- Research
- Open Access

# How many replicates of arrays are required to detect gene expression changes in microarray experiments? A mixture model approach

- Wei Pan
^{1}Email author, - Jizhen Lin
^{2}and - Chap T Le
^{1}

**3**:research0022.1

https://doi.org/10.1186/gb-2002-3-5-research0022

© Pan et al., licensee BioMed Central Ltd 2002

**Received: **27 December 2001

**Accepted: **11 March 2002

**Published: **22 April 2002

## Abstract

### Background

It has been recognized that replicates of arrays (or spots) may be necessary for reliably detecting differentially expressed genes in microarray experiments. However, the often-asked question of how many replicates are required has barely been addressed in the literature. In general, the answer depends on several factors: a given magnitude of expression change, a desired statistical power (that is, probability) to detect it, a specified Type I error rate, and the statistical method being used to detect the change. Here, we discuss how to calculate the number of replicates in the context of applying a nonparametric statistical method, the normal mixture model approach, to detect changes in gene expression.

### Results

The methodology is applied to a data set containing expression levels of 1,176 genes in rats with and without pneumococcal middle-ear infection. We illustrate how to calculate the power functions for 2, 4, 6 and 8 replicates.

### Conclusions

The proposed method is potentially useful in designing microarray experiments to discover differentially expressed genes. The same idea can be applied to other statistical methods.

## Keywords

## Background

Microarrays are used to measure the (relative) expression levels of thousands of genes (or expressed sequence tags). A comparison of gene expression in cells or tissues from two conditions may provide useful information on important biological processes or functions [1,2]. The challenge now is how to detect those genuine changes from noisy data. It is now known that simply using fold changes, as in the earlier days, is unreliable and inefficient [3,4]. More sophisticated statistical methods are called for. Many proposals have appeared in the literature [3,4,5,6,7,8,9,10]. In particular, it has been noticed that it may be necessary to design an experiment that uses multiple arrays (or multiple spots on each array) containing multiple measurements for each gene under each condition. One reason is that because of a high noise-to-signal ratio, a single array may not provide enough information that can be reliably extracted [11]. More important, multiple measurements from each gene make it possible to assess the potentially different variability of genes. The problem then seems to fall within the traditional two-sample comparison in statistics. Two of the best known two-sample statistical tests are the two-sample *t*-test and the Wilcoxon test (or equivalently, Mann-Whitney test). The *t*-test is parametric and is based on the assumption that the gene-expression levels have normal distributions. In contrast, the Wilcoxon test is nonparametric and is based on the ranks of observed gene-expression levels. Although the *t*-test is robust to departures from normality and the Wilcoxon test does not depend on the normality assumption, the problem is that under non-normal situations the *t*-test may be too conservative, and hence, as with the Wilcoxon test, may have too low power, especially when the sample size is small, which is the case for most microarray experiments. These points have been verified in two case studies using real data [8,12]. In a class of nonparametric approaches [5,9,10], a version of the two-sample *t*-statistic is used but its null distribution is estimated nonparametrically, rather than directly assumed to be a *t*-distribution. In addition, some earlier studies have suggested that the variability of gene expression may be related to the mean expression [3,4,6]. Therefore, it implies that the *t*-statistic being used should be based on unequal variances for the two samples.

An important and natural question often asked by biologists is how many replicates are required. For microarray experiments, unlike many other experimental contexts, this issue has rarely been discussed in the literature. To our knowledge, the only exception is the work by Black and Doerge [13], which, however, is for the situation where parametric statistical methods are applied to detect expression changes. In this paper, we discuss the problem when a nonparametric method, the normal mixture model approach [10], is used to detect differential expression. But to facilitate calculations of sample size, the formulation is slightly changed from their original one. Nonparametric methods of microarray data analysis have been pioneered by Efron and Tibshirani and co-workers [5,9]. They take advantage of the presence of replicates and thus can impose much weaker modeling assumptions. For instance, the parametric methods of Black and Doerge [13] depend on the assumption on the log-normal or gamma distribution of gene-expression levels, whereas the mixture model approach does not have such a distributional assumption and directly estimates distributions related to random errors. Note that modeling the distribution of random errors has advantages over direct modeling of expression levels, and is a common practice in applied statistics. For example, gene-expression levels may be correlated (for example, as a result of coexpression of some genes) whereas random errors can be more reasonably assumed to be independent. This is similar to modeling longitudinal data using a linear mixed-effects model [14]: the responses from each subject (corresponding to a group of coregulated genes here) are in general correlated, but the measurement errors from the same subject can be considered to be independent after incorporating a random-subject effect in the model. Note that the random effect will be canceled out from the *t*-statistic for each gene. Our proposal here also shows an attractive feature of the mixture model approach, as compared to the other two nonparametric approaches [5,9], because it is still unclear how the sample size/power calculation can be done in the other two approaches.

The problem of calculating the number of replicates required in a microarray experiment is similar to that of sample size/power calculations in clinical trials and other experiment designs; the (to-be-determined) sample size in microarray experiments refers to the number of replicates, whereas the number of genes is not an issue here. As usual, we assume that the replicates are (approximately) independent with each other, whether they are drawn from the same individual or multiple individuals. In general, the required sample size depends on several factors: the true magnitude of the change of gene expression (say, *d*), the desired statistical power (that is, probability) (
) to detect the change, and the specified Type I error rate (
). The problem of how to calculate the number of replicates for any given triplet (*d*,
,
) is equivalent to that of how the power
depends on the pair (*d*,
) and the number of replicates, which we consider in the paper.

The proposed method is not restricted to any specific microarray technology. From now on, the expression level can refer to a summary measure of relative red-to-green-channel intensities in a fluorescence-labeled cDNA array, a radioactive intensity of a radiolabeled cDNA array (as used in the example later), or a summary difference of the perfect match (PM) and mismatch (MM) scores from an oligonucleotide array. The gene-expression levels may have been suitably preprocessed, including dimension reduction, data normalization and data transformation [5,15,16,17,18].

## Results and discussion

### A statistical model

We consider a generic situation that, for each gene *i, I* = 1,2,..., *N,* we have (relative) expression levels *X*_{1i},..., *X*_{
mi
} from *m* microarrays under condition 1, and *Y*_{1i},..., *Y*_{
mi
} from *m* arrays under condition 2. We need to assume that *m* is an even integer. A general statistical model is assumed for gene expression data:

*X*_{
ji
} =
_{(1),i} +
_{
ji
}, *Y*_{
li
} =
_{(2),i} + *e*_{
li
},

_{(1),i}and

_{(2),i}are the mean expression levels for gene

*i*under the two conditions respectively, and

_{ ji }and

*e*

_{ li }are independent random errors with means and variances

for any *j* = 1,..., *m*, *l* = 1,..., *m* and *i* = 1,..., *N*. It is assumed that random errors
_{
ji
}/
_{(1),i} and
are randomly taken respectively from one of two (not necessarily equal) distributions that are symmetric about their mean 0. Note that the above assumption on the distributions of random errors, not on that of gene expression levels (that is, *X*_{
ji
} and *Y*_{
li
}), is often reasonable, and similar assumptions are common in other statistical applications. In addition, we do not assume that the expression levels of all the genes have an equal variance, because some previous studies [3,4,6] have found that the variance
^{2}_{(c),i} (for *c* = 1,2) of gene-expression levels may depend on the mean expression
_{(c),i}. Also, we do not even need to assume that
^{2}_{(1),i} =
^{2}_{(2),i} unless
_{(1),i} =
_{(2),i}.

### A test statistic

To test the null hypothesis *H*_{0}:
_{(1),i} =
_{(2),i}, we use a *t*-type test statistic or score

Note that the mean and variance of *Z*_{
i
} are

whereas the mean *E*(*Z*_{
i
}) = 0 under *H*_{0}. Hence, it can be seen that a large absolute value of *Z*_{
i
}, |*Z*_{
i
}|, gives evidence against *H*_{0}. As the number of arrays (that is, *m*) increases, the variance of the test statistic *Z*_{
i
} decreases. Hence, it is possible to reject *H*_{0} (that is, detect differential expression for gene *i*) with any *E*(*Z*_{
i
}) ≠ 0 if *m* is large enough. In other words, if the Type I error rate and other parameters are fixed, then the statistical power of the test will increase as *m* increases. This is the key point that motivates the discussion on sample size calculations.

To determine the cut-off point for |*Z*_{
i
}| to reject *H*_{0}, we need to know or estimate the distribution of *Z*_{
i
} under *H*_{0}, the null distribution *f*_{0}. In a parametric approach, based on some full distributional assumptions for *X*_{
ji
} and *Y*_{
ji
}, one may derive the null distribution *f*_{0}, such as in a two-sample *t*-test. However, the validity of such a parametric method critically depends on the correctness of assumed distributions, which of course is not guaranteed. Here, we consider a nonparametric approach: a finite normal mixture model is used to estimate *f*_{0} nonparametrically.

### Estimating the null distribution

There may be various ways to estimate the null distribution *f*_{0}. For instance, using expression levels of some housekeeping genes that are known to have non-differential expression, one can construct their *Z*_{
i
} scores and then estimate *f*_{0} using the obtained *Z*_{
i
} scores. In practice, however, there may be only a small number of or no housekeeping genes in a given experiment. Here, following the basic idea in a class of nonparametric methods [5,9,10], we construct a null score *z*_{
i
} for each gene and then use these null scores to estimate *f*_{0} nonparametrically. The null score is constructed from the same observed gene expression data as used in *Z*_{
i
}:

_{ ji }and

*e*

_{ ji }have symmetric distributions, then

_{ ji }and -

_{ ji }have the same distribution, and

*e*

_{ ji }and -

*e*

_{ ji }have the same distribution. Thus, by comparing the form of

*z*

_{ i }with that of

*Z*

_{ i }, we know that the distribution of

*z*

_{ i }is exactly

*f*

_{0}, the null distribution for

*Z*

_{ i }(under

*H*

_{0}). Note that under

*H*

_{0},

_{(1),i}=

_{(2),i}, and hence

_{(1),i}=

_{(2),I}(since we assume that

_{(c),i}only depends on

_{(c),i}), then

Thus *z*_{
i
} and *Z*_{
i
} have the same distribution *f*_{0} under *H*_{0}. We use all *z*_{
i
} values across all genes to estimate *f*_{0}.

_{(c),i}(for

*c*= 1, 2) are unknown, and can be estimated using the sample standard deviations (SDs)

*s*

_{(c),i}. Although the sample SD

*s*

_{(c),i}is asymptotically unbiased, if

*m*and

*n*are small,

*s*

_{(c),i}may not be stable, and some modifications may be necessary. In any case, substituting

_{(c),i}by any suitable estimates, we can calculate the scores

*z*

_{ i }values and

*Z*

_{ i }values, on the basis of which we can estimate

*f*

_{0}and

*f*respectively. By comparing

*f*

_{0}and

*f*, we can gain insight about genes with altered expression (that is,

_{(1),i}≠

_{(2),i}).

We assume that all the *z*_{
i
} values for *i* = 1,..., *N* are a random sample from *f*_{0}; thus we can use the observed *z*_{
i
} values to estimate *f*_{0}. Pan *et al*. [10] proposed estimating *f*_{0} using a finite normal mixture model [19]. Specifically, it is assumed that

*z*;

*a*

_{ r },

*V*

_{ r }) denotes the density function of a normal distribution

*N*(

*a*

_{ r },

*V*

_{ r }) with mean

*a*

_{ r }and variance

*V*

_{ r }, and

_{ r }values are mixing proportions. Ω

_{g0}represents all unknown parameters {

_{ r },

*a*

_{ r },

*V*

_{ r }) :

*r*= 1,...

*g*

_{0}} in a

*g*

_{0}-component mixture model. Among others, a normal mixture is essentially nonparametric and flexible, and easy to use with stable tail probabilities.

A mixture model can be fitted by maximum likelihood using the expectation-maximization (EM) algorithm [19,20,21]. The number of components can be selected adaptively using the Akaike Information Criterion (AIC) [22] or the Bayesian Information Criterion (BIC) [23]. In using the AIC or BIC, one first fits a series of models with various values of *g*_{0}, then picks up the *g*_{0} corresponding to the first local minimum of AIC or BIC [24]. Some empirical studies seem to favor the use of BIC [24].

### Determining the cut-off point

Once we obtain an estimate of the null distribution *f*_{0}, we can determine the cut-off point of the rejection region for testing *H*_{0}. In general, as for a two-sample test, the rejection region can be selected in the tails of *f*_{0} because, under the null hypothesis, *Z*_{
i
} should be close to the center of *f*_{0}, whereas if there is differential expression for gene *i*, *Z*_{
i
} is likely to be in one of the two tails of *f*_{0}. The specific choice may depend on the goal of the analysis. For example, if we are only interested in detecting upregulated genes, we can choose the rejection region at the right-tail of *f*_{0}. Our proposed method works for any specified way of determining the rejection region. As *f*_{0} should be symmetric about its mean 0, and often we are interested in both up- and downregulated genes, we propose to take the rejection region at the two tails of *f*_{0}, {*z* : *f*_{0}(*z*) <*C*
}, where the constant *C*
> 0 is the cut-off point and depends on the specified (gene-specific) Type I error rate
. As usual, *C*
> 0 is chosen such that the rejection rate under *H*_{0} is exactly
:

*a*,

*V*) is the corresponding cumulative distribution function for (.;

*a*,

*V*). Using a numerical algorithm, such as the bisection method [25], we can solve the above equation to obtain

*C*for any given .

For microarray data, because we are testing *H*_{0} for each gene, the multiple test problem arises and some control on it is necessary. Usually we can use Bonferroni's method. For instance, if we want to maintain the genome-wide Type I error rate at the usual 5% level, then the Bonferroni-adjusted gene-specific (that is, test-specific) Type I error rate is
= 0.05/*N*, where *N* is the total number of genes to be tested.

Once *C*
is determined, we can calculate the power as a function of *d*, the magnitude of the expression change targeted to be detected. Note that

_{(1),i}=

_{(2),i},

*d*can be interpreted as the change of the mean expression levels from condition 1 to condition 2. Otherwise, it can be regarded as the difference of (variation) standardized mean expression levels. Specifically, we have the power function

### Calculation of replicate numbers

Now we describe how to calculate replicate numbers based on some pilot data taken from earlier studies. We use *z*_{m,i} to explicitly denote the *z*_{
i
} scores in (2) with *m* replicates. Based on the data we can estimate the density function *f*_{0,m} (*z*;Ω_{g0}) of *z*_{m,i}values as a normal mixture

From now on, we treat *f*_{0,m} as known in Equation (5).

With estimated *f*_{0,m}, we want to estimate the density function *f*_{0,mk} for *z*_{mk,i}, the *z*_{
i
} scores based on *mk* replicates (with *k* > 1). If we can have an estimate of *f*_{0,mk}, then we can obtain the corresponding power function
(*d*,
) for *mk* replicates in the same way as described earlier for m replicates. Of course, we assume that our pilot data are drawn from only *m* arrays under each of the two experimental conditions, and thus we do not observe any *z*_{mk,i} based on *mk* arrays. However, we show next that it is possible to generate *z*_{mk,i} values from *z*_{m,i} values. Note that we can draw random realizations of *z*_{m,i} from the estimated *f*_{0,m} (see Pan *et al*. [10] or the example below). Suppose *z*_{m,i}^{(j)} values (for *j* = 1,2,..., *k*) are *k* independent realizations of *z*_{m,i}, then it is easy to show that

have the distribution *f*_{0,mk}. Thus, the density function for *z*_{mk,i} values is

For example, if we triple the number of replicates, the resulting density function is

The number of components of *f*_{0,mk} may be too large. For example, if the number of components is *g*_{0} = 3 for *m* = *n* = 2, the corresponding numbers of components for *m* = *n* = 4, *m* = *n* = 6 and *m* = *n* = 8 are, respectively, *g*_{0}^{2} = 9, *g*_{0}^{3} = 27 and *g*_{0}^{4} = 81. In fact, some of these components maybe very similar or have a negligible role, hence the form of *f*_{0,mk}, may be simplified. In the extreme situation, as *mk* → ∞, by the Central Limit Theorem, the mixture model will reduce to a single-component normal distribution. Hence, we propose a simulation-based method to select a more parsimonious model for *f*_{0,mk}.

On the basis of the mixture model *f*_{0,m} in Equation (5), we can generate a random sample of *z*_{m,i}^{(j)} values [10], from which we can calculate *z*_{mk,i} values using Equation (6). Using *z*_{mk,i} values we can fit a normal mixture model for *f*_{0,mk}. As we shall show later, we find such a fitted mixture model often contains a smaller number of components than *g*^{
k
}_{0}, as dictated in Equation (7), leading to a simplified form of *f*_{0,mk}.

### Summary of the proposed method

In summary, our proposed method of calculating the required replicate number works in the following steps.

*Step 1.* Suppose that we have pilot gene expression data *X*_{
ji
} and *Y*_{
ij
} from *m* arrays under each condition. Use formula (2) to calculate the scores *z*_{i,m}.

*Step 2.* Use *z*_{i,m} and the normal mixture model (5) to estimate *f*_{0,m}.

*Step 3.* For a specified Type I error rate
, determine the cutoff point *C*
for the rejection region using formula (3), in which *f*_{0} is replaced with the estimated *f*_{0,m}.

*Step 4.* For any specified *d*, calculate the power function
(*d*,
) using formula (4), in which *f*_{0} is replaced with the estimated *f*_{0,m}.

*Step 5.* For any given *k* > 1, use formula (7) or (6) to estimate *f*_{0,mk}.

*Step 6.* For a specified Type I error rate
, determine the cutoff point *C*
for the rejection region using formula (3), in which *f*_{0} is replaced with the estimated *f*_{0,mk}.

*Step 7.* For any specified *d*, calculate the power function
(*d*,
) using formulae (4), in which *f*_{0} is replaced with the estimated *f*_{0,mk}.

*Step 8.* Repeat Steps 5 to 7 until all *k* > 1 of interest have been tried.

After the power functions for many possible *mk* replicates have been obtained, we can determine an appropriate number of replicates by considering all the factors involved, the desired power and Type I error rate, the targeted expression changes and other experimental constraints.

### An example

To understand the pathogenesis of otitis media, a study was conducted to identify genes involved in response to pneumococcal middle-ear infection and to study their roles in otitis media. Radioactively labeled DNA microarrays were applied to the mRNA analysis of 1,176 genes in middle-ear mucosa of rats with and without subacute pneumococcal middle-ear infection [26]. The data are available for the control group and for the pneumococcal middle-ear infection group. A more detailed description of how the data were collected and their public availability was provided in Pan *et al.* [26]. For the purpose of sample size calculations and to mimic many practical situations with only a small number of replicates, we only use *m* = *n* = 2 arrays from each group. We first take a natural logarithm transformation for all the observed gene-expression levels (that is, radioactive intensities) so that the resulting distributions are less skewed (which will reduce the number of components of a fitted mixture model). Then, for each microarray, we standardize the transformed gene-expression levels by subtracting their median.

*m*= 2, the sample SDs may not be stable. One way is to add a small constant as suggested by Efron

*et al*. [5]. Here we follow the idea of Lin

*et al*. [27] and use a loess smoother [28] to nonparametrically model the sample SDs in terms of the mean expression levels (Figure 1). Then we plug in the smoothed SD to calculate

*z*

_{2,i}. Note that an alternative use of SD or its modification in calculating

*z*

_{2,i}values will not change the basic idea and the following steps in sample size calculations.

*f*

_{0,2}with

*g*

_{0}ranging from 1 to 3. Table 1 summarizes the model-fitting results.

*g*

_{0}= 1 was selected as both AIC and BIC achieve their minima there. So the fitted

*f*

_{0}is a normal distribution,

*N*(-0.0013, 0.1278). However, for the purposes of general illustration, we choose

*g*

_{0}= 2 as the fitted model:

AIC and BIC for fitted mixture models with various number of components *g*_{0}

Two replicates | Four replicates | Six replicates | Eight replicates | |||||
---|---|---|---|---|---|---|---|---|

| AIC | BIC | AIC | BIC | AIC | BIC | AIC | BIC |

1 | 3928.10 | 3938.24 | 3111.75 | 3121.89 | 2612.98 | 2623.12 | 2322.85 | 2332.99 |

2 | 3928.54 | 3953.89 | 3116.40 | 3141.75 | 2617.65 | 2643.00 | 2327.03 | 2352.38 |

3 | 3932.67 | 3973.23 | 3122.20 | 3162.76 | 2622.61 | 2663.17 | 2331.92 | 2372.48 |

*f*_{0,2}(*z*) = 0.76
(*z*;-0.0415, 1.3117) + 0.24
(*z*;0.0700, 2.6970).

*z*

_{ i }values and the fitted

*f*

_{0}with

*g*

_{0}= 1 and 2. There is not much difference between the two fitted

*f*

_{0,2}, both of which fit the data well. In particular,

*f*

_{0,2}does not look like a

*t*-distribution with small degrees of freedom, as predicted from the

*t*-test.

A realization of *z*_{2,i} can be simulated in the following two steps. First, we draw a random number *p*_{
i
} from {1, 2} with probability 0.76 and 0.24 respectively. Second, if the drawn *p*_{
i
} = 1, *z*_{
i
} is randomly drawn from a normal distribution
(*z*; -0.0415, 1.3117); otherwise, it is drawn from
(*z*; 0.0700, 2.6970). From the generated *z*_{2,i} values, following expression (6) we generated three simulated data sets: *z*_{2k,i} values, *I* = 1,..., 1,176 for *k* = 2, 3 and 4. Then a normal mixture model was fitted to each data set. From Table 1, it can be seen that a single-component normal distribution was selected in each case. In Figure 2, each of the fitted normal distributions, *N*(-0.0494, 0.8226), *N*(-0.0644, 0.5383) and *N*(-0.0438, 0.4206), is compared with its theoretically derived mixture model in Equation (7); they are all very close. Here we see that using simulated data to fit a mixture model results in a much-simplified model. For example, for *k* = 4, it is a fitted single-component model versus a 2^{4} = 16-component model in Equation (7). Note that, as predicted, all the means of the fitted models are all essentially 0, and their variances decrease as *k* increases.

*f*

_{0,2k}, the cut-off points

*C*are determined. Then the power functions (

*d*, ) are drawn in Figure 3, which may help make a decision on the required number of replicates. For instance, if we want to detect an expression change

*d*= 3 with probability at least 80% and with = 0.09%, then six replicates are needed. Also, with just two replicates, the power to detect a change as high as 4 is very low, smaller than 30%. Note that the choice of

*d*may depend on some prior knowledge. For instance, based on the pilot data, we can estimate the

*d*values for some selected genes (with the sample means and sample SDs substituting the true means and SDs in the formula for

*d*), from which one can determine a range of

*d*values of interest.

*N*= 1,000, 5,000 and 10,000 genes, respectively, while controlling the genome-wide Type I error rate at the usual 5% level. It can be seen that as

*N*increases, we also need a larger number of arrays to maintain the power of the statistical test when other parameters are fixed. For instance, for

*N*= 10,000 (Figure 6), even eight replicates cannot detect a change as large as

*d*= 3 with 80% power, but six replicates can detect a change

*d*= 4 with 80% power.

## Conclusions

We have described a method for calculating the number of replicates in microarray experiments. This method is designed for the situation where the mixture approach is going to be taken to analyze the data. Note that any method for sample size/power calculations has to depend on a specific statistical test to be used in data analysis; this explains why there is a huge literature on the topic for clinical trials. However, because of the close relation between the mixture approach and the other two recently proposed nonparametric approaches - the empirical Bayes method [5] and the statistical analysis of microarray (SAM) method [9] - our proposed method can be also applied to provide some useful guideline for designing microarray experiments even when one of the latter two approaches (or other approaches) is planned to be used for data analysis in a later stage. For instance, even though the null distribution *f*_{0} is estimated using the null scores *z*_{
i
} in our proposal, there maybe alternative ways of estimating *f*_{0}, such as using an alternative nonparametric method (for example, kernel or local likelihood), rather than the finite normal mixture model, to estimate *f*_{0} or using the test statistics, *Z*_{
i
}, of a large number of housekeeping genes to estimate *f*_{0}. Some modifications to the test statistic *Z*_{
i
} and the null statistic *z*_{
i
} are also possible, especially when we consider differential gene expression across more than two conditions. These are all interesting topics we are investigating now.

In most sample size/power calculations, some pilot data are needed to provide reasonable estimates of some parameters needed for subsequent calculations. An alternative is to obtain reasonable estimates from other similar studies in the literature. However, because of the rapid development of microarray technology, the latter is not likely and we expect a researcher will have to do his or her own pilot study. This was the situation we considered in the example. A particular challenge is how to obtain good estimates of the variances of gene expression levels from a small number of replicates. In our example, we considered a nonparametric method to smooth sample variances. Some alternative smoothing methods have also appeared in the literature. But it is not clear which one is the most desirable. This is a topic for future study.

The proposed method is straightforward to statisticians and can be implemented in many existing statistical packages. Our sample S-Plus program and data are available at [29].

## Declarations

### Acknowledgements

This research was partially supported by NIH.

## Authors’ Affiliations

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