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For transcription to start, the RNA polymerase bound to the promoter needs to undergo a conformational change to the so-called open complex. The rate of open complex formation is often much smaller than the rates for the polymerase binding and falling off the promoter. Here, we investigate within a simple model how this state of affairs might justify the equilibrium assumption underlying thermodynamic models of gene regulation, namely that the equilibrium probability that the promoter is occupied by the RNA polymerase determines the level of gene expression. (a) Write down the chemical kinetics equation for this situation. Consider three states: RNA polymerase bound nonspecifically on the DNA (N); RNA polymerase bound to the promoter in the closed complex (C); and RNA polymerase bound to the promoter in the open complex (O). To simplify matters, take both the rate for \(\mathrm{N} \rightarrow \mathrm{C}\) and the rate for \(C \rightarrow N\) to be \(k\). Assume that the transition \(C \rightarrow O\) is irreversible, with rate \(\Gamma\) (b) For \(\Gamma=0,\) show that in the steady state there are equal numbers of RNA polymerases in the \(\mathrm{N}\) and \(\mathrm{C}\) states. What is the steady state in the case \(\Gamma \neq 0 ?\) (c) For the case \(\Gamma \neq 0\), show that for times \(1 / k \ll t \ll 1 / \Gamma\), the numbers of RNA polymerases in the \(\mathrm{N}\) and \(\mathrm{C}\) states are equal, as would be expected in equilibrium.

Short Answer

Expert verified
The chemical kinetics equations for these reactions are given by d[N]/dt = -\(k_1[N]\) + \(k_{-1}[C]\), d[C]/dt = \(k_1[N]\) - \(k_{-1}[C]\) - \(\Gamma[C]\) and d[O]/dt = \(\Gamma[C]\). In the equilibrium state, there are equal numbers of RNA polymerases in N and C states for \(\Gamma=0\) and at times significantly greater than \(1/k_1\) and significantly less than \(1/\Gamma\) when \(\Gamma \neq 0\). At steady state when \(\Gamma \neq 0\), all RNA polymerases will transition to the O state.

Step by step solution

01

Represent the Biochemical Reactions

The biochemical reactions are represented as:\nN \(\xrightleftharpoons[k_{-1}]{k_1}\) C \(\xrightarrow[{}]{\Gamma}\) O\nwhere N represents RNA polymerase bound nonspecifically to the DNA, C represents RNA polymerase in the closed complex, and O represents RNA polymerase in the open complex. Here, \(k_1\) and \(k_{-1}\) represent the forward and backward rate constants.
02

Write Down the Chemical Kinetics Equations

The corresponding rate of change equations are:\n\nFor N: d[N]/dt = -\(k_1[N]\) + \(k_{-1}[C]\)\nFor C: d[C]/dt = \(k_1[N]\) - \(k_{-1}[C]\) - \(\Gamma[C]\)\nFor O: d[O]/dt = \(\Gamma[C]\)
03

Show Steady-State Conditions for \(\Gamma=0\)

At steady state, we have: d[N]/dt = 0 and d[C]/dt = 0. From these equations, one can derive that there are equal numbers of RNA polymerases in the N and C states.
04

Show Steady-State Conditions for \(\Gamma \neq 0\)

When \(\Gamma \neq 0\), in the steady state, all RNA polymerases will end up in the O state. To show that the numbers of RNA polymerases in the N and C states are equal at equilibrium for times significantly greater than \(1/k_1\) and significantly less than \(1/\Gamma\), one must take the appropriate limits of the equations written in step 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

RNA Polymerase
RNA polymerase is a crucial enzyme involved in the process of transcription, where DNA is transcribed into RNA. This enzyme binds to a specific region of the DNA known as the promoter. The binding goes through several stages:
  • Nonspecific binding to DNA, where RNA polymerase associates randomly with DNA.
  • Formation of the closed complex, where RNA polymerase binds more tightly to the promoter sequence.
  • Transition to the open complex, which is essential for the initiation of transcribing the DNA template into RNA.
Transitioning from the closed to the open complex is often the rate-limiting step in transcription initiation. This means it's usually slower than the steps leading to polymerase binding or disassociating. The irregular rate of these transitions is a focus when applying thermodynamic models in gene expression, as it reflects how equilibrium probabilities can predict levels of gene expression.
Chemical Kinetics
Chemical kinetics studies the rates of chemical reactions and factors affecting them. When looking at RNA polymerase's transitions, chemical kinetics provides an approach to model and understand these rates. In our context, the following reactions take place:- RNA polymerase binding nonspecifically to DNA, transitioning to the closed complex with rate constant \( k \).- Transition from the closed to the open complex, an irreversible step with rate constant \( \Gamma \).The steps involve:
  • First order kinetics for each state, which is typical of enzyme reactions.
  • At \( \Gamma = 0 \), steady state means equal distribution in the nonspecific (\( N \)) and closed (\( C \)) complex states.
  • For \( \Gamma eq 0 \), at equilibrium and within specific timeframes, enzymes distribute equally between \( N \) and \( C \), before proceeding to the open complex \( O \) state.
By taking limits based on the kinetics equations, we illustrate conditions for equilibrium, balancing enzyme behaviors over given time frames.
Thermodynamic Models
Thermodynamic models in gene expression consider the concept of equilibrium greatly. These models assume that the gene expression level is determined by the probability that RNA polymerase occupies the promoter in an equilibrium state. In essence, even before transcription starts, the likelihood of having RNA polymerase bound to a promoter can indicate potential gene expression levels.
  • Thermodynamic models use equilibrium statistics to simplify gene regulation, reducing complexity from kinetics.
  • They are particularly effective when kinetic transitions, like \( C \rightarrow O \), are much slower. This state of affairs encourages the simplification to equilibrium assumptions.
  • These models reveal that thermodynamic equilibrium can simplify predictions about biological systems when detailed kinetic data are impractical.
Understanding gene regulation via thermodynamic models assists in acknowledging how biological systems prioritize stability and equilibrium, thus offering a fundamental framework for gene expression study in cellular biology.

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Most popular questions from this chapter

An important concept in gene regulation is the sensitivity, that is, how steep is the change in gene expression (for example, the steepness of the transition from the "off" to the "on" state in activation) in response to a change in the number of transcription factors. It can be quantified by obtaining the slope on a log-log plot of the level of gene expression versus the number of transcription factors at this transition. Using thermodynamic models of gene regulation, determine how the sensitivity depends on the relevant parameters for the following regulatory motifs in the case of a weak promoter: (a) Simple activation. (b) Simple repression. (c) Two binding sites where the same species of repressor can bind. They can recruit each other and repress RNA polymerase independently. What happens when the interaction is turned off? For simplicity, assume that both binding sites have the same binding energy. (d) Repression in the presence of DNA looping.

In the last section of the chapter, we considered the action of N-WASP using a simple one-dimensional random walk model to treat the statistical mechanics of looping. Redo that analysis by using the Gaussian model of a polymer chain. First, assume that the loop has to close on itself and then account for the finite size of the protein domain. Compare your results with those obtained in the chapter.

In this problem, we work out an expression for the repression for the case in which there are \(N\) plasmids, each harboring the same promoter subjected to repression by the simple repression motif. (a) Write a partition function for \(P\) RNA polymerase molecules that can bind to the plasmids, resulting in expression of our gene of interest. Take into account the cell's nonspecific reservoir and assume that \(P \gg N\) Calculate the mean number of plasmids occupied by RNA polymerase, \(\langle N\rangle .\) Could you just have predicted this result based on what you know about the \(N=1\) case? (b) Work out an expression for the repression defined as $$\text { repression }=\frac{\langle N\rangle(R=0)}{\langle N\rangle(R \neq 0)}$$ Make sure to take into account the distinct cases where \(NR,\) where \(R\) is the number of repressors, and assume that you are dealing with a weak promoter, namely \(\left(P / N_{\mathrm{NS}}\right) \mathrm{e}^{-\beta \Delta \varepsilon_{\mathrm{pd}}} \ll 1\) (c) Show that your result yields the same expression for simple repression in the case where \(N=1\) that we found in the chapter. (d) Consider the case where there are two plasmids (that is, \(N=2)\) and work out the repression as a function of the number of repressors and make a corresponding plot.

The model of the Poisson promoter considered in the chapter assumed that the number of copies of the gene of interest was fixed at one. However, as a result of the replication of the chromosomal DNA, during some part of the cell cycle there will be two (or even more for rapidly dividing cells) copies of the gene of interest. In this problem, we imagine that during a fraction \(f\) of the cell cycle, there is one copy of our gene of interest and during the rest of the cell cycle there are two such copies. (a) Write down the appropriate distribution \(p(m)\) for \(m\) mRNA molecules as a function of the parameter \(f\) (b) Find \(\langle m\rangle\) (c) Find \(\left\langle m^{2}\right\rangle\) and use it to find the Fano factor. (d) Plot the Fano factor as a function of \(f\) for different choices of the mean mRNA copy number for a single promoter. How "Poissonian" do you expect an unregulated promoter to be? (Problem courtesy of Rob Brewster and Daniel Jones.)

In chemotaxis experiments, a source of nutrient molecules can be introduced into the medium containing bacteria via a micropipette. The outward diffusion of the nutrient molecules creates a position-dependent concentration gradient, and the chemotactic response of the bacteria can be observed under a microscope. (a) Estimate the nutrient gradient in steady state as a function of the distance from the micropipette \(r\) by assuming that it keeps the concentration fixed at \(c_{0}\) for distances \(r

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