Missing Heritability

1
Instructions for use
ICAR-Indian Agricultural Statistical Research
Institute
Missing Heritability
Shashank kshandakar Ph.D.(Agricultural
Statistics), Roll No.10684
2
Contents

• Introduction
• Dissolving the of Missing Heritability Problem
• Estimation of Heritability from Common Variants
• HasemanElston Regression Method
• Heritability from Case-control Study
• Liability Threshold Model
• Illustration
• Conclusions
• References

3
Introduction

• Heritability is a key genetic parameter that can
  help to understand the genetic architecture of
  complex traits.
• Heritability (in narrow sense) defined as the
  proportion of total phenotypic variation that is
  due to additive genetic factors
• In GWAS, statistical significance variants
  explain only a small fraction of the heritability
  and the heritability estimates obtained in GWAS
  are much lower than those of traditional
  quantitative methods.

4
Introduction

• Estimation and knowledge of missing heritability
  is important because disease susceptibility is
  known to be due to genetic factors, and
  understanding this genetic variation may
  contribute to better prevention, diagnosis and
  treatment of disease.
• Knowledge of Missing heritability help in
  planning of research strategies to uncover the
  genetic risk factors (Maher, 2008)

5
Heritability and number of loci for several traits (Manolio et al.? 2009)  Heritability and number of loci for several traits (Manolio et al.? 2009)  Heritability and number of loci for several traits (Manolio et al.? 2009) 
Disease Number of loci Proportion of heritability explained (GWASs)
ARMD 5 50
Crohns disease 32 20
Fasting glucose 4 1.5
HDL cholesterol 7 5.2
Height 40 5
Myocardial infarction 9 2.8
Type 2 diabetes 18 6
6
Probable Reasons for Missing heritability

• Nadeau (2009), pointing that the missing
  heritability is due to epigenetic factors
• Epigenetics is the study of heritable changes in
  gene expression that are not caused by changes in
  DNA sequence
• Epigenetic modifications are influenced by the
  environmental factors, such as smoking, stress,
  and nutrients
• Trans-generational epigenetic inheritance
  that contribute to disease risk would not be
  detectable in GWAS but may contribute to average
  risk and to similarities among relatives.
  (Richards, 2006)

7
Probable Reasons for Missing heritability

• Manolio et al. (2009) termed Missing Heritability
  as a dark matter of GWAS, dark matter in the
  sense that one is sure it exists, can detect its
  influence, but simply cannot see it (yet)
• Kong (2009), suggest that the rare variants
  contribute to missing heritability in two ways
  first, they are more difficult to discover and
  second, even if discovered, their contribution to
  heritability would be underestimated when
  evaluated under models that do not take parental
  origin into account

8
Probable Reasons for Missing heritability

• Eichler et al. (2010), explore the impact of
  large variants (deletions, duplications and
  inversions) that are individually rare but
  collectively common in the population
• Rare variants that can potentially affect many
  genes and various biological pathways in an
  individual organism are inaccessible by most
  existing genotyping and sequencing technologies
  (Zuk et al. 2014)

9
Dissolving the Missing Heritability Problem by
Traditional Method

• Heritability estimation from family study
• Twin study
• Parents-offspring regression
• Heritability estimation from GWAS study

10
Estimation of Missing Heritability

• In quantitative genetics, we assume that there is
  absence of gene-environment interaction and
  correlation (Falconer et al. 1996)
• VP VG VE VG VA VD VI
• H2 VG / VP h2 VA / VP
• where VP is the phenotypic variance of a
  population
• VG is the genotypic variance
• VE is the environmental variance
• VA is the additive genetic variance
• VD is the dominance variance
• VI is the epistasis variance

11
Estimation

• Cov (P1, P2) Cov (A1D1I1E1, A2D2I2E2)
• Cov (A1, A2) Cov (D1, D2)
  Cov (I1, I2) Cov (E1, E2)
• Twin studies is the most commonly used
  traditional methods for estimating heritability
  (Pierrick et al. 2016).
• Twin studies are a special type of
  epidemiological studies designed to measure the
  contribution of gene as opposed to the
  environment, for a given trait.
• Monozygotic and dizygotic twins share almost 100
  and 50 of their genetic material respectively.

12
Estimation

• The environment is typically divided into
• Shared Environment (C) - the part of the
  environment that affects both twins in the same
  way
• Unique Environment (U) - the part of the
  environment that affects one twin but not the
  other
• (Silventoinen et al. 2003)
• In the absence of interaction and correlation
  between C and U, we have
• E C U

13
Estimation

• Assuming epistasis effects to be negligible
  (assumption in twin studies), then
• Cov (PT1, PT2) Cov (AT1DT1CT1UT1 ,
  AT2DT2CT2UT2)
• Cov (AT1, AT2) Cov (DT1, DT2) Cov (CT1, CT2)
  Cov (UT1, UT2)
• where indexes T1 and T2 represent the two
  twins for each twin pair studied
• Cov (UT1, UT2) is zero for both monozygotic and
  dizygotic twins as each twins unique environment
  by definition is independent of that of the other
  twin.

14
Estimation

• Variance is a special case of covariance when the
  two variables are identical, and that for
  monozygotic twins AT1, DT1, and CT1 equal to AT2,
  DT2, and CT2 respectively, then
• CovMT (PT1, PT2) VAVDVC
  CovDT (PT1, PT2) 1 2 VA 1 4 VD
  VC
• ?? ???? 2 2 ?????? ???? ?? ??1, ??
  ??2 -2 ( ?????? ???? ?? ??1, ?? ??2 ??
  ?? ?? ?? ?? ?? 3 2 ?? ?? ?? ??
• where ?? ???? 2 is the broad-sense heritability
  from twin studies, because the resulting estimate
  provides an accurate estimate of neither H2 nor
  h2, although it is closer to H2 than to h2
  (Falconer and Mackay, 1996).

15
Estimation

• The covariance between the traits of parents (one
  or the mean of both) and the mean of their
  offspring (Falconer and Mackay 1996)
• Cov (PP, PO) Cov (AP DP IP EP, AO DO
  IO EO)
• Cov(AP, AO) Cov(DP, DO)
  Cov(IP, IO) Cov(EP, EO)
• Doolittle, 2012 assumes that Cov (DP, DO) and Cov
  (EP, EO) are zero
• Environments experienced by individuals are
  likely to be more similar within a family line,
  so Cov(EP,EO) might be some value (Guo et al.
  2014)

16
Estimation

• Covariance of the parents and their offspring is
  equal to half of additive genetic variance, and a
  variance term representing effects due to
  dominance and similarities between environments
• Cov (PP,PO) Cov (AP,AO) Cov (DP,DO) Cov
  (EP,EO) 1 2 VA VDEC
• ?? ???? 2 2 ?????? ( ?? ??, ?? ??
  ) ?? ?? ?? ?? ?? ?? ?? ??????
  ?? ??
• Heritability estimates in both twin studies and
  parent-offspring regression include an extra term
  when compared to h2, but they do not correspond
  to H²
• ?? 2 h 2 h ????h???? 2
• where h ????h???? 2 is the part of heritability
  contributed by the extra component(s)
  representing non-additive variance.

17
Estimation

• Some epigenetic factors can lead to additive
  genetic effects (Pierrick et al. 2017), the
  additive variance of them ( ?? ?? ?????? )
  should be added to the additive variance of DNA
  sequences ( ?? ?? ?????? ) to obtain VA,
  assuming there is no interaction between ?? ??
  ?????? and ?? ?? ?????? then,
• VA ?? ?? ?????? ?? ?? ??????
• h2 ?? ?? ?????? ?? ?? ?? ??
  ?????? ?? ??
• h ?????? 2 h ?????? 2
• h ?????? 2 h2 - h ?????? 2

18
Estimation

• Missing heritability (MH) equals to difference
  between the estimates obtained by traditional
  quantitative methods (H2) and the estimates
  obtained by GWAS ( h ?????? 2 ). Thus,
• MH H2 - h ?????? 2
• h ????h???? 2 h ?????? 2
• Missing heritability results from the part of
  heritability originating from epigenetic factors
  stably transmitted across generations, plus the
  part of heritability originatingfrom
  non-additives factors.

19
Estimation of Heritability from Common Variants
20
Estimation of heritability from common variants

• SNPs identified by GWAS explain only a small
  fraction of the heritability
• Genome-wide complex trait analysis (Yang et al.
  2011) estimates the variance explained by all the
  SNPs to solve the missing heritability problem.
• The basic concept behind GCTA, is to fit the
  effects of all the SNPs as a random effects by
  using linear mixed model (Hayes et al. 2009) and
  H-E regression method (Haseman et al. 1972)

21
HasemanElston regression method

• The unbiased estimator of ?? ?? 2 is provided
  by HasemanElston regression method
• Let ,Yj (x1j-x2j)2 be the squared pair
  difference for jth sib pair
• x1jµ g1j e1j x2j µ g2j e2j
• gij a, d, -a for BB, Bb and bb individuals
• ?? ?? (0, 1 2 ???? 1) is the proportion of
  gene IBD for jth sib pair
• Conditional expectation of the squared pair
  differences
• E(Yj ?? ?? ) (?? ?? 2 2?? ?? 2 ) -(2?? ??
  2 ) ?? ?? a ß ?? ??
• where a ?? ?? 2 2?? ?? 2 ß -2?? ?? 2
  ?? ?? 2 -ß/2

22
Estimation of heritability from common variants

• In GWAS the associations between individual SNPs
  and the trait are represented by following simple
  regression model
• yj µ xijai ej
• ejN(0, ?? ?? 2 )
  
• where yj is the phenotypic value of the jth
  individual µ is the general mean ai is the
  allele substitution effect of ith SNP xij is an
  indicator variable that takes a value of 0, 1 or
  2 if the genotype of the j th individual at ith
  SNP is bb, Bb or BB respectively and ej is the
  residual effect.

23
Estimation of heritability from common variants

• Let, m causal variants are genotyped, then the
  model is
• yj µ gj ej and gj ??1 ?? ?? ????
  ?? ??
• where gj is the total genetic effect of jth
  individual m is the number of causal loci ui is
  the additive effect of the ith causal variant
  zij is the design matrix allocating casual allele
  to trait
• -2pi / 2?? ?? (1-?? ?? )
  if the genotype of the jth
• zij
  individual at ith locus is qq
• (1-2pi) / 2?? ?? (1-?? ?? )
  Qq
• 2(1-pi ) / 2?? ?? (1-?? ?? )
  QQ

24
Estimation of heritability from common variants

• In matrix notation,
• y µ1 g e and g Zu
• variance-covariance matrix of y (the vector of
  observations) can be expressed as
• var(y) ZZ ?? ?? 2 I ?? ?? 2 ????' ??
  ?? 2 ?? I ?? ?? 2 G ?? ?? 2 I ?? ?? 2
• u N(0, I ?? ?? 2 ) gj N(0, ?? ?? 2 m ??
  ?? 2 )
• where ?? ?? 2 is the variance of causal
  (random) effects ?? ?? 2 is the variance of
  total additive genetic effects I is an n x n
  identity matrix, G is the genetic relationship
  matrix between pairs of individuals at causal
  loci.

25
Estimation of heritability from common variants

• The number and positions of the causal variants
  are exactly not known, so G matrix is not
  directly obtained
• The Genome-wide relationship matrix (A) is
  obtained from a genome-wide sample of SNPs
• A ?? ?? ' ?? wij ?? ????
  -???? ?? ?? ?? ?? (??- ?? ?? )
• where W is a standardized genotype matrix with
  the ijth element xij is the number of copies of
  the allele for the ith SNP of the jth individual
  and pi is the frequency of the allele

26
Estimation of heritability from common variants

• The Genome-wide relationship matrix (A) between
  individual j and k can be estimated by the
  following equation
• Ajk 1 ?? ??1 ?? ( ?? ???? -2?? ??
  )( ?? ???? -2?? ?? ) 2?? ?? (1-?? ?? )
  when j ? k
• 1 1 N i1 N ?? ???? 2 - 1
  2p i x ik 2?? ?? 2 ) 2p i (1-p i )
  when jk
• Gjk is not known, so to fit model and estimate
  the genetic variance ( ?? ?? 2 ), A is used
  i.e. estimate of relationship matrix based on the
  Genome-wide relationship matrix (A) .

27
Estimation of heritability from common variants

• Randomly sample 2N SNPs from all the SNPs across
  the genome and randomly split them into two
  groups (N SNPs in each group).
• Calculate Ajk using all the SNPs in the first
  group.
• Calculate Gjk using SNPs with MAF ? in the
  second group
• Regress Gjk on Ajk for j k (use Gjk - 1 and Ajk
  - 1 when j k). the regression coefficient is
• ß ?????? ( ?? ???? , ?? ???? ) ??????( ?? ????
  )
• Repeat the procedure using different numbers of
  SNPs
•  

28
Estimation of heritability from common variants

• Yjk (z1j-z2j)2 squared z-score difference
  between individual
• Gjk is not known we replace it by an estimate A
  jk such that
• E(Gjk A jk) A jk
• E(Yjk) E(a ßGjk) a ß A jk
• Yjk is plotted against the A jk i.e. regression
  of Yjk on A jk -2 ?? ?? 2
• ?? ?? 2 - ß/2
• The relationship at causal loci is predicted with
  error by the observed SNPs, and the error is c 
  1/N
• ?? 1- (?? 1 ?? ) ??????( ?? ???? )

29
Estimation of Heritability from Case-control
Study
30
Liability Threshold Model

• Liability describe all the genetic and
  environmental factors that contribute to the
  development of a multi-factorial disorder
• The level of liability at which we distinguish
  population into case or control is referred as
  the threshold level.

31
Liability Threshold Model

• Liability is best represented as a standard
  normal distribution curve as most individual who
  is affected or unaffected will possess some
  degree of liability
• li gi ei
• where li is an unknown liability of ith
  individual and a person is assumed to be a case
  if his liability exceeds a threshold t
• gi is a genetic random effect, which can be
  correlated across individuals
• ei is the environmental random effect, which is
  assumed to be independent of each other and of
  the genetic effects.

32
Estimation of heritability from case-control study

• The advantages of working on the scale of
  liability are that the, population parameters
  such as variance components and heritability are
  independent of prevalence
• l µ1N g e
• where lN(0, 1)and g N(0, ?? ?? 2 )
• Mean of the distribution of liability is zero (µ
  0) when there is no ascertainment
• The total phenotypic variance ( ?? ?? ?? ) on the
  scale of liability is per definition equal to 1
• The heritability on the liability scale is
• ?? ?? ?? ?? ?? ??

33
Estimation of heritability from case-control
study

• Applying the properties of truncated normal
  distributions, the mean liability is
• i E(ly1)z/K for case
  and
• i2 E(ly0) -z/(1-K)
  for control
• Squared mean liability
• E(l2y1)1it for case
• E(l2y0) 1i2t for
  control
• The covariance between y (unaffected/affected
  status) and l (liability) to describe the
  relationship between the phenotypes on the two
  scales
• Cov(y,l) E(y.l)- E(y)E(l) K1i (1-K)0i2 Ki
  z

34
Estimation of heritability from case-control
study

• The genetic value on the observed 01 risk scale
  for an individual (u), defined in Equation , as
• u c bg
• where c is a constant, The linear regression
  coefficient (b) that links the two scales is
  derived from the regression of the phenotype on
  the observed scale (y) on the additive genetic
  effect on the scale of liability (g)
• b cov(y, g) / ?? ?? 2 E(y.g) -
  E(y)E(g)/ h ?? 2
• Ki h ?? 2 / h ?? 2
• z
• u c bg c zg
• ?? ?? 2 var(zg) z2 ?? ?? 2

35
Estimation of heritability from case-control
study

• The proportion of the total variance of 01
  observations, which is the Bernoulli distribution
  variance K(1 - K) and can be written as
• h ?? 2 ?? ?? 2 /
  K(1-K) ?? ?? 2 cov (y, g)/ ?? ?? 2
  2/K(1-K)
• s g 2 b2/K(1-K) h l 2 z2/K(1-K)
• h l 2 h ?? 2 K(1-K)
  /z2
• The mean of the estimated genetic values is
• E( ?? y 1) zi s g 2 for case
• E( ?? y 0) z ?? 2 s g 2 for control

36
Selection probabilities

• When the study is observational, the probability
  of being included in the study is independent of
  the phenotype.
• In case-control study, the proportion of cases is
  usually greatly ascertained
• ?? ?? ???????? (1-??) ?? ?????????????? ??
  1-?? ?? ?????????????? ??(1-??)
  ??(1-??) ?? ????????
• where Pcase and Pcontrol are the probabilities
  that a case and a control would be selected for
  the study respectively
• K is the prevalence of a condition in the
  population
• P is the prevalence in condition the study

37
Non-normality of the liability

• When the proportions of cases and controls are
  not a random sample from the population.
• The mean and variance for case and control
  disease status (ycc), disease liability (lcc),
  and genetic liability (gcc) are
• E(ycc) P
• (usually, P 1/2)
• var(ycc) P (1-P) which is the phenotypic
  variance on the observed scale in the
  case-control sample
• where P is the proportion of cases in the
  case-control study sample

38
Non-normality of the liability

• E(lcc) Pi (1-P)i2 i(P-K)/(1-K)
• i? where ? is, (??-??)
  (1-??)
• var(lcc) ?? ?????? 2 E( ?? ???? 2 )
  E(lcc)2
• P(1 it )(1 - P)(1 i2t) - i2 ?? 2
  1Pit-(i-P)tik/(1-K) - i2 ?? 2
• 1i ?(t-i ?)
• 1?
• ? ??? ??-???
• var(lcc) gt 1, in a case-control study because
  individuals from the tails of the distribution of
  liability have been selected.

39
Non-normality of the liability

• The mean of genetic liability depends on the mean
  liability phenotype of case-control sample and
  the heritability of liability
• E(gcc) h l 2 E(lcc) h l 2 Pi (1-P)i2
• h l 2 i?
• Variance of genetic liability as
• var(gcc) s gcc 2 E( ?? ???? 2 ) E(gcc)2
• h l 2 E ?? ???? 2 h l 2 E(lcc)2
• h l 2 P(1it)(1-P)(1i2t)- h l 4 i2?2
  
• h l 2 1 h l 2
  ??

40
Non-normality of the liability

• The regression of phenotype on the observed risk
  scale on genetic liability in the case-control
  study
• bcc cov(ycc,gcc)/var(gcc)
• E(ycc.gcc)-E(ycc)E(gcc)/var(gc
  c)
• h ?? 2 iP- h ?? 2 i?/ ?? ?????? 2
  
  ??h ?? 2 i(1-?)/ ?? ?????? 2
• z ??(1-??) ?? ?? 2 ??(1-??) ??
  ?????? 2 ?? ? where ? ??(1-??) ?? ?? 2
  ??(1-??) ?? ?????? 2
• where, ? quantifies the change of the regression
  coefficient due to ascertainment in a regression
  of phenotype on the observed risk scale onto
  genetic factors on the scale of liability

41
Non-normality of the liability

• The genetic value on the observed scale (ucc) for
  an individual in a case-control study is
• ucc c bccgcc
• c z?gcc
• c z ??(1-??) ?? ?? 2 ?? 1-?? ??
  ?????? 2 gcc
• and,
• var (ucc) ?? ?????? 2 ?? ???? 2 ?? ?????? 2
  
• z ??(1-??) ?? ?? 2 ?? 1-??
  ?? ?????? 2 2 ?? ?????? ??

42
Non-normality of the liability

• The mean of the estimated genetic values, when
  samples are ascertained
• E( ?? ccycc1) ??(1-??)(1-??) ??(1-??)(1-??)
  E( ?? y1) for case
• E( ?? ccycc0) ??(1-??)?? ??(1-??)?? E( ??
  y0) for control
• ?? ???? 2 is a squared regression coefficient
  that transform the estimate of genetic factor on
  the observed risk scale to liability scale
• ?? ?? 2 ??(1-??) ??(1-??) ?? 2 ??
  ?????? 2 ??(1-??) ??(1-??) ?? ???? 2
  ?? ???? 2 ?? 2 ?? ?????? 2 ?? ?? 2
  ?? 2 ??(1-??) ??(1-??)

43
Non-normality of the liability

• The mean genetic liability for cases transformed
  the observed scale by
• bcci ?? ?????? 2 bcci ??(1-??)
  ??(1-??) ?? ?? 2
• bcci2 ?? ?????? 2 bcci2 ??(1-??)
  ??(1-??) ?? ?? 2
• ?? ?????? 2 ??(1-??) ??(1-??) ?? ?? 2
• i (1-??) (1-??) ?? ???? ?? and
• i2 i2 ?? ?? - ?? ???? 1-??

44
Non-normality of the liability

• h l 2 h ?????? 2 ??(1-??) ?? 2 ????
  
• h ?? 2 h ?????? 2 1 ?? ??(1-??)
  ??(1-??) 2
• var (h ?? 2 ) ?????? ( h ?????? 2 ) 1 ??
  ??(1-??) ??(1-??) 4

45
Illustration

• Estimate the heritability of a trait from GWAS
  study when fitting significant SNPs and all SNPs
  simultaneously.
• Sol-Phenotypic and marker data was simulated
  with the help of R-package and the dimension of
  phenotypic data and marker data is 200X1 and
  500X200 respectively. The model is represented
  as
• y µ1 g e
• rrBLUP is used to estimate the effect of random
  additive genetic variance from SNPs information.
  From 500 SNPs, 10 most significant SNPs are
  selected by LASSO. The heritability of a trait
  from GWAS study when fitting significant SNPs and
  all SNPs was 0.2998 and 0.3755 respectively.

46
Illustration

• 2. To estimate the heritability on the liability
  scale along with standard error from ascertained
  case-control data (K 0.1, l gt 1.282 s ?? )
• Sol-
• The phenotypic data of 2500 case and 2500 control
  is simulated (mvrnorm function in the R package).
  The heritability estimated from observed data is
  0.1021 and heritability estimated in liability
  scale is
• h l 2 ?? ?? 2 0.1032
• var ( h l 2 ) 0.00017

47
Conclusions

• Heritability is not missing but hidden
• In the form of common variants of small effect
  scattered across the genome
• In the form of low frequency variants only
  partially tagged by common variants
• Estimates of heritability from traditional method
  are inflated
• If there is physical material (epigenetic
  factors), other than DNA pieces, that can affect
  the phenotype and be transmitted stably across
  generations, then it should also be thought to
  play the role that contributes to additive
  genetic effects.

48
Conclusions

• There are many character of biological or
  economic interest which vary in discontinuous
  manner but are not inherited in a simple
  Mendelian manner, for this type of traits, the
  estimation of heritability based on liability and
  threshold model provide an unbiased effect.
• The missing heritability of complex traits can be
  resolved by estimates of heritability explained
  by all genotyped SNPs.
• The general framework for heritability
  estimation, called GCTA based on HE regression
  method provides the unbiased estimates of
  heritability

49
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52
References

• Visscher, P.M., et al. (2006). Assumption-Free
  Estimation of Heritability from Genome-Wide
  Identity-by-Descent Sharing Between Full
  Siblings. PLoS Genet.2e41.
• Yang, J., Benyamin, B., McEvoy, B.P., Gordon, S.,
  Henders, A. K., Nyholt, D. R. and Madden, P. A.
  (2010). Common SNPs Explain a Large Proportion of
  the Heritability for Human Height. Nature
  Genetics. 42(7)565569.
• Yang, J., Lee, S.H., Goddard, M.E. and Visscher,
  P.M. (2011). GCTA A Tool for Genome-Wide Complex
  Trait Analysis. American Journal of Human
  Genetics. 88(1)7682.
• Zuk O, Hechter, E., Sunyaev, S.R., Lander, E.S.
  (2012). The Mystery of Missing Heritability
  Genetic Interactions Create Phantom Heritability.
  Proceedings of the National Academy of
  Sciences.109(4)11931198.
• Zuk, O., Schaffner, S. F., Samocha, K., Do, R.,
  Hechter, E., Kathiresan, S. and Lander, E. S.
  (2014). Searching for Missing Heritability
  Designing Rare Variant Association Studies.
  Proceedings of the National Academy of Sciences.
  111(4)455464

53
Instructions for use
54
LASSO

• Data train
• Input y x1-x500
• Cards
• Data test
• Input y x1-x500
• Cards
• proc glmselect datatrain valdatatest
• plotscoefficients
• model y x1-x500/
• selectionLASSO(steps10 choosevalidate)
• run

55
Estimation of heritability

• xlt-read.csv(file.choose("g1"))
• ylt-read.csv(file.choose("p1"))
• library(rrBLUP)
• xlt-as.matrix(x)
• ylt-as.matrix(y)
• ans lt- mixed.solve(y,x)
• betalt-ansu
• glt-xbeta
• sigma2glt-var(g)
• sigma2plt-var(y)
• h2lt-sigma2g/sigma2p
• h2
• 0.3755903

• xlt-read.csv(file.choose("g2"))
• ylt-read.csv(file.choose("p1"))
• library(rrBLUP)
• xlt-as.matrix(x)
• ylt-as.matrix(y)
• ans lt- mixed.solve(y,x)
• betalt-ansu
• glt-xbeta
• sigma2glt-var(g)
• sigma2plt-var(y)
• h2lt-sigma2g/sigma2p
• h2
• 0.2998782

56
Illustration No. 2 (code)

• library(MASS)
• library(pps)
• sigmag1
• sigmae3
• m1matrix(sigmag,100,100)
• m2diag(sigmag,100,100)
• m10.05(m1-m2)
• sigmam1m2
• murep(0,100)
• gvlt-mvrnorm(n 50, mu, sigma, tol 1e-6,
  empirical FALSE, EISPACK FALSE)
• gvlt-t(matrix(gv,nrow1,byrowT))

• evefflt-rnorm(5000,0,sigmae)
• llt-gveveff
• sigmallt-sd(l)
• threslt-1.283sigmal
• cc1lt-ifelse(lgtthres,1,0)
• idlt-rep(c(150),each100)
• cclt-as.data.frame(cbind(id,cc1))
• names(cc)lt-c("id","cc1")
• csumlt-tapply(cc,2,cc,1,sum)
• s1lt-cccc,21,
• s2lt-cccc,20,
• s3lt-stratsrs(s2,1,csum)
•  

57
If y follows a standard normal distribution with
a truncation point at t, with t gt 0, so that the
fraction of y that is larger than t is K, then
the mean value of y above the truncation point
is E(y y gt t) i ?? ?? E(y y lt t)
?? ?? - ???? ??-?? Var (y y gt t) 1-
i(i-t) Var (y y lt t) 1 - ?????? (??-??)
?? ?? ???? (?? - ??) where z the height
of the normal curve at point t
Fig. 2
58
Fig .3
var(lcc) gt 1, in a case-control study because
individuals from the tails of the distribution of
liability have been selected.