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An Unconditional Exact Test for the Hardy-Weinberg Equilibrium Law: Sample-Space Ordering Using the Bayes Factor
Luis E. Montoya-Delgadoa, Telba Z. Ironyb, Carlos A. de B. Pereirac, and Martin R. Whittleda Universidad del Cauca, Popayan, Cauca, Colombia,
b Division of Biostatistics, Center for Devices and Radiological Health, Food and Drug Administration, Rockville, Maryland 20850,
c Instituto de Matematica e Estatistica, Universidade de São Paulo, 05008-090 São Paulo, SP, Brazil
d Genomic Engenharia Molecular Ltda, 01332-903 São Paulo, SP, Brazil
Corresponding author: Martin R. Whittle, Genomic Engenharia Molecular Ltda, Rua Itapeva 500, cj 5AB, 01332-903 São Paulo, SP, Brazil., mwhittle{at}genomic.com.br (E-mail)
Communicating editor: G. A. CHURCHILL
| ABSTRACT |
|---|
Much forensic inference based upon DNA evidence is made assuming that the Hardy-Weinberg equilibrium (HWE) is valid for the genetic loci being used. Several statistical tests to detect and measure deviation from HWE have been devised, each having advantages and limitations. The limitations become more obvious when testing for deviation within multiallelic DNA loci is attempted. Here we present an exact test for HWE in the biallelic case, based on the ratio of weighted likelihoods under the null and alternative hypotheses, the Bayes factor. This test does not depend on asymptotic results and minimizes a linear combination of type I and type II errors. By ordering the sample space using the Bayes factor, we also define a significance (evidence) index, P value, using the weighted likelihood under the null hypothesis. We compare it to the conditional exact test for the case of sample size n = 10. Using the idea under the method of
2 partition, the test is used sequentially to test equilibrium in the multiple allele case and then applied to two short tandem repeat loci, using a real Caucasian data bank, showing its usefulness.
ONE of the major uses of data from human multiallelic DNA loci is forensic inference. Because of the increasing use of variable number of tandem repeats (VNTR) and short tandem repeats (STR) loci, the importance of the Hardy-Weinberg equilibrium (HWE) has been reinforced (![]()
![]()
![]()
![]()
![]()
To test the HWE, usually the
2 test, the conditional exact test, and the likelihood-ratio test are used. For a complete discussion on these procedures, including some comparisons, see, for instance, ![]()
![]()
![]()
![]()
![]()
![]()
![]()
![]()
![]()
In cases where the number of alleles per locus or the sample size is large, a technique to generate a Markov chain is used. The objective is to estimate the p values for the exact conditional test. For more details see ![]()
![]()
The
2 test is highly dependent on asymptotic results. In addition to being inefficient when the sample is insufficiently large, it may fail whenever there are categories (genotypes) of low expected frequencies. In the problem we consider, these genotypes are present in large numbers because of the inherent genetic structure and so it is not unusual to find an allele that appears only once or twice in a database, even when there are a considerable total number (
5000) of points.
Being a nonasymptotic test, the conditional exact test is, on the other hand, useful in cases where small samples are dealt with. It depends on a multinomial distribution and requires the ranking of all the possible samples with the same frequencies of alleles and same sample size. However, this ranking of the possible samples is problematic as we enlarge the size of the sample. Another disadvantage of this test is that it imposes unrealistic probabilities on the data points whose probabilities had already been determined from conditioning. Recall that, by conditioning, the sample space is severely reduced and the probabilities may be drastically increased.
![]()
![]()
![]()
The problem of estimating the allelic frequencies, under the Bayesian perspective, was considered by ![]()
![]()
The aim of this study is to develop an exact test on the basis of the comparison between weighted likelihoods (![]()
![]()
The weight used to calculate the weighted likelihoods under each of the hypotheses is based on the a priori preferences, which are derived from the choice of an a priori distribution over the parametric space. The dimension of the subspace defined by the hypothesis of HWE is smaller than that of the original parametric space. Therefore, to calculate the weighted likelihood under the null hypothesis, we use line integrals. In the same manner, type I and type II errors are weighted average errors (![]()
1 and ß1 are the errors associated with the Neyman-Pearson test of simple hypotheses H0,
=
0 vs. H1,
=
1, where
0 is a parametric element of the equilibrium curve and
1 is outside the curve. Consider now the set of all such pairs. The weighted errors for the unconditional exact test are the weighted averages of these Neyman-Pearson errors.
Usually, a test of hypothesis consists of comparing the supremum of the likelihood in the subset of the parameter space corresponding to the null hypothesis with the supremum in the whole parameter space. The test presented in this article consists of comparing the averages of the likelihoods of these sets. That is, instead of comparing suprema the test compares averages. The most important property of the test is that it minimizes linear combinations of the two kinds of errors. For instance, if
and ß are errors of the first and second types, the test presented here minimizes
+ ß (![]()
, it maximizes (1 - ß) -
and consequently also the power of the test.
We define the significance level of the test by ordering the sample space using the Bayes factor. The BF, following ![]()
![]()
![]()
![]()
The significance level, the P value, takes into consideration the alternative hypothesis in its calculus, which controls the type II error. Recall that the Bayes factor is the ratio between the probabilities under the two hypotheses. Usually this is not considered for the standard P value, which can cause problems such as rejecting the null hypothesis even when, under the alternative hypothesis, the observed sample has a lower probability (![]()
A program to compute our P value is available for the MatLab environment at the website http://www.ime.usp.br/~cpereira/signifpr.html.
A more complex test environment where two loci are considered simultaneously is presented by ![]()
| METHODS |
|---|
To exemplify the use of the test (see Examples), data from two STR loci were analyzed: D17S250 (![]()
![]()
-32P]dCTP, separated on DNA sequencing gels by electrophoresis, and visualized by autoradiography. Allele sizing was done by running adjacent M13 sequence ladders.
Definition and presentation of the exact unconditional test
Consider a single autosomal biallelic locus, comprising alleles a1 and a2, which does not undergo mutation. Let

be the genotypes at this locus, and we denote p1, p2, and p3 as the respective proportions of these genotypic classes in the population (p1 + p2 + p3 = 1). Let us suppose that the system is codominant; that is, distinct genotypic classes define distinct phenotypic classes. In this way, in a sample of size n, the frequencies of members in each class n1, n2, and n3, satisfying the condition n1 + n2 + n3 = n, can be observed.
In a panmictic population obeying Mendelian rules, equilibrium is attained in one reproductive generation and this assures the existence of a real number p
(0, 1), such that the genotypic proportions satisfy the relations
![]() |
(1) |
Hence, to decide as to the existence or not of equilibrium, it is necessary to test the null hypothesis H0,
= (p1, p2, p3)
0, where

against the alternative hypothesis H1,
1 =
-
0, where

Consequently, the statistical problem of interest is the construction of a procedure to test the following two alternative hypotheses,

and

where
,
0, and
1 are defined as above (see Fig 1).
|
Assuming that the sample elements are obtained independently, by using a Bernoulli multivariate process, prefixing the sample size n, and representing the data by d = (n1, n2, n3), we have that the likelihood function is given by
![]() |
(2) |
where
.
Let us represent the researcher prior preferences by a Dirichlet density function (see ![]()
1,
2,
3),
i > 0. That is, if
[·] represents the Gamma function,
![]() |
(3) |
is the function that defines the prior preferences of the possible parameter points
= (p1, p2, p3). From the Bayesian perspective, to choose this density is to choose a conjugate prior since the posterior density will also be a Dirichlet density with parameters (A1, A2, A3), where Ai =
i + ni, i = 1, 2, or 3.
Considering now (3) as the weighing system, the weighted likelihood average over
is given by
![]() |
(4) |
Also note that
0 is a line inside the simplex
and hence the weighted likelihood average over
0 is the ratio of two line integrals as follows:
![]() |
(5) |
Let us suppose that the a priori density over
is uniform. Thus, making
i = 1 for i = 1, 2, 3 in Equation 4 and 6, this means that the exact values of the weighted likelihoods over
1 and
0 are given, respectively, by
![]() |
(7) |
and

Consequently if we assume that a priori the two hypotheses have equal probabilities, 0.5, we obtain
![]() |
(8) |
A test for the hypothesis of HWE consists of comparing BF [d] with unity. In this case we have a test that minimizes the sum of the average of the two types of errors.
Sometimes the exact calculation of Equation 8 is not feasible and so it is useful to show approximations to its determination. An approximation to Equation 8 (using Taylor's expansion) is given by

Note that although we have no closed form for general Dirichlet priors, using Equation 5 and 6, and numerical integration, we can easily compute the Bayes factor for any choice of the prior parameters. In computing the P value using the program mentioned above for general Dirichlet priors, one needs only to adjust the data input. Instead of inputing the vector (x1, x2, x3), one must use (A1 - 1, A2 - 1, A3 - 1).
In this discussion we emphasize the use of uniform priors only for the purpose of a fair comparison with the alternative classical methods. Recall that with the uniform prior, the posterior is the normalized likelihood function. In the next section we provide a comparison of this proposed test for HWE to the conditional exact test.
Comparison between the unconditional and the conditional exact tests
In this section we compare the exact unconditional test proposed in the previous section to the traditional conditional exact test. Considering all possible samples of size n = 10, we calculate the P value for each of these samples. To pinpoint the two different ways of computing the probabilities, we refer to p value in the conditional test and to P value in the unconditional one (see ![]()
|
|
|
From Table 2 and Table 3 it can be seen that the conditional exact test is more conservative than the unconditional exact test. The unconditional test is observed to minimize the sum of the average of the two types of errors. For a sample where (n1, n2, n3) = (1, 8, 1) the unconditional exact test rejects the hypothesis of equilibrium (P = 0.036744
0.04). Meanwhile the conditional exact test will not reject this hypothesis (p = 0.20). The p value of the conditional exact test can be obtained from the weighted likelihood ratios. For example, let us fix the sample size at n = 10 and suppose that the total number of observed elements that show the allele a1 is 9. That is, T = 2n1 + n2 = 9. To compute the p value of the conditional exact test, it is enough to consider all possible sample points for which T = 9. To determine the conditional probability of a sample point (n1, n2, n3) given that 2n1 + n2 = 9 we have only to divide the BF obtained in this point by the sum of the BF of all points having T = 9. With these probabilities calculated we compute the p values in the usual manner, adding to the probability of each point all the smaller ones. Table 4 illustrates this calculation. Note that column 4, which is equal to column 3, is the second column divided by its sum.
|
In the general case, we maintain the sample size at n and the total number of observed elements that have the allele a1 at T = t. Considering the statistic T defined by T = 2n1 + n2, where n1 and n2 are random variables that denote the number of individuals observed in the sample who have the genotype (a1, a1) and (a1, a2), respectively, this means that for each d = (n1, n2, n3) with 2n1 + n2 = t and
3i=1ni = n, whatever the value of p at (0, 1), the conditional probability Pr[d|T = t] is given by the equation

where

Note that the sample space for n = 10 has a total of 66 sample points, in contrast to the conditional test, where T = 10, which considers only a sample set of 6 points: (0, 10, 0), (1, 8, 1), (2, 6, 2), (3, 4, 3), (4, 2, 4), and (5, 0, 5).
Hierarchical sequential testing for multiple alleles
The ideal situation to build the significance test for the multiallelic case would be to consider an ordering in the whole space. However, the dimensionality of the parameter space is incredibly large. For instance, consider a locus with 20 alleles. For this example, the parameter space will have dimension 210. Hence the number of possible sample points increases drastically. Theoretically, substituting line integrals with surface integrals, we can proceed exactly as in the biallelic case but at an extremely high computational cost. Next we present a sequential procedure that loses in precision in benefit of cost.
Let us consider a single autosomal locus with multiple codominant alleles. Let (ai, aj) be the genotype referring to the alleles ai and aj, i = 1, · · ·, m; j = i, · · ·, m, and

is the proportion of the genotypic class (ai, aj) in the population.
Because the system is codominant, in a sample of size n, the frequencies of the elements in each genotypic class n1, n2, · · ·, nm(m+1)/2, with 
i=1ni = n can be identified.
Assuming that the sample elements are obtained independently, by using a Bernoulli (multivariate) process, we see that the likelihood function is proportional to

where

In a manner analogous to the case of the biallelic locus, the condition of equilibrium given in Equation 1 is characterized in the general case by Equation 9, because of the following statement: in a panmictic population that obeys Mendelian laws, equilibrium is attained in one reproductive generation and there are u1, u2, · · ·, um
(0, 1) with
mj=1 uj = 1, such that they obey the relation
![]() |
(9) |
Although we can define the exact unconditional test using the BF (as in the case with two alleles), great difficulties arise when calculating the surface integrals in this case. Upon examining the population HWE determined by Equation 9, we can prove that the following statement, a property of the multinomial distribution, is true. This comprises the basis of the procedure that we propose to test the hypothesis of equilibrium in a situation of multiple alleles.
Statement:
Let a1, a2, · · ·, am be the alleles under consideration. If for any i1 = 1, 2, · · ·, n, the condition of HWE given in Equation 1, is satisfied by the biallelic system ai1, AI1, where

and if for every j = 2, · · ·, m - 1, the condition of HWE is satisfied by the biallelic system aij, AIj, whatever ij in Ij-1, where

then the system with alleles a1, a2, · · ·, am satisfies the condition of HWE given by Equation 9.
Thus if a system with n alleles, A0 = {a1, a2, · · ·, am}, obeys the law of HWE, this law is obeyed by "any system" obtained by partitioning A0 into at least two nonempty subsets and upon considering each element of this partition as "an allele."
The idea under the HWE test for the multiallelic case is based on the chi-square partition in contingency tables (see ![]()
The sequential procedure, to test the hypothesis of HWE, is as follows.
Procedure:
- Step 1: Without loss of generality, call a1 the least frequent allele in sample S.
- Step 2: Divide the sample S into three mutually exclusive sets:
- S11, all individuals with genotype (a1, a1); S1., all individuals with genotype (a1, a1), i
1; and S.., all individuals not having the allele a1. - Step 3: Apply the unconditional exact test for the biallelic case in the partition (S11, S1., S..). If HWE is rejected stop and declare the population to be in disequilibrium. If HWE is accepted go to step 4.
- Step 4: If S.. is composed of elements with only one allele involved, stop and declare the population to be in equilibrium. If more than one allele is involved in the elements of S.., rename S.. as S and go to step 1.
Examples
In the examples we illustrate the sequence in which the tests for equilibrium were performed and present the values of the Bayes factors with respective p values and the values of the p values for the corresponding
2 tests.
Let ma be the number of alleles present in the locus being studied. For each i = 1, · · ·, ma - 1, we define
- ai, the allele chosen to carry out the ith test;
- (ai)c, the set made up of the remaining alleles to carry out the ith test;
- ni1, the number of genotypes (ai, ai);
- ni2, the number of genotypes (ai, (ai)c);
- ni3, the number of genotypes ((ai)c, (ai)c); and
- BFi, the Bayes factor corresponding to the ith test.
EXAMPLE 1: The following data were obtained from the STR locus MYC in which 19 alleles are observed and na = 5714 (Table 5).
|
In this case, therefore, the hypothesis of HWE should be rejected.
EXAMPLE 2: Here the data were obtained from the STR locus D17S250 in which 21 alleles are seen and na = 5592 (Table 6).
|
The hypothesis is not rejected in this example.
| DISCUSSION |
|---|
A Bayesian test of hypothesis was presented in this article. However, its evaluation and comparison with the alternative conditional test are done under a classical perspective. To start this discussion we recall that the generalized Neyman-Pearson (GNP) test is an optimal test under both perspectives, classical and Bayesian (![]()
) k f1, then the null hypothesis is accepted. On the other hand, if f0
(<) k f1, then the null hypothesis is rejected. Note that, in fact, we compare the values of the two likelihoods at the sample point that effectively occurred, to make the decision. If
and ß are, respectively, the probabilities of the two kinds of errors, the GNP is the test that minimizes the linear combination
- kß. Considering adequately the choice of k as a function of losses and prior probabilities, this linear combination is the minimum expected loss, which makes the test optimal under the Bayesian perspective. ![]()
The HWE case is different in that both hypotheses are composite. That is, each hypothesis is represented as a set of probability functions. One of the difficulties is that the two sets have different dimensions. The alternative hypothesis is bidimensional although the null hypothesis is unidimensional. The idea of the test, under the classical point of view, is to define as the two probability functions, f0 and f1, the averages of the likelihood over the parametric sets defined by the null and the alternative hypotheses, respectively. Having now two probability functions, we apply the GNP procedure to define the critical region. To compute the average p value we have to order the sample space. We say that a sample point xi is higher than xj, denoted xj
xi, if f0(xi)/f1(xi) > f0(xj)/f1(xj). To define the P value (not p value) we consider the sum of f0(xj) over all sample points xj
xo, where xo is the sample point effectively observed. Since we cover the whole sample space, we have called the test an unconditional exact test.
The unconditional test is opposite to the one that considers as the likelihood under the null hypothesis the conditional probability function given the observed value of T = 2n1 + n2, which is a sufficient statistic under the null hypothesis (![]()
![]()
![]()
![]()
![]()
![]()
![]()
Turning here to our procedure, from the Bayesian point of view, where a posterior density is defined over the parametric space, one could say that the test is fully conditional. The reason is obvious because we compute a conditional density for the parameter given the observed sample point. Considering the uniform prior in the parametric space and 0.5 as the prior probability of the null hypothesis to be true, the ratio of the average likelihoods, as presented, is the posterior odds, which compared with a chosen k would define the testing procedure presented. This is a test that minimizes
+ k
, where
and
are the average of the probabilities of errors of types I and II, respectively (![]()
As far as we know, ![]()
![]()
![]()
![]()
The hierarchical sequential procedure described in this article is based on intuition. Recall that multinomial likelihoods can be factorized using partitions on the set of categories. Also, the HWE is a special kind of association among alleles at a specific locus. Whenever we conclude that a specific allele is in HWE association with all the others, we believe we do not have to use it again when testing the remaining alleles. It could be argued that, by using this procedure, the probability of rejecting HWE may increase as alleles are being eliminated. However, since the sample size is decreasing, the power of the tests will decrease. Hence, it is reasonable to believe that there is a compensation and that the procedure will do the job fairly. Note also that the sequence order depends on the rarity of the alleles in such a way that the sample size reduction occurs as slowly as possible.
Today, the use of Bayesian ideas in genetics is a reality. More recently, an interesting article by ![]()
| ACKNOWLEDGMENTS |
|---|
The views expressed are those of the authors and not necessarily those of the FDA.
Manuscript received September 17, 1999; Accepted for publication March 1, 2001.
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