covariance F+Q+E | F _
F | F+Q+E _ ! IBD 0 Covariance matrix
F+Q+E | _
| F+Q+E _ ! IBD 1 Covariance matrix
F+Q+E | F+Q _
F+Q | F+Q+E; ! IBD 2 Covariance matrix
Weights B ;
The means matrix contains corrections for age and sex – it is repeated 3 times and vertically stacked
Tells Mx to weight each of the means and var/cov matrices by the IBD prob. which we placed in the B matrix

31. Your job Pick a number location that you ran with the pihat script
Change the script to run linkage at the location you picked
Test for linkage

33. Summary Weighted likelihood approach more powerful than pi-hat
Quickly becomes unfeasible
3 models for sibship size 2
27 models for sibship size 4
Q: How many models for sibship size 3?
For larger sib-ships/arbitrary pedigrees pi-hat approach is method of choice

34. Break – and then association

35. Association
Introduction

36. Association
Simplest design possible
Correlate phenotype with genotype

37. The equivalent for a quantitative trait - run a regression
Yi = a + bXi + ei
where
Yi = trait value for individual i
Xi = 1 if allele individual i has allele ‘A’
0 otherwise
i.e., test of mean differences between ‘A’ and ‘not-A’ individuals
Play with Association.xls 1

38. Biometrical model Genotype Genetic Value BB Bb bb a d -a
midpoint Bb BB
Genotype Genetic Value BB Bb bb a d -a
Va (QTL) = 2pqa2 (no dominance)

39. How can we do this in Mx?

40. Where this goes wrong… Chose a phenotype – chopstick use
Collect 2 groups - cases and controls
Unrelated individuals – university students
Matched for relevant covariates – age sex 1

41. Where this goes wrong…
Pick your candidate gene and type it – SUSHI (Successful use of selected hand instruments gene)
Count the number of cases and controls with each genotype – highly significant and replicatable 1

42. Obviously this is a spurious example
The cases were of Asian decent, the controls were of Caucasian decent
SUSHI – was a histocompatability antigen gene with different allele frequencies in different racial groups
This confound is known as population stratification
(Yes this is a fairy tale but it is based on a real asthma study)

43. Practical – Find a gene for sensation seeking:
Two populations (A & B) of 100 individuals in which sensation seeking was measured
In population A, gene X (alleles 1 & 2) does not influence sensation seeking
In population B, gene X (alleles 1 & 2) does not influence sensation seeking
Mean sensation seeking score of population A is 90
Mean sensation seeking score of population B is 110
Frequencies of allele 1 & 2 in population A are .1 & .9
Frequencies of allele 1 & 2 in population B are .5 & .5

44. Population B scores higher than population A
Genotypic freq.
Sensation seeking score is the same across genotypes, within each population.
Population B scores higher than population A
Differences in genotypic frequencies

45. Suppose we are unaware of these two populations and have measured 200 individuals and typed gene X
The mean sensation seeking score of this mixed population is 100
What are our observed genotypic frequencies and means?

46. Calculating genotypic frequencies in the mixed population
Genotype 11:
1 individual from population A, 25 individuals from population B on a total of 200 individuals: (1+25)/200=.13
Genotype 12: (18+50)/200=.34
Genotype 22: (81+25)/200=.53

47. Calculating genotypic means in the mixed population
Genotype 11:
1 individual from population A with a mean of 90, 25 individuals from population B with a mean of 110 = ((1*90) + (25*110))/26 =109.2
Genotype 12: ((18*90) + (50*110))/68 = 104.7
Genotype 22: ((81*90) + (25*110))/106 = 94.7

48. Gene X is the gene for sensation seeking!
Genotypic freq.
Now, allele 1 is associated with higher sensation seeking scores, while in both populations A and B, the gene was not associated with sensation seeking scores…
FALSE ASSOCIATION

49. False positives and false negatives
Reversal effects
Overestimation
Underestimation -3 -2 -1 1 2 3 4 5
0.49
0.42
0.35
0.28
0.21
0.14
0.07
0.00
-0.07
-0.14
-0.21
-0.28
-0.35
-0.42
-0.49
Estimated value of allelic effect
=-10
=-5
=5
=10
Genuine allelic
effect=+2
Difference in gene frequency in subpopulations
Dm =
Difference in subpopulation mean
Posthuma et al., Behav Genet, 2004

50. How to avoid spurious association?
True association is detected in people coming from the same genetic stratum
Can check that individuals come from the same population using a large set of highly polymorphic genes – genomic control
Can use family members as controls – family based association

51. Biometrical model Genotype Genetic Value BB Bb bb a d -a 2a d bb Bb BB
midpoint Bb BB
Genotype Genetic Value BB Bb bb a d -a

52. Fulker (1999) between/within association model
Fulker et al, AJHG, 1999

55. BTW – this is on top of a linkage model

56. So… 4 tests for the price of one Test for pop-stratification aw=ab
Robust test for association aw=0
Test so see if linked loci is the functional variant QTL≠ 0 in the presence of aw it is in LD with the variant but is not the casual variant
Test for dominance effects if dominance is also modeled

57. Combined Linkage & association Implemented in QTDT (Abecasis et al
Combined Linkage & association Implemented in QTDT (Abecasis et al., 2000) and Mx (Posthuma et al., 2004)
Association and Linkage modeled simultaneously:
Association is modeled in the means
Linkage is modeled in the (co)variances
QTDT: simple, quick, straightforward, but not so flexible in terms of models
Mx: less simple, but highly flexible

58. Implementation in Mx link_assoc.mx
#define n 3 ! number of alleles is 3, coded 1, 2, 3 …
G1: calculation group between and within effects
Data Calc
Begin matrices;
A Full 1 n free ! additive allelic effects within
C Full 1 n free ! additive allelic effects between
D Sdiag n n free ! dominance deviations within
F Sdiag n n free ! dominance deviations between
I Unit 1 n ! one's
End matrices;
Specify A
Specify C
Specify D
Specify F
The locus has 3 alleles
These 1*3 vector matrices contain the b/n & w/n additive effects of each of the 3 alleles
These 3*3 off-diagonal matrices contain the b/n & w/n dominance effects of each of the 3 alleles

59. Sticking it together… Begin algebra;
K = + ; ! Within effects, additive
L = D + D' ; ! Within effects, dominance
W = K+L ; ! Within effects - additive and dominance in one matrix
M = + ; ! Between effects, additive
N = F + F' ; ! Between effects, dominance
B = M+N ; ! Between effects - additive and dominance in one matrix
End algebra ;

60. K = (A'@I) + (A@I') ; A matrix contains additive w/n effects
Allele 1
Allele 2
Allele 3 a1
100 a2
101 a3
102
Parameter numbers
I matrix is unit 1*3 so … 1|1|1

61. K = + ; = a1 a2 a3

62. K = + ; =
2*a1
a1+a2
a1+a3
2*a2
a2+a3
2*a3

63. L = D + D' ;
This is the deviation of the 1/2 genotype mean from the simple additive effects of allele 1 + allele 2
D+D’ =
d12
d13
d23

64. W = K+L ; K+L = 2*a1 a1+a2+ d12 a1+a3+ d13 2*a2 a2+a3+ d23 2*a3
This is the 1/2 genotype mean composed of the simple additive effects of allele 1 + allele 2 and any deviation from these simple additive effects (dominance effects)
K+L =
2*a1
a1+a2+ d12
a1+a3+ d13
2*a2
a2+a3+ d23
2*a3

65. W = K+L ; K+L (parameter numbers) =
Between effects stick together in the same way

66. M = (C'@I) + (C@I') ; ! Between effects, additive
K = + ; ! Within effects, additive
L = D + D' ; ! Within effects, dominance
W = K+L ; ! Within effects total
I = [ 1 1 1], A = [a1 a2 a3]
D = d
d31 d32 0
K = + =
a1 1
[1 1 1] + [a1 a2 1 =
a3 1
a1 a1 a1 a1 a2 a a1a1 a1a2 a1a3
a2 a2 a2 + a1 a2 a3 = a2a1 a2a2 a2a3
a3 a3 a3 a1 a2 a a3a1 a3a2 a3a3
L = D + D' =
d21 d d21 d31
d d32 = d d32
d31 d d31 d32 0
W = K+L =
a1a1 a1a2 a1a d21 d a1a a1a2d21 a1a3d31
a2a1 a2a2 a2a3 + d d32 = a2a1d21 a2a a2a3d32
a3a1 a3a2 a3a3 d31 d a3a1d31 a3a2d32 a3a3
M = + ; ! Between effects, additive
N = F + F' ; ! Between effects, dominance
B = M+N ; ! Between effects - total

67. We have a sibpair with genotypes 1,1 and 1,2.
B = c1c c1c2f21 c1c3f31
c2c1f21 c2c c2c3f32
c3c1f31 c3c2f32 c3c3
W = a1a a1a2d21 a1a3d31
a2a1d21 a2a a2a3d32
a3a1d31 a3a2d32 a3a3
We have a sibpair with genotypes 1,1 and 1,2.
μ1 = Grand mean + pair mean + ½ pair difference
μ2 = Grand mean + pair mean - ½ pair difference
To calculate the between-pairs effect, or the mean genotypic effect of this pair, we need matrix B: ((c1c1) + (c1c2f21)) / 2
To calculate the within-pair effect we need matrix W and the between pairs effect:
For sib1: (a1a1) + ((c1c1) + (c1c2f21)) / 2
For sib2: (a1a2d21) - ((c1c1) + (c1c2f21)) / 2

68. G3: datagroup: sibship size two, DZ … Definition_variables tw1a1 tw1a2
Begin Matrices;
G Full 1 nvar = G2 ! grand mean
B Computed n n = B1 ! spurious and genuine genotypic effects (between)
W Computed n n = W1 ! genuine genotypic effects (within)
K Full 1 4 Fix ! Will contain first and second allele of twin1
L Full 1 4 Fix ! Will contain first and second allele of twin2
S Full 1 1 Fix ! Will contain 2 (for two individuals per family)
End Matrices;
Matrix K
Matrix L
Specify K tw1a1 tw1a2 tw1a1 tw1a2
Specify L tw2a1 tw2a2 tw2a1 tw2a2
Alleles 1 and 2 for twins 1 and 2 respectively
We are going to put the allele numbers here
These matrices must be initialized –given default values
Alleles 1 and 2 for twins 1 and 2 respectively

69. Sticking it together using part
For sibpairs 1,1 and 1,2
To calculate the between-pairs effect, or the mean genotypic effect of this pair, we need matrix B: ((c1c1) + (c1c2f21)) / 2
We can use the part function to draw out a specified element of a matrix
B = c1c c1c2f21 c1c3f31
c2c1f21 c2c c2c3f32
c3c1f31 c3c2f32 c3c3

70. Sticking it together using part
We can use the part function to draw out a specified element of a matrix
So to draw out c3c1f31 we could say:
\part(B, K)
where K is a matrix containing
B = c1c c1c2f21 c1c3f31
c2c1f21 c2c c2c3f32
c3c1f31 c3c2f32 c3c3

71. Sibpair with genotypes: 1,1 and 1,2
W = a1a a1a2d21 a1a3d31
a2a1d21 a2a a2a3d32
a3a1d31 a3a2d32 a3a3
B = c1c c1c2f21 c1c3f31
c2c1f21 c2c c2c3f32
c3c1f31 c3c2f32 c3c3
Sibpair with genotypes: 1,1 and 1,2
Specify K tw1a1 tw1a2 tw1a1 tw1a2 =
Specify L tw2a1 tw2a2 tw2a1 tw2a2 =
V = (\part(B,K) + \part(B,L) ) %S ; (c1c1 + c1c2f21)/2
C = (\part(W,K) + \part(W,L) ) %S ; (a1a1 + a1a2d21)/2
Means G + F*R '+ V + (\part(W,K)-C) | G + I*R' + V +(\part(W,L)-C); =
G + F*R’ + (c1c1 + c1c1f21)/2 + (a1a1 - (a1a1 + a1a2d21)/2) |
G + I*R' + (c1c1 + c1c1f21)/2 + (a2a1 - (a1a1 + a1a2d21)/2)

72. Constrain sum additive allelic within effects = 0
Constraint ni=1
Begin Matrices;
A full 1 n = A1
O zero 1 1
End Matrices;
Begin algebra;
B = \sum(A) ;
End Algebra;
Constraint O = B ;
end
Constrain sum additive allelic between effects = 0
C full 1 n = C1 !
B = \sum(C) ;

73. Variance/covariance matrix
AEQ model using pi-hat formulation
So 4 tests for the price of one
Test for pop-stratification aw=ab
Robust test for association aw=0
Test so see if linked loci is the functional variant
QTL≠ 0 in the presence of aw it is in LD with the variant but is not the casual variant
Test for dominance effects if dominance is also modeled

74. !1.test for linkage in presence of full association
Drop D 2 1 1
end
!2.Test for population stratification:
!between effects = within effects.
Specify 1 A
Specify 1 C
Specify 1 D
Specify 1 F
!3.Test for presence of dominance
!4.Test for presence of full association
!5.Test for linkage in absence of association
Free D 2 1 1

75. Model
Test
-2ll df
Vs model
Chi^2
Df-diff
P-value - 1
Linkage in presence of association 2
B=W 3
Dominance 4
Full association 5
Linkage in absence of association