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  • Writer's pictureKatie Galloway

Lesson 1: Modeling gene expression with differential equations

These lessons introduce how to model gene expression based on defined species reactions, the law of mass action, and differential equations.

Note: These lessons are adapted from material generously supplied by Professor Mary Dunlop, Boston University, and Professor Elisa Franco, UCLA, experts in modeling with extensive experience in training students in the fields of synthetic biology and quantitative biology.

Why model gene expression?

Modeling gene expression enables quantitative study of transcriptional networks both native and synthetic to extract systems-level principles and parameter sensitivity. Systems-level principles such as how cell division rate, degradation rate, or transcription rate impact steady-state and dynamic responses can guide the design of synthetic circuits as well as suggest interventions in native pathways. Additionally, identifying how parameters impact system performance can direct experiments even when the absolute values of parameters remain unknown. Finally, models codify understandings of system behaviors. Defining these principles enables better communication of research findings, informs experimental approach, and offers predictions that can be more easily understood across researchers of various backgrounds.

Constitutive gene expression

(e.g. constant, no regulation)

Figure 1. Model of constitutive gene expression includeing mRNA and protein with defined synthesis and degration rates.

Shown above we have a constitutive promoter which means that the promoter produces a mRNA (m) at a constant rate defined as alpha-m. Other inducible or tunable promoters will output mRNA at rates dependent on input of small molecules or transcription factors. We will treat these types of promoters later in our lessons.

Following the transcription of the gene to make mRNA, protein is produced via translation which is defined by an alpha-p.

Both mRNA and protein are degraded at some rate beta-m and beta-p.

Introducing Ordinary Differentially Equations (ODEs)

We would like to be able to turn our model of gene expression into a mathematical expression that allows us to solve how each species varies in time and understand how different parameters impact dynamics as well as steady state output.

We want to define the rate of change of X over time by writing an ODE for species x. That takes the form of dX/dt. X can be any chemical species

Static: In the case that there is no change in X over time (left graph), dX/dt=0. In this static case, the initial condition defines X. Steady state solutions are also obtained by setting non-static cases to 0 and solving.

Figure 2. Solutions to ODEs of different orders indicate the expected rate of change over time.

Zeroth order: When the rate of change of X does not depend on the concentration of X, the reaction is zeroth order. This can occur for several reasons. Either X does not participate in a reaction that impacts it generation or destruction or it's concentration is irrelevant to the rate of that reaction. For example, the rate may depend on another limited species in which case the concentration of X is "saturated". Similarly, X may be so low as to not significantly impact the rate and the rate is dominated by other factors. The result is that X increases linearly with time.

First Order: When the rate of change of X linearly depends on the concentration of X, the reaction is first order. The solution takes the form of the exponential.

So how do we know how to define dX/dt? What can we use to inform us?

Writing out the reactions and species that impact X is the first step. From there, we will use the law of mass action to define dX/dt.

Law of Mass Action

The rate of reaction if proportional to the product of the concentrations of the reactants.

For the reaction X + Y -> Z, we define the the rates of change for species X, Y, and Z using the rate coefficients K1 and K2 as shown below.

Figure 3. Law of mass action defines ODES for X+Y-> Z.

For dimers, X+ X -> Y, we can define dX/dt and dY/dt as follows.

Figure 4. Law of mass action defines ODES for dimerization of X, X+X-> Y.

Applying Mass Action to Gene Expression

So now we want to draw out the process of gene expression, define the constants for each reaction, and apply the law of mass action to define the rates of change for mRNA (m) and protein (p).

For mRNA, dm/dt increases as a zeroth order reaction with a synthesis rate of alpha-m (e.g. synthesis constant) and is degraded at a rate of beta-m (degradation constant) times it's concentration. Note: The sign of the degradation term is negative while synthesis is positive. Also note that the rate of degradation slows downs and approaches zero as the concentration of mRNA approaches zero. In the limit that makes sense since you cannot degrade what does not exist. Also, note that the degradation constant beta-m can be a function of other parameters. Most notably degradation rate can be influenced by cell division rate when the division rate is faster than the time constant of halving the concentration. In other words, if cell division cuts the concentration faster than other mechanisms, than it dominates degradation rate. For some systems, which we will discuss in the future, this is very important.

Figure 4. Using the law of mass action to convert species and rates into ODEs for mRNA and protein rates of change over time.

For protein, we have a similar equation for dp/dt where synthesis is also a function of mRNA concentration and degradation is proportional to protein concentration.

So the final model for the system we defined in the beginning gives us two differential equations.

Figure 5. Final model of ODEs for constitutive gene expression defined in fig 1.

Before we can solve this system, we need 1. parameter values and 2. initial conditions. These may be straight-forward or require a lot of experimental or literature research to define. Once we have those values we can use ODE solvers to solve the system and precisely define the how mRNA and protein change in time.

In the absence of a solver, we can very easily understand how parameters impact steady-state solutions using pencil and paper. By setting each ODE to 0, indicating the system has come to steady state, we can solve for the relationship between the parameters and the steady state concentrations of mRNA (mss) and protein (pss) as shown below.

Figure 6. Steady state solutions for constitutive gene expression highlight how the rates of synthesis and degradation define the steady-state set points for mRNA and protein.

Note: How the steady state concentration of mRNA is simply the ratio of synthesis and degradation parameters. As synthesis rate increases or degradation rate decreases, the steady state concentration of mRNA increases.

For protein the result is slightly more complex since the protein concentration depends on the mRNA concentration. However, the general form remains. As synthesis rate increases or degradation rate decreases, the steady state concentration of protein increases (assuming no change in mRNA).

Ok, so now you know how to model a single gene expressed from a constitutive promoter. In our next lesson, we will go over how to convert these equations into MATLAB scripts that can solved.

Notes without commentary available here.

Test your understanding

1. What would the set of ODEs be for the following reactions?

A+ B-> C with rate constant k1

C+B->A with rate constant k2

2. What is the steady state concentration of A set by?

3. Does the steady state concentration of A depend on the the steady state concentration of B?

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