1. Heritability, prediction and the genomic
architecture of complex traits
David Balding and Doug Speed
EMBO Practical Course on Genotype to Phenotype Mapping of Complex Traits
Hinxton, July 30, 2014
Any questions,

2. Datafiles data.bed, data.bim, data.fam –
contain genotypic data for 1500 individuals and 60k SNPs (chr 1, 6 & 10)
phen.pheno – contains three phenotypes
sex.covar – contains gender
mlist1, mlist2 – lists of kinship files
genelist.txt – gene annotations
weights.txt – SNP weightings
train.fam – (randomly drawn) list of individuals used to train the prediction model
test.fam – list of individuals used to test the prediction model
Note, all arguments are preceded by two dashes (even if it appears as one long dash!)

3. Why use allelic correlations A = XXT/N?
The linear mixed model with A=XXT/N is equivalent to a random effects regression model:
Y = α
+ β1 X β2 X β3 X β4 X β5 X β6 X β7 X β8 X8
+ β9 X β10X β11X β12X β13X β14X β15X β16X16
+ β17X17 + β18X β19X β20X β21X β22X β23X β24X24
β60,000X60,000
+ e
Suppose βj ~ N ( 0, σg2 / N ) and ei ~ N ( 0, σe2 )
Then g = Σ Xj βj ~ N ( 0, Aσa2 ) where A= XXT / N
We obtain the mixed model: Y = α + g + e ~ N( α, A σg2 + I σe2 )
Variance explained by all SNPs is σa2 / (σa2 + σe2 )

4. Using GCTA (Genome-wide Complex Trait Analysis)
1 – calculate similarity matrices separately across Chromosomes 1, 6 and 10
[because XXT = X1X1T + X6X6T + X10X10T]
gcta –bfile data –make-grm –chr 1 –out chr1
gcta –bfile data –make-grm –chr 6 –out chr6
gcta –bfile data –make-grm –chr 10 –out chr10
Other options: –extract, –maf or –keep, or for dosage data –dosage-mach
2 – join these three similarity matrices together
[mlist1.txt contains the prefixes of the matrices]
gcta –mgrm mlist1 –make-grm –out chrALL
Elements of chrALL.grm.bin reflect “genome-wide” similarity for pairs of individuals
GRM is stored in binary format, but can view it using
gcta –mgrm mlist1 –make-grm-gz –out chrALL
less –S chrALL.grm.gz
gunzip –c chrALL.grm.gz | awk ‘{if($1!=$2 && $4>0.25) print $0}’
E.g. individual pairs 314 & 236 have kinship 0.404, while 433 & 264 have kinship 0.999

5. Using GCTA (Genome-wide Complex Trait Analysis)
3 – identify and remove closely related individuals (lose 18 individuals)
gcta –bfile data –grm chrALL –grm-cutoff 0.05 –make-grm –out filter
4 – perform PCA (used to identify population outliers which should be removed)
gcta –bfile data –grm chrALL –keep filter.grm.id –pca 2 –out filter

6. Using GCTA (Genome-wide Complex Trait Analysis)
5 – estimate total variance explained by SNPs for Phenotype 1
gcta –reml –grm chrALL –pheno phen.pheno –mpheno 1 --keep filtered.grm.id –out p1
Could add covariates with option –covar
Find that total variance explained is 93%
6a – estimate total variance explained by SNPs for Phenotype 3 (binary trait)
gcta –reml –grm chrALL –pheno phen.pheno –mpheno 1 --keep filtered.grm.id –out p3
Find that estimate on the observed scale is 63%
6b – to convert to the liability scale provide the prevalence (assumed to be 1 in 100)
gcta –reml –grm chrALL–pheno phen.pheno –mpheno 3 --keep filtered.grm.id –out binary –prevalence 0.01
On the liability scale the variance explained is 36%

7. Genome partitioning
Y = α
+ β1 X β2 X β3 X β4 X β5 X β6 X β7 X β8 X8
+ β9 X β10X β11X β12X β13X β14X β15X β16X16
+ β17X17 + β18X β19X β20X β21X β22X β23X β24X24
+ β25X25 + β26X β27X β28X β29X β30X β31X β32X32
+ e
Suppose: β1, β2, β3, ..., β16 ~ N ( 0, σa12/ N ) and β17, β18, β19, ..., β32 ~ N ( 0, σa22/N )
Now we have the mixed model: Y = α + g + e ~ N( α, A1 σa12 + A2 σa22 + I σe2 )
where: A1 = X1X1T / N1 and A2 = X2X2T / N2
Variance explained by first 16 SNPs is σa12 / ( σa12 + σa22 + σe2 )
by last 16 SNPs is σa22 / ( σa12 + σa22 + σe2 )

8. a) height, b) BMI, c) vWF, d) QTi
Genome partitioning
Genome partitioning has been used to estimate the proportion of a trait’s phenotypic variance explained by individual chromosomes
a) height, b) BMI, c) vWF, d) QTi
Used to demonstrate traits are polygenic

9. Testing for inflation due to cryptic relatedness
Divide the SNPs X into left half X1 and right half X2
(e.g. X1 contains SNPs in Chr 1-8; X2 contains those in Chr 9-22)
Estimate total h2 mixed model with K = XXT / N
Estimate left half h2L mixed model with K1 = X1X1T / N1
Estimate right half h2R mixed model with K2 = X2X2T / N2
If cryptic relatedness is not inflating estimates, then would expect h2 = h2L + h2R
In which case, estimates of h2L + h2R from the joint model Y ~ N ( α, K σg2 + K σg2 + I σe2 )
should match those from individual models
Y ~ N ( α, K σg2 + I σe2 ) and Y ~ N( α, K σg2 + I σe2 ) [5]
h2L
h2R
h2R

10. Genome partitioning
7a – estimate variance explained for Phenotype 1 by each chromosome jointly
gcta –reml –mgrm mlist1 –pheno phen.pheno –mpheno 1 --keep filtered.grm.id –out gp
Get breakdown as 47%, 23% & 24% (sum to 95%, slightly higher than combined)
7b – can compare this with variance explained by each chromosome separately
gcta –reml –grm chr1 –pheno phen.pheno –mpheno 1 --keep filtered.grm.id –out chr1
gcta –reml –grm chr6 –pheno phen.pheno –mpheno 1 --keep filtered.grm.id –out chr6
gcta –reml –grm chr10 –pheno phen.pheno –mpheno 1 --keep filtered.grm.id –out chr10
Estimates are 45%, 22% & 27% (sum to 93%)
The only slight difference between sums (93% and 95%) indicates only a mild contribution from inflation due to residual relatedness and population structure (a useful check).

11. Bivariate analysis
SNP-based heritability analysis can be used to assess the concordance between traits
Y = α β1 X β2 X β3 X β4 X β5 X β6 X e1
Y = α λ1 X λ 2 X λ 3 X λ 4 X λ 5 X5 + λ 6 X e2
The individual analyses suppose βj ~ N ( 0, σg12 / N) and λj ~ N ( 0, σg22/N )
But we can additionally specify cor( βj , λj ) = τ τ takes value in [-1, 1]
Similar to univariate analysis, except now we analyse Traits 1 and 2 jointly
using a generalized version of REML which estimates σ2g1, σg22 and τ [9]
For different traits, interested in testing whether τ is significantly non-zero (pleiotropy)
Can also test whether for sub-phenotypes τ is significantly less than one (heterogeneity)

12. Bivariate analysis
Mainly used to test concordance between traits (pleiotropy)
e.g. demonstrate overlap between
Bipolar Disorder and schizophrenia
But can also be used to test concordance within traits
e.g., how similar are focal and
non-focal epilepsy?

13. Bivariate analysis 8a – bivariate analysis between Phenotypes 1 and 2
gcta –reml-bivar 1 2 –grm both –pheno phen.pheno --keep filtered.grm.id –out bivar
Estimate the correlation between Traits 1 and 2 to be 53%
8b – bivariate analysis – test whether significantly different to 0 (pleiotropy)
gcta –reml-bivar 1 2 –grm both –pheno phen.pheno --keep filtered.grm.id –out bivar0 –reml-bivar-lrt-rg 0
The correlation is significantly greater than 0 (P < 1e-16, chi squared test stat 527)
8c – bivariate analysis – test whether significantly lower than 1 (heterogeneity)
gcta –reml-bivar 1 2 –grm both –pheno phen.pheno --keep filtered.grm.id –out bivar1 –reml-bivar-lrt-rg 1
The correlation is significantly lower than 1 (P < 1e-16, chi squared test stat 82)

14. Using LDAK (Linkage Disequilibrium Adjusted Kinships)
The aim of the SNP weightings is to equalise the signal variance across the genome.
If SNPs are weighted equally, then signal in high LD (well tagged) regions will be overrepresented when calculating similarity, and vice versa.
E.g., suppose two SNPs are identical, then could give both weighting 0.5.
In practice, many SNPs get weighting zero, which means that their signal is completely explained by the SNPs which remain.

15. Using LDAK (Linkage Disequilibrium Adjusted Kinships)
1 – calculate SNP weightings (will not carry out these during the practical)
ldak3 –cut-weights sections –bfile data
ldak3 –calc-weights sections –bfile data –section 1
...
ldak3 –calc-weights sections –bfile data –section 21
ldak3 –join-weights sections
Weightings (prepared earlier) are stored in weights.txt
For computational reasons, it is necessary to divide the genome into sections (default size 3000 SNPs).
For sparse genotyping, 3000 SNPs ~ 30Mb, which is sufficiently large
For dense (sequence / imputed) data, 3000 SNPs ~ 1Mbp, so we suggest calculating weightings “twice”, the second time using only SNPs with non-zero weights from the first run. This will increase the window size about 10-fold.

16. Using LDAK (Linkage Disequilibrium Adjusted Kinships)
2 – calculate kinships
ldak3 –calc-kins partitions –bfile data –by-chr YES
ldak3 –calc-kins partitions –bfile data –weights weights.txt –partition 1
ldak3 –calc-kins partitions –bfile data –weights weights.txt –partition 2
ldak3 –calc-kins partitions –bfile data –weights weights.txt –partition 3
ldak3 –join-kins partitions –kinship-matrix YES
Other options: --min-maf, --min-var, --max-maf and --min-obs
To facilitate genome partitioning, can provide lists of SNPs
Can examine kinships using
less –S partitions/kinshipALL.grm.raw

17. Without weightings After weightings
Using LDAK (Linkage Disequilibrium Adjusted Kinships)
3 – decompose kinship
ldak3 –decompose partitions/kinshipALL.eigen –grm partitions/kinshipALL –keep filter.grm.id
Will save eigenvalues and eigenvectors in partitions/kinshipALL.eigen
Without weightings After weightings

18. Without weightings After weightings
Using LDAK (Linkage Disequilibrium Adjusted Kinships)
4a – estimate total variance explained by SNPs for Phenotype 1
ldak3 –reml outp1 –grm partitions/kinshipALL –pheno phen.pheno –mpheno 1 –keep filter.grm.id
As before, could add covariates with option –covar
Find that total variance explained is 88% (compared to 93% without weightings)
4b – estimate total variance explained by SNPs for Phenotype 3
ldak3 –reml outp3 –grm partitions/kinshipALL –pheno phen.pheno –mpheno 3 –keep filter.grm.id
Conversion from observed to liability scale not yet implemented, sorry
Without weightings After weightings

19. Using LDAK (Linkage Disequilibrium Adjusted Kinships)
5a – genome partitioning for Phenotype 1
ldak3 –reml outgp –mgrm mlist2 –pheno phen.pheno –mpheno 1 –keep filter.grm.id
The breakdown is 41%, 18% and 27% (sum to 86%)
5b – estimate total variance explained by SNPs for Phenotype 3
ldak3 –reml outchr1 –grm partitions/kinship1 –pheno phen.pheno –mpheno 1 –keep filter.grm.id
ldak3 –reml outchr6 –grm partitions/kinship2 –pheno phen.pheno –mpheno 1 –keep filter.grm.id
ldak3 –reml outchr10 –grm partitions/kinship3 –pheno phen.pheno –mpheno 1 –keep filter.grm.id
Individual estimates are 44%, 18%, 32% (sum to 94%)

20. Gene-based / regional tests
Y = α
+ β1 X β2 X β3 X β4 X β5 X β6 X β7 X β8 X8
+ β9 X β10X β11X β12X β13X β14X β15X β16X16
+ β17X17 + β18X β19X β20X β21X β22X β23X β24X24
β500,000X500,000
+ e
Can focus in on very small regions or individual genes (tens of SNPs)
Suppose we want to compute variance explained by SNPs 14 to 18
Naive approach is to compute K from these SNPs and perform REML
But because N=5 < # individuals, fast computational tricks can be used (FastLMM)

21. Gene-based analysis 6a – divide genome into genes
ldak3 –cut-genes genes –bfile data –genefile genelist.txt –weights weights.txt –gene-buffer 10000
3356 of the 4580 genes are found (have at least one non-zero weight SNP)
Can split the genes into partitions, and analyse each in parallel, but typically not necessary
6b – test each gene for association with Phenotype 3
ldak3 –calc-genes-reml genes –bfile data –weights weights.txt –pheno phen.pheno –mpheno 3 –keep filter.grm.id –covar sex.covar –partition 1
6c – join the partitions (here, only one partition)
ldak3 –join-genes-reml genes

22. Heritability LRT Pvalue Score Pvalue
Gene-based analysis
Heritability LRT Pvalue Score Pvalue
LRT Pvalue can be conservative, although suffices for our purposes
Listgarten and Lippert implement a permutation-based p-value

23. Chunk-based analysis Restricting only to genes can miss a lot!
7a – divide into 75,000 base pair chunks
ldak3 –cut-genes chunks –bfile data –chunks-bp weights weights.txt
Can also divide according to weighting using --chunks
7b – now test each chunk
ldak3 –calc-genes-reml chunks –bfile data –weights weights.txt –pheno phen.pheno –mpheno 3 –keep filter.grm.id –covar sex.covar –partition 1

24. BLUP Standard BLUP using a single (genetic) random effect
8a – perform REML analysis (this time restricted to training individuals)
ldak3 –reml blupALL –grm partitions/kinshipALL –pheno phen.pheno ---mpheno 3 –keep train.fam –covar sex.covar
8b – now get the blup estimates of effect sizes and project onto these
ldak3 –calc-blups blupALL –remlfile blupALL.reml –grm partitions/kinshipALL –bfile data
blupALL.blup contains effect sizes blupALL.pred contains predictions
Can measure prediction performance
by correlation between predicted and observed phenotypic values
With standard BLUP performance is 2%

25. MultiBLUP Standard BLUP model:
Y = α + g + e g ~ N(0, σg2A) e ~ N(0,σe2I)
MultiBLUP model
Y = α + g1 + g gM + e gm ~ N(0, σm2Am) e ~ N(0,σe2I)
Each similarity matrix Am corresponds to a subset of SNPs
Prediction improved when subset of SNPs have distinct effect size variances
Performance of MultiBLUP depends on how well SNP subsets are chosen
We provide an algorithm for finding suitable SNP subsets

26. Adaptive MultiBLUP
Step 1: Divide genome into (say) 75kb overlapping chunks (--chunks-bp)
Step 2: Test each chunk for association (--calc-genes-reml)
Step 3: Identify all significant chunks (say, P< )
Create regions by merging these with neighbouring chunks with P<0.001
( join-genes-reml –sig1 1e-6 –sig )
Step 4: Run MultiBLUP using these local regions plus the background region
(--reml with –grm, –region-prefix and –region-number)

27. Adaptive MultiBLUP 9a - Collect results of 7a
ldak3 --join-genes-reml chunks –bfile data –sig1 1e-6 –sig2 1e-3
This creates six regions, chunks/region1, ... , chunks/region6

28. Adaptive MultiBLUP
9b – Run MultiBLUP using these 6 regions and the background kinship matrix
ldak3 –reml outmb –grm partitions/kinshipALL –region-prefix chunks/region –region-number 6 –pheno phen.pheno –mpheno 3 –keep train.fam –weights weights.txt –bfile data
9c – compute effect size estimates and predictions
ldak3 –calc-blups outmb –remlfile outmb.reml –bfile data –grm partitions/kinshipALL
Prediction improves
From 2% to 25%