linkage analysis

LINKAGE ANALYSIS

Recombination Fraction

During synapsis, crossing-over may occur between any two non-sister chromatids

If there are allelic differences at the site of crossing over, the genetic result is recombination

Genes on the same chromosome are connected physically (syntenic)

At least a thousand to several thousand for each human chromosome

Recombination fraction

Prophase of the first meiotic cell division Homologous pairing of chromosomes (synapsis) Each chromosome consists of two fully formed

chromatids joined at the centromere; four chromatids for each pair of chromosomes

Each chromatid represents a separate DNA molecule


If two syntenic genes are close enough that a crossover occurs between them less than once per meiosis, on the average, the two genes are genetically linked

Recombination fraction (RF or ) is 1/2 the frequency of crossovers (a single crossover involves two of four chromatids in a synapsed pair of chromosomes)

If two loci are so far apart that on average there is at least one crossover between them in every meiosis, then = 50%, the loci are unlinked

can not be greater than 50% Morton et al. (1982) used cytological preps of

spermatocytes and reported an average of 52 crossovers per male meiosis


Expressed in map units or centiMorgans (cM) 1 mu = 1 cM; of 1% over small distances One crossover on average implies a genetic map length of 50 cM If two loci are separated by a distance such that an average of

one crossover occurs between them in every meitotic cell, then those loci are 50 cM apart

52 crossovers implies a total genetic map length of 2600 cM in humans; thus, 1 cM equals approximately 1 megabase of sequence

Not additive over long distances due to multiple crossovers (positive or negative interference); mapping functions have been developed to address this phenomenon


gametes total

gametest recombinan ofnumber

)(fraction ion Recombinat

Linkage

Linkage describes the phenomenon whereby allele at neighbouring loci are close to one another on the same chromosome, they will be transmitted together more frequently than chance.

= 0 : no recombination => complete linkage < 0.5 : partial linkage = 0.5 : no linkage

Linkage Analysis

For a couple of which the genotypes at the A and B are known, the probability of observing the genotypes of the offspring depends on the value of

Let us assume the following crossing:

Therefore, such a couple can have 4 types of offspring

Coupling/Repulsion

There are two possible situations: The alleles A1 and B1 may be on the same

chromosome within the pair, in which case A1 and B1 are said to be "coupled";

They may be on different chromosomes, in which case A1 and B1 are said to be in a state of "repulsion".

Linkage analysis

Assuming that there is gamete equilibrium at the A and B loci, in parent 1 there is a probability of 1/2 that alleles A1 and B1 will be coupled, and a probability of 1/2 that they will be in repulsion.

(1) A1 and B1 are coupled, The probability that parent (1) provides the gametes A1B1 and A2B2 is (1-

)/2 and the probability that this parent provides gametes A1B2 and A2B1 is /2. The probability that the couple will have child of type (1) or (2) is (1-)/2, and that of their having a type (3) or type (4) child is /2.

The probability of finding n1 children of type (1), n2 of type (2), n3 of type (3) and n4 of type (4) is therefore

[(1- )/2]n1+n2 x (/2)n3+n4 (2) A1 and B1 are in a state of repulsion

The probability that parent (1) provides the gametes A1B2 and A2B1 is (1-)/2 and the probability that this parent provides gametes A1B1 and A2B2 is /2.

The probability of the previous observation is therefore: (/2)n1+n2 x[(1-)/2]n3+n4

Linkage analysis With no additional information about the A1 and B1 phase, and

assuming that the alleles at the A and B loci are in a state of coupling equilibrium, the probability of finding n1, n2, n3 and n4 children in categories (1), (2), (3), (4) is: p(n1,n2,n3,n4/)=1/2{[(1 -)/2]n1+n2 x (/2)n3+n4 + (/2) n1+n2 x [(1-)/2] n3+n4}

So the liklihood of for an observation n1, n2, n3, n4 can be written:

L(q/n1,n2,n3,n4)=1/2 {[(1-)/2]n1+n2 (/2)n3+n4 + (/2) n1+n2 [(1-)/2] n3+n4}

Special case: number of children n= 1

Regardless of the category to which this child belongs L() = 1/2 [(1-)/2] + 1/2 [/2] = 1/4 The likelihood of this observation for the family does not depend on

. We can say that such a family is not informative for .

Informative families

An "informative family" is a family for which the liklihood is a variable function of .

One essential condition for a family to be informative is, therefore, that it has more than one child. Furthermore, at least one of the parents must be heterozygotic.

Definition: if one of the parents is doubly heterozygotic and the other is A double homozygote, we have a backcross A single homozygote, we have a simple backcross A double heterozygote, we have a double intercross

Definition Of The "Lod Score" Of A Family

Take a family of which we know the genotypes at the A and B loci of each of the members.

Let L() be the likelihood of a recombination fraction 0 and < 1/2 L(1/2) is the likelihood of = 1/2, that is of independent segregation into

A and B. The lod score of the family in is:

Z() = log10 [L()/L(1/2)]

Z can be taken to be a function of defined over the range [0,1/2]. The likelihood of a value of for a sample of independent families is the

product of the likelihoods of each family, and so the lod score of the whole sample will be the sum of the lod scores of each family.

Test For Linkage Several methods have been proposed to detect linkage: "U

scores", were suggested by Bernstein in 1931, "the sib pair test" by Penrose in 1935, "likelihood ratios" by Haldane and Smith in 1947, "the lod score method" proposed by Morton in 1955 (1). Morton’s method is the one most commonly used at present.

The test procedure in the lod score method is sequential (Wald, 1947 (2)). Information, i.e. the number of families in the sample, is accumulated until it is possible to decide between the hypotheses H0 and H1 :

H0 : genetic independence = 1/2 H1: linkage of 1 0 < 1 < 1/2 The lod score of the 1 sample Z(1) = log10 [L(1)/L(l/2)] indicates the relative probabilities of finding that the sample is

H1 or H0. Thus, a lod score of 3 means that the probability of finding that the sample is H1 is 1000 times greater than of finding that it is H0 ("lod = logarithm of the odds").

Test For Linkage

The decision thresholds of the test are usually set at -2 and +3, so that if: Z(1) > 3 H0 is rejected, and linkage is accepted. Z(1) < -2 linkage of 1 is rejected. -2 < Z(1) < 3 it is impossible to decide between

H0 and H1. It is necessary to go on accumulating information.

For the thresholds chosen, -2 and +3, we can show that:

The first degree error, (False negative) < 10-3 The second degree error (false positive), < 10-2 The reliability, 1- > 0.95 1 is the probability that we conclude that H1 is true

given H0

Significance of results In fact, what is being

tested is not a single value of 1 relative to 1 = 1/2, but a whole set of values between 0 and 1/2, with a step of various size (0.01 or 0.05).

If there is a value of 1 such that Z(1) >3: linkage is concluded to exist.

If there is a value of 1 such that

Z(1) = -2

The linkage is excluded for any 1 1

If 2 < Z() < 3, no conclusion can be drawn, the sample is not sufficiently informative.

Criticism

The proposed test has the advantage of being very simple, and of providing protection against falsely concluding linkage.

However, some criticisms can be leveled, not only against the criteria chosen , but also against the entire principle of using a sequential procedure .

The number of families typed is, indeed, rarely chosen in the light of the test results.

Estimation Of The Recombination Fraction

If the test, on a sample of the family, has demonstrated linkage between the A and B loci, then one may want to estimate the recombination fraction for these loci.

The estimated value of is the value which maximizes the function of the lod score Z, and this is equivalent to taking the value of for which the probability of observing linkage in the sample is greatest.

Recombination Fraction For A Disease Locus And A Marker Locus

Let us assume we are dealing with a disease carried by a single gene, determined by an allele, g0, located at a locus G (g0: harmful allele, G0: normal allele).

We would like to be able to situate locus G relative to a marker locus T, which is known to occupy a given locus on the genome. To do this, we can use families with one or several individuals affected and in which the genotype of each member of the family is known with regard to the marker T.

In order to be able to use the lod scores method described above, what is needed is to be able to extrapolate from the phenotype of the individuals (affected, not affected) to their genotype at locus G (or their genotypical probability at locus G).

Disease and Marker Locus What we need to know is:

the frequency, g0 the penetration vector f1, f2,f3 f1 = Pr (affected /g0g0) f2 = Pr (affected /g0G0) f3 = Pr (affected /G0G0)

It will often happen that the information available for the marker is not also genotypic, but phenotypic in nature. Once again, all possible genotypes must be envisaged.

As a general rule, the information available about a family concerns the phenotype. To calculate the likelihood of we must envisage all the possible genotype configurations at each of the loci, for this family, writing the likelihood of for each configuration, weighting it by the probability of this configuration, and knowing the phenotypes of individuals in A and B.

Knowledge of the genetic parameters at each of the loci (gene frequency, penetration values) is therefore necessary before we can estimate

D d dd

21

Estimation of L as a function of and f

Allele distribution. If the frequency of D is .01, H-W equilibrium is

Pr(Dd ) = 2x.01x.99 Genotypes of founder couples are (usually)

treated as independent. Pr(Dd , dd ) = (2x.01x.99)x(.99)2

Pedigree analyses usually suppose that, given the genotype at all loci, and in some cases age and sex, the chance of having a particular phenotype depends only on genotype at one locus, and is independent of all other factors: genotypes at other loci, environment, genotypes and phenotypes of relatives, etc.

Complete penetrance:

pr(affected | DD ) = 1

Incomplete penetrance:

pr(affected | DD ) = .8

DD

DD


Assume penetrances pr(affected | dd ) = .1, pr(affected | Dd ) = .3 pr(affected | DD ) = .8, and that allele D has frequency .01.

The probability of this pedigree is the product:

(2 x .01 x .99 x .7) x (2 x .01 x .99 x .3) x (1/2 x 1/2 x .9) x (2 x 1/2 x 1/2 x .7) x (1/2 x 1/2 x .8)

D d D d

D dd d D D

1

4 53

2


3

21

D d

T t

d d

t t

D D

T T

3

T t

D (1-)/2 /2 1/2

d /2 (1-)/2 1/2

1/2 1/2


Two-locus founder probabilities are typically calculated assuming linkage equilibrium, i.e. independence of genotypes across loci.

If D and d have frequencies .01 and .99 at one locus, and T and t have frequencies .25 and .75 at a second, linked locus, this assumption means that DT, Dt, dT and dt have frequencies .01 x .25, .01 x .75, .99 x .25 and .99 x .75 respectively. Together with Hardy-Weinberg, this implies that

pr(DdTt ) = (2 x .01 x .99) x (2 x .25 x .75)

= 2 x (.01 x .25) x (.99 x .75)

+ 2 x (.01 x .75) x (.99 x .25).

This last expression adds haplotype pair probabilities.

Dd

Tt


D d

T t

d d

t t

D d

T t

Initially, this must be done with haplotypes, so that account can be taken of recombination. Then terms like that below are summed over possible phases. Here only the father can exhibit recombination: mother is uninformative.

pr(kid DT/dt | pop DT/dt & mom dt/dt )

= pr(kid DT | pop DT/dt ) x pr(kid dt | mom dt/dt )

= (1-)/2 x 1.


Two Loci: Penetrance

In all standard linkage programs, different parts of phenotype are conditionally independent given all genotypes, and two-loci penetrances split into products of one-locus penetrances. Assuming the penetrances for DD, Dd and dd given earlier, and that T,t are two alleles at a co-dominant marker locus.

Pr( affected & Tt | DD, Tt ) = Pr(affected | DD, Tt ) Pr(Tt | DD, Tt ) = 0.8 1

Estimation of L as a function of and f We assume below pop is as likely to be DT / dt as Dt / dT.

d d

t t

D d

T t

D d

T tD d

t t

d d

t t

D d

T t

Pr (all data | ) = pr(parents' data | ) pr(kids' data | parents' data, )= pr(parents' data) {[((1-)/2)3 /2]/2+ [(/2)3 (1-)/2]/2}

Lod Scores and LA software It is obvious that calculating the lod scores, despite being simple in

theory, is in fact a lengthy and tedious business. In 1955, Morton provided a set of tables giving the lod scores for various values of q for a disease locus and a marker locus for nuclear families with sibling sizes of 2 to 7. However, the situations envisaged were very restrictive. In particular, it was assumed that the disease was determined by a dominant or recessive completely penetrating rare gene (f1=1,f2,f3=0 or f1,f2,f3=1)

. "LIPED" written by Ott in 1974 was the pioneering software in linkage

analysis. It is able to carry out this calculation, in an extensive pedigree for any values of , f1, f2, f3 and for penetration as a function of age.

The "Linkage" program of Lathrop et al, 1984 is the one most often used for gene mapping. It can be used to carry out multipoint analysis.

All the software we have described is based on the same recursive algorithm, r (Elston and Stewart), which means that it can be used to investigate pedigrees of any size, but that it envisages all the possible haplotypical combinations of markers, and is therefore limited by the number of markers to be taken into account.

LA software

In contrast, "Genehunter", which is based on a Markov chain principle, is limited not by the number of markers taken into consideration in the analysis, but by the size of the family structure.

The very recently developed software package "Allegro" can apply information from a large number of markers and extended family structures.

Analysis of gene linkage has made it possible to construct a gene map by locating the new polymorphisms relative to one other on the genome. The measurement used on the gene map is not the recombination fraction, which is not an additive datum, but the gene distance, which we will define below.

Linkage Analysis For Three Loci : Interference

Now let us consider three loci A, B and C. Let the recombination fraction between A and B be 1, that between B and C be 2 and that between A and C be 3.

Let us consider the double recombinant event, firstly between A and B, and secondly between B and C. Let R12 be the probability of this event. If the crossings-over occur independently in segments AB and BC, then:

R12

Interference

If this is not the case, an interference phenomenon is occurring and

R12 = C 1 2 where C1 If C < 1 the interference is said

to be positive; and crossings-over in segment AB inhibit those in segment BC.

If C >1 the interference is said to be negative; and crossings-over in segment AB promote those in segment BC.

Let us consider the case of a triple heterozygotic individual.

Such an individual can provide 8 types of gametes. 1221

222

111

3

122122

211

121112

221

12212

121

1 RCBA

CBA

RCBA

CBA

RCBA

CBA

RCBA

CBA

Interference

We can write that :

RC1

If C = 1

= 1 + - 21 The recombination fraction is a non-additive measurement.

However, we can write (1-2 ) = (1-2 )(1-2 ) if x() = k Log (1-2) then we have x() = x() + x() and for k = -1/2, x()~ for small values of . x() = -1/2 Log (1-2

) is an additive measurement. It is known as the genetic distance, and is measured in

Morgans. It can be shown that x measures the mean number of crossings-over.

Test for the presence of interference

Let us consider a sample of families with the genotypes A, B and C. Let Lc be the greatest likelihood for 1, 2, and L1 the greatest likelihood when we impose the constraint C=1

(i.e. = 1 + - 21 ) Then -2 Log (L1/Lc ) follows a 2 pattern, with

one degree of freedom.

Reminder 2 distribution

The critical value depends on the degrees of freedom

E

EO 2

2 )( Where ‘O’ is observed, ‘E’ is expected valueTest Statistic

The Chi-Square distribution is a skewed distribution

0 4 8 12 16 202

H 0 rejectedH 0 accepted

The critical region

Genetic Heterogeneity Of Localization

The analysis of genetic linkage can be complicated by the fact that mutations of several genes, located at different places on the genome, can give rise to the same disorder. This is known as genetic heterogeneity of localization.

One of the following two tests is used to identify heterogeneity of this type: The "Predivided sample test" The "Admixture Test".

The first test is usually only appropriate if there is a good family stratification criterion or if each family individually has high informativity.

II- 1. The Predivided Sample Test

This test is intended to demonstrate linkage heterogeneity in different sub-groups of a sample of families. The aim is to test whether the genetic linkage between a disease and its marker(s) is the same in all sub-groups. These groups are formed ad hoc on the basis of clinical or geographical criteria etc....

Let us assume that the total sample of families has been divided into n sub-groups (it is possible to test for the existence of as many sub-groups as families). i denotes the true value of the recombination fraction of sub-group i.

We want to test the null hypothesis H0: 1= 2 = 3 = …= n against the alternative hypothesis H1: the values of i are not all equal.

Therefore, the quantity

Follows a distribution with (n-1) degrees of freedom. The homogeneity of the sample for linkage with a type-I error of the sample for linkage with a type I error equal to a if Q is above the critical threshold 2

(n-l) corresponding to .

iiii ZZQ )ˆ()(log2 10

The Admixture Test The "admixture test" is not based on an ad hoc subdivision of the families. It is

assumed that among all the families studied genetic linkage between the disease and the marker is found only in a proportion of the families, with a recombination fraction < 1/2. In the remaining (l-) families, it is assumed that there is no linkage with the marker (=1/2).

For each family i of the sample, the likelihood is calculated

Li() = Li() + (1-) Li(1/2),

where Li() is the likelihood of for family i. The likelihood of the couple (a, ) is defined by the product of the likelihoods associated with all the families :

L(, )= Li(, )

We test to find out whether a is significantly different from 1 by comparing Lmax( = 1, ), the maximized likelihood for assuming homogeneity, and Lmax(, ), the maximized likelihood for the two parameters and (nested models). Then variable Q =2[Ln Lmax (, ) —Ln Lmax (= 1, )] follows a distribution with one degree of freedom.

Generalization Of The Admixture Test

In some single-gene diseases, several genes have been shown to exist at different locations. This is true, for example of multiple exostosis disease, for which 3 genes have been identified successively on 3 different chromosomes.

The "admixture test" is then extended to determine the proportion of families in which each of the three genes is implicated , and the possibility that there is a fourth gene.

The three locations on chromosomes 8, 19 and 11 were reported as El, E2 and E3, and the proportions of families concerned as 1, 2 and 3 respectively. 4 was used to represent the proportion of the families in which another location was involved.

For each family i of the sample, the likelihood was calculated using the observed segregation within the family of the markers available in each of the three regions, according to the clinical status of each of its members.

Li(El, E2, E3, |Fi) = (L(E1|Fi)/L(E1=1/2 |Fi)] + (L(E2|Fi)/L(E2=1/2 |Fi)] + 3 [L(E3|Fi)/L(E3=1/2 | Fi)]+ 4

For all the families Li(El, E2, E3, |Ft) = i Li(El, E2, E3, | Fi) Each i can be tested to see if it is equal to 0, and then the corresponding non nulla i and Ei values are estimated.

Generalization of The Admixture Test -Results

It is also possible to calculate the probability that the gene implicated is at El, E2 or E3 for each of the families in the sample. The post hoc probability makes use of the estimated i proportions, but also the specific observations in this family.

The sample investigated has been shown to consist of three types of families: in 48% of families, the gene is located on chromosome 8, in 24% of them on chromosome 19, and in 28% of families the gene is located on chromosome 11. There was no evidence of a fourth location in this sample.

The post hoc probabilities of belonging to one of these 3 sub-groups were then estimated: the probability that the gene implicated would be on chromosome 8 was over 90% for 5 families, that it would be on chromosome 19 for 3 of them, and that it would be on chromosome 11 for 4 families. For the other families, the situation was less clear-cut: the post-hoc probabilities are similar to the ad hoc probabilities because of the paucity of information provided by the markers used.

Gamete Disequilibrium Between Alleles At The Disease Locus And At

The Marker Locus An association between a susceptibility gene and a marker can

lead to bias in the estimation of the recombination fraction. In particular, the "lod scores" method specifies that there must be no selection for the marker in the sample.

However, in a context of an association, selection based on the status of the patient implicitly involves selection for a marker.

Furthermore, the calculation assumes that the probability for each genetic combination is equal in the parents, and this is not true if there is an association.

In the analysis, failing to take into account the disequilibrium existing between disease alleles and marker alleles, induces a very great under-estimation of the "lod score" (in other terms, a marked reduction in the power of the linkage test) and a very slight under-estimation of the recombination fraction.

The Problem Of Multiple Tests One of the difficulties encountered in the statistical interpretation of the

analyses of the genetic linkage of complex diseases arises in fact from the fact that in general and with a varying degree of explicitness, the data are subjected to multiple tests: several clinical classifications, several genetic markers, several models, several samples.

It is quite clear that the discontinuation criteria usually used in the lod score test no longer have the same statistical significance when several tests are applied simultaneously to the same sample or to several samples.

The situation is much more complex for multifactorial diseases, because the multiplicity of the tests has several types of impact and these are not independent.

Multiple tests could be taken into account by readjusting the discontinuation criterion of the lod scores test. However, on the one hand, it is not always clear which tests have actually been carried out, and on the other, this can make the test too conservative.

Replication strategy- If a positive result is replicated for a new sample (using the same classification, the same marker, the same transmission model) this provides a reliable threshold of significance.

linkage analysis

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