Lecture notes. Written from the material of MIT 8.592J, licensed CC BY-NC-SA 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

10. Protein Search Kinetics on DNA¶

What this covers¶

A repressor protein finds its one target sequence among millions of DNA base pairs fast enough to shut off a gene within a fraction of a second. This chapter asks how, quantitatively. It starts from the measured on-rate for a real protein-DNA pair, builds the classical diffusion-limited theory of that rate, shows the theory is two orders of magnitude too slow, and then works through the fix — a protein that alternates between sliding along the DNA and jumping through solution. It assumes the diffusion equation, elementary mass-action kinetics, and the geometric and exponential distributions.

The kinetics of a protein-DNA complex¶

Consider a protein $P$ that can bind a specific site on a DNA molecule. The concentration of the bound complex changes for two reasons: complexes form, at a rate proportional to how likely a free site and a free protein are to be at the same place, and complexes break up, at a rate proportional to how many there already are. At the low concentrations relevant to a cell, both proportionalities are exact (this is ordinary mass-action kinetics), giving

\[\frac{d}{dt}[P{\cdot}\text{DNA}] = k_a [P][\text{DNA}] - k_d [P{\cdot}\text{DNA}],\]

where $k_a$ is the on-rate and $k_d$ the off-rate. For the lac repressor binding its operator on E. coli DNA, in vitro measurements give $k_a \sim 10^{10}\ \text{M}^{-1}\text{s}^{-1}$.

The question. Suppose lactose is abundant, so the repressor sits inactive and no complexes have formed. At $t = 0$ lactose vanishes, the repressor activates, and it must now find the operator to switch the gene off. How long does that take?

There are only a handful of operator sequences per cell. Taking the cell volume as $1\ \mu\text{m}^3$ gives an operator number density of about $1/\mu\text{m}^3$, which converted to molar units ($1\ \text{M} = 6\times 10^{26}\ \text{m}^{-3}$) is $[\text{DNA}] \sim 10^{-9}\ \text{M}$. Because this concentration barely changes while the search is still going on, the rate equation linearises at early times:

\[\frac{d}{dt}[P{\cdot}\text{DNA}] \approx \big(k_a[\text{DNA}]\big)[P] = [P]/\tau, \qquad \tau \equiv \frac{1}{k_a[\text{DNA}]}.\]

$\tau$ is the characteristic time for one free repressor to locate the operator. Plugging in the measured $k_a$ and the estimated $[\text{DNA}]$ gives $\tau \sim 0.1\ \text{s}$: the repressor finds its target in about a tenth of a second. The rest of the chapter asks where the number $k_a\sim 10^{10}\ \text{M}^{-1}\text{s}^{-1}$ that made this possible actually comes from.

Diffusion-limited kinetics: the Debye-Smoluchowski rate¶

The classical answer is that binding is limited only by how fast diffusion can bring the two partners together. Model the cell as a sphere of radius $R$ with the target site at the centre, and let $C(\vec r, t)$ be the concentration field of free protein, obeying the diffusion equation $\partial C/\partial t = D_3\nabla^2 C$ with $D_3$ the protein's diffusion constant in the cytoplasm.

The protein must find the exact target sequence, so represent the target as a small absorbing sphere of radius $b\approx 0.34\ \text{nm}$ (one base pair) at the origin: whenever a diffusing protein touches it, it disappears (it has become a bound complex). This is a good approximation as long as the target is usually unoccupied. Hold the concentration fixed at $C_R$ on the outer boundary, as if protein is constantly resupplied there. Under these conditions a time-independent current of protein flows steadily inward, so instead of the full diffusion equation it is enough to solve Laplace's equation, $\nabla^2 C = 0$, with $C(R) = C_R$ and $C(b) = 0$.

The Debye-Smoluchowski setup: concentration is pinned at $C_R$ on the outer sphere of radius $R$ and at zero on the absorbing target of radius $b$ at the centre. The steady state of Laplace's equation carries a constant inward current $I$.

The spherically symmetric solution is $C(r) = C_0[1 - b/r]$, with $C_0 = C_R/(1-b/R)\approx C_R$ since $b\ll R$. Fick's law gives the radially inward current density $J = -D_3\,\partial C/\partial r = -D_3 b C_0/r^2$, and the total current through any sphere is

\[I = J(r)\cdot 4\pi r^2 = -4\pi D_3 b\, C_0.\]

The number of complexes forming per second — the left side of the linearised rate equation above — must equal (minus) this incoming current, with $C_0$ identified as the bulk free-protein concentration $[P]$. Matching the two gives the Debye-Smoluchowski rate:

\[k_a = 4\pi D_3 b.\]

Using the target size $b\approx 0.34\ \text{nm}$ and the measured diffusion constant of the lac repressor in water, $D_3 \approx 3\times 10^{-11}\ \text{m}^2\text{s}^{-1}$, gives $k_a \approx 10^8\ \text{M}^{-1}\text{s}^{-1}$ — and the real value should if anything be smaller still, because a protein also needs time to align itself with the target and because $D_3$ in the crowded cytoplasm is lower than in water. Either way this is two orders of magnitude below the measured $k_a \sim 10^{10}\ \text{M}^{-1}\text{s}^{-1}$. Plain three-dimensional diffusion cannot account for how fast the repressor actually finds its target — something else must be speeding the search up.

Facilitated diffusion: sliding and jumping (Berg-von Hippel theory)¶

Berg and von Hippel's proposal is that the protein does not diffuse only in three dimensions. It can bind any site on the DNA, not just the specific one, and once bound it diffuses along the DNA backbone (sliding, one-dimensional diffusion with constant $D_1$). Eventually it detaches and diffuses freely in solution (a jump) until it lands on the DNA again, somewhere else, and resumes sliding. It keeps alternating rounds of sliding and jumping until, purely by chance, a sliding event carries it across the specific target — at which point it stays, because the specific site binds far more strongly ($E_s \sim 20$–$25\,k_BT$) than a generic site does ($E_{ns} \sim 5$–$10\,k_BT$): non-specific binding is weak enough to let the protein keep exploring, while the target is a deep trap.

The protein alternates rounds of one-dimensional sliding along the DNA (thick segments) with three-dimensional jumps through solution (dashed arcs), until a slide carries it across the specific target site.

Why jumps can be treated as landing uniformly at random. DNA is packed into a compact coil, so two sites that are close together in real space can be very far apart along the sequence. It is therefore reasonable to treat a jump as landing on a uniformly random site among the $M$ total sites, independently of where the previous slide ended. Each sliding event then independently has some probability $q = n/M$ of passing over the target, where $n$ is the number of sites visited during that slide.

A fixed sliding time, first¶

Suppose, unrealistically, that every slide takes the same time $\tau_1$ and visits the same number of sites $n$. The number of rounds $N_R$ needed to hit the target is then geometric: the protein misses in each of the first $N_R - 1$ rounds and hits on the $N_R$-th,

\[p(N_R) = q(1-q)^{N_R - 1}, \qquad \overline{N_R} = \sum_{N_R=1}^\infty N_R\, q(1-q)^{N_R-1} = \frac{1}{q} = \frac{M}{n}.\]

Each round costs the sliding time plus the average jump time $\tau_3$, so the average search time is

\[\overline{t_s} = \overline{N_R}(\tau_1 + \tau_3) = \frac{M}{n}(\tau_1+\tau_3).\]

Sliding times are actually random¶

Real sliding events end when the protein happens to detach, at rate $k_d^{(ns)} = 1/\overline{\tau_1}$, so the duration of a slide is exponentially distributed, $\rho(\tau_1) = \exp(-\tau_1/\overline{\tau_1})/\overline{\tau_1}$.

The number of sites visited in a slide of duration $\tau_1$ now needs care. What matters is not where the slide ends but every site it passes over, since the protein would be trapped the instant it crossed the target during the slide, not only if it happened to stop there. The relevant quantity is therefore the range of the one-dimensional random walk — the span between its leftmost and rightmost visited site — rather than its end-to-end displacement:

\[n(\tau_1) = \sqrt{\frac{16 D_1 \tau_1}{\pi b^2}} \qquad \text{(range)}, \qquad \text{as opposed to} \qquad \sqrt{\frac{2D_1\tau_1}{b^2}} \qquad \text{(end-to-end displacement)},\]

with $b$ the base-pair distance. The range is the larger of the two, which matters, since it is what sets how quickly the protein actually explores new ground.

Averaging $n(\tau_1)$ over the exponential distribution of sliding times gives $\langle n\rangle = 2\sqrt{D_1\overline{\tau_1}}/b$. Because successive slides are still independent draws (each one's outcome depends only on its own random duration), the same geometric-distribution argument applies with $q$ replaced by its average $\langle q\rangle = \langle n\rangle /M$:

\[\langle N_R\rangle = \frac{1}{\langle q\rangle} = \frac{Mb}{2\sqrt{D_1\overline{\tau_1}}}.\]

Computing the average search time is more delicate, because a round that ends in success is, on average, a longer round than one that fails — but working through the sum over rounds (weighting each round's duration by the probability that the search ends exactly there) the correlation cancels and the compact form survives:

\[\langle t_s\rangle = \langle N_R\rangle\left(\overline{\tau_1} + \tau_3\right) = \frac{Mb}{2\sqrt{D_1\overline{\tau_1}}}\left(\overline{\tau_1}+\tau_3\right).\]

The optimal sliding time¶

Treat $\overline{\tau_1}$ as a free parameter and minimise $\langle t_s\rangle$ over it:

\[0 = \frac{\partial \langle t_s\rangle}{\partial \overline{\tau_1}} = \frac{Mb}{2\sqrt{D_1\overline{\tau_1}}}\left(\frac{1}{2} - \frac{\tau_3}{2\overline{\tau_1}}\right) \quad\Longrightarrow\quad \overline{\tau_1}^{(\text{opt})} = \tau_3.\]

The search is fastest when the average time spent sliding equals the average time spent jumping. Slide too long ($\overline{\tau_1} > \tau_3$) and the protein wastes time re-covering ground it has already ruled out, since one-dimensional diffusion explores new sites only as $\sqrt{\text{time}}$; jump too often ($\overline{\tau_1} < \tau_3$) and it wastes time in solution making no progress along the sequence at all.

The sliding-off rate $k_d^{(ns)} = 1/\overline{\tau_1}$ depends strongly on the non-specific binding energy $E_{ns}$, which in turn depends on the salt concentration in the cytoplasm, since the non-specific interaction is predominantly electrostatic: more salt screens the interaction and lowers $E_{ns}$. One might expect a cell to sit at the optimum, but at physiological salt concentrations the measured sliding time for the lac repressor, $\overline{\tau_1}\sim 10^{-3}\ \text{s}$, is noticeably longer than the estimated jump time, $\tau_3\sim V/(D_3 L)\sim 10^{-4}\ \text{s}$ (using cell volume $V\sim 1\ \mu\text{m}^3$, DNA length $L = Mb \sim 1\ \text{mm}$, and $D_3\approx 3\times 10^{-11}\ \text{m}^2\text{s}^{-1}$): real proteins slide somewhat longer than the strict optimum. Feeding these numbers, together with $D_1\approx 5\times 10^{-14}\ \text{m}^2\text{s}^{-1}$, into the search-time formula gives

\[\langle t_s\rangle = \frac{L}{2\sqrt{D_1\overline{\tau_1}}}(\overline{\tau_1}+\tau_3) \sim 10\text{--}100\ \text{s}.\]

This is the search time for one repressor molecule doing facilitated diffusion — and it is far longer than the $\sim 0.1\ \text{s}$ inferred at the very start of the chapter from the measured on-rate. The gap is closed by noting that a cell does not search with only one copy.

Many searchers: the fastest of $n_p$ proteins¶

Take the single-protein search time to be exponentially distributed, $\rho_1(t_s) = \exp(-t_s/\langle t_s\rangle)/\langle t_s\rangle$. Gene expression shuts off as soon as the first of $n_p$ independently searching copies binds the target, so what matters is the distribution of the minimum of $n_p$ independent draws from $\rho_1$. That distribution is standard: the minimum exceeds $t$ only if every one of the $n_p$ searches has not yet succeeded by time $t$, giving

\[\rho_{n_p}(t_s) = n_p\,\rho_1(t_s)\left(1 - \int_0^{t_s}\rho_1(t)\,dt\right)^{n_p - 1} = \frac{n_p}{\langle t_s\rangle}\exp\!\left(-\frac{t_s\, n_p}{\langle t_s\rangle}\right),\]

itself exponential, with mean $\langle t_s\rangle/n_p$. With of order $n_p\sim 100$ copies of the repressor typically present in the cell, this brings the effective search time down from $10$–$100\ \text{s}$ for a lone protein to roughly $0.1$–$1\ \text{s}$ — matching, in order of magnitude, the search time inferred from the measured on-rate at the start of the chapter. Searching in parallel, not a faster mechanism per molecule, is what reconciles facilitated diffusion with the observed switching speed.

Exercises¶

The non-specific dissociation rate $k_d^{(ns)} = 1/\overline{\tau_1}$ depends on the non-specific binding energy $E_{ns}$. Work out this dependence, and explain why increasing the salt concentration in the cytoplasm — which screens the (electrostatic) non-specific interaction — changes $\overline{\tau_1}$, and hence the optimal balance between sliding and jumping.

Sources¶

Slides: lectures/19-slides/01-1-1-reaction-kinetics.md ("1.1 Reaction Kinetics"), 02-1-2-debye-smoluchowski-theory.md ("1.2 Debye-Smoluchowski theory"), and 03-1-3-berg-von-hippel-theory.md ("1.3 Berg-von Hippel theory"), from MIT OCW 8.592J/HST.452J Statistical Physics in Biology, Spring 2011, Lecture 22 (4/30/09). No transcript, written notes, or exercise sheet were supplied for this lecture; the one exercise above is the "(homework)" aside named explicitly in the Berg-von Hippel slide.
The slides themselves note that the first two sections (reaction kinetics and Debye-Smoluchowski theory) closely follow R. F. Bruinsma, Physica A 313, 211-237 (2002) — a paper the lecture points to but that was not itself supplied here.
These slide files are flagged in their own front matter as a model's reconstruction of a PDF with no extractable text layer ("fidelity: reconstructed"), with every equation marked unverified. One apparent inconsistency was resolved silently for readability: the slide text defines $\tau \equiv 1/(k_a[P{\cdot}\text{DNA}])$ immediately after deriving $d[P{\cdot}\text{DNA}]/dt \approx (k_a[\text{DNA}])[P]$, which is self-consistent only as $\tau = 1/(k_a[\text{DNA}])$; the latter is what is used here. Given the reconstruction caveat, the numbered equations in this chapter should be checked against the original PDF slides before being cited elsewhere.

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