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Using Fano factors to determine certain types of gene autoregulation

Yue Wang Department of Computational Medicine, University of California, Los Angeles, California, United States of America Institut des Hautes Études Scientifiques, Bures-sur-Yvette, Essonne, France E-mail address: yuew@g.ucla.edu (Y. W.). ORCID: 0000-0001-5918-7525 Siqi He Simons Center for Geometry and Physics, Stony Brook University, Stony Brook, New York, United States of America
Abstract

The expression of one gene might be regulated by its corresponding protein, which is called autoregulation. Although gene regulation is a central topic in biology, autoregulation is much less studied. In general, it is extremely difficult to determine the existence of autoregulation with direct biochemical approaches. Nevertheless, some papers have observed that certain types of autoregulations are linked to noise levels in gene expression. We generalize these results by two propositions on discrete-state continuous-time Markov chains. These two propositions form a simple but robust method to infer the existence of autoregulation in certain scenarios from gene expression data. This method only depends on the Fano factor, namely the ratio of variance and mean of the gene expression level. Compared to other methods for inferring autoregulation, our method only requires non-interventional one-time data, and does not need to estimate parameters. Besides, our method has few restrictions on the model. We apply this method to four groups of experimental data and find some genes that might have autoregulation. Some inferred autoregulations have been verified by experiments or other theoretical works.

Keywords.

inference; gene expression; autoregulation; Markov chain.

Frequently used abbreviations:

GRN: gene regulatory network.

VMR: variance-to-mean ratio

1 Introduction

In general, genes are transcribed to mRNAs and then translated to proteins. We can use the abundance of mRNA or protein to represent the expression levels of genes. Both the synthesis and degradation of mRNAs/proteins can be affected (activated or inhibited) by the expression levels of other genes [42], which is called (mutual) gene regulation. Genes and their regulatory relations form a gene regulatory network (GRN) [16], generally represented as a directed graph: each vertex is a gene, and each directed edge is a regulatory relationship. See Fig. 1 for an example of a GRN.

PLCGPIP3AKTPIP2RAFMEKPKCPKAERKJNKP38

Figure 1: An example of a GRN in human T cells [91]. Each vertex is a gene. Each arrow is a regulatory relationship. Notice that it has no directed cycle.

The expression of one gene could promote/repress its own expression, which is called positive/negative autoregulation [11]. Autoregulation is very common in E. coli [63]. Positive autoregulation is also called autocatalysis or autoactivation, and negative autoregulation is also called autorepression [4, 21]. For instance, HOX proteins form and maintain spatially inhomogeneous expression of HOX genes [64]. For genes with position-specific expressions during development, it is common that the increase of one gene can further increase or decrease its level [81]. Autoregulation has the effect of stabilizing transposons in genomes [6], which can affect cell behavior [41, 79]. Autoregulation can also stabilize the cell phenotype [2], which is related to cancer development [96, 52, 15, 90].

While countless works infer the regulatory relationships between different genes (the GRN structure) [86], determining the existence of autoregulation is an equally important yet less-studied field. Due to technical limitations, it is difficult and sometimes impossible to directly detect autoregulation in experiments. Instead, we can measure gene expression profiles and infer the existence of autoregulation. In this paper, we consider a specific data type: measure the expression levels of certain genes without intervention for a single cell (which reaches stationarity) at a single time point, and repeat for many different cells to obtain a probability distribution for expression levels. Such single-cell non-interventional one-time gene expression data can be obtained with a relatively low cost [48].

With such single-cell level data for one gene VV, we can calculate the ratio of variance and mean of the expression level (mRNA or protein count). This quantity is called the variance-to-mean ratio (VMR) or the Fano factor. Many papers that study gene expression systems with autoregulations have found that negative autoregulation can decrease noise (smaller VMR), and positive autoregulation can increase noise (larger VMR) [70, 68, 30, 51, 25, 18, 17]. This means VMR can be used to infer the existence of autoregulation.

We generalize the above observation and develop two mathematical results that use VMR to determine the existence of autoregulation. They apply to some genes that have autoregulation. For genes without autoregulation, these results cannot determine that autoregulation does not exist. We apply these results to four experimental gene expression data sets and detect some genes that might have autoregulation.

We start with some setup and introduce our main results (Section 2). Then we cite some previous works on this topic and compare them with our results (Section 3). For a single gene that is not regulated by other genes (Section 4) and multiple genes that regulate each other (Section 5), we develop mathematical results to identify the existence of autoregulation. These two mathematical sections can be skipped. We summarize the procedure of our method and apply it to experimental data (Section 6). We finish with some conclusions and discussions (Section 7).

2 Setup and main results

One possible mechanism of “the increase of one gene’s expression level further increases its expression level” is a positive feedback loop between two genes [31]. Here V1V_{1} and V2V_{2} promote each other, so that the increase of V1V_{1} increases V2V_{2}, which in return further increases V1V_{1}. We should not regard this feedback loop as autoregulation. When we define autoregulation for a gene VV, we should fix environmental factors and other genes that regulate VV, and observe whether the expression level of VV can affect itself. If VV is in a feedback loop that contains other genes, then those genes (which regulate VV and are regulated by VV) cannot be fixed when we change VV. Therefore, it is essentially difficult to determine whether VV has autoregulation in this scenario. In the following, we need to assume that VV is not contained in a feedback loop that involves other genes.

The actual gene expression mechanism might be complicated. Besides other genes/factors that can regulate a gene, for a gene VV itself, it might switch between inactivated (off) and activated (on) states [10]. These states correspond to different transcription rates to produce mRNAs. When mRNAs are translated into proteins, those proteins might affect the transition of gene activation states, which forms autoregulation [23]. See Fig. 2 for an illustration. Therefore, for a gene VV, we should regard the gene activation state, mRNA count, and protein count as a triplet of random variables (G,M,P)(G,M,P), which depend on each other.

inactivated genemRNAproteinactivated gene

Figure 2: The mechanism of gene expression. A gene might switch between inactivated state and activated state, which correspond to different transcription rates. Gene is transcribed into mRNAs, which are translated into proteins. Proteins might (auto)regulate the state transition of the corresponding gene.

When we fix environmental factors and other genes that affect VV, the triplet (G,M,P)(G,M,P) should follow a continuous-time Markov chain. A possible state is the gene activation state on/off (for GG), the mRNA count on \mathbb{Z} (for MM), and the protein count on \mathbb{Z} (for PP). Thus the total space is {0,1}××\{0,1\}\times\mathbb{Z}\times\mathbb{Z}. When we consider the expression level MM or PP (but have no access to the value of GG), sometimes itself is Markovian (its dynamics can be fully determined by itself, without the knowledge of GG), and we call this scenario “autonomous”. In other cases, MM or PP itself is no longer Markovian (its dynamics explicitly depends on GG), and we call this scenario “non-autonomous”. We need to consider the triplet (G,M,P)(G,M,P) in the non-autonomous scenario. This is similar to a hidden Markov model, where a two-dimensional Markov chain is no longer Markovian if projected to one dimension (since this dimension depends on the other dimension).

For the autonomous scenario, we can fully classify autoregulation for a gene VV. Assume environmental factors and other genes that affect the expression of VV are kept at constants. Define the expression level (mRNA count for example) of one cell to be X=nX=n, the mRNA synthesis rate at X=n1X=n-1 to be fnf_{n}, and the degradation rate for each mRNA molecule at X=nX=n to be gng_{n}. This is a standard continuous-time Markov chain on \mathbb{Z} with transition rates

1Δt[X(t+Δt)=nX(t)=n1]=fn,\frac{1}{\Delta t}\mathbb{P}[X(t+\Delta t)=n\mid X(t)=n-1]=f_{n},
1Δt[X(t+Δt)=n1X(t)=n]=ngn.\frac{1}{\Delta t}\mathbb{P}[X(t+\Delta t)=n-1\mid X(t)=n]=ng_{n}.

Define the relative growth rate hn=fn/gnh_{n}=f_{n}/g_{n}. If there is no autoregulation, then hnh_{n} is a constant. Positive autoregulation means hn>hn1h_{n}>h_{n-1} for some nn, so that fn>fn1f_{n}>f_{n-1} and/or gn<gn1g_{n}<g_{n-1}; negative autoregulation means hn<hn1h_{n}<h_{n-1} for some nn, so that fn<fn1f_{n}<f_{n-1} and/or gn>gn1g_{n}>g_{n-1}. Notice that we can have hn>hn1h_{n}>h_{n-1} for some nn and hn<hn1h_{n^{\prime}}<h_{n^{\prime}-1} for some other nn^{\prime}, meaning that positive autoregulation and negative autoregulation can both exist for the same gene, but occur at different expression levels. Such non-monotonicity in regulating gene expression often appear in reality [1].

For the non-autonomous scenario, we can still define autoregulation. Consider the expression level XX of VV (mRNA count or protein count) and its interior factor II. If XX is the mRNA count, then II is the gene state; if XX is the protein count, then II is the gene state and the mRNA count. If there is no autoregulation, then XX cannot affect II, and for each value of II, the relative growth rate hnh_{n} of XX is a constant. If XX can affect II, or hnh_{n} is not a constant, then there is autoregulation. When XX can affect II, there is a directed cycle (XIXX\to I\to X), and the change of XX can affect itself through II. In this case, it is not always easy to distinguish between positive autoregulation and negative autoregulation.

Quantitatively, for the autonomous scenario, when we fix other factors that might regulate this gene VV, if VV has no autoregulation, then hn=fn/gnh_{n}=f_{n}/g_{n} is a constant hh for all nn. In this case, the stationary distribution of VV satisfies (X=n)/(X=n1)=h/n\mathbb{P}(X=n)/\mathbb{P}(X=n-1)=h/n, meaning that the distribution is Poissonian with parameter hh, (X=n)=hneh/n!\mathbb{P}(X=n)=h^{n}e^{-h}/n!, and VMR=1\text{VMR}=1. If there exists positive autoregulation of certain forms, VMR>1\text{VMR}>1; if there exists negative autoregulation of certain forms, VMR<1\text{VMR}<1. However, such results are derived by assuming that fn,gnf_{n},g_{n} take certain functional forms, such as linear functions [54, 57], quadratic functions [24], or Hill functions [67]. There are other papers that consider Markov chain models in gene expression/regulation [32, 61, 65, 14, 62, 45], but the role of VMR is not thoroughly studied.

In this paper, we generalize the above result of inferring autoregulation with VMR by dropping the restrictions on parameters. Consider a gene VV in a known GRN, and assume it is not regulated by other genes, or assume other factors that regulate VV are fixed. Assume we have the autonomous scenario, meaning that its expression level X=nX=n satisfies a general Markov chain with synthesis rate fnf_{n} and per molecule degradation rate gng_{n}. We do not add any restrictions on fnf_{n} and gng_{n}. Use the single-cell non-interventional one-time gene expression data to calculate the VMR of VV. Proposition 1 states that VMR>1\text{VMR}>1 or VMR<1\text{VMR}<1 means the existence of positive/negative autoregulation.

Nevertheless, the autonomous condition requires some assumptions, and often does not hold in reality [5, 39, 33, 32]. Consider a gene VV that is not regulated by other genes, and has no autoregulation. The mRNA count or the protein count is regulated by the gene activation state (an interior factor), which cannot be fixed. Due to this non-controllable factor, there might be transcriptional bursting [60, 19] or translational bursting [8], where transcription or translation can occur in bursts, and we have VMR>1\text{VMR}>1. This does not mean that Proposition 1 is wrong. Instead, it means that the expression level itself is not Markovian, and the scenario is non-autonomous. In this scenario, we should apply Proposition 2, described below, which states that no autoregulation means VMR1\text{VMR}\geq 1.

We extend the idea of inferring autoregulation with VMR to a gene that is regulated by other genes, or with non-autonomous expression. Consider a gene VV^{\prime} in a known GRN. Denote other genes that regulate VV^{\prime} and the interior factors (gene state and/or mRNA count) of VV^{\prime} by 𝑭\boldsymbol{F}. Denote the values of V,𝑭V^{\prime},\boldsymbol{F} as X,𝒀X,\boldsymbol{Y}. Assume VV^{\prime} is not contained in a feedback loop, and assume gng_{n}, the per molecule degradation rate of VV^{\prime} at X=nX=n, is not regulated by other genes or its interior factors (gene state and/or mRNA count). We do not add any restrictions on the synthesis rate fnf_{n}. Proposition 2 states that if VV^{\prime} has no autoregulation, then VMR(X)1\text{VMR}(X)\geq 1. Therefore, VMR(X)<1\text{VMR}(X)<1 means autoregulation for VV^{\prime}.

Proposition 2 is derived in a “one-step” Markov chain model, where at one time point, only transitions to the nearest neighbors are allowed: (X=n,𝒀=𝒂)(X=n+1,𝒀=𝒂)(X=n,\boldsymbol{Y}=\boldsymbol{a})\to(X=n+1,\boldsymbol{Y}=\boldsymbol{a}), (X=n,𝒀=𝒂)(X=n1,𝒀=𝒂)(X=n,\boldsymbol{Y}=\boldsymbol{a})\to(X=n-1,\boldsymbol{Y}=\boldsymbol{a}), and (X=n,𝒀=𝒂)(X=n,𝒀=𝒂)(X=n,\boldsymbol{Y}=\boldsymbol{a})\to(X=n,\boldsymbol{Y}=\boldsymbol{a}^{\prime}). This one-step Markov chain model is the most common approach in stochastic representations of gene regulation [70, 30, 54, 51, 17]. Recently, there are some studies that consider “multi-step” Markov chain models, where at one time point, the change of mRNA/protein count can be accompanied with the change of other factors, such as the gene state [7, 43, 74]. For example, the following transition is allowed: (G,M=n)(G,M=n+1)(G^{*},M=n)\to(G,M=n+1). In this multi-step model, Proposition 2 is no longer valid: even without autoregulation, it is possible that VMR(X)<1\text{VMR}(X)<1. Consider an example that the production of one mRNA molecule needs many steps of gene state transition, and the gene returns to the initial step after producing one mRNA molecule: G1G2GkG1+MG_{1}\to G_{2}\to\cdots\to G_{k}\to G_{1}+M, MM\to\emptyset. Since there are many steps, the total time for one cycle of G1GkG1+MG_{1}\to\cdots\to G_{k}\to G_{1}+M can be highly deterministic, such as 11 second. Assume the degradation probability for each mRNA molecule in 11 second is 0.010.01. Then the mRNA count is highly concentrated near 100100, and VMR(X)<1\text{VMR}(X)<1 (close to 0.50.5 in numerical simulations).

Since multi-step models allow more transitions, they are more general than one-step models. However, it is still a question that whether such generalizations are necessary, since one-step models have good fitting with experimental data [38, 18, 9]. Proposition 2 provides a method to verify this problem: If a gene has VMR(X)<1\text{VMR}(X)<1, but we use other methods to determine that it has no autoregulation, then Proposition 2 states that one-step models deviate from reality, and multi-step models should be adopted. Therefore, when one-step models hold, Proposition 2 is a valid method to determine the existence of autoregulation; when one-step models do not hold, combined with other methods to determine autoregulation, Proposition 2 can detect the failure of one-step models.

In the scenario that Proposition 2 may apply, if VMR1\text{VMR}\geq 1, Proposition 2 cannot determine whether autoregulation exists. In fact, with VMR, or even the full probability distribution, we might not distinguish a non-autonomous system with autoregulation from a non-autonomous system without autoregulation, which both have VMR1\text{VMR}\geq 1 [9]. In the non-autonomous scenario, we only focus on the less complicated case of VMR<1\text{VMR}<1, and derive Proposition 2 that firmly links VMR and autoregulation.

In reality, Proposition 1 and Proposition 2 can only apply to a few genes (which are not regulated by other genes or have VMR<1\text{VMR}<1), and they cannot determine negative results. Thus the inference results about autoregulation are a few “yes” and many “we do not know”. Besides, for the results inferred by Proposition 1, especially those with VMR>1\text{VMR}>1 (positive autoregulation), we cannot verify whether their expression is autonomous, and the inference results are less reliable.

Current experimental methods can hardly determine the existence of autoregulation, and to determine that a gene does not have autoregulation is even more difficult. Therefore, about whether genes in a GRN have autoregulation, experimentally, we do not have “yes” or “no”, but a few “yes” and many “we do not know”. Thus there is no gold standard to thoroughly evaluate the performance of our inference results. We can only report that some genes inferred by our method to have autoregulation are also verified by experiments or other inference methods to have autoregulation. Besides, if the result by Proposition 2 does not match with other methods, it is possible that the one-step model fails. Instead, in Section A, we test our methods with numerical simulations, and the performances of both Propositions are satisfactory.

3 Related works

There have been some results of inferring autoregulation with VMR [70, 68, 30, 51, 25, 18, 17]. However, these VMR-based methods have various restrictions on the model, and some of them are derived by applying linear noise approximations, which are not always reliable in gene regulatory networks [72].

Besides VMR-based methods, there are other mathematical approaches to infer the existence of autoregulation in gene expression [59, 92, 22, 73, 37, 97, 35, 34]. We introduce some works and compare them with our method. (A) Sanchez-Castillo et al. [59] considered an autoregressive model for multiple genes. This method (1) needs time series data; (2) requires the dynamics to be linear; (3) estimates a group of parameters. (B) Xing et al. [92] applied causal inference to a complicated gene expression model. This method (1) needs promoter sequences and information on transcription factor binding sites; (2) requires linearity for certain steps; (3) estimates a group of parameters. (C) Feigelman et al. [22] applied a Bayesian method for model selection. This method (1) needs time series data; (2) estimates a group of parameters. (D) Veerman et al. [73] considered the probability-generating function of a propagator model. This method (1) needs time series data; (2) estimates a group of parameters; (3) needs to approximate a Cauchy integral. (E) Jia et al. [37] compared the relaxation rate with degradation rate. This method (1) needs interventional data; (2) only works for a single gene that is not regulated by other genes; (3) requires that the per molecule degradation rate is a constant.

Compared to other more complicated methods, VMR-based methods (including ours) have two advantages: (1) VMR-based methods use non-interventional one-time data. Time series data require measuring the same cell multiple times without killing it, and interventional data require some techniques to interfere with gene expression, such as gene knockdown. Therefore, non-interventional one-time data used in VMR-based methods are much easier and cheaper to obtain. (2) VMR-based methods do not estimate parameters, and only calculate the mean and variance of the expression level. Some other methods need to estimate many parameters or approximate some complicated quantities, meaning that they need large data size and high data accuracy. Therefore, our method is easy to calculate, and need lower data accuracy and smaller data size.

Compared to other VMR-based methods, our method has few restrictions on the model, making them applicable to various scenarios with different dynamics. Besides, our derivations do not use any approximations.

In sum, compared to other VMR-based methods, our method is universal. Compared to other non-VMR-based methods, our method is simple, and has lower requirements on data quality.

Compared to other non-VMR-based methods, our method has some disadvantages: (1) The GRN structure needs to be known. (2) Our method does not work for certain genes, depending on regulatory relationships. Proposition 1 only works for a gene that is not regulated by other genes, and we require its expression to be autonomous; Proposition 2 only works for a gene that is not in a feedback loop. (3) Proposition 2 requires the per molecule degradation rate to be a constant, and it cannot provide information about autoregulation if VMR1\text{VMR}\geq 1. (4) Our method only works for cells at equilibrium. Thus time series data that contain time-specific information cannot be utilized other than treated as one-time data. With just the stationary distribution, sometimes it is impossible to build the causal relationship (including autoregulation) [85]. Thus with this data type, some disadvantages are inevitable. Such impossibility results might be generalized to other data types or even other fields [77].

4 Scenario of a single isolated gene

4.1 Setup

We first consider the expression level (e.g., mRNA count) of one gene VV in a single cell. At the single-cell level, gene expression is essentially stochastic, and we do not further consider differential equation approaches [80] dynamical system approaches with deterministic [82] or stochastic [95] operators. We use a random variable XX to represent the mRNA count of VV. We assume VV is not in a feedback loop. We also assume all environmental factors and other genes that can affect XX are kept at constant levels, so that we can focus on VV alone. This can be achieved if no other genes point to gene VV in the GRN, such as PIP3 in Fig. 1. Then we assume that the expression of VV is autonomous, thus XX satisfies a time-homogeneous Markov chain defined on \mathbb{Z}^{*}.

Assume that the mRNA synthesis rate at X(t)=n1X(t)=n-1, namely the transition rate from X=n1X=n-1 to X=nX=n, is fn0f_{n}\geq 0. Assume that with nn mRNA molecules, the degradation rate for each mRNA molecule is gn>0g_{n}>0. Then the overall degradation rate at X(t)=nX(t)=n, namely the transition rate from X=nX=n to X=n1X=n-1, is gnng_{n}n. The associated master equation is

d[X(t)=n]dt=[X(t)=n+1]gn+1(n+1)+[X(t)=n1]fn[X(t)=n](fn+1+gnn).\begin{split}\frac{\mathrm{d}\mathbb{P}[X(t)=n]}{\mathrm{d}t}=&\mathbb{P}[X(t)=n+1]g_{n+1}(n+1)+\mathbb{P}[X(t)=n-1]f_{n}\\ &-\mathbb{P}[X(t)=n](f_{n+1}+g_{n}n).\end{split} (1)

When fn,gnf_{n},g_{n} take specific forms, this master equation also corresponds to a branching process, so that related techniques can be applied [40]. Define the relative growth rate hn=fn/gnh_{n}=f_{n}/g_{n}. We assume that as time tends to infinity, the process reaches equilibrium, where (1) the stationary probability distribution Pn=limt[X(t)=n]P_{n}=\lim_{t\to\infty}\mathbb{P}[X(t)=n] exists, and Pn=Pn1hn/nP_{n}=P_{n-1}h_{n}/n; (2) the mean limt𝔼[X(t)]\lim_{t\to\infty}\mathbb{E}[X(t)] and the variance limtσ2[X(t)]\lim_{t\to\infty}\sigma^{2}[X(t)] are finite. Such requirements can be satisfied under simple assumptions, such as assuming hnh_{n} has a finite upper bound [53, 83].

If hn>hn1h_{n}>h_{n-1} for some nn, then there exists positive autoregulation. If hn<hn1h_{n}<h_{n-1} for some nn, then there exists negative autoregulation. If there is no autoregulation, then hnh_{n} is a constant hh, and the stationary distribution is Poissonian with parameter hh. In this setting, positive autoregulation and negative autoregulation might coexist, meaning that hn+1<hnh_{n+1}<h_{n} for some nn and hn+1>hnh_{n^{\prime}+1}>h_{n^{\prime}} for some nn^{\prime}.

4.2 Theoretical results

With single-cell non-interventional one-time gene expression data for one gene, we have the stationary distribution of the Markov chain XX. We can infer the existence of autoregulation with the VMR of XX, defined as VMR(X)=σ2(X)/𝔼(X)\text{VMR}(X)=\sigma^{2}(X)/\mathbb{E}(X). The idea is that if we let fnf_{n} increase/decrease with nn, and control gng_{n} to make 𝔼(X)\mathbb{E}(X) invariant, then the variance σ2(X)\sigma^{2}(X) increases/decreases [76, Section 2.5.1]. We shall prove that VMR>1\text{VMR}>1 implies the occurrence of positive autoregulation, and VMR<1\text{VMR}<1 implies the occurrence of negative autoregulation. Notice that VMR>1\text{VMR}>1 does not exclude the possibility that negative autoregulation exists for some expression level. This also applies to VMR<1\text{VMR}<1 and positive autoregulation.

We can illustrate this result with a linear model:

Example 1.

Consider a Markov chain that satisfies Eq. 1, and set fn=k+b(n1)f_{n}=k+b(n-1), gn=cg_{n}=c. Here bb (can be positive or negative) is the strength of autoregulation, and cc satisfies c>0c>0 and cb>0c-b>0. We can calculate that VMR=1+b/(cb)\text{VMR}=1+b/(c-b) (see Appendix B.1 for details). Therefore, VMR>1\text{VMR}>1 means positive autoregulation, b>0b>0; VMR<1\text{VMR}<1 means negative autoregulation, b<0b<0; VMR=1\text{VMR}=1 means no autoregulation, b=0b=0.

Lemma 1.

Consider a Markov chain X(t)X(t) that follows Eq. 1 with general transition coefficients fn,gnf_{n},g_{n}. Here X(t)X(t) models the mRNA/protein count of one gene whose expression is autonomous. Calculate VMR(X)\text{VMR}(X) at stationarity. (1) Assume hn+1hnh_{n+1}\geq h_{n} for all nn. We have VMR(X)1\text{VMR}(X)\geq 1; moreover, VMR(X)=1\text{VMR}(X)=1 if and only if hn+1=hnh_{n+1}=h_{n} for all nn. (2) Assume hn+1hnh_{n+1}\leq h_{n} for all nn. We have VMR(X)1\text{VMR}(X)\leq 1; moreover, VMR(X)=1\text{VMR}(X)=1 if and only if hn+1=hnh_{n+1}=h_{n} for all nn.

We can take negation of Lemma 1 to obtain the following proposition.

Proposition 1.

In the setting of Lemma 1, (1) If VMR(X)>1\text{VMR}(X)>1, then there exists at least one value of nn for which hn+1>hnh_{n+1}>h_{n}; thus this gene has positive autoregulation. (2) If VMR(X)<1\text{VMR}(X)<1, then there exists at least one value of nn for which hn+1<hnh_{n+1}<h_{n}; thus this gene has negative autoregulation. (3) If VMR(X)=1\text{VMR}(X)=1, then either (A) hn+1=hnh_{n+1}=h_{n} for all nn, meaning that this gene has no autoregulation; or (B) hn+1<hnh_{n+1}<h_{n} for one nn and hn+1>hnh_{n^{\prime}+1}>h_{n^{\prime}} for another nn^{\prime}, meaning that this gene has both positive and negative autoregulation (at different expression levels).

Remark 1.

Results similar to Proposition 1 have been proven by Jia et al. in another model of expression for a single gene [38]. However, they require that gi=gjg_{i}=g_{j} for any i,ji,j. Proposition 1 can handle arbitrary gig_{i}, thus being novel.

Proof of Lemma 1.

Define λ=logP0\lambda=-\log P_{0}, so that P0=exp(λ)P_{0}=\exp(-\lambda). Define dn=i=1nhi>0d_{n}=\prod_{i=1}^{n}h_{i}>0 and stipulate that d0=1d_{0}=1. We can see that

dndn+2dn+12=hn+2hn+1.\frac{d_{n}d_{n+2}}{d_{n+1}^{2}}=\frac{h_{n+2}}{h_{n+1}}.

Also,

Pn=Pn1fn/(gnn)=Pn1hn/n==P0(i=1nhi)/n!=eλdnn!.P_{n}=P_{n-1}f_{n}/(g_{n}n)=P_{n-1}h_{n}/n=\cdots=P_{0}(\prod_{i=1}^{n}h_{i})/n!\ =e^{-\lambda}\frac{d_{n}}{n!}.

Then

𝔼(X2)𝔼(X)=n=1(n2n)Pn=eλn=1(n2n)dnn!=eλn=2dn(n2)!=eλn=0dn+2n!,\begin{split}\mathbb{E}(X^{2})-\mathbb{E}(X)&=\sum_{n=1}^{\infty}(n^{2}-n)P_{n}=e^{-\lambda}\sum_{n=1}^{\infty}(n^{2}-n)\frac{d_{n}}{n!}\\ &=e^{-\lambda}\sum_{n=2}^{\infty}\frac{d_{n}}{(n-2)!}=e^{-\lambda}\sum_{n=0}^{\infty}\frac{d_{n+2}}{n!},\end{split}
[𝔼(X)]2=(n=1nPn)2=e2λ(n=1ndnn!)2=e2λ(n=0dn+1n!)2.[\mathbb{E}(X)]^{2}=\left(\sum_{n=1}^{\infty}nP_{n}\right)^{2}=e^{-2\lambda}\left(\sum_{n=1}^{\infty}n\frac{d_{n}}{n!}\right)^{2}=e^{-2\lambda}\left(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}\right)^{2}.

Besides,

1=n=0Pn=eλn=0dnn!.1=\sum_{n=0}^{\infty}P_{n}=e^{-\lambda}\sum_{n=0}^{\infty}\frac{d_{n}}{n!}.

Now we have

𝔼(X2)𝔼(X)[𝔼(X)]2=e2λ(n=0dnn!)(n=0dn+2n!)e2λ(n=0dn+1n!)2.\mathbb{E}(X^{2})-\mathbb{E}(X)-[\mathbb{E}(X)]^{2}=e^{-2\lambda}\left(\sum_{n=0}^{\infty}\frac{d_{n}}{n!}\right)\left(\sum_{n=0}^{\infty}\frac{d_{n+2}}{n!}\right)-e^{-2\lambda}\left(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}\right)^{2}.

(1) Assume hn+1hnh_{n+1}\geq h_{n} for all nn. Then

𝔼(X2)𝔼(X)[𝔼(X)]2e2λ(n=0dndn+2n!)2e2λ(n=0dn+1n!)20.\begin{split}&\mathbb{E}(X^{2})-\mathbb{E}(X)-[\mathbb{E}(X)]^{2}\\ \geq&e^{-2\lambda}\left(\sum_{n=0}^{\infty}\frac{\sqrt{d_{n}d_{n+2}}}{n!}\right)^{2}-e^{-2\lambda}\left(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}\right)^{2}\geq 0.\end{split} (2)

Here the first inequality is from the Cauchy-Schwarz inequality, and the second inequality is from dndn+2dn+12d_{n}d_{n+2}\geq d_{n+1}^{2} for all nn. Then VMR(X)={𝔼(X2)[𝔼(X)]2}/𝔼(X)1\text{VMR}(X)=\{\mathbb{E}(X^{2})-[\mathbb{E}(X)]^{2}\}/\mathbb{E}(X)\geq 1. Equality holds if and only if dn/dn+2=dn+1/dn+3d_{n}/d_{n+2}=d_{n+1}/d_{n+3} for all nn (the first inequality of Eq. 2) and dndn+2=dn+12d_{n}d_{n+2}=d_{n+1}^{2} for all nn (the second inequality of Eq. 2). The equality condition is equivalent to hn+1=hnh_{n+1}=h_{n} for all nn.

(2) Assume hn+1hnh_{n+1}\leq h_{n} for all nn. Then dn+2/dn+1dn+1/dnd_{n+2}/d_{n+1}\leq d_{n+1}/d_{n}, and dnh1nd_{n}\leq h_{1}^{n} for all nn. Define

H(t)=n=0dnn!tn.H(t)=\sum_{n=0}^{\infty}\frac{d_{n}}{n!}t^{n}.

Since 0<dnh1n0<d_{n}\leq h_{1}^{n}, this series converges for all tt\in\mathbb{C}, so that H(t)H(t) is a well-defined analytical function on \mathbb{C}, and

H(t)=n=0dn+1n!tn, and H′′(t)=n=0dn+2n!tn.H^{\prime}(t)=\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}t^{n},\ \text{ and }\ H^{\prime\prime}(t)=\sum_{n=0}^{\infty}\frac{d_{n+2}}{n!}t^{n}.

In the following, we only consider H(t),H(t),H′′(t)H(t),H^{\prime}(t),H^{\prime\prime}(t) as real functions for tt\in\mathbb{R}.

To prove VMR(X)1\text{VMR}(X)\leq 1, we just need to prove 𝔼(X2)𝔼(X)[𝔼(X)]2=e2λ{H(1)H′′(1)[H(1)]2}0\mathbb{E}(X^{2})-\mathbb{E}(X)-[\mathbb{E}(X)]^{2}=e^{-2\lambda}\{H(1)H^{\prime\prime}(1)-[H^{\prime}(1)]^{2}\}\leq 0. However, we shall prove H′′(t)H(t)[H(t)]2H^{\prime\prime}(t)H(t)\leq[H^{\prime}(t)]^{2} for all tt\in\mathfrak{I}, where =(a,b)\mathfrak{I}=(a,b) is a fixed interval in \mathbb{R} with 0<a<10<a<1 and 1<b<1<b<\infty. Thus t=1t=1 is an interior point of \mathfrak{I}. Since H(t),H(t),H′′(t)H(t),H^{\prime}(t),H^{\prime\prime}(t) have positive lower bounds on \mathfrak{I}, the following statements are obviously equivalent: (i) H′′(t)H(t)[H(t)]2H^{\prime\prime}(t)H(t)\leq[H^{\prime}(t)]^{2} for all tt\in\mathfrak{I}; (ii) {log[H(t)/H(t)]}0\{\log[H^{\prime}(t)/H(t)]\}^{\prime}\leq 0 for all tt\in\mathfrak{I}; (iii) log[H(t)/H(t)]\log[H^{\prime}(t)/H(t)] is non-increasing on \mathfrak{I}; (iv) H(t)/H(t)H^{\prime}(t)/H(t) is non-increasing on \mathfrak{I}. To prove (i), we just need to prove (iv).

Consider any t1,t2t_{1},t_{2}\in\mathfrak{I} with t1t2t_{1}\leq t_{2} and any p,qp,q\in\mathbb{N} with pqp\geq q. Since dp+1/dpdq+1/dqd_{p+1}/d_{p}\leq d_{q+1}/d_{q}, and t1pqt2pqt_{1}^{p-q}\leq t_{2}^{p-q}, we have

dpdqt1qt2q(dp+1dpdq+1dq)(t1pqt2pq)0,d_{p}d_{q}t_{1}^{q}t_{2}^{q}(\frac{d_{p+1}}{d_{p}}-\frac{d_{q+1}}{d_{q}})(t_{1}^{p-q}-t_{2}^{p-q})\geq 0,

which means

dp+1dqt1pt2q+dq+1dpt1qt2pdp+1dqt2pt1q+dq+1dpt2qt1p.d_{p+1}d_{q}t_{1}^{p}t_{2}^{q}+d_{q+1}d_{p}t_{1}^{q}t_{2}^{p}\geq d_{p+1}d_{q}t_{2}^{p}t_{1}^{q}+d_{q+1}d_{p}t_{2}^{q}t_{1}^{p}.

Sum over all p,qp,q\in\mathbb{N} with pqp\geq q to obtain

H(t1)H(t2)=(n=0dn+1n!t1n)(n=0dnn!t2n)(n=0dn+1n!t2n)(n=0dnn!t1n)=H(t2)H(t1).\begin{split}H^{\prime}(t_{1})H(t_{2})&=(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}t_{1}^{n})(\sum_{n=0}^{\infty}\frac{d_{n}}{n!}t_{2}^{n})\\ &\geq(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}t_{2}^{n})(\sum_{n=0}^{\infty}\frac{d_{n}}{n!}t_{1}^{n})=H^{\prime}(t_{2})H(t_{1}).\end{split}

Thus H(t1)/H(t1)H(t2)/H(t2)H^{\prime}(t_{1})/H(t_{1})\geq H^{\prime}(t_{2})/H(t_{2}) for all t1,t2t_{1},t_{2}\in\mathfrak{I} with t1t2t_{1}\leq t_{2}. This means H′′(t)H(t)[H(t)]2H^{\prime\prime}(t)H(t)\leq[H^{\prime}(t)]^{2} for all tt\in\mathfrak{I}, and VMR(X)1\text{VMR}(X)\leq 1.

About the condition for the equality to hold, assume hn+1<hnh_{n^{\prime}+1}<h_{n^{\prime}} for a given nn^{\prime}. Then

dndn1t1n1t2n1(dn+1dndndn1)(t1t2)C(t2t1)d_{n^{\prime}}d_{n^{\prime}-1}t_{1}^{n^{\prime}-1}t_{2}^{n^{\prime}-1}(\frac{d_{n^{\prime}+1}}{d_{n^{\prime}}}-\frac{d_{n^{\prime}}}{d_{n^{\prime}-1}})(t_{1}-t_{2})\geq C(t_{2}-t_{1})

for all t1,t2t_{1},t_{2}\in\mathfrak{I} with t1t2t_{1}\leq t_{2} and a constant CC that does not depend on t1,t2t_{1},t_{2}. Therefore,

[H(t1)/H(t1)H(t2)/H(t2)][H(t1)H(t2)]=(n=0dn+1n!t1n)(n=0dnn!t2n)(n=0dn+1n!t2n)(n=0dnn!t1n)dndn1t1n1t2n1(dn+1dndndn1)(t1t2)C(t2t1).\begin{split}&[H^{\prime}(t_{1})/H(t_{1})-H^{\prime}(t_{2})/H(t_{2})]\cdot[H(t_{1})H(t_{2})]\\ =&(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}t_{1}^{n})(\sum_{n=0}^{\infty}\frac{d_{n}}{n!}t_{2}^{n})-(\sum_{n=0}^{\infty}\frac{d_{n+1}}{n!}t_{2}^{n})(\sum_{n=0}^{\infty}\frac{d_{n}}{n!}t_{1}^{n})\\ \geq&d_{n^{\prime}}d_{n^{\prime}-1}t_{1}^{n^{\prime}-1}t_{2}^{n^{\prime}-1}(\frac{d_{n^{\prime}+1}}{d_{n^{\prime}}}-\frac{d_{n^{\prime}}}{d_{n^{\prime}-1}})(t_{1}-t_{2})\\ \geq&C(t_{2}-t_{1}).\end{split}

Since H(t)H(t) has a finite positive upper bound AA and a positive lower bound BB on \mathfrak{I}, we have

H(t1)/H(t1)H(t2)/H(t2)C(t2t1)/A2,H^{\prime}(t_{1})/H(t_{1})-H^{\prime}(t_{2})/H(t_{2})\geq C(t_{2}-t_{1})/A^{2},

meaning that

t,[H(t)/H(t)]={H(t)H′′(t)[H(t)]2}/[H(t)]2C/A2,\forall t\in\mathfrak{I},\,\,[H^{\prime}(t)/H(t)]^{\prime}=\{H(t)H^{\prime\prime}(t)-[H^{\prime}(t)]^{2}\}/[H(t)]^{2}\leq-C/A^{2},

and thus

t,H(t)H′′(t)[H(t)]2CB2/A2<0.\forall t\in\mathfrak{I},\,\,H(t)H^{\prime\prime}(t)-[H^{\prime}(t)]^{2}\leq-CB^{2}/A^{2}<0.

Therefore, 𝔼(X2)𝔼(X)[𝔼(X)]2=e2λ{H(1)H′′(1)[H(1)]2}<0\mathbb{E}(X^{2})-\mathbb{E}(X)-[\mathbb{E}(X)]^{2}=e^{-2\lambda}\{H(1)H^{\prime\prime}(1)-[H^{\prime}(1)]^{2}\}<0, and VMR(X)<1\text{VMR}(X)<1.

We have proved in (1) that if hn+1=hnh_{n+1}=h_{n} for all nn, then VMR(X)=1\text{VMR}(X)=1. Thus when hn+1hnh_{n+1}\leq h_{n} for all nn, VMR(X)=1\text{VMR}(X)=1 if and only if hn+1=hnh_{n+1}=h_{n} for all nn. ∎

In sum, for the Markov chain model of one gene (by assuming the expression to be autonomous), when we have the stationary distribution from single-cell non-interventional one-time gene expression data, we can calculate the VMR of XX. VMR(X)>1\text{VMR}(X)>1 means the existence of positive autoregulation (while negative autoregulation might still be possible at different expression levels), and VMR(X)<1\text{VMR}(X)<1 means the existence of negative autoregulation (while positive autoregulation might still be possible at different expression levels). VMR(X)=1\text{VMR}(X)=1 means either (1) no autoregulation exists; or (2) both positive autoregulation and negative autoregulation exist (at different expression levels). In reality, many genes are non-autonomous, and transcriptional/translational bursting can make the VMR to be larger than 100100 [54]. Since Proposition 1 does not apply to non-autonomous cases, such genes might not have autoregulations.

5 Scenario of multiple entangled genes

5.1 Setup

We consider mm genes V1,,VmV_{1},\ldots,V_{m} for a single cell. Denote their expression levels by random variables X1,,XmX_{1},\ldots,X_{m}. The change of XiX_{i} can depend on XjX_{j} (mutual regulation) and XiX_{i} itself (autoregulation). Since these genes regulate each other, and their expression levels are not fixed, we cannot consider them separately. If the expression of gene VkV_{k} is non-autonomous, we also need to add its interior factors (gene state and/or mRNA count) into X1,,XmX_{1},\ldots,X_{m}.

We can use a continuous-time one-step Markov chain on ()m(\mathbb{Z}^{*})^{m} to describe the dynamics. Each state of this Markov chain, (X1=n1,,Xi=ni,,Xm=nm)(X_{1}=n_{1},\ldots,X_{i}=n_{i},\ldots,X_{m}=n_{m}), can be abbreviated as 𝒏=(n1,,ni,,nm)\boldsymbol{n}=(n_{1},\ldots,n_{i},\ldots,n_{m}). For gene ViV_{i}, the transition rate of ni1nin_{i}-1\to n_{i} is fi(𝒏)f_{i}(\boldsymbol{n}), and the transition rate of nini1n_{i}\to n_{i}-1 is gi(𝒏)nig_{i}(\boldsymbol{n})n_{i}. Transitions with more than one step are not allowed. The master equation of this process is

d(𝒏)dt=i(n1,,ni+1,,nm)gi(n1,,ni+1,,nm)(ni+1)+i(n1,,ni1,,nm)fi(𝒏)(𝒏)i[fi(n1,,ni+1,,nm)+gi(𝒏)ni].\begin{split}\frac{\mathrm{d}\mathbb{P}(\boldsymbol{n})}{\mathrm{d}t}=&\sum_{i}\mathbb{P}(n_{1},\ldots,n_{i}+1,\ldots,n_{m})g_{i}(n_{1},\ldots,n_{i}+1,\ldots,n_{m})(n_{i}+1)\\ +&\sum_{i}\mathbb{P}(n_{1},\ldots,n_{i}-1,\ldots,n_{m})f_{i}(\boldsymbol{n})\\ -&\mathbb{P}(\boldsymbol{n})\sum_{i}[f_{i}(n_{1},\ldots,n_{i}+1,\ldots,n_{m})+g_{i}(\boldsymbol{n})n_{i}].\end{split} (3)

Define 𝒏i¯=(n1,,ni1,ni+1,,nm)\boldsymbol{n}_{\bar{i}}=(n_{1},\ldots,n_{i-1},n_{i+1},\ldots,n_{m}). Define hi(𝒏)=fi(𝒏)/gi(𝒏)h_{i}(\boldsymbol{n})=f_{i}(\boldsymbol{n})/g_{i}(\boldsymbol{n}) to be the relative growth rate of gene ViV_{i}. Autoregulation means for some fixed 𝒏i¯\boldsymbol{n}_{\bar{i}}, hi(𝒏)h_{i}(\boldsymbol{n}) is (locally) increasing/decreasing with nin_{i}, thus fi(𝒏)f_{i}(\boldsymbol{n}) increases/decreases and/or gi(𝒏)g_{i}(\boldsymbol{n}) decreases/increases with nin_{i}. For the non-autonomous scenario, another possibility for autoregulation is that ViV_{i} can affect its interior factors (gene state and/or mRNA count).

5.2 Theoretical results

With expression data for multiple genes, there are various methods to infer the regulatory relationships between different genes, so that the GRN can be reconstructed [86]. In the GRN, if there is a directed path from gene ViV_{i} to gene VjV_{j}, meaning that ViV_{i} can directly or indirectly regulate VjV_{j}, then ViV_{i} is an ancestor of VjV_{j}, and VjV_{j} is a descendant of ViV_{i}.

Fix a gene VkV_{k} in a GRN. We consider a simple case that VkV_{k} is not contained in any directed cycle (feedback loop), which means no gene is both an ancestor and a descendant of VkV_{k}, such as PIP2 in Fig. 1. This means VkV_{k} itself is a strongly connected component of the GRN. This condition is automatically satisfied if the GRN has no directed cycle. If the expression of VkV_{k} is non-autonomous, we need to add the interior factors (gene state and/or mRNA count) of VkV_{k} into V1,,VmV_{1},\ldots,V_{m}, and it is acceptable that VkV_{k} regulates its interior factors. In this case, if the one-step model holds, we can prove that if VkV_{k} does not regulate itself, meaning that hk(𝒏)h_{k}(\boldsymbol{n}) is a constant for fixed 𝒏k¯\boldsymbol{n}_{\bar{k}} and different nkn_{k}, and XkX_{k} does not affect its interior factors (if non-autonomous), then VMR(Xk)1\text{VMR}(X_{k})\geq 1. The reason is that VMR<1\text{VMR}<1 requires either a feedback loop or autoregulation. Certainly, VMR<1\text{VMR}<1 might also mean that the one-step model fails. One intuition is to assume the transitions of Vk¯V_{\bar{k}} are extremely slow, so that VkV_{k} is approximately the average of many Poisson variables. It is easy to verify that the average of Poisson variables has VMR1\text{VMR}\geq 1. We need to assume that the per molecule degradation rate gk()g_{k}(\cdot) for VkV_{k} is not affected by V1,,VmV_{1},\ldots,V_{m}, which is not always true in reality [42]. With this result, when VMR<1\text{VMR}<1, there might be autoregulation.

Proposition 2.

Consider the one-step Markov chain model for multiple genes, described by Eq. 3. Assume the GRN has no directed cycle, or at least there is no directed cycle that contains gene VkV_{k}. Assume gk()g_{k}(\cdot) is a constant for all 𝐧\boldsymbol{n}. If VkV_{k} has no autoregulation, meaning that hk()h_{k}(\cdot) and fk()f_{k}(\cdot) do not depend on nkn_{k}, and VkV_{k} does not regulate its interior factors (gene state and/or mRNA count), then VkV_{k} has VMR1\text{VMR}\geq 1. Therefore, VkV_{k} has VMR<1\text{VMR}<1 means VkV_{k} has autoregulation, or the one-step model fails.

Paulsson et al. study a similar problem [29, 93], and they state Proposition 2 in an unpublished work. Proposition 2 also appears in a preprint by Mahajan et al. [49], but the proof is based on a linear noise approximation, which requires that fk()f_{k}(\cdot) is linear with 𝒏k¯\boldsymbol{n}_{\bar{k}}. We propose a rigorous proof independently.

Proof.

Denote the expression level of VkV_{k} by WW. Assume the ancestors of VkV_{k} are V1,,VlV_{1},\ldots,V_{l}. For simplicity, denote the expression levels of V1,,VlV_{1},\ldots,V_{l} by a (high-dimensional) random variable YY. Assume VkV_{k} has no autoregulation. Since VkV_{k} does not regulate V1,,VlV_{1},\ldots,V_{l}, WW does not affect YY. Denote the transition rate from Y=iY=i to Y=jY=j by qij0q_{ij}\geq 0. Stipulate that qii=jiqijq_{ii}=-\sum_{j\neq i}q_{ij}. When Y=iY=i, the transition rate from W=nW=n to W=n+1W=n+1 is FiF_{i} (does not depend on nn), and the transition rate from W=nW=n to W=n1W=n-1 is GG.

The master equation of this process is

d[W(t)=n,Y(t)=i]dt=[W(t)=n1,Y(t)=i]Fi+[W(t)=n+1,Y(t)=i]G(n+1)+ji[W(t)=n,Y(t)=j]qji[W(t)=n,Y(t)=i](Fi+Gn+jiqij).\begin{split}&\frac{\mathrm{d}\mathbb{P}[W(t)=n,Y(t)=i]}{\mathrm{d}t}\\ =&\mathbb{P}[W(t)=n-1,Y(t)=i]F_{i}+\mathbb{P}[W(t)=n+1,Y(t)=i]G(n+1)\\ &+\sum_{j\neq i}\mathbb{P}[W(t)=n,Y(t)=j]q_{ji}-\mathbb{P}[W(t)=n,Y(t)=i](F_{i}+Gn+\sum_{j\neq i}q_{ij}).\end{split}

Assume there is a unique stationary probability distribution Pn,i=limt[W(t)=n,Y(t)=i]P_{n,i}=\lim_{t\to\infty}\mathbb{P}[W(t)=n,Y(t)=i]. This can be guaranteed by assuming the process to be irreducible. Then we have

Pn,i[Fi+Gn+jqij]=Pn1,iFi+Pn+1,iG(n+1)+jPn,jqji.P_{n,i}\Big{[}F_{i}+Gn+\sum_{j}q_{ij}\Big{]}=P_{n-1,i}F_{i}+P_{n+1,i}G(n+1)+\sum_{j}P_{n,j}q_{ji}. (4)

Define Pi=nPn,iP_{i}=\sum_{n}P_{n,i}. Sum over nn for Eq. 4 to obtain

Pijqij=jPjqji,P_{i}\sum_{j}q_{ij}=\sum_{j}P_{j}q_{ji}, (5)

meaning that PiP_{i} is the stationary probability distribution of YY.

Define WiW_{i} to be WW conditioned on Y=iY=i at stationarity. Then (Wi=n)=(W=nY=i)=Pn,i/Pi\mathbb{P}(W_{i}=n)=\mathbb{P}(W=n\mid Y=i)=P_{n,i}/P_{i}, and 𝔼(Wi)=nnPn,i/Pi\mathbb{E}(W_{i})=\sum_{n}nP_{n,i}/P_{i}. Multiply Eq. 4 by nn and sum over nn to obtain

(G+jqij)Pi𝔼(Wi)=FiPi+jqjiPj𝔼(Wj).\Big{(}G+\sum_{j}q_{ij}\Big{)}P_{i}\mathbb{E}(W_{i})=F_{i}P_{i}+\sum_{j}q_{ji}P_{j}\mathbb{E}(W_{j}). (6)

Here and in the following, we repeatedly apply the tricks of splitting nn and shifting the index of summation. For example,

n=1Pn1,iFinn=1Pn,iFin=n=1Pn1,iFi(n1)+n=1Pn1,iFin=1Pn,iFin=n1=0Pn1,iFi(n1)+n1=0Pn1,iFin=0Pn,iFin=n=0Pn,iFin+Fin=0Pn,in=0Pn,iFin=FiPi.\begin{split}&\sum_{n=1}^{\infty}P_{n-1,i}F_{i}n-\sum_{n=1}^{\infty}P_{n,i}F_{i}n\\ =&\sum_{n=1}^{\infty}P_{n-1,i}F_{i}(n-1)+\sum_{n=1}^{\infty}P_{n-1,i}F_{i}-\sum_{n=1}^{\infty}P_{n,i}F_{i}n\\ =&\sum_{n-1=0}^{\infty}P_{n-1,i}F_{i}(n-1)+\sum_{n-1=0}^{\infty}P_{n-1,i}F_{i}-\sum_{n=0}^{\infty}P_{n,i}F_{i}n\\ =&\sum_{n=0}^{\infty}P_{n,i}F_{i}n+F_{i}\sum_{n=0}^{\infty}P_{n,i}-\sum_{n=0}^{\infty}P_{n,i}F_{i}n=F_{i}P_{i}.\end{split}

Sum over ii for Eq. 6 to obtain

GiPi𝔼(Wi)=iFiPi.G\sum_{i}P_{i}\mathbb{E}(W_{i})=\sum_{i}F_{i}P_{i}. (7)

Multiply Eq. 4 by n2n^{2} and sum over nn to obtain

(2G+jqij)Pi𝔼(Wi2)=FiPi+(2Fi+G)Pi𝔼(Wi)+jqjiPj𝔼(Wj2).\Big{(}2G+\sum_{j}q_{ij}\Big{)}P_{i}\mathbb{E}(W_{i}^{2})=F_{i}P_{i}+(2F_{i}+G)P_{i}\mathbb{E}(W_{i})+\sum_{j}q_{ji}P_{j}\mathbb{E}(W_{j}^{2}). (8)

Sum over ii for Eq. 8 to obtain

2GiPi𝔼(Wi2)=iFiPi+2iFiPi𝔼(Wi)+GiPi𝔼(Wi).2G\sum_{i}P_{i}\mathbb{E}(W_{i}^{2})=\sum_{i}F_{i}P_{i}+2\sum_{i}F_{i}P_{i}\mathbb{E}(W_{i})+G\sum_{i}P_{i}\mathbb{E}(W_{i}). (9)

Multiply Eq. 6 by 𝔼(Wi)\mathbb{E}(W_{i}) and sum over ii to obtain

GiPi[𝔼(Wi)]2+i,jPiqij[𝔼(Wi)]2=iFiPi𝔼(Wi)+i,jPjqji𝔼(Wi)𝔼(Wj).\begin{split}&G\sum_{i}P_{i}[\mathbb{E}(W_{i})]^{2}+\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{i})]^{2}\\ =&\sum_{i}F_{i}P_{i}\mathbb{E}(W_{i})+\sum_{i,j}P_{j}q_{ji}\mathbb{E}(W_{i})\mathbb{E}(W_{j}).\end{split} (10)

Then we have

iFiPi𝔼(Wi)GiPi[𝔼(Wi)]2=i,jPiqij[𝔼(Wi)]2i,jPjqji𝔼(Wi)𝔼(Wj)=12{i,jPiqij[𝔼(Wi)]2+i[𝔼(Wi)]2jPiqij2i,jPiqij𝔼(Wi)𝔼(Wj)}=12{i,jPiqij[𝔼(Wi)]2+i[𝔼(Wi)]2jPjqji2i,jPiqij𝔼(Wi)𝔼(Wj)}=12{i,jPiqij[𝔼(Wi)]2+i,jPiqij[𝔼(Wj)]22i,jPiqij𝔼(Wi)𝔼(Wj)}=12i,jPiqij[𝔼(Wi)𝔼(Wj)]20.\begin{split}&\sum_{i}F_{i}P_{i}\mathbb{E}(W_{i})-G\sum_{i}P_{i}[\mathbb{E}(W_{i})]^{2}\\ =&\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{i})]^{2}-\sum_{i,j}P_{j}q_{ji}\mathbb{E}(W_{i})\mathbb{E}(W_{j})\\ =&\frac{1}{2}\Big{\{}\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{i})]^{2}+\sum_{i}[\mathbb{E}(W_{i})]^{2}\sum_{j}P_{i}q_{ij}-2\sum_{i,j}P_{i}q_{ij}\mathbb{E}(W_{i})\mathbb{E}(W_{j})\Big{\}}\\ =&\frac{1}{2}\Big{\{}\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{i})]^{2}+\sum_{i}[\mathbb{E}(W_{i})]^{2}\sum_{j}P_{j}q_{ji}-2\sum_{i,j}P_{i}q_{ij}\mathbb{E}(W_{i})\mathbb{E}(W_{j})\Big{\}}\\ =&\frac{1}{2}\Big{\{}\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{i})]^{2}+\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{j})]^{2}-2\sum_{i,j}P_{i}q_{ij}\mathbb{E}(W_{i})\mathbb{E}(W_{j})\Big{\}}\\ =&\frac{1}{2}\sum_{i,j}P_{i}q_{ij}[\mathbb{E}(W_{i})-\mathbb{E}(W_{j})]^{2}\geq 0.\end{split} (11)

Here the first equality is from Eq. 10, the third equality is from Eq. 5, and other equalities are equivalent transformations.

Now we have

𝔼(W2)𝔼(W)[𝔼(W)]2=iPi𝔼(Wi2)iPi𝔼(Wi)[iPi𝔼(Wi)]2=1GiFiPi𝔼(Wi)+iPi𝔼(Wi)iPi𝔼(Wi)[iPi𝔼(Wi)]2iPi[𝔼(Wi)]2[iPi𝔼(Wi)]2=(iPi)iPi[𝔼(Wi)]2[iPi𝔼(Wi)]20,\begin{split}&\mathbb{E}(W^{2})-\mathbb{E}(W)-[\mathbb{E}(W)]^{2}\\ =&\sum_{i}P_{i}\mathbb{E}(W_{i}^{2})-\sum_{i}P_{i}\mathbb{E}(W_{i})-\Big{[}\sum_{i}P_{i}\mathbb{E}(W_{i})\Big{]}^{2}\\ =&\frac{1}{G}\sum_{i}F_{i}P_{i}\mathbb{E}(W_{i})+\sum_{i}P_{i}\mathbb{E}(W_{i})-\sum_{i}P_{i}\mathbb{E}(W_{i})-\Big{[}\sum_{i}P_{i}\mathbb{E}(W_{i})\Big{]}^{2}\\ \geq&\sum_{i}P_{i}[\mathbb{E}(W_{i})]^{2}-\Big{[}\sum_{i}P_{i}\mathbb{E}(W_{i})\Big{]}^{2}\\ =&\Big{(}\sum_{i}P_{i}\Big{)}\sum_{i}P_{i}[\mathbb{E}(W_{i})]^{2}-\Big{[}\sum_{i}P_{i}\mathbb{E}(W_{i})\Big{]}^{2}\geq 0,\\ \end{split} (12)

where the first equality is by definition, the second equality is from Eqs. 7,9, the first inequality is from Eq. 11, the third equality is from iPi=1\sum_{i}P_{i}=1, and the second inequality is the Cauchy-Schwarz inequality.

Since 𝔼(W2)[𝔼(W)]2𝔼(W)\mathbb{E}(W^{2})-[\mathbb{E}(W)]^{2}\geq\mathbb{E}(W), VMR(W)={𝔼(W2)[𝔼(W)]2}/𝔼(W)1\text{VMR}(W)=\{\mathbb{E}(W^{2})-[\mathbb{E}(W)]^{2}\}/\mathbb{E}(W)\geq 1. ∎

Remark 2.

In gene expression, the total noise (σ2(X)/(𝔼X)2\sigma^{2}(X)/(\mathbb{E}X)^{2}) can be decomposed into intrinsic (cellular) noise and extrinsic (environmental) noise [3, 71, 27, 47, 75]. Inspired by that, we can decompose the VMR into intrinsic and extrinsic components. Denote intrinsic and extrinsic stochastic factors as I,EI,E, and the expression level XX is a deterministic function of these factors: X=X(I,E)X=X(I,E). Then

VMRint=𝔼E(𝔼IEX2)𝔼E(𝔼IEX)2𝔼X,\text{VMR}_{\text{int}}=\frac{\mathbb{E}_{E}(\mathbb{E}_{I\mid E}X^{2})-\mathbb{E}_{E}(\mathbb{E}_{I\mid E}X)^{2}}{\mathbb{E}X},
VMRext=𝔼E(𝔼IEX)2[𝔼E(𝔼IEX)]2𝔼X,\text{VMR}_{\text{ext}}=\frac{\mathbb{E}_{E}(\mathbb{E}_{I\mid E}X)^{2}-[\mathbb{E}_{E}(\mathbb{E}_{I\mid E}X)]^{2}}{\mathbb{E}X},

where 𝔼IE\mathbb{E}_{I\mid E} is the expectation conditioned on EE. This decomposition might lead to further understanding of Proposition 2.

We hypothesize that the requirement for gk()g_{k}(\cdot) in Proposition 2 can be dropped:

Conjecture 1.

Assume VkV_{k} is not contained in a directed cycle in the GRN, and VkV_{k} does not regulate its interior factors (gene state and/or mRNA count). If VkV_{k} has no autoregulation, meaning that hk()h_{k}(\cdot) does not depend on nkn_{k} (but might depend on 𝐧k¯\boldsymbol{n}_{\bar{k}}), then VkV_{k} has VMR1\text{VMR}\geq 1.

The main obstacle of proving this conjecture is that the second equality in Eq. 12 does not hold. The reason is that GiG_{i} cannot be extracted from the summation, and we cannot link iPi𝔼(Wi2)\sum_{i}P_{i}\mathbb{E}(W_{i}^{2}) and iGiPi𝔼(Wi2)\sum_{i}G_{i}P_{i}\mathbb{E}(W_{i}^{2}).

If the GRN has directed cycles, there is a result by Paulsson et al. [29, 93], which is proved under first-order approximations of covariances. The general case (when the approximations do not apply) has been numerically verified but not proved yet:

Conjecture 2.

Assume for each ViV_{i}, gi()g_{i}(\cdot) does not depend on 𝐧\boldsymbol{n}, and fi()f_{i}(\cdot) does not depend on nin_{i} (no autoregulation). Then for at least one gene VjV_{j}, we have VMR1\text{VMR}\geq 1 [29, 93].

Due to the existence of directed cycles, one gene can affect itself through other genes, and we cannot study them separately.

Notice that Conjecture 2 does not hold if gig_{i} depends on 𝒏i¯\boldsymbol{n}_{\bar{i}}:

Example 2.

Consider a one-step Markov chain that satisfies Eq. 3, where m=2m=2, f1(n2)=g1(n2)=1f_{1}(n_{2})=g_{1}(n_{2})=1 for n2=2n_{2}=2, f1(n2)=g1(n2)=0f_{1}(n_{2})=g_{1}(n_{2})=0 for n22n_{2}\neq 2, and f2(n1)=g2(n1)=1f_{2}(n_{1})=g_{2}(n_{1})=1 for n1=2n_{1}=2, f2(n1)=g2(n1)=0f_{2}(n_{1})=g_{2}(n_{1})=0 for n12n_{1}\neq 2. The initial state is (n1=2,n2=2)(n_{1}=2,n_{2}=2). Then VMR=2e/(4e1)0.55\text{VMR}=2e/(4e-1)\approx 0.55 for both genes (see Appendix B.2 for details).

Assume Conjecture 2 is correct. For mm genes, if we find that VMR for each gene is less than 11, then we can infer that autoregulation exists, although we do not know which gene has autoregulation. Another possibility is that the one-step model fails.

6 Applying theoretical results to experimental data

 
  1. 1.

    Input

    Single-cell non-interventional one-time expression data for genes V1,,VmV_{1},\ldots,V_{m}

    The structure of the GRN that contains V1,,VmV_{1},\ldots,V_{m}

  2. 2.

    Calculate the VMR of each VkV_{k}

  3. 3.

    If VkV_{k} is not in a directed cycle (like PIP2 in Fig. 1) and VMR<1\text{VMR}<1

    Output VkV_{k} has autoregulation

    // Assume the degradation of VkV_{k} is not regulated by V1,,VmV_{1},\ldots,V_{m}

    Else

    If VkV_{k} has no ancestor in the GRN (like PIP3 in Fig. 1) and VMR>1\text{VMR}>1

    Output VkV_{k} has autoregulation

    //Assume the expression of VkV_{k} is autonomous

    Else

    Output We cannot determine whether VkV_{k} has autoregulation

    End of if

    End of if

Algorithm 1 Detailed workflow of inferring autoregulation with gene expression data.

We summarize our theoretical results into Algorithm 1. Proposition 1 applies to a gene that has no ancestor in the GRN. However, it requires the corresponding gene has autonomous expression (or the transition rates of gene states are high enough, so that the non-autonomous process is close to an autonomous process), which is difficult to validate and often does not hold in reality. Thus the inference result by Proposition 1 for VMR>1\text{VMR}>1 (positive autoregulation) is not very reliable. When VMR<1\text{VMR}<1 and Proposition 1 could apply, we should instead apply Proposition 2 to determine the existence of autoregulation, since Proposition 2 does not require the expression to be autonomous, thus being much more reliable, although it may fail if the one-step model does not hold. Proposition 2 applies when the gene is not in a feedback loop and has VMR<1\text{VMR}<1. Notice that our result cannot determine that a gene has no autoregulation.

For a given gene without autoregulation, its expression level satisfies a Poisson distribution, and VMR is 11. If we have nn samples of its expression level, then the sample VMR (sample variance divided by sample mean) asymptotically satisfies a Gamma distribution Γ[(n1)/2,2/(n1)]\Gamma[(n-1)/2,2/(n-1)], and we can determine the confidence interval of sample VMR [20]. If the sample VMR is out of this confidence interval, then we know that VMR is significantly different from 11, and Propositions 1,2 might apply.

We apply our method to four groups of single-cell non-interventional one-time gene expression data from experiments, where the corresponding GRNs are known. Notice that we need to convert indirect measurements into protein/mRNA count. See Table 1 for our inference results and theoretical/experimental evidence that partially validates our results. See Appendix C for details. There are 186 genes in these four data sets, and we can only determine that 12 genes have autoregulation (7 genes determined by Proposition 1, and 5 genes determined by Proposition 2). Not every VMR is less than 11, so that Conjecture 2 does not apply. For the other 174 genes, (1) some of them are not contained in the known GRN, and we cannot determine if they are in directed cycles; (2) some of them are in directed cycles; (3) some of them have ancestors, and we cannot reject the hypothesis that VMR1\text{VMR}\geq 1; (4) some of them have no ancestors, and we cannot reject the hypothesis that VMR=1\text{VMR}=1. Therefore, Proposition 1 and Proposition 2 do not apply, and we do not know whether they have autoregulation.

In some cases, we have experimental evidence that some genes have autoregulation, so that we can partially validate our inference results. Nevertheless, as discussed in the Introduction, there is no gold standard to evaluate our inference results. Besides, Proposition 2 requires that the one-step model holds, which we cannot verify.

In the data set by Guo et al. [26], Sanchez-Castillo et al. [59] inferred that 17 of 39 genes have autoregulation, and 22 genes do not have autoregulation. We infer that 5 genes have autoregulation, and 34 genes cannot be determined. Here 3 genes are shared by both inference results to have autoregulation. Consider a random classifier that randomly picks 5 genes and claims they have autoregulation. Using Sanchez-Castillo et al. as the standard, this random classifier has probability 62.55%62.55\% to be worse than our result, and 10.17%10.17\% to be better than our result. Thus our inference result is better than a random classifier, but the advantage is not substantial.

Source
Propo-
sition 1
Propo-
sition 2
Theory Experiment
Guo
et al. [26]
FN1
HNF4A
TCFAP2C
BMP4
CREB312
BMP4 [59]
HNF4A [59]
TCFAP2C [59]
BMP4 [55]
HNF4A [12]
TCFAP2C [44]
Psaila
et al. [56]
BIM
CCND1
ECT2
PFKP
ECT2 [28]
Moignard
et al. [50]
EIF2B1
HOXD8
Sachs
et al. [58]
PIP3
Table 1: The autoregulation inference results by our method on four data sets. Source column is the paper that contains this data set. Proposition 1 column is the genes that can be only inferred by Proposition 1 to have autoregulation. Proposition 2 column is the genes that can be inferred by Proposition 2 to have autoregulation. Theory column is the genes inferred by both our method and other theoretical works to have autoregulation. Experiment column is the genes inferred by both our method and other experimental works to have autoregulation. Bold font means the inferred gene with autoregulation is validated by other results. Details can be found in Appendix C.

7 Conclusions

For a single gene that is not affected by other genes, or a group of genes that form a connected GRN, we develop rigorous theoretical results (without applying approximations) to determine the existence of autoregulation. These results generalize known relationships between autoregulation and VMR by dropping restrictions on parameters. Our results only depend on VMR, which is easy to compute and more robust than other complicated statistics. We also apply our method to experimental data and detect some genes that might have autoregulation.

Our method requires independent and identically distributed samples from the exact stationary distribution of a fully observed Markov chain, plus a known GRN. Proposition 1 requires that the expression is autonomous. Proposition 2 requires that the Markov chain model is one-step, the GRN has no directed cycle, and degradation is not regulated. If our inference fails, then some requirements are not met: (1) cells might affect each other, making the samples dependent; (2) cells are heterogeneous; (3) the measurements have extra errors; (4) the cells are not at stationarity; (5) there exist unobserved variables that affect gene expression; (6) the GRN is inferred by a theoretical method, which can be interfered by the existence of autoregulation; (7) the expression is non-autonomous; (8) the Markov chain is multi-step; (9) the GRN has unknown directed cycles; (10) the degradation rate is regulated by other genes. Such situations, especially the unobserved variables, are unavoidable. Therefore, current data might not satisfy these requirements, and our inference results should be interpreted as informative findings, not ground truths.

There are some known methods that overcome the above obstacles, and there are also some possible solutions that might appear in the future. (1) The dependency can be solved by better measurements for isolated cells that do not affect each other. In fact, the relationship between autoregulation and cell-cell interaction has been studied [46]. (2) About cell heterogeneity, we prove a result in Appendix D that if several cell types have VMR1\text{VMR}\geq 1, then for a mixed population of such cell types, we still have VMR1\text{VMR}\geq 1. Therefore, cell heterogeneity does not fail Proposition 2, since VMR<1\text{VMR}<1 for the mixture of several cell types means VMR<1\text{VMR}<1 for at least one cell type. (3) With the development of experimental technologies, we expect that the measurement error can decrease. (4) Some works study autoregulation in non-stationary situations [10, 69, 66, 36]. (5) Since hidden variables hurt any mechanism-based models, we can develop methods (especially with machine learning tools) that determine autoregulation based on similarities between gene expression profiles [87, 94, 78, 88, 89]. (6) Some GRN inference methods can also determine the existence of autoregulation [59]. (7) Many methods (including our Proposition 2) work in non-autonomous situations. (8) Some works study multi-step models [7, 43, 74]. (9) We expect the appearance of more advanced GRN inference methods. (10) If probabilists can prove Conjecture 1, then the restriction on degradation rate can be lifted.

In fact, other theoretical works that determine gene autoregulation, or general gene regulation, also need various assumptions and might fail. Nevertheless, with the development of experimental technologies and theoretical results, we believe that some obstacles will be lifted, and our method will be more applicable in the future. Besides, our method can be further developed and combined with other methods.

Acknowledgments

Y.W. would like to thank Jiawei Yan for fruitful discussions, and Xiangting Li, Zikun Wang, Mingtao Xia for helpful comments. The authors would like to thank some anonymous reviewers for their wise suggestions.

Declarations

  • Funding: This research was partially supported by NIH grant R01HL146552 (Y.W.).

  • Competing interests: The authors declare no conflict of interest.

  • Data and code availability: All code files are available in

    https://github.com/YueWangMathbio/Autoregulation.

Appendix A Simulation results

A.1 Test Proposition 1 without autoregulation

Simulation 1: Consider the Markov chain in Example 1 with k=1,b=0,c=1k=1,b=0,c=1. The stationary distribution is Poissonian with parameter 11. This process has no autoregulation with the true VMR=1\text{VMR}=1. For sample sizes n=100n=100, n=1000n=1000, n=10000n=10000, we repeat the experiment for 1000010000 times and calculate the rate that the sample VMR falls in the 95%95\% confidence interval. See Table 2 for results. Proposition 1 has about 95%95\% probability to produce the correct result that VMR=1\text{VMR}=1 (no autoregulation), since the confidence interval is 95%95\%.

VMR<1\text{VMR}<1 VMR=1\text{VMR}=1 (true) VMR>1\text{VMR}>1
n=100n=100 2.5%2.5\% 95.3%95.3\% 2.2%2.2\%
n=1000n=1000 2.3%2.3\% 95.2%95.2\% 2.5%2.5\%
n=10000n=10000 2.5%2.5\% 95.1%95.1\% 2.4%2.4\%
Table 2: Success rates for determining VMR in Simulation 1. For different sample sizes nn, calculate the rate that VMR is different from 11.

A.2 Test Proposition 1 with autoregulation

Simulation 2: Consider the Markov chain in Example 1 with k=1,b=1,c=2k=1,b=1,c=2. The stationary distribution is geometric with parameter 0.50.5. This process has positive autoregulation with the true VMR=2\text{VMR}=2. For sample sizes n=100n=100, n=1000n=1000, n=10000n=10000, we repeat the experiment for 1000010000 times and calculate the rate that the sample VMR falls in the 95%95\% confidence interval. See Table 3 for results. When nn is not too small, Proposition 1 always produces the correct result that VMR>1\text{VMR}>1 (positive autoregulation).

VMR<1\text{VMR}<1 VMR=1\text{VMR}=1 VMR>1\text{VMR}>1 (true)
n=100n=100 0%0\% 1.9%1.9\% 98.1%98.1\%
n=1000n=1000 0%0\% 0%0\% 100%100\%
n=10000n=10000 0%0\% 0%0\% 100%100\%
Table 3: Success rates for determining VMR in Simulation 2. For different sample sizes nn, calculate the rate that VMR is different from 11.

A.3 Test Proposition 2 without autoregulation

Simulation 3: Consider a Markov chain (G,M)(G,M) that satisfies Eq. 3. GG can take 0 and 11, and MM can take values in \mathbb{Z}. GG does not depend on MM, and transition rates are both 101010^{-10} for G=0G=1G=0\to G=1 and G=1G=0G=1\to G=0. Restricted on G=0G=0, MM is the same as Example 1 with k=1,b=0,c=1k=1,b=0,c=1. Restricted on G=1G=1, MM is the same as Example 1 with k=2,b=0,c=1k=2,b=0,c=1. The stationary distribution is the average of two Poisson distributions with parameters 11 and 22. This process has no autoregulation with the true VMR=1.167\text{VMR}=1.167. For sample sizes n=100n=100, n=1000n=1000, n=10000n=10000, we repeat the experiment for 1000010000 times and calculate the rate that the sample VMR falls in the 95%95\% confidence interval. See Table 4 for results. When nn increases, we are very likely to obtain the correct VMR>1\text{VMR}>1, but Proposition 2 cannot determine whether autoregulation exists.

VMR<1\text{VMR}<1 VMR=1\text{VMR}=1 VMR>1\text{VMR}>1 (true)
n=100n=100 0.2%0.2\% 81.6%81.6\% 18.2%18.2\%
n=1000n=1000 0%0\% 7.7%7.7\% 92.3%92.3\%
n=10000n=10000 0%0\% 0%0\% 100%100\%
Table 4: Success rates for determining VMR in Simulation 3. For different sample sizes nn, calculate the rate that VMR is different from 11.

A.4 Test Proposition 2 with autoregulation

Simulation 4: Consider a Markov chain (G,M)(G,M) that satisfies Eq. 3. GG can take 0 and 11, and MM can take values in \mathbb{Z}. GG does not depend on MM, and transition rates are both 101010^{-10} for G=0G=1G=0\to G=1 and G=1G=0G=1\to G=0. Restricted on G=0G=0, MM is the same as Example 1 with k=10,b=1,c=1k=10,b=-1,c=1. Restricted on G=1G=1, MM is the same as Example 1 with k=10,b=1,c=2k=10,b=-1,c=2. This process has autoregulation with the true VMR=0.733\text{VMR}=0.733. For sample sizes n=100n=100, n=1000n=1000, n=10000n=10000, we repeat the experiment for 1000010000 times and calculate the rate that the sample VMR falls in the 95%95\% confidence interval. See Table 5 for results. When nn is not too small, Proposition 2 always produces the correct result that VMR<1\text{VMR}<1 (autoregulation).

VMR<1\text{VMR}<1 (true) VMR=1\text{VMR}=1 VMR>1\text{VMR}>1
n=100n=100 56.6%56.6\% 43.4%43.4\% 0%0\%
n=1000n=1000 100%100\% 0%0\% 0%0\%
n=10000n=10000 100%100\% 0%0\% 0%0\%
Table 5: Success rates for determining VMR in Simulation 4. For different sample sizes nn, calculate the rate that VMR is different from 11.

Appendix B Details of examples

B.1 Details of Example 1

In Example 1, the stationary distribution PnP_{n} exists, and satisfies

[b(n1)+k]Pn1=cnPn.[b(n-1)+k]P_{n-1}=cnP_{n}. (13)

Taking summation for Eq. (13), we have

n=1[b(n1)+k]Pn1=n=1cnPn,bn=0nPn+kn=0Pn=cn=0nPn,b𝔼X+k=c𝔼X.\begin{split}&\sum_{n=1}^{\infty}[b(n-1)+k]P_{n-1}=\sum_{n=1}^{\infty}cnP_{n},\\ \Rightarrow&b\sum_{n=0}^{\infty}nP_{n}+k\sum_{n=0}^{\infty}P_{n}=c\sum_{n=0}^{\infty}nP_{n},\\ \Rightarrow&b\mathbb{E}X+k=c\mathbb{E}X.\end{split}

Thus 𝔼X=k/(cb)\mathbb{E}X=k/(c-b).

Also, multiplying Eq. (13) by n1n-1 and taking summation, we have

n=1[b(n1)2+k(n1)]Pn1=n=1cn2Pnn=1cnPn,bn=0n2Pn+kn=0nPn=cn=0n2Pncn=0nPn,b𝔼(X2)+k𝔼X=c𝔼(X2)c𝔼X.\begin{split}&\sum_{n=1}^{\infty}[b(n-1)^{2}+k(n-1)]P_{n-1}=\sum_{n=1}^{\infty}cn^{2}P_{n}-\sum_{n=1}^{\infty}cnP_{n},\\ \Rightarrow&b\sum_{n=0}^{\infty}n^{2}P_{n}+k\sum_{n=0}^{\infty}nP_{n}=c\sum_{n=0}^{\infty}n^{2}P_{n}-c\sum_{n=0}^{\infty}nP_{n},\\ \Rightarrow&b\mathbb{E}(X^{2})+k\mathbb{E}X=c\mathbb{E}(X^{2})-c\mathbb{E}X.\end{split}

Thus 𝔼(X2)=(c+k)k/(cb)2\mathbb{E}(X^{2})=(c+k)k/(c-b)^{2}, and the variance σ2(X)=ck/(cb)2\sigma^{2}(X)=ck/(c-b)^{2}. Then VMR(X)=1+b/(cb)\text{VMR}(X)=1+b/(c-b).

B.2 Details of Example 2

In Example 2, the process is restricted to two lines: n1=2n_{1}=2 and n2=2n_{2}=2. Since this process has no cycle, it is detailed balanced [84], and the stationary distribution P(n1,n2)P(n_{1},n_{2}) satisfies

P(k,2)=(k+1)P(k+1,2)P(k,2)=(k+1)P(k+1,2)

and

P(2,k)=(k+1)P(2,k+1).P(2,k)=(k+1)P(2,k+1).

Restricted on n1=2n_{1}=2 or n2=2n_{2}=2, the stationary distribution is Poissonian, P(k,2)=c/(ek!)P(k,2)=c/(ek!). After normalization, we find that c=2e/(4e1)c=2e/(4e-1). Thus

P(k,2)=P(2,k)=2k!(4e1).P(k,2)=P(2,k)=\frac{2}{k!(4e-1)}.

Besides,

P(2,)=k=0P(2,k)=k=02k!(4e1)=2e4e1.P(2,\cdot)=\sum_{k=0}^{\infty}P(2,k)=\sum_{k=0}^{\infty}\frac{2}{k!(4e-1)}=\frac{2e}{4e-1}.

Then for the first gene XX, we have

𝔼X=k2kP(k,2)+2P(2,)=k=0kP(k,2)+2[P(2,)P(2,2)]=k=02kk!(4e1)+4e24e1=6e24e1,\begin{split}\mathbb{E}X&=\sum_{k\neq 2}kP(k,2)+2P(2,\cdot)=\sum_{k=0}^{\infty}kP(k,2)+2[P(2,\cdot)-P(2,2)]\\ &=\sum_{k=0}^{\infty}\frac{2k}{k!(4e-1)}+\frac{4e-2}{4e-1}=\frac{6e-2}{4e-1},\end{split}

and

𝔼(X2)=k2k2P(k,2)+4P(2,)=k=0k2P(k,2)+4[P(2,)P(2,2)]=k=02k(k1)k!(4e1)+k=02kk!(4e1)+8e44e1=12e44e1.\begin{split}\mathbb{E}(X^{2})&=\sum_{k\neq 2}k^{2}P(k,2)+4P(2,\cdot)=\sum_{k=0}^{\infty}k^{2}P(k,2)+4[P(2,\cdot)-P(2,2)]\\ &=\sum_{k=0}^{\infty}\frac{2k(k-1)}{k!(4e-1)}+\sum_{k=0}^{\infty}\frac{2k}{k!(4e-1)}+\frac{8e-4}{4e-1}=\frac{12e-4}{4e-1}.\end{split}

Thus

VMR(X)=𝔼(X2)𝔼X𝔼X=26e24e1=24e10.55.\text{VMR}(X)=\frac{\mathbb{E}(X^{2})}{\mathbb{E}X}-\mathbb{E}X=2-\frac{6e-2}{4e-1}=\frac{2}{4e-1}\approx 0.55.

Due to symmetry, the other gene also has VMR(Y)0.55\text{VMR}(Y)\approx 0.55.

Appendix C Details of applications on experimental data

In experiments, the expression levels of genes are not directly measured as mRNA or protein counts. Rather, they are measured as cycle threshold (Ct) values or fluorescence intensity values. Such indirect measurements need to be converted. Related details can be found in other papers [38].

Guo et al. [26] measured the expression (mRNA) levels of 48 genes for mouse embryo cells at different developmental stages. We consider three groups (16-cell stage, 32-cell stage, 64-cell stage) that have more than 50 samples. Sanchez-Castillo et al. [59] used such data to infer the GRN structure, including autoregulation, but the inferred GRN only contains 39 genes. We cannot guarantee that the other 9 genes have no ancestors in the true GRN (to apply Proposition 1) or these genes are not contained in directed cycles (to apply Proposition 2). Thus we ignore those 9 genes not in this GRN. In the inferred GRN, genes BMP4, CREB312, and TCFAP2C are not contained in directed cycles. In the 16-cell stage group with 75 samples, if there is no autoregulation, then the 95%95\% confidence interval of VMR is [0.7041,1.3470][0.7041,1.3470]. BMP4 (VMR=0.2139\text{VMR}=0.2139), CREB312 (VMR=0.1971\text{VMR}=0.1971), and TCFAP2C (VMR=0.3468\text{VMR}=0.3468) have significantly small VMR, and we can apply Proposition 2 to infer that BMP4, CREB312, and TCFAP2C might have autoregulation. In the other two groups, these genes do not have VMR<1\text{VMR}<1, and the results are relatively weak. Besides, in the inferred GRN, genes FN1 and HNF4A have no ancestors. For the 16-cell stage with 75 samples, the VMR of FN1 and HNF4A are 3.45223.4522 and 1.35991.3599, outside of the 95%95\% confidence interval [0.7041,1.3470][0.7041,1.3470]; for the 32-cell stage with 113 samples, the VMR of FN1 and HNF4A are 93.107093.1070 and 46.768846.7688, outside of the 95%95\% confidence interval [0.7554,1.2784][0.7554,1.2784]; for the 64-cell stage with 159 samples, the VMR of FN1 and HNF4A are 117.3059117.3059 and 93.958993.9589, outside of the 95%95\% confidence interval [0.7917,1.2322][0.7917,1.2322]. Thus we can apply Proposition 1 to infer that FN1 and HNF4A (VMR>1\text{VMR}>1 for all three cell groups) might have positive autoregulation. Nevertheless, it is more likely that the expressions of FN1 and HNF4A are non-autonomous, and there is no autoregulation. Sanchez-Castillo et al. [59] inferred that BMP4, HNF4A, TCFAP2C have autoregulation. Besides, there is experimental evidence that BMP4 [55], HNF4A [12], TCFAP2C [44] have autoregulation. Therefore, our inference results are partially validated.

Psaila et al. [56] measured the expression (mRNA) levels of 90 genes for human megakaryocyte-erythroid progenitor cells. Chan et al. [13] inferred the GRN structure (autoregulation not included). In the inferred GRN, genes BIM, CCND1, ECT2, PFKP have no ancestors. BIM has 214 effective samples, and VMR is 187.7187.7, outside of the 95%95\% confidence interval [0.8191,1.1987][0.8191,1.1987]. CCND1 has 68 effective samples, and VMR is 111.3111.3, outside of the 95%95\% confidence interval [0.6905,1.3660][0.6905,1.3660]. ECT2 has 56 effective samples, and VMR is 8.28.2, outside of the 95%95\% confidence interval [0.6618,1.4069][0.6618,1.4069]. PFKP has 134 effective samples, and VMR is 82.182.1, outside of the 95%95\% confidence interval [0.7742,1.2543][0.7742,1.2543]. Thus we can apply Proposition 1 to infer that BIM, CCND1, ECT2, PFKP might have positive autoregulation. Nevertheless, it is more likely that the expressions of these four genes are non-autonomous, and there is no autoregulation. There is experimental evidence that ECT2 has autoregulation [28], which partially validates our inference results. No other gene fits the requirement of Proposition 2.

Moignard et al. [50] measured the expression (mRNA) levels of 46 genes for mouse embryo cells. Chan et al. [13] inferred the GRN structure (autoregulation not included). Gene EIF2B1 has 3934 effective samples, and VMR is 0.660.66, outside of the 95%95\% confidence interval [0.9563,1.0447][0.9563,1.0447]. Gene EIF2B1 has 12 effective samples, and VMR is 0.240.24, outside of the 95%95\% confidence interval [0.3469,1.9927][0.3469,1.9927]. We can apply Proposition 2 to infer that EIF2B1 and HOXD8 might have autoregulation. No other gene fits the requirement of Proposition 1.

Sachs et al. [58] measured the expression (protein) levels of 11 genes in the RAF signaling pathway for human T cells. The measurements were repeated for 14 groups of cells under different interventions. Werhli et al. [91] inferred the GRN structure (autoregulation not included). In the inferred GRN (Fig. 1), PIP3 gene has no ancestor, and its VMRs in all 14 groups are larger than 55, while the 95%95\% confidence intervals for all 14 groups are contained in [0.8,1.2][0.8,1.2]. Therefore, we can apply Proposition 1 and infer that PIP3 might have positive autoregulation. Nevertheless, it is more likely that the expression of PIP3 is non-autonomous, and there is no autoregulation. No other gene fits the requirement of Proposition 2.

Appendix D Heterogeneity and VMR

Proposition 3.

Consider nn independent random variables X1,,XnX_{1},\ldots,X_{n} and probabilities p1,,pnp_{1},\ldots,p_{n} with pi=1\sum p_{i}=1. Consider an independent random variable RR that equals ii with probability pip_{i}. Construct a random variable ZZ that equals XiX_{i} when R=iR=i. If each XiX_{i} has VMR1\text{VMR}\geq 1, then ZZ has VMR1\text{VMR}\geq 1.

Proof.

We only need to prove this for n=2n=2. The case for general nn can be proved by mixing two variables iteratively.

Consider random variables X,YX,Y and construct ZZ that equals XX or YY with probability pp or 1p1-p. Since VMR(X)1\text{VMR}(X)\geq 1, VMR(Y)1\text{VMR}(Y)\geq 1, we have 𝔼(X2)[𝔼(X)]2𝔼(X)\mathbb{E}(X^{2})-[\mathbb{E}(X)]^{2}\geq\mathbb{E}(X) and 𝔼(Y2)[𝔼(Y)]2𝔼(Y)\mathbb{E}(Y^{2})-[\mathbb{E}(Y)]^{2}\geq\mathbb{E}(Y). Then

VMR(Z)=p𝔼(X2)+(1p)𝔼(Y2)p𝔼(X)+(1p)𝔼(Y)+p2[𝔼(X)]22p(1p)𝔼(X)𝔼(Y)(1p)2[𝔼(Y)]2p𝔼(X)+(1p)𝔼(Y)=p𝔼(X2)p[𝔼(X)]2+(1p)𝔼(Y2)(1p)[𝔼(Y)]2p𝔼(X)+(1p)𝔼(Y)+p(1p)[𝔼(X)]22p(1p)𝔼(X)𝔼(Y)+p(1p)[𝔼(Y)]2p𝔼(X)+(1p)𝔼(Y)p𝔼(X)+(1p)𝔼(Y)p𝔼(X)+(1p)𝔼(Y)+p(1p)[𝔼(X)𝔼(Y)]2p𝔼(X)+(1p)𝔼(Y)1.\begin{split}&\text{VMR}(Z)\\ =&\frac{p\mathbb{E}(X^{2})+(1-p)\mathbb{E}(Y^{2})}{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}\\ &+\frac{-p^{2}[\mathbb{E}(X)]^{2}-2p(1-p)\mathbb{E}(X)\mathbb{E}(Y)-(1-p)^{2}[\mathbb{E}(Y)]^{2}}{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}\\ =&\frac{p\mathbb{E}(X^{2})-p[\mathbb{E}(X)]^{2}+(1-p)\mathbb{E}(Y^{2})-(1-p)[\mathbb{E}(Y)]^{2}}{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}\\ &+\frac{p(1-p)[\mathbb{E}(X)]^{2}-2p(1-p)\mathbb{E}(X)\mathbb{E}(Y)+p(1-p)[\mathbb{E}(Y)]^{2}}{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}\\ \geq&\frac{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}+\frac{p(1-p)[\mathbb{E}(X)-\mathbb{E}(Y)]^{2}}{p\mathbb{E}(X)+(1-p)\mathbb{E}(Y)}\\ \geq&1.\end{split}

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