lecture 18: association studies i

Lecture 18: Association Studies I

Date: 10/24/02 A mathematical formalism for linkage disequilibrium Allelic association in random populationsAllelic association in case-control populations

Where We Are

We will be covering chapter 4 of Sham. This material is also covered in chapter 8 of

Liu’s Statistical Genomics.

The Limitations of Per-Cross Linkage Analysis

Family-based linkage analysis is limited by the family unit.

When restricted to natural populations where families are small (e.g. humans), it is difficult to obtain an adequate sample size to detect small differences in linkage A and B or equivalently small .

B

A

A B

How Many to Detect Recombination with =

0.005? Suppose you run a coupling linkage two-point

backcross with recombination fraction = 0.005. You expect a proportion (1-) of the offspring to be

nonrecombinant (e.g. AB and ab). Suppose you want to guarantee that you have

= 0.99 probability offinding at least one recombinant among the offspring scored. You seek N such that N11

How Many to Detect Recombination with =

0.005?

7.918

1log

1log

1log1log

11

N

N

N

N

How Many to Achieve Resolution of d = 0.005?

1

1

1

1

1

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unit

22

222

2

IN

N

nn-

d

ldI

nn

d

dlS

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NRR

NRR

NRR

How Many to Achieve Resolution of d = 0.005?

004.0449,2

04.0602,23

4.0518,147

192.3

192.3

96.12

1

2

2

2

N

dN

dN

dN

Association Studies

Suppose instead of sampling families, which only allow a single generation of recombination to take place, you now let nature takes its course and you allow many generations to pass before checking for linkage.

You are now performing an association study and you have the power to detect tight linkage that you could not previously detect because of a phenomenon known as allelic association.

Allelic Association

A

A

recombinantwith respect

to marker loci.

nonrecombinantwith respect

to marker loci.

Consider two loci A and B. Suppose there are m alleles at locus A, denoted A1,

A2, ..., Am with allele frequencies pi.

Suppose there are n alleles at locus B, denoted B1, B2, ..., Bn with allele frequencies qj.

There are mn possible haplotypes constructed by combining one A allele and one B allele. Denote their frequencies by hij.

Allelic Association – Mathematical Definition

Allelic Association – Mathematical Definition

If the two loci are independent (i.e. there is no association of their alleles), then

If hij > piqj, then there is a positive association between Ai and Bj.

If hij < piqj, then there is a negative association between Ai and Bj.

jijiijji qpBBAAhBBAA PP,P

Mathematical Formulation for Linkage Disequilibrium

One cause of allelic association is linkage disequilibrium, which we can formalize in a mathematical treatment by defining the recombination fraction between two loci.

What happens in a single generation of random mating?

Mathematical Formulation for Linkage Disequilibrium

001

01 1

ijjiijij

jiijij

hqphh

qphh

Linkage equilibrium: hij0 = piqj.

Linkage disequilibrium: hij0 piqj.

The amount of change from generation to generation is proportional to .

Rate of Approach to Linkage Equilibrium

Linkage equilibrium is approached at a geometric rate.

Dij = hij – piqj is termed linkage disequilibrium, rather inappropriately.

jiij

kkjiij

jiijjiij

qphqph

qphqph

01

01

1

1

Other Causes of Allelic Association

Random genetic drift: Each generation is created by (randomly) sampling from the alleles present in the preceding generation. Because populations are finite, there will be sampling variation introduced in allele and haplotype frequencies.

Mutation: Mutation is another random process that changes allele and haplotype frequencies.

Other Causes of Allelic Association

Founder effect: If a population is initiated by a small group of individuals, this group will tend to be in substantial linkage disequilibrium which will take some time to dissipate.

In addition, if the population subsequently expands, the linkage disequilibrium will take even longer to dissipate. In fact, theory provides that the linkage disequilibrium will dissipate at the rate that it would for a population with effective size Ne approximately equal to the harmonic mean of the population size over all generations.

Other Causes of Allelic Association

Selection: When the genotype affects the reproductive fitness of individuals in the population, then alleles at two loci which act synergistically to improve the individual’s phenotype will be on the over-represented haplotype in the population.

Other Causes of Allelic Association

Population Subdivision: When the population is divided into subdivisions which do not interact (mate) for cultural or geographical reasons, then genetic drift will differentiate these populations. If the subpopulations are merged in the analysis, then artificial allelic associations will be generated.

Subpopulations - Example

N A B AB

1000 0.2 0.8 0.16

1500 0.5 0.6 0.30

5000 0.01 0.4 0.004

7500 0.13 0.49 0.08

Expected: 0.06

What is Allelic Association Then?

With so many other causes of allelic association, it is very difficult to determine the cause of a significant allelic association.

One must understand the population in order to use association studies.

If the contributions of random drift, mutation, selection, and population subdivision can be eliminated (or bounded below), then one can take any (or substantial) allelic association as evidence of linkage.

Linkage Disequilibrium in Human Populations

Human populations are usually assumed to be in approximate linkage equilibrium for unselected marker loci. Therefore, only very tight linkage would be sustained over many generations.

Association studies are used for fine mapping of human diseases and traits. Start with family-based linkage analysis. Finish with an association study.

Association Study on Random Population Sample

Because it is a random sample, it is infeasible to perform such an analysis on rare alleles or traits.

Usually used to analyze linkage of markers. Verify suspected linkage Obtain more accurate maps Test a population for the level of linkage

detectable in association study

Random Population – General Model

Using the same framework described earlier, there are m(m+1)/2 genotypes possible at locus A, n(n+1)/2 at B. The total number of joint genotypes is the product.

We observe the counts nijkl of each of the possible joint genotypes, which follow a multinomial distribution with parameters n++++ and the genotypes frequencies gijkl.

1

2

1

2

1d.f. ˆ

loglog

3

,,,2

nnmm

n

ng

gngL

ijklijkl

lkjiijklijklijkl

Random Population – HWE and LE Model

A submodel of the general model given above is that the population is in Hardy-Weinberg and linkage equilibrium.

2d.f. 2

ˆ

log,log

or

,,,0

mnn

n

n

np

qqppnqpL

qqppg

jisjsiij

sss

lkjilkjiijkl

lkjiijkl

Random Population – Testing HWE & LE

2(logL2 – logL0) ~ 2 with m(m+1)n(n+1)/4 – n – m + 1 degrees of freedom.

Jointly tests the assumption of HWE and linkage equilibrium.

Random Population – HWE Model

A submodel of the general model is that the population is in Hardy-Weinberg but not linkage equilibrium.

1d.f. ??ˆ

loglog

ow22

or 2or 2

,

,,,1

2

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hhhh

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lkjih

g

ij

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kjiljlik

jkikilik

ik

ijkl

Random Population – Testing LE | HWE

2(logL1 – logL0) ~ 2 with (m – 1)(n – 1) degrees of freedom.

Tests the assumption of linkage equilibrium conditional on the the assumption of HWE.

Random Population – Estimating hij

There is missing information because we cannot tell whether the doubly heterozygous genotype gijkl is made up of haplotypes hik and hjl or haplotypes hil and hjk. All other genotypes have identifiable haplotypes.

The stage is set for using the EM algorithm to obtain the hij maximum likelihood estimates.

Start by choosing initial values, reasonable to assume linkage equilibrium.

Random Population – Conditional Probabilities

jkiljlij

jlij

jkilijkljlikijkl

jlikijklijklik

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Random Population – E Step

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Random Population – M Step

iikkijkkjlikik nnnnnh 2

1

2

11 /

Random Population - Example

Sham, example 4.4 in section 4.5. 2(logL2 – logL1) = 9.40, df = 5, p = 0.09

2(logL1 – logL0) = 8.89, df = 1, p = 0.003

Conclusion: There may be linkage disequilibrium (test 2), but there is no substantial evidence for nonrandom mating (test 1).

Association Study on Case-Control Sample

When a trait is rare, random sampling of the population is inefficient. It is more appropriate to sample affected individuals in the population and compare them to a random sample of unaffected members from the same population.

The approach is to derive the conditional probabilities of marker genotypes given the disease state, thus permitting genotype comparisons between the sample of affected individuals (cases) and the unaffected individuals (control).

Case-Control – Derivation I

Suppose the rare trait is controlled by locus D with alleles D1 and D2.

Suppose that the penetrance parameters are

And the allele frequencies are

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Case-Control – Derivation II

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122121

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Case-Control – Derivation III

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1 and

1

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Let

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Case-Control – Derivation IV

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Case-Control – Derivation V

Unless we are very lucky though, we are not genotyping at the disease/trait locus, rather at a marker which may or may not be linked to the disease/trait locus.

Thus, we must continue the derivation to determine the conditional genotype probabilities at the marker locus, not the disease/trait locus.

Let the marker locus be called B and suppose it has n possible alleles with frequencies qj.

Case-Control – Derivation VI

The key in the analysis of association is that the trait locus D and marker locus B may not be in linkage equilibrium. This deviation from linkage equilibrium is quantitated through the haplotype frequencies hij and the fact that they are not just multiplicative functions of the allele frequencies.

Case-Control – Derivation VII

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qq

hhfhhhhfhhfBBA

q

hfhhfhfBBA

2

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Case-Control – Derivation VIII

Q

hhshhhhshhsUBB

Q

hshhshsUBB

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hhfhhhhfhhfABB

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hfhhfhfABB

jiijjijiji

iiiiii

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22222112

2111

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2P

Case-Control – Hypothesis Testing

Full Model (no assumptions): n(n+1)-1 degrees of freedom.

Restricted Model 1 (LE): 2n – 2 Restricted Model 2 (LE and HWE): n - 1

Case-Control – Multiple Loci

Suppose that you have genotyped multiple loci and there is one locus underlying the trait of interest.

Then one can set up a model where all the loci are in linkage disequilibrium and a nested model where all the loci are in linkage disequilibrium except the locus of interest.

Test these nested models with G statistic to test for evidence of linkage of the loci to the disease.

Case-Control – Unknown Genetic Model

When the penetrance parameters and allele frequencies are unknown, the above model gets you nowhere.

Treat the data as a traditional contingency table and test for significant difference in the two populations.

Allele Case Control

B1 1n1 2n1

B2 1n2 2n2

n-1degreesoffreedom

Case-Control – Polymorphic Loci – The Problem

When the makers are very polymorphic, there are many, many possible genotypes. Not all will be observed in the data and some cells will have very low counts, even if we consider just allele frequencies.

Case-Control – Polymorphic Loci - Solutions

Group alleles together until the counts are high enough to perform chi-squared test. Unfortunately, power is lost.

Compare each allele against the other alleles, resulting in a 2x2 table. There are multiple tests and the significance level must be adjusted. May also reduce power.

Case-Control – Polymorphic Loci - Solutions

Monte Carlo simulation or exact tests to determine significance of Pearson chi-squared statistic conditional on the marginal totals in the table.

There are other methods. Perhaps you’ll see them in homework...