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DATA ANALYSIS Module Code: CA660 Lecture Block 7

DATA ANALYSIS Module Code: CA660 Lecture Block 7

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Page 1: DATA ANALYSIS Module Code: CA660 Lecture Block 7

DATA ANALYSIS

Module Code: CA660

Lecture Block 7

Page 2: DATA ANALYSIS Module Code: CA660 Lecture Block 7

2

Examples in Genomics and Trait Models

• Genetic traits may be controlled by No.genes-usually unknown

Taking “genetic effect” as one genotypic term, a simple model for

where yij is the trait value for genotype i in replication j, is the mean, Gi the genetic effect for genotype i and ij the errors.

• If assume Normality (and want Random effects) + assume

zero covariance between genetic effects and error

Note: If same genotype replicated b times in an experiment, with phenotypic means used, error variance averaged over b.

ijiij Gy

222egp

),0(~),,0(~),,(~ 222egp NIDNIDGNIDy

Page 3: DATA ANALYSIS Module Code: CA660 Lecture Block 7

3

Example - Trait Models contd.• What about Environment and GE interactions? Extension to

Simple Model.

ANOVA Table: Randomized Blocks within environment and within sets/blocks in environment = b = replications. Focus - on genotype effect

Source dof Expected MSQ

Environment e-1 know there are differences

Blocks (b-1)e again – know there are differences

Genotypes g-1

GE (g-1)(e-1)

Error (b-1)(g-1)eNote: individuals blocked within each of multiple environments, so environmental

effect intrinsic to error. Model form is standard, but only meaningful comparisons are within environment, hence form of random error = population variance = ; so random effects of interest from additional variances & ratios

222ggee beb

22gee b

2e

ijkijjiijk GEEGy )(

2e

Genotypic effects measured within blocks

Page 4: DATA ANALYSIS Module Code: CA660 Lecture Block 7

4

Example contd.

• HERITABILITY = Ratio genotypic to phenotypic variance

• Depending on relationship among genotypes, interpretation of genotypic variance differs. May contain additive, dominance, other interactions, variances

(Above = heritability in broad terms).

• For some experimental or mating schemes, an additive genetic variance may be calculated. Narrow/specific sense heritability then

• Again, if phenotypic means used, obtain a mean-based heritability for b replications.

22

2

2

2

eg

g

p

gH

2222

2

2

2

elda

a

p

aH

bH

eg

g

p

g

/22

2

2

2

222 ,, lda

Page 5: DATA ANALYSIS Module Code: CA660 Lecture Block 7

5

Extended Example- Two related traits

Have

where 1 and 2 denote traits, i the gene and j an individual in population. Then ‘y’ is the trait value, overall mean, G genetic effect, = random error.

To quantify relationship between the two traits, the variance- covariance matrices for phenotypic, p genetic g and environmental effects e

So correlations between traits in terms of phenotypic, genetic and environmental effects:

jij

jiij

Gy

Gy

2222

111

2221

1221

22

1221

2221

1221

21 ee

ee

gg

gg

eg

pp

pp

p

22

21

12

22

21

12

22

21

12 ;;ee

ee

gg

gg

pp

pp

Page 6: DATA ANALYSIS Module Code: CA660 Lecture Block 7

6

MAXIMUM LIKELIHOOD ESTIMATION

• Recall general points: Estimation, definition of Likelihood function for a vector of parameters and set of values x.

Find most likely value of = maximise the Likelihood fn.

Also defined Log-likelihood (Support fn. S() ) and its derivative, the Score, together with Information content per observation, which for single parameter likelihood is given by

• Why MLE? (Need to know underlying distribution).

Properties: Consistency; sufficiency; asymptotic efficiency (linked to variance); unique maximum; invariance and, hence most convenient parameterisation; usually MVUE; amenable to conventional optimisation methods.

)(log2

2)(log)(2

xLExLEI

Page 7: DATA ANALYSIS Module Code: CA660 Lecture Block 7

7

VARIANCE, BIAS & CONFIDENCE

• Variance of an Estimator - usual form or

for k independent estimates• For a large sample, variance of MLE can be approximated by

can also estimate empirically, using re-sampling* techniques.

• Variance of a linear function (of several estimates) – (common need in genomics analysis), e.g. heritability.

• Recall Bias of the Estimator

then the Mean Square Error is defined to be:

expands to

so we have the basis for C.I. and tests of hypothesis.

)ˆ(E2)ˆ( EMSE

2

11

22 ˆ1ˆˆ

k

i

i

k

i

i k

)(

1ˆ 2

nI

22ˆ

2 ])ˆ([]})ˆ([)]ˆ(ˆ{[ EEEE

Page 8: DATA ANALYSIS Module Code: CA660 Lecture Block 7

8

COMMONLY-USED METHODS of obtaining MLE

• Analytical - solving or when simple solutions exist

• Grid search or likelihood profile approach

• Newton-Raphson iteration methods

• EM (expectation and maximisation) algorithm

N.B. Log.-likelihood, because max. same value as Likelihood

Easier to compute

Close relationship between statistical properties of MLE

and Log-likelihood

0ddL 0d

dS

Page 9: DATA ANALYSIS Module Code: CA660 Lecture Block 7

9

METHODS in brief

Analytical : - recall Binomial example earlier

• Example : For Normal, MLE’s of mean and variance, (taking derivatives w.r.t mean and variance separately), and equivalent to sample mean and actual variance (i.e. /N), -unbiased if mean known, biased if not.

• Invariance : One-to-one relationships preserved

• Used: when MLE has a simple solution

0)(

xnx

d

dSScore

n

x

Page 10: DATA ANALYSIS Module Code: CA660 Lecture Block 7

10

Methods for MLE’s contd.

Grid Search – Computational

Plot likelihood or log-likelihood vs parameter. Various features

• Relative Likelihood =Likelihood/Max. Likelihood (ML set =1).

Peak of R.L. can be visually identified /sought algorithmically. e.g.

Plot likelihood and parameter space range - gives 2 peaks, symmetrical around likelihood profile for the well-known mixed linkage phase problem in linkage analysis.

If e.g. constrain MLE = R.F. between genes (possible mixed linkage phase).

])1()1([)( 20808020 LogS

10

5.0ˆ5.00

2.0ˆ

Page 11: DATA ANALYSIS Module Code: CA660 Lecture Block 7

11

contd.

• Graphic/numerical Implementation - initial estimate of , direction of search determined by evaluating likelihood at both sides of .

Search takes direction giving increase. Initial search increments large, e.g. 0.1, then when likelihood change starts to decrease or become negative, stop and refine increment.

• Multiple peaks – can miss global maximum, computationally intensive

• Multiple Parameters - grid search. Interpretation of Likelihood profiles can be difficult.

Page 12: DATA ANALYSIS Module Code: CA660 Lecture Block 7

12

Example

• Recall Exs 2, Q. 8.

Data used to show a linkage relationship between marker and a “rust-resistant”gene.

Escapes = individuals who are susceptible, but show no disease (rust) phenotype under experimental conditions. So define as proportion escapes and R.F. respectively.

is penetrance for disease trait, i.e. P{ that individual with susceptible genotype has disease phenotype}.

Purpose of expt.-typically to estimate R.F. between marker and gene.

• Use: Support function = Log-Likelihood

,

1

)1log(163)log(52)log(3)1log(168),( S

Page 13: DATA ANALYSIS Module Code: CA660 Lecture Block 7

13

Example contd.

• Setting 1st derivatives (Scores) w.r.t = 0. Expected value of Score (w.r.t. is zero, (see analogies in classical sampling/hypothesis testing). Similarly for . Here, however, No simple analytical solution, so can not solve directly for either.

• Using grid search, likelihood reaches maximum at • In general, this type of experiment tests H0: Independence between

marker and gene and H0: no escapes Uses Likelihood Ratio Test statistics. (MLE 2 equivalent)

• N.B: Moment estimates solve slightly different problem, because no info. on expected frequencies, - (not same as MLE)

,

22.0ˆ,02.0ˆ

)5.0( )0(

Page 14: DATA ANALYSIS Module Code: CA660 Lecture Block 7

14

MLE Estimation Methods contd.Newton-Raphson Iteration

Have Score () = 0 from previously. N-R consists of replacing Score by linear terms of its Taylor expansion, so if ´´ a solution, ´=1st guess

Repeat with ´´ replacing ´ Each iteration - fits a parabola to

Likelihood Fn.

• Problems - Multiple peaks, zero Information, extreme estimates • Multiple parameters – need matrix notation, where S matrix e.g. has

elements = derivatives of S(, ) w.r.t. and respectively. Similarly, Information matrix has terms of form

Estimates are

0)]([

)()()(

2

2

d

Sd

d

dS

d

dS

22 )(

)]([

dSd

dSd

.),(),(2

2

2

etcSESE

)()(1 1

SIN

L.F.

2nd

1st

Variance of Log-L i.e.S()

Page 15: DATA ANALYSIS Module Code: CA660 Lecture Block 7

15

Methods contd.

Expectation-Maximisation Algorithm - Iterative. Incomplete data

(Much genomic data fits this situation e.g. linkage analysis with marker genotypes of F2 progeny. Usually 9 categories observed for 2-locus, 2-allele model, but 16 = complete info., while 14 give info. on linkage. Some hidden, but if linkage parameter known, expected frequencies can be predicted – as you know - and the complete data restored using expectation).

• Steps: (1) Expectation estimates statistics of complete data, given observed incomplete data.

• -(2) Maximisation uses estimated complete data to give MLE. • Iterate till converges (no further change)

Page 16: DATA ANALYSIS Module Code: CA660 Lecture Block 7

16

E-M contd.

Implementation

• Initial guess, ´, chosen (e.g. =0.25 say = R.F.). • Taking this as “true”, complete data is estimated, by distributional

statements e.g. P(individual is recombinant, given observed genotype) for R.F. estimation.

• MLE estimate ´´ computed. • This, for R.F. sum of recombinants/N.

• Thus MLE, for fi observed count,

Convergence ´´ = ´ or

)(1

GRPfN ii

)00001.0(tolerance

Page 17: DATA ANALYSIS Module Code: CA660 Lecture Block 7

17

LIKELIHOOD : C.I. and H.T.• Likelihood Ratio Test – c.f. with 2.

• Principal Advantage of G is Power, as unknown parameters involved in hypothesis test.

Have : Likelihood of taking a value A which maximises

it, i.e. its MLE and likelihood under H0 : N , (e.g. N = 0.5)

• Form of L.R. Test Statistic

or, conventionally

- choose; easier to interpret.• Distribution of G ~ approx. 2 (d.o.f. = difference in dimension of

parameter spaces for L(A), L(N) )

• Goodness of Fit : notation as for 2 , G ~ 2n-1 :

• Independence: notation again as for 2

)(

)(2

xL

xLLogG

N

A

)(

)(2

xL

xLLogG

A

N

i

i

n

i

i E

OLogOG

1

2

ij

ij

r

i

c

j

ij E

OLogOG

1 1

2

Page 18: DATA ANALYSIS Module Code: CA660 Lecture Block 7

18

Power-Example extended• Under H0 :

• At level of significance =0.05, suppose true = 1 = 0.2, so if n=25

(e.g. in genomics might apply where R.F. =0.2 between two genes (as opposed to 0.5). Natural logs. used, though either possible in practice. Hence, generic form “Log” rather than Ln here. Assume Ln throughout for genetic/genomic examples unless otherwise indicated)

• Rejection region at 0.05 level is

• If sketch curves, P{LRTS falls in the acceptance region} = 0.13,

= Prob.of a false negative when actual value of = 0.2

• If sample size increased, e.g. n=50, E{G} = 19 and easy to show that P{False negative} = 0.01

• Generally: Power for these tests given by

0}5.05.05.05.05.0{2}{ LogLogLognGE

6.9}5.08.08.02.02.0{50}{ LogLogLogGE

84.321

}{ 22}{,

unitGnEdfP

Page 19: DATA ANALYSIS Module Code: CA660 Lecture Block 7

19

Likelihood C. I.’s - method • Example: Consider the following Likelihood function is the unknown parameter ; a, b observed counts• For 4 data sets observed, A: (a,b) = (8,2), B: (a,b)=(16,4) C: (a,b)=(80, 20) D: (a,b) = (400, 100)

• Likelihood estimates can be plotted vs possible parameter values, with MLE = peak value.

e.g. MLE = 0.2, Lmax=0.0067 for A, and Lmax=0.0045 for B etc.

Set A: Log Lmax- Log L=Log(0.0067) - Log(0.00091)= 2 gives 95% C.I. so =(0.035,0.496) corresponding to L=0.00091, 95% C.I. for A.

Similarly, manipulating this expression, Likelihood value corresponding to 95% confidence interval given as L = 7.389Lmax

Note: Usually plot Log-likelihood vs parameter, rather than Likelihood. As sample size increases, C.I. narrower and symmetric

baL )1()(

Page 20: DATA ANALYSIS Module Code: CA660 Lecture Block 7

20

Multiple Populations: Extensions to G -Example• Recall Mendel’s data - earlier and Extensions to 2 for same In brief Round Wrinkled Plant O E O E G dof p-value 1 45 42.75 12 14.25 0.49 1 0.49 2 0.09 1 0.77 3 0.10 1 0.75

4 1.30 1 0.26 5 0.01 1 0.93 6 0.71 1 0.40 7 0.79 1 0.38 8 0.63 1 0.43 9 1.06 1 0.30 10 0.17 1 0.68 Total 336 101 5.34 10 Pooled 336 327.75 101 109.25 0.85 1 0.36Heterogeneity 4.50 9 0.88

Page 21: DATA ANALYSIS Module Code: CA660 Lecture Block 7

21

Multiple Populations - summary

• Parallels

• Partitions therefore

and Gheterogeneity = Gtotal - GPooled (n=no. classes, p = no.populations)

Example: Recall Backcross (AaBb x aabb)- Goodness of fit (2- locus model). For each of 4 crosses, a Total GoF statistic can be calculated according to

expected segregation ratio 1:1:1:1 – (assumes no segregation distortion for both loci and no linkage between loci).

For each locus GoF calculated using marginal counts, assuming each genotype segregates 1:1.

Difference between Total and 2 individual locus GoF statistics is L-LRTS (or chi-squared statistic) contributed by association/linkage between 2 loci.

2

p

i

n

j ij

ijiTotal E

OOG

1 1

log2

n

j

p

i

p

i

ij

p

i

ij

ijPooled

E

O

OG1 1

1

1log2

Page 22: DATA ANALYSIS Module Code: CA660 Lecture Block 7

22

Example: Marker Screening

Screening for Polymorphism - (different detectable alleles) – look at stages involved.

Genomic map –based on genome variation at locations (from molecular assay or traditional trait observations).

(1) Screening polymorphic genetic markers is Exptal step 1 - usually assay a large number of possible genetic markers in

small progeny set = random sample of mapping population.

If a marker does not show polymorphism for set of progeny, then marker non-informative ; will not be used for data analysis).

Page 23: DATA ANALYSIS Module Code: CA660 Lecture Block 7

23

Example contd.

(2) Progeny size for screening – based on power, convenience etc.,

e.g. False positive = monomorphic marker determined to be polymorphic. Rare since m-m cannot produce segregating genotypes if these determined accurately.

False negatives high particularly for small sample. e.g. for markers segregating 1:1 – (i)Backcross, recombinant inbred lines, doubled haploid lines, or (ii)F2 with codominant markers,

So, e.g. (i) P{sampling all individuals with same genotype) = 2(0.5)n

(ii) P{false negative for single marker, n=5} = 2(0.25)5+0.55=0.0332 Hence Power curves as before.

Page 24: DATA ANALYSIS Module Code: CA660 Lecture Block 7

24

Example contd. S.R 1:1 vs 3:1- use LRTS

• Detection of departure from S.R. of 1:1

n = sample size, O1, O2 observed counts of 2 genotypic classes.

• For true S.R. 3:1, O1 genotypic frequency of dominant genotype, T.S. parametric value is approx.

n

OO

n

OOG

5.0log

5.0log2 2

21

1

)]5.0([2 2211 nnLogLogOOLogOO

n

OELogOE

n

OELogOEGE 5.0

)()(

5.0

)()(2 2

21

1

n

nnLog

n

nnLog

5.0

25.025.0

5.0

75.075.02

n2616.0

Page 25: DATA ANALYSIS Module Code: CA660 Lecture Block 7

25

Example contd.

• To reject a S.R. of 1:1 at 0.05 significance level, a LogLRTS of at least 3.84 (critical value for rejection) is required.

• Statistical Power

• For n=15 then, power is

• For a power of 90%, n 40 needed

• If problem expressed other way. i.e. calculating Expected LRTS (for rejecting a 3:1 S.R. when true value is S.R. 1:1), this is 0.2877n and n 35 needed.

}84.3{ 21, EGP

}84.3{ 21,924.3 P

Page 26: DATA ANALYSIS Module Code: CA660 Lecture Block 7

26

Maximum Likelihood Benefits

• Good Confidence Intervals Coverage probability realised and interval biologically

meaningful • MLE Good estimator of a CI MSE consistent Absence of Bias - does not “stand-alone” – minimum variance important

Asymptotically Normal Precise – large sample Biological inference valid Biological range realistic

0)ˆ( 2 ELimn

)ˆ(E

nasN )1,0(~ˆ