Analysis of Single Locus Trajectories for Extracting In Vivo Chromatin Tethering Interactions

Is it possible to extract tethering forces applied on chromatin from the statistics of a single locus trajectories imaged in vivo? Chromatin fragments interact with many partners such as the nuclear membrane, other chromosomes or nuclear bodies, but the resulting forces cannot be directly measured in vivo. However, they impact chromatin dynamics and should be reflected in particular in the motion of a single locus. We present here a method based on polymer models and statistics of single trajectories to extract the force characteristics and in particular when they are generated by the gradient of a quadratic potential well. Using numerical simulations of a Rouse polymer and live cell imaging of the MAT-locus located on the yeast Saccharomyces cerevisiae chromosome III, we recover the amplitude and the distance between the observed and the interacting monomer. To conclude, the confined trajectories we observed in vivo reflect local interaction on chromatin.


Supplementary Information
In this Supplementary Information, we detail the analysis, calculations and results presented in the main text.

Extracting the strength of a potential well for a Rouse polymer
A Rouse polymer which has one monomer interacting with a infinite potential well, can be described by the energy potential where R = (R 1 , R 2 , ..., R N ) is the collection of the monomers, connected by a spring constant κ = dk B T /b 2 , b is the standard-deviation of the distance between adjacent monomers [21], k B the Boltzmann coefficient, T the temperature and d the dimensionality (dim 2 or 3), k is the strength of the external harmonic well acting on monomer n, located at the origin.
To extract the strength of the potential well applied on monomer n from the measured velocity of locus c (n < c), we will compute ∫ Ω dR N (∇ Rc Φ)P (R|R c = x). (2) The force acting on monomer c, when its position is x is given by The equilibrium probability distribution function is the Boltzmann distribution, conditioned to R c = x: with the normalization factor The matrix A is a matrix that can be decomposed into d blocks A i , each of size (N − 1) × (N − 1). A i is also a block matrix of with which is of size (c − 1) × (c − 1) and The vector B is composed of d blocks, each given by We solve again the recurrence relation Eq.(18) has the same characteristic polynomial (eq.12), thus and with the new initial conditions (eq.18), we find Thus, since detA i 1 = θ c−1 [34]: and using eq.(15) we find The term in the exponential in eq.(5) is given by Since A i 1 is a tridiagonal matrix [34] ( while [34] ( Substituting (24) and (25) into (23) we find . (26) Substituting (26) into (5) we find We can now we calculate the conditional expectation of the measured velocity of monomer c (eq.2). where For a Rouse polymer the force depends only on the distance along the chain |c − n|. Interestingly, the restoring force decays inversely proportional to the distance along the chain.

The variance of monomer c position
We now compute the variance of a monomer position ⟨R 2 c ⟩ with respect to its average position (mean zero) where the potential Φ is given by (1) and G is a matrix composed of d blocks (G i ), each of size N × N : Thus, rewriting eq.(31) where [34] ( and ϕ l = b l θ l+1 − a l+1 c l θ l+2 for n < l ≤ N, Since c > n, solving the recurrence relation 35, we find Thus, using the value θ c−1 (eq.20) while Finally, the determinant is given by and the inverse element Finally, substituting (40) into eq.(33), we find the variance for the position to be where we use the definition 30 for k cn .

Extracting the strength of a potential well for a βpolymer
We now derive an expression for the measured velocity of a monomer c when monomer n further away along the chain interacts with an harmonic potential well for the β-polymer model [24]. We recall that for the β-model, the polymer potential is giveñ where When a localized interaction acts on monomer n, the polymer energy becomes which can be represented as where The probability distribution function is the Boltzmann distribution while the conditional probability distribution function is The normalization factor N is given by The matrixC is the reduced matrix C (eq.46) when the row and column c is removed. It is a block matrix with d blocks each of size (N − 1) × (N − 1), where each block i given bỹ and the vector F is compose of d blocks, each of size (N − 1): The force acting on monomer c is given by The conditional expectation of the velocity of monomer c at small time steps is where the conditional force is given by (52) and the conditional probability by (48) and (49). Thus We conclude that where While for a Rouse polymer the force depends only on the distance along the chain |c − n|, for a β-polymer, the effective spring coefficient depends on all monomers.

Extracting the strength of two potential wells for a Rouse polymer
For a Rouse polymer subjected to two gradient forces, the energy U ext (R) can now described bỹ The external potentials represent the interaction of two monomers. We compute here the average position ⟨R c ⟩ of monomer c, which is situated between monomers n and m. For that goal, we rewrite the potentialΦ(R) in a quadratic formΦ where the matrix A is made of d blocks, each of size N × N , where each block i is given by and the vector B is compose of d blocks, each given by Since A i is a tridiagonal matrix, we use the algorithm proposed in [34] to calculate its determinant and inverse of matrix. For a, b < m, A i a,b has the same structure as the matrix 7. Thus We next find the series solution at position m: and Again, we solve the recurrence relation Eq.(66) has the characteristic polynomial (eq.12), thus using the initial conditions (eq.66), we find Using that we obtain that the determinant is given by The inverse of the matrix A i is given by [34] ( where Solving the recurrence relation 71 we find The normalization factor is Substituting eq.(60) in the exponential term of the last equation, we get We now estimate these elements of (A i ) −1 : Finally, we compute the average position of monomer c Thus We now estimate the conditional expectation of the velocity for the observed monomer c, in the small time step regime The conditional expectation of the velocity of monomer c is whereÑ is the normalization as computed in 73 when the c column is removed.
which is of size (c − 1) × (c − 1) and which is of size (N − c) × (N − c). The vectorB is compose of d blocks, each given byB To find the determinant detÃ i 1 , we use eq.(63): and using eq.(72) The elements of the inverse matrix are given by , , Now, we estimate the term in the parenthesis in eq.(80) by substituting eqs.(87) We introduce into expression eq.(88) the position of monomer c with respect to its mean position (x = x − ⟨R c ⟩) (eq.78) .

Variance of monomer c position
We now find the variance of a monomer position ⟨R 2 c ⟩ with respect to its average position We begin by calculating where Substituting (94), (95) and (78) into eq.(93) we find

Two potential wells located on the left side of the observed locus (n < m < c)
Finally, we solve for the case n < m < c. We being with the average position of monomer c where A i is given in eq.(59). Thus , .
The average position of c is The conditional expectation of the velocity of monomer c for a small time step is defined as where U i is also a block matrix of size N − 1 × N − 1 with which is of size (c − 1) × (c − 1) and The vector V is composed of d blocks, each given by To find the determinant detŨ i 1 , we use eq.(67): and using eq.(61), we get The (c-1,c-1) term of the inverse matrix is given by where we used eqs.(72),(67). For simplicity we name the indices of the matrix and We estimate the term in the parenthesis in eq.(100) by substituting eqs.(107)-(110) where k c,nm = κ κ(k n + k m ) + (m − n)k n k m κ 2 + ((c − n)k n + (c − m)k m )κ + (m − n)(c − m)k n k m .
for t 2 > t 1 , where we approximated for long times and introduced back the expression for b p . Similarly, for the center of mass we have Thus, the auto-correlation function of monomer c is where (α n 0 ) 2 = 1 N .