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Nucleosomes in a box Repeat the derivation of nucleosome accessibility using the canonical distribution. This means that you should imagine a nucleosome in a box with \(L\) DNA-binding proteins and then work out the probability that the nucleosome will be occupied by one of those proteins as a function of their concentration and as a function of the position of the binding site on the DNA.

Short Answer

Expert verified
The probability of a nucleosome being occupied by a protein depends exponentially on the concentration of the protein and the position of the binding site on the DNA. It was determined by applying the canonical distribution to this biological system.

Step by step solution

01

Define the Problem Variables

First, define the variables for the problem. Let \( p \) be the probability of a nucleosome being occupied by a protein, \( L \) be the total number of proteins in the system, and \( x \) be the position of the binding site on the DNA.
02

Apply Canonical Distribution

Next, apply the concept of the canonical distribution. According to this distribution, the relative number of particles (proteins) that have a certain level (position on the DNA) is given by an exponential factor of the energy of the state at that level. In this case, the energy is assumed to be a function of the position \( x \). The canonical distribution, hence, gives the probability that a nucleosome at position \( x \) is occupied by a protein as \( p(x) = \frac{1}{Z} e^{-E(x)} \), where \( E(x) \) is the energy at position \( x \) and \( Z \) is the partition function for normalizing the probability, given by \( Z = \int_{0}^{L} e^{-E(x)} dx \).
03

Calculate Protein Concentration Effect

To find the probability as a function of the protein concentration, assume that the energy required for a protein to bind to the DNA is inversely proportional to the concentration of the protein. Therefore, the energy can be written as \( E(x) = k/n \), where \( k \) is a constant and \( n \) is the concentration of the protein. Substituting into the earlier equation gives: \( p(x) = \frac{1}{Z} e^{-k/n} \).
04

Final Result

Thus, from the above steps, we have the probability function as a function of the concentration of proteins and position on the DNA. The main conclusion here is that the probability of a nucleosome being occupied by a protein depends on the concentration of the protein and the position of the binding site on the DNA. The higher the protein concentration, the more likely it is that a nucleosome will be occupied by a protein. Similarly, the location of the binding site on the DNA also influences this probability.

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Key Concepts

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

Canonical Distribution
The canonical distribution is a principle used to understand how particles, such as proteins, distribute themselves among various energy states. In the context of nucleosome accessibility, it helps us predict how likely a DNA nucleosome is to be occupied by a protein. Imagine each nucleosome as a box where proteins can "live."

In this scenario, each "state" or energy level depends on the position of the nucleosome, which is crucial since proteins may bind more readily to certain sites. The canonical distribution formula represents the probability of a state being occupied by a protein. This probability depends on the energy of the state and is expressed as: \[ p(x) = \frac{1}{Z} e^{-E(x)} \]Here, \(E(x)\) denotes the energy associated with a position \(x\), and \(Z\) is the partition function that ensures all probabilities across possible states sum to one.
DNA Binding Proteins
DNA binding proteins are essential molecules that interact specifically with DNA sequences. They play a critical role in regulating gene expression and maintaining cellular functions. These proteins can latch onto particular sequences of DNA known as binding sites.

Their affinity, or how strongly they bind to DNA, can depend heavily on the sequence of the binding site. Some proteins find it easier to attach to certain sequences, which can influence where and how often they bind. This behavior is integral to determining nucleosome accessibility because not every protein will bind with the same strength to every site available along the DNA. Thus, understanding which proteins are available and how selective they are provides insights into which sites on the DNA are accessible at any given time.
Protein Concentration
The concentration of proteins in a given cell environment can dramatically affect nucleosome accessibility. Protein concentration refers to how many proteins are present in a specific volume. Higher concentrations often mean more proteins are available to bind to DNA.

Imagine standing in a room full of balloons: the denser the balloon population, the likelier it is you will grab one easily. Similarly, with more DNA binding proteins floating around, the chance increases that they will find and occupy their respective binding sites on the DNA.

In our formula, as the protein concentration \(n\) increases, the effective energy \(E(x)\) necessary for a protein to bind reduces, making it easier for proteins to attach. This dynamic showcases the direct relationship between concentration and binding probability: \[ p(x) = \frac{1}{Z} e^{-k/n} \]where \(k\) is a constant indicating energy-related constraints.
Binding Site Position
The position of the binding site on the DNA plays a pivotal role in determining the likelihood of a protein binding there. DNA has a long sequence, and each segment can present different "sites" with varying affinity.

Certain positions might be energetically more favorable for protein binding, thus often being occupied. Conversely, some areas may be less accessible due to structural hindrances or lower affinity for the proteins in question.

The position of these sites dictates changes in the potential energy \(E(x)\) in our probability formula. Therefore, understanding these spatial preferences involves mapping these energetically favorable sites where proteins naturally gravitate. This concept helps explain why protein binding doesn't happen uniformly throughout the DNA, but rather, follows a distribution influenced by specific "hotspots" or preferred sites on the DNA.

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Most popular questions from this chapter

Nucleosome formation and assembly (a) Repeat the derivation given in the chapter for the nucleosome formation energy, but now assuming that there is a discrete number of contacts \((N=14)\) between the DNA and the histone octamer. (b) Use the discrete model to calculate the equilibrium accessibility of binding sites wrapped around nucleosomes. Apply this model to the data by Polach and Widom (1995) and fit the adhesive energy per contact \(\gamma\) discrete (c) Reproduce Figure 10.25 and compare your results for the equilibrium accessibility versus burial depth from (a) and (b) with the continuum model. (d) Look at some of the binding affinities of different DNA sequences to histones reported by Lowary and Widom \((1998) .\) Once again, assume that the electrostatic interaction between the histone and the different DNA molecules does not vary, that is, it is not sequence-dependent. This is equivalent to saying that the difference between each sequence lies in its flexibility, in its bending energy. What would one expect the difference in their persistence lengths to be? (e) Model the case of having two binding sites for the same DNA-binding protein on a DNA molecule that is wrapped around a histone octamer. How does the equilibrium accessibility depend on the protein concentration and the relative position of the binding site? How does the problem change if the two binding sites correspond to two different DNA-binding proteins? Relevant data for this problem is provided on the book's website.

Flexural rigidity of biopolymers (a) Recall from \(\mathrm{p} .389\) that when treating macromolecules as elastic beams it is the combination \(K_{\mathrm{eff}}=E I,\) the flexural rigidity, that dictates the stiffness of that molecule. The flexural rigidity is a product of an energetic ( \(E,\) Young modulus) and geometric \((I,\) areal moment of inertia) factor. Reproduce the argument given in the chapter that culminated in Equation 10.8 for the bending energy of a beam and show that the flexural rigidity enters as claimed above. (b) Using what you know about the geometry of DNA, actin filaments, and microtubules, determine the areal moment of inertia \(I\) for each of these molecules. Be careful and remember that microtubules are hollow. Make sure that you comment on the various simplifications that you are making when you replace the macromolecule by some simple geometry. (c) Given that the elastic modulus of actin is \(2.3 \mathrm{GPa}\), take as your working hypothesis that \(E\) is universal for the macromolecules of interest here and has a value 2 GPa. In light of this choice of modulus, compute the stress needed to stretch both actin and DNA with a strain of \(1 \%\). Convert this result into a pulling force in piconewtons. (d) Using the results from (b) and (c), compute the persistence lengths of all three of these molecules. (e) Given that the measured persistence length of DNA is \(50 \mathrm{nm},\) and using the areal moment of inertia you computed in (b), compute the Young modulus of DNA. How well does it agree with our 2 GPa rule of thumb from above?

Persistence length and Fourier analysis In the chapter, we computed the tangent-tangent correlation function for a polymer, which we modeled as an elastic beam undergoing thermal fluctuations. The calculation was carried out in the limit of small fluctuations and it led to an expression for the persistence length in terms of the flexural rigidity of the polymer. Here we reexamine this problem for a two-dimensional polymer, but without the assumption of small fluctuations. (a) For a polymer confined to a plane, the tangent vector \(\mathbf{t}(s)\) can be written in terms of the polar angle \(\theta(s)\) as \(\mathbf{t}(s)=(\cos \theta(s), \sin \theta(s)) .\) Rewrite the beam bending energy Equation \(10.9,\) in terms of the polar angle \(\theta(s)\) (b) Expand the polar angle \(\theta(s)\) into a Fourier series, taking into account the boundary conditions \(\theta(0)=0\) and \(\mathrm{d} \theta / \mathrm{d} s=0\) for \(s=L\). The first boundary condition comes about by choosing the orientation of the polymer so that the tangent vector at \(s=0\) is always along the \(x\) -axis. Convince yourself that the second is a consequence of there being no force acting on the end of the polymer. (c) Rewrite the bending energy in terms of the Fourier amplitudes \(\tilde{\theta}_{n},\) introduced in (b), and show that it takes on the form equivalent to that of many independent harmonic oscillators. Use equipartition to compute the thermal average of each of the Fourier amplitudes. (d) Make use of the identity \(\langle\cos X\rangle=\mathrm{e}^{-X^{2} / 2},\) which holds for a Gaussian distributed random variable \(X,\) to obtain the equation for the tangent-tangent correlation function, \((\mathbf{t}(s) \cdot \mathbf{t}(0)\rangle=\mathrm{e}^{-\theta(s)^{2} / 2} .\) Then compute \(\left\langle\theta(s)^{2}\right)\) by using the Fourier series representation of \(\theta(s)\) and the average values of the Fourier amplitudes \(\bar{\theta}_{n}\) obtained in (c). Convince yourself either by plotting or Fourier analysis that on the \\[\text { interval } 0

The J-factor connection Derive the relation between the \(J\) -factor and the free energy of cyclization described in Section \(10.3 .3 .\) In particular. construct a lattice model of the cyclization process by imagining a box of \(\Omega\) lattice sites each with volume \(v\). We consider three species of DNA: monomers with sticky ends that are complementary to each other; dimers, which reflect two monomers sticking together; and DNA circles in which the two ends on the same molecule have stuck together; The number of molecules of each species is \(N_{1}, N_{2},\) and \(N_{C}\) respectively. (a) Use the lattice model to write down the free energy of this assembly and attribute an energy \(\varepsilon_{\mathrm{b}}\) to the binding of complementary ends, \(z_{\text {loop }}\) to the looped configurations, and \(\epsilon_{\text {?ol }}\) as the energy associated with DNA-solvent interactions when the length of the DNA corresponds to one monomer. Further, assume that the solvent energy for a dimer is \(2 x_{\text {sol }}\) and for a looped configuration is identical to that of a monomer. Use a Lagrange multiplier \(\mu\) to impose the constraint that \(N_{\mathrm{tot}}=\) \(N_{1}+2 N_{2}+N_{c}\) (b) Minimize the free energy with respect to \(N_{1}, N_{2},\) and \(N_{c}\) to find expressions for the concentrations of the three species. Then use the fact that \(J\) is the concentration of monomers at which \(N_{2}=N_{\mathrm{c}}\) to solve for the unknown Lagrange multiplier \(\mu\) and to obtain an expression for the J-factor in terms of the looping free energy.

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