Background & Motivation

At the moment, there does not exist an assay which reports directly on the occupancies of transcription factor binding sites across the genome in single cells. However, we can measure chromatin accessibility in single cells via single-cell ATAC-seq, which reports a 1-dimensional projection of TFBS occupancies. This blog post explores an approach to lift single cell ATAC signals onto the space of TFBS occupancies, i.e. to deconvolve ATAC signals into TF binding tracks.

Our approach relies on an orthogonal suite of experiments characterizing transcription factor binding in vitro. In High-Throughput SELEX experiments, the...

Equilibrium Binding Model

There exist two mathematical formalisms to describe the probabilities of molecular configurations at equilibrium: thermodynamics and kinetics. The kinetics formalism is more general: it describes a system's trajectory to equilibrium, and can describe non-equilibrium systems. Tthe thermodynamics formalism only describes equilibrium, but requires fewer parameters to do so. We'll derive an expression for the probability of a single TFBS' occupancy with both formalisms, but drop the kinetic description for more elaborate configurations.

Single TF

$$\newcommand{\kon}{k_{\mathrm{on}}} \newcommand{\koff}{k_{\mathrm{off}}}$$

Most derivations of the probability of TFBS occupancy at equilibrium employ a kinetics formalism, so we'll walk through that first, and then explore the analog in the thermodynamics description. In the kinetic description, the parameters are rates.

$$ \mathrm{TF} + \mathrm{TFBS} \underset{\koff}{\overset{\kon}{\rightleftarrows}} \mathrm{TF\colon TFBS} $$

Note: We will assume the Law of Mass Action for well-mixed solutions throughout.

The natural rates are the rate of TF binding $\kon$ and unbinding $\koff$. Equilibrium is reached when binding and unbinding are balanced:

$$\frac{d[\mathrm{TF\colon TFBS}]}{dt} = k_{\mathrm{on}}[\mathrm{TF}][\mathrm{TFBS}] - k_{\mathrm{off}}[\mathrm{TF\colon TFBS}] = 0 \text{ at equilibrium}$$ $$k_{\mathrm{on}}[\mathrm{TF}]_{\mathrm{eq}}[\mathrm{TFBS}]_{\mathrm{eq}} = k_{\mathrm{off}}[\mathrm{TF\colon TFBS}]_{\mathrm{eq}}$$ $$\text{(dropping eq subscript) }[\mathrm{TF\colon TFBS}] = \frac{k_{\mathrm{on}}[\mathrm{TF}][\mathrm{TFBS}]}{k_{\mathrm{off}}} = \frac{[\mathrm{TF}][\mathrm{TFBS}]}{k_{d}}$$

where $k_{d} = \frac{\koff}{\kon}$ is called the dissociation constant (or equilibrium constant). We'd like to determine the probability of finding the TFBS occupied, i.e. the fraction of time it spends in the bound state. That fraction is $\frac{[\mathrm{bound}]}{([\mathrm{unbound}] + [\mathrm{bound}])} = \frac{[\mathrm{TF\colon TFBS}]}{([\mathrm{TFBS}] + [\mathrm{TF\colon TFBS}])}$. Define the denominator as $[\mathrm{TFBS}]_{0} = [\mathrm{TFBS}] + [\mathrm{TF\colon TFBS}]$ so that $[\mathrm{TFBS}] = [\mathrm{TFBS}]_{0} - [\mathrm{TF\colon TFBS}]$ and substitute:

$$[\mathrm{TF\colon TFBS}] = \frac{[\mathrm{TF}]([\mathrm{TFBS}]_{0} - [\mathrm{TF\colon TFBS}])}{k_{d}}$$ $$[\mathrm{TF\colon TFBS}](k_d + [\mathrm{TF}]) = [\mathrm{TF}][\mathrm{TFBS}]_{0}$$ $$\frac{[\mathrm{TF\colon TFBS}]}{[\mathrm{TFBS}]_{0}} = \frac{[\mathrm{TF}]}{k_d + [\mathrm{TF}]}$$

Note: We could also ask for this expression in terms of $[\mathrm{TF}]_0 = [\mathrm{TF}] + [\mathrm{TF\colon TFBS}]$ however, since we’re considering a single TFBS, $[\mathrm{TF\colon TFBS}]$ is at most 1, and so $[\mathrm{TF}]_0 \approx [\mathrm{TF}]$. In instances of ligand-receptor binding in which that approximation cannot be made, the fraction bound is a messy quadratic. Derivation here.

In the thermodynamic description, the parameters are Gibbs free energies $\Delta G$. Let's follow the derivation from Physical Biology of the Cell (pp. 242) and consider the number microstates underlying each of the of bound and unbound macrostates, and their energies.

In order to count microstates, we imagine distributing $L$ TF molecules across a space-filling lattice with $\Omega$ sites. The energy of a TF in solution is $\varepsilon_{\mathrm{solution}}$ and the energy of a bound TF is $\varepsilon_{\mathrm{bound}}$. $\beta$ is the constant $1/k_b T$ where $k_b$ is Boltzmann's constant and $T$ is the temperature.

In our case, the number of microstates in the unbound macrostate is $\frac{\Omega !}{L!(\Omega -L)!}\approx \frac{\Omega^L}{L!}$ and they each have energy $L \cdot \varepsilon_s$. The number of microstates in the bound macrostate is $\frac{\Omega !}{(L-1)!(\Omega -(L+1))}\approx \frac{\Omega^{(L-1)}}{(L-1)!}$ and they each have energy $(L-1) \varepsilon_s + \varepsilon_b$.

The Boltzmann distribution describes the probability of a microstate as a function of its energy: $p(E_i) = e^{-\beta E_i}/Z$ where $Z$ is the "partition function" or simply $\sum_i e^{-\beta E_i}$ the sum of the weights of the microstates, which normalizes the distribution. In our case:

$$Z(L,\Omega)=\left(\colorbox{LightCyan}{$ \frac{\Omega^L}{L!} e^{-\beta L \varepsilon_s}$}\right) + \left(\colorbox{Seashell}{$\frac{\Omega^{(L-1)}}{(L-1)!} e^{-\beta [(L-1) \varepsilon_s + \varepsilon_b]}$}\right)$$

With that expression in hand, we can express the probability of the bound macrostate, $p_b$:

$$p_b=\frac{ \colorbox{Seashell}{$\frac{\Omega^{(L-1)}}{(L-1)!} e^{-\beta [(L-1) \varepsilon_s + \varepsilon_b]}$}}{\colorbox{LightCyan}{$\frac{\Omega^L}{L!} e^{-\beta L \varepsilon_s}$} + \colorbox{Seashell}{$\frac{\Omega^{(L-1)}}{(L-1)!} e^{-\beta [(L-1) \varepsilon_s + \varepsilon_b]}$}} \cdot \color{DarkRed}\frac{\frac{\Omega^L}{L!}e^{\beta L \varepsilon_s}}{\frac{\Omega^L}{L!}e^{\beta L \varepsilon_s}} \color{black} = \frac{(L/\Omega)e^{- \beta \Delta \varepsilon}}{1+(L/\Omega)e^{- \beta \Delta \varepsilon}} $$

Where we have defined $\Delta \varepsilon = \varepsilon_b - \varepsilon_s$. $L/\Omega$ is really just the concentration $[\mathrm{TF}]$, which leaves us with an expression suspiciously similar to the one we derived from the kinetics formalism:

$$p_b = \frac{[\mathrm{TF}]e^{-\beta \Delta \varepsilon}}{1+[\mathrm{TF}]e^{-\beta \Delta \varepsilon}} \cdot \color{DarkRed}\frac{e^{\beta \Delta \varepsilon}}{e^{\beta \Delta \varepsilon}} \color{black} = \frac{[\mathrm{TF}]}{e^{\beta \Delta \varepsilon}+[\mathrm{TF}]}$$

From which we recapitulate an important correspondence between kinetics and thermodynamics at equilibrium: $ k_d = e^{\beta \Delta \varepsilon} = e^{\Delta \varepsilon / k_bT} $ more commonly written for different units as $k = e^{-\Delta G / RT}$.

The takeaway is that both the kinetic and thermodynamic formalisms produce an equivalent expression for the probabilities of each of the bound and unbound configurations, parameterized respectively by $k_d$ and $\Delta G$.

Direct Cooperativity

Suppose now that two TFs bind adjacent segments of DNA in such a way that the binding of either facilitates the binding of the other. We call this direct cooperativity.

As before, we enumerate the configurations, their multiplicities, and their energies. We'll call the TFs A and B. We'll denote their counts as $A$ and $B$. The energy of a TF in solution will once more be $A_s$ and bound to its cognate TFBS $A_b$. The energy of cooperativity will be $C_{AB}$.

StateEnergyMultiplicityWeight
$A \cdot A_s + B \cdot B_s$ $\frac{\Omega!}{(\Omega - A - B)!A!B!} \approx \frac{\Omega^{A+B}}{A!B!}$ $\frac{\Omega^{A+B}}{A!B!} \cdot e^{-\beta \left[ A \cdot A_s + B \cdot B_s \right]}$
$(A - 1) A_s + A_b + B \cdot B_s$ $\frac{\Omega!}{(\Omega - (A - 1) - B)!(A-1)!B!} \approx \frac{\Omega^{A+B-1}}{(A-1)!B!}$ $\frac{\Omega^{A+B-1}}{(A-1)!B!} \cdot e^{-\beta \left[ (A - 1) A_s + A_b + B \cdot B_s \right]}$
$A \cdot A_s + (B - 1) B_s + B_b$ $\frac{\Omega!}{(\Omega - A - (B - 1))!A!(B-1)!} \approx \frac{\Omega^{A+B-1}}{A!(B-1)!}$ $\frac{\Omega^{A+B-1}}{A!(B-1)!} \cdot e^{-\beta \left[ A \cdot A_s + (B - 1) B_s + B_b \right]}$
$(A - 1) A_s + A_b + (B - 1) B_s + B_b + C_{AB}$ $\frac{\Omega!}{(\Omega - (A - 1) - (B-1))!(A-1)!(B-1)!} \approx \frac{\Omega^{A+B-2}}{(A-1)!(B-1)!}$ $\frac{\Omega^{A+B-2}}{(A-1)!(B-1)!} \cdot e^{-\beta \left[ (A - 1) A_s + A_b + (B - 1) B_s + B_b + C_{AB} \right]}$

The partition function is the sum of the weights:

$$ Z = \frac{\Omega^{A+B}}{A!B!} \cdot e^{-\beta \left[ A \cdot A_s + B \cdot B_s \right]} + \frac{\Omega^{A+B-1}}{(A-1)!B!} \cdot e^{-\beta \left[ (A - 1) A_s + A_b + B \cdot B_s \right]} + \frac{\Omega^{A+B-1}}{A!(B-1)!} \cdot e^{-\beta \left[ A \cdot A_s + (B - 1) B_s + B_b \right]} + \frac{\Omega^{A+B-2}}{(A-1)!(B-1)!} \cdot e^{-\beta \left[ (A - 1) A_s + A_b + (B - 1) B_s + B_b + C_{AB} \right]}$$

Which we can greatly simplify by multiplying the entire expression by the reciprocal of the "base state" weight, $\color{DarkRed}\frac{A!B!}{\Omega^{A+B}} \cdot e^{\beta \left[ A \cdot A_s + B \cdot B_s \right]}$, normalizing that weight to 1:

$$ Z = 1 + \frac{A}{\Omega} \cdot e^{-\beta \left[ A_b-A_s \right]} + \frac{B}{\Omega} \cdot e^{-\beta \left[ B_b-B_s \right]} + \frac{A \cdot B}{\Omega^2} \cdot e^{-\beta \left[ A_b-A_s+B_b-B_s+C_{AB} \right]}$$

Taking the definition $[A] = A/\Omega$ and $\Delta G_A = A_b-A_s$ produces:

$$ Z = 1 + [A] e^{-\beta \left[ \Delta G_A \right]} + [B] e^{-\beta \left[ \Delta G_B \right]} + [A][B] e^{-\beta \left[ \Delta G_A+\Delta G_B+C_{AB} \right]}$$

Then the probability of any state is just the weight of that state divided by the partition function expression $Z$.

Competitive binding

Indirect cooperativity

Putting it all together

TFBSs = [{'name':'A','start':10,'end':18,'dG':100},
         {'name':'B','start':22,'end':30,'dG':100}]

TF_concs = {'A': 100, 'B': 100}
 
def enumerate_configs(TFBSs):
    pass

def Z(TFBSs, TF_concs):

    return [P_config(config) for config in enumerate_configs(TFBSs)]

def P_config(config):

    return multiplicity(config, TF_concs) * exp(-energy(TFBSs, TF_concs)/kbT)


def energy(TFBSs, TF_concs):
    pass
    # energy of binding of each occupied TFBS
    # energy of cooperativity between each bound TFBS
    # energy of competition between each bound TFBS
    # energy of competition between each bound TFBS and the nucleosome
    # energy of every unbound TF in solution


def multiplicity(TFBSs, TF_concs):
    pass
# supposing I have concentrations rather than molecule counts, what is the continuum limit of the energy expressions?
# TFBSs is a list of structs with fields 'TF_name', 'pos_start', 'pos_end', 'dG'

# we also need a way to start nucleosome positions
config = [1., 1.]
TFs = list(TF_concs.keys())
genomic_range = [0, 50]
df = pd.DataFrame(TFBSs)
def c(TF_name):
    return {'A': 'blue', 'B': 'orange'}[TF_name]
def draw_config(TFBSs, config):

    plt.rcParams['figure.figsize'] = [12, int(np.log2(len(TFs)))]

    fig, ax = plt.subplots()

    ax.scatter(df.start.values, df.name.values,  alpha=0)
    ax.set(ylabel='TFs', yticks=range(len(TFs)), yticklabels=TFs, ylim=[-1, len(TFs)+1], xlabel='Genome', xlim=genomic_range)

    errorboxes = []

    for i, tfbs in enumerate(TFBSs):
        ax.add_patch(Rectangle((tfbs['start'], i), tfbs['end']-tfbs['start'], 0.9, fc=c(tfbs['name']), alpha=0.05))
        ax.add_patch(Rectangle((tfbs['start'], i), tfbs['end']-tfbs['start'], 0.1, fc=c(tfbs['name']), alpha=1))

    ax.grid(axis='y', lw=0.1)
    for side in ["right","top","left","bottom"]: ax.spines[side].set_visible(False)

    return ax
ax = draw_config(TFBSs, config)

Conditioning on ATAC