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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 under Iying thermodynamic models of gene regulation, namely that the equilibrium probability that the promoter is occu pied 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 cormplex (C), and RNA poly. merase bound to the promoter in the open complex (0). 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 \(C \rightarrow 0\) is irreversible, with rate \(r\) (b) For \(\Gamma=0,\) show that in the steady state there are equal numbers of RNA polymerases in the N and C states. What is the steady state in the case \(r \neq 0 ?\) (c) For the case \(r \neq 0,\) show that for times \(1 / k

Short Answer

Expert verified
The chemical kinetic equations that represent the transitions of the RNA polymerase are \(\mathrm{N} \xrightarrow{k} \mathrm{C}, \mathrm{C} \xrightarrow{k} \mathrm{N}\) and \(\mathrm{C} \xrightarrow{r} \mathrm{O}\). When \(\Gamma = 0\), there are equal amounts of \(\mathrm{N}\) and \(\mathrm{C}\) complexes ([N]=[C]). However when \(r≠0\), the \(\mathrm{O}\) increases due to the irreversible reaction of C to O, causing less C. For a time of \(1/k < t < 1/r\), [N]=[C] representing an equilibrium condition.

Step by step solution

01

Write the chemical kinetics equation

The transitions given can be represented by the equations:\(\mathrm{N} \xrightarrow{k} \mathrm{C}, \mathrm{C} \xrightarrow{k} \mathrm{N}, \mathrm{C} \xrightarrow{r} \mathrm{O}\)
02

Evaluate the steady state condition for Γ=0

For \(\Gamma = 0\), the steady state condition implies \(\mathrm{d[N]}/\mathrm{dt} = \mathrm{d[C]}/\mathrm{dt} = 0\). This means that, for the reactions \(\mathrm{N} \xrightarrow{k} \mathrm{C}\) and \(\mathrm{C} \xrightarrow{k} \mathrm{N}\), the rate of the forward reaction is equal to the rate of the reverse reaction. Therefore, the system has equal amounts of \(\mathrm{N}\) and \(\mathrm{C}\) complexes which implies [N]=[C].
03

Steady State for r ≠ 0

Now, if \(r≠0\), the third reaction comes into play where \(\mathrm{C} \rightarrow \mathrm{O}\) which is an irreversible reaction. Due to this, the level of the \(\mathrm{O}\) (open promoter) will start to increase, creating less \(\mathrm{C}\) (closed promoter).
04

Equilibrium for r ≠ 0

For \(r ≠ 0\) and for time \(1/k < t < 1/r\), the system is as if it is in equilibrium implying [N]=[C]. During these times the polymerase has reached C state, but not yet transitioned to O due to the slower rate of C to O.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the study of reaction rates and the steps involved in a chemical reaction. It provides insights into how molecules interact, change, or degrade under certain conditions. In the context of gene expression, chemical kinetics helps explain how quickly RNA polymerase can bind to the DNA promoter site, undergo conformational changes, and initiate transcription.

Key aspects include:
  • Reaction Rates: These describe how quickly reactants convert to products. In our exercise, the rate constant \( k \) determines the speed of the RNA polymerase's movement between different states like the nonspecific DNA binding (N), closed complex (C), and open complex (O).
  • Equilibrium State: This is the condition when the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of reactants and products. At equilibrium, the amounts of RNA polymerase in the N and C states are equal when \( \Gamma = 0 \).
  • Irreversibility: The transition from the closed complex (C) state to the open complex (O) is represented as irreversible with a rate \( r \), indicating a commitment to gene expression once the RNA polymerase reaches the open state.
RNA Polymerase
RNA polymerase is an essential enzyme in the transcription process, facilitating the synthesis of RNA from a DNA template. It plays a pivotal role in gene expression by transcribing genes into messenger RNA (mRNA), which is then translated into proteins.

Features you should know:
  • Function: RNA polymerase reads the DNA template strand and assembles RNA nucleotides to form a complementary RNA strand.
  • Steps: Initially, RNA polymerase binds nonspecifically to DNA (N state). With chemical kinetics at play, it proceeds to bind specifically at the promoter region as a closed complex (C), before forming the open complex (O).
  • Types: In prokaryotes, a single RNA polymerase synthesizes all types of RNA. In eukaryotes, there are three types of RNA polymerases (I, II, III) each responsible for transcribing different types of RNA.
Promoter Binding
Promoter binding is a critical step in the initiation of transcription. The promoter is a special DNA sequence that signals RNA polymerase where to start transcription. Right at this spot, RNA polymerase must attach to begin reading the gene.

Important aspects of promoter binding include:
  • Specificity: Promoter regions are specific sequences that ensure RNA polymerase binds at the correct location on the DNA for accurate gene transcription.
  • Closed to Open Complex Transition: Initially, RNA polymerase forms a closed complex (C), indicating that it is bound but not yet ready to transcribe. This changes to an open complex (O) when the DNA strands unwind, granting access for transcription to start.
  • Rate-Limiting Step: In many cases, the formation of the open complex is slower compared to initial binding and falling off. This means it is the rate-limiting step in transcription initiation.
Transcription Regulation
Transcription regulation determines how genes are expressed by controlling when and how often a gene is transcribed into RNA. It ensures that cells produce the right amount of proteins at the right time, adapting to internal and external signals.

Crucial points in regulation include:
  • Control Elements: These may include enhancers, silencers, and promoters that influence the binding of RNA polymerase to the DNA and the rate of transcription.
  • Regulatory Proteins: Transcription factors and other proteins bind to specific DNA sequences to modulate the transcription rate, either enhancing or repressing gene expression.
  • Feedback Mechanisms: Gene expression levels can be fine-tuned through feedback loops, ensuring homeostasis within the cell.
  • Dynamic Process: The regulation of transcription is often dynamic, responding to signals such as nutrient availability, stress, and cellular differentiation status.

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Most popular questions from this chapter

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.

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 vs. 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 oft? For simplicity, assume that both binding sites have the same binding energy. (d) Repression in the presence of DNA looping.

In the thermodynamic models of gene regulation discussed in the chapter the RNA polymerase is treated as a single molecular species. While this might be a reasonable assumption for transcription in prokaryotes, in eukary. otes tens of different molecules need to come together in order to form the transcriptional machinery. The objective of this problem is to develop intuition about the requirements for our simple model to apply in such a com plex case by assuming that the transcriptional machinery is made out of two different subunits. \(X\) and \(Y\), that come together at the promoter. (a) Calculate the probability of finding the complex \(x+Y\) bound to the promoter in the case where unit \(x\) binds to DNA and unit \(Y\) binds to \(X\). Can you reduce this to an effective one molecule problem such as in the bacterial case? (b) Calculate the fold change in gene expression for \(\operatorname{sim}\) ple repression using transcriptional machinery such as that proposed in part (a). Explore the weak promoter assumption in order to reduce the expression to that corresponding to the bacterial case. Repeat this for the case where an activator can contact \(Y\) (c) Repeat parts \((a)\) and \((b)\) for a case when \(Y\) binds to a site on the DNA which is near the \(X\) binding site, and there is an interaction energy between \(X\) and \(Y\)

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