Lecture notes. Written from the material of MIT 8.591J 2014, 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.

17. Binding, Kinetics, and Ultrasensitivity¶

What this covers¶

This chapter follows one lecture across four connected pieces: the equilibrium binding of two molecules, what changes when the bound complex can also react (Michaelis–Menten kinetics), how an inhibitor's effect on that kinetics gives away its mechanism, and the dynamics of a simple transcription-factor-to-target gene circuit — ending with two distinct ways to make that circuit's response ultrasensitive. It assumes only the definitions of a reaction rate and an equilibrium constant, and enough calculus to solve $\dot Y = \beta - \alpha Y$.

Binding equilibrium: two molecules, one complex¶

Two molecules $E$ and $S$ associate at rate $k_f$ and their complex $ES$ falls apart at rate $k_r$: the forward flux is $k_f[E][S]$, the reverse flux is $k_r[ES]$. Their ratio defines the dissociation constant, $$K_d \equiv \frac{k_r}{k_f}.$$

A first thing worth being careful about: $k_f$ and $k_r$ look like the same kind of object — both just "a $k$ with a subscript" — but they do not have the same units, because the forward rate carries an extra factor of concentration that the reverse rate does not. It is $K_d$, the ratio, that comes out with units of concentration, once the two rates are worked out consistently. This is an easy point to lose track of later, precisely because the two symbols look so similar.

Fraction bound, and four limits worth thinking through before any algebra¶

Define the fraction of $E$ that is in complex, $$F_b \equiv \frac{[ES]}{[E] + [ES]},$$ on the assumption that free and bound are the only two states available to $E$. Before writing down a formula for $F_b$, it pays to fix intuition against a few limits: if the algebra later disagrees with the limit, one of the two — the intuition or the algebra — has to be wrong, and finding out which is how you catch mistakes rather than just making them.

$S_{\text{total}} \to 0$. Nothing to bind, so $F_b \to 0$.
$S_{\text{total}} \to \infty$. Enough substrate saturates every enzyme molecule, so $F_b \to 1$.
$E_{\text{total}} \to 0$ at fixed $S$. This is the subtle one. Written as a population fraction, $[ES]/([E]+[ES])$ looks like $0/0$ as both go to zero together, and one honest answer is that the ratio is genuinely ill-defined without more care. But there is a cleaner way to see it: stop thinking about a population and imagine being the one enzyme molecule in the tube. You bind a substrate molecule at some rate set by $[S]$, stay bound for a while, and fall apart again at rate $k_r$. $F_b$ is then the time-average fraction of your own life spent bound — a perfectly well-defined quantity for a single molecule, and it settles at $$F_b = \frac{S}{K_d + S}.$$ This is the limit in which one enzyme does not deplete the pool of substrate it swims in, so the free concentration $S$ in that formula is safely the same as $S_{\text{total}}$.
$E_{\text{total}} \to \infty$ at fixed $S_{\text{total}}$. Now there is so much enzyme that essentially all of the substrate ends up bound — but that substrate is spread across an ever-growing enzyme population, so $F_b = [ES]/E_{\text{total}} \to S_{\text{total}}/E_{\text{total}} \to 0$. Almost every enzyme molecule is free simply because there are so many of them chasing a fixed amount of substrate.

Setting $d[ES]/dt = k_f[E][S] - k_r[ES] = 0$ gives, in one line, the same formula found above, $$F_b = \frac{S}{K_d + S}.$$

This expression is easy to misread. The $S$ in it is the free substrate concentration, not $S_{\text{total}}$ — and free $S$ itself is a function of $S_{\text{total}}$, $E_{\text{total}}$, and $K_d$ together. Treating $S \approx S_{\text{total}}$ is only safe in the limit where enzyme is scarce relative to substrate (the $E_{\text{total}} \to 0$ limit above), which is exactly the regime in which real enzyme–substrate systems are usually studied — and is the reason the Michaelis–Menten curve below has this same shape.

Plotting $F_b$ against $E_{\text{total}}$ at some fixed $S_{\text{total}}$ comparable to $K_d$ is a good way to check this understanding: the curve starts at $1/2$ as $E_{\text{total}} \to 0$ (from the third limit above, with $S \approx S_{\text{total}} \approx K_d$), falls monotonically, and goes to $0$ as $E_{\text{total}} \to \infty$ (the fourth limit) — with $K_d$, the only concentration scale in the problem, setting where the curve bends.

From binding to catalysis: Michaelis–Menten kinetics¶

Add a third step to the scheme: the bound complex can also react forward, $ES \to E + P$, at rate $k_{\text{cat}}$. From the standpoint of the enzyme, the complex $ES$ can now fall apart two ways — back to $E + S$, or forward to $E + P$ — so its effective lifetime is shorter than before. The two loss rates simply add, giving the Michaelis constant $$K_m = \frac{k_r + k_{\text{cat}}}{k_f},$$ which plays exactly the role $K_d$ played above. The whole earlier discussion of fraction bound carries over unchanged with $K_d$ replaced by $K_m$, and the reaction velocity (rate of product formation) is $$V = V_{\max}\,\frac{S}{K_m + S}, \qquad V_{\max} = k_{\text{cat}}[E]_{\text{total}},$$ valid, as before, once $[E]_{\text{total}}$ is small enough that $S \approx S_{\text{total}}$. Two conditions have to hold for this steady-state formula to apply in practice: enough time must have passed for the initial transient — building up the $ES$ pool from nothing — to have died away, but not so much time that the substrate has been noticeably used up.

A limitation worth stating precisely, not just noting. As written, this scheme has no way back from $P$ to $S$, so if it is allowed to run forever, all the substrate becomes product. That is not just a detail missing from one particular model — it is a way of getting the physics of a catalyst wrong in general. A catalyst does not change the equilibrium ratio between substrate and product; it only speeds up the forward and reverse reactions equally. Drop invertase into a test tube of sucrose and it will reach the same equilibrium between sucrose and its hydrolysis products that the tube would reach on its own after being left for a million years — the enzyme just gets there many orders of magnitude faster. The reason is thermodynamic: the enzyme itself is present, unchanged, on both sides of the reaction (bound at the start, bound at the end), so from that standpoint it cannot appear in the free-energy difference between substrate and product, and cannot shift where that equilibrium sits. The one-way $E + S \to ES \to E + P$ scheme is therefore a good local approximation while the reaction is far from its own equilibrium — which is usually the regime an experiment is run in — but it is not a globally valid model of what an enzyme does.

Reading a mechanism off $V_{\max}$ and $K_m$¶

Different inhibitors change the Michaelis–Menten curve differently, and this is the practical payoff of having derived it carefully.

An inhibitor $I$ that binds $E$ reversibly, forming an $EI$ complex that competes with $S$ for the enzyme, leaves $V_{\max}$ unchanged. At saturating substrate, there is enough $S$ to pull essentially every enzyme molecule away from $I$ and into $ES$ — you just need more substrate to get there than you would without the inhibitor. So this kind of inhibitor raises $K_m$ but does not touch $V_{\max}$.
An inhibitor that instead binds the $ES$ complex itself, blocking it from turning over into product, does the opposite: it lowers $V_{\max}$ (some fraction of complexes are stuck, unable to react, no matter how much substrate is present) without changing $K_m$.

This is why titrating in a candidate inhibitor and measuring how it moves $V_{\max}$ and $K_m$ is a standard way enzymologists diagnose which of these two mechanisms — competition for the enzyme, or blocking of the catalytic step — a small molecule is actually using.

Gene expression: separation of time scales and the dilution clock¶

Now consider a transcription factor $X$ that is activated by some signal into $X^*$, which binds a promoter and drives expression of $Y$. Two of the steps here happen on very different time scales. Activation of $X$ into $X^*$ — often something like an allosteric switch triggered by a small molecule — can be extremely fast, well under a second; whatever is rate-limiting about the response to an environmental change is usually getting the signal into the cell in the first place, not the activation step itself. Equilibration of $X^*$ on the promoter is also fast on the time scale of gene expression, though not quite as fast — more like seconds. Against those, transcription and translation take minutes. So on the time scale that matters for $Y$, $X^*$ can be treated as switching essentially instantaneously in response to the signal, and the real dynamics is all in $Y$.

The dynamics is written $$\dot Y = f(X^*) - \alpha Y,$$ where $\alpha$ is the sum of two physically distinct contributions: active degradation of the protein, and dilution from cell growth. Even a protein that is never actively degraded still has $\alpha > 0$, because a cell that is growing and dividing while holding its number of protein copies fixed is diluting that protein's concentration. Real growth is not perfectly uniform across the cell cycle, but averaged over it, a first-order dilution rate is a reasonable description.

If $f(X^*)$ switches on to a constant $\beta$ at $t = 0$, the solution approaches $Y \to \beta/\alpha$ with characteristic time $1/\alpha$, and the time to reach the halfway point is $T_{1/2} = \ln 2/\alpha$. For a stable protein — no active degradation, $\alpha$ entirely from dilution — $T_{1/2}$ is exactly the cell division time.

The relaxation is symmetric. If the signal is switched off again, $Y$ decays back down on the same characteristic time $1/\alpha$ that it took to rise — not faster, not slower. That is not obvious in advance, but it follows directly from $\dot Y = \beta - \alpha Y$ being linear: the same $\alpha$ governs the approach to any target, whether that target is $\beta/\alpha$, $0$, or any other constant.

One consequence: since $1/\alpha$ sets the response time in both directions, the only way to make the circuit genuinely faster — on or off — is to increase $\alpha$, i.e., to actively degrade $Y$. That has a cost. Holding the same steady-state level $\beta/\alpha$ while raising $\alpha$ requires raising $\beta$ to match, which means synthesizing more protein to sustain the same concentration. This cost is smaller for a protein that is kept at low copy number to begin with — which is precisely how many transcription factors are expressed, since they only need to be present in amounts sufficient to occupy binding sites, not to build cellular structure. That combination — needing a fast response, and being cheap to overexpress — is a plausible reason active degradation of transcription factors is common, while it would be far more costly for a structural protein made in bulk.

Ultrasensitivity: making a small input change into a large output change¶

A default input–output relationship, with a single $X$ monomer binding the promoter, has exactly the same hyperbolic shape derived above: $$\text{rate of } Y \text{ expression} = \beta\,\frac{X}{K_d + X}.$$ This is never ultrasensitive: doubling $X$ always produces less than a doubling in the response, because the curve is already bending over. The ideal ultrasensitive response would instead be a step — nothing until $X$ reaches some threshold, then all of $\beta$ at once.

Cooperative binding steepens the curve¶

One way to approach the step is cooperative binding — dimerization or multimerization of the transcription factor before or during promoter binding, or cooperative interactions among multiple binding sites at the promoter. Such curves are commonly summarized with a Hill function, $$\text{rate} = \beta\,\frac{X^n}{K^n + X^n},$$ where $K$ is still the point of half-maximal response, but $n > 1$ makes the transition around $K$ sharper. As $n$ increases the curve becomes progressively more step-like.

Hill functions with the same half-maximal point K but growing cooperativity n: the curve does not move, it steepens, approaching the ideal step response as n increases.

For many real genes, measured input–output curves are reasonably well described by such a curve with $n$ somewhere between $1$ and $4$ — moderate cooperativity is common, extreme cooperativity is not.

Molecular titration: sliding the curve instead of steepening it¶

A second route to ultrasensitivity needs no cooperativity at all. Suppose $X$ still binds the promoter to activate $Y$ with dissociation constant $K_d$, but there is also some other protein $W$ that binds $X$ reversibly, with dissociation constant $K_w$, forming an inert complex $XW$. $W$ acts as a sponge or decoy: while $W$ is present in excess, most of $X$ is siphoned off into $XW$ and essentially none is free to activate the promoter, so expression stays near zero. Only once $X_{\text{total}}$ exceeds $W_{\text{total}}$ does free $X$ start to rise substantially, and expression switches on. The effect on the curve is different in kind from cooperative binding — the response does not steepen in place, the whole curve slides sideways so that essentially nothing happens until $X_{\text{total}}$ passes $W_{\text{total}}$.

Without the titrating protein W, Y turns on near X = 0 (faint curve). With W present in excess, expression is suppressed until X total exceeds W total, then rises — the response threshold moves to sit at W total rather than the curve becoming steeper where it already was.

This mechanism was motivated in the lecture by work from Nick Buchler, who helped work out how molecular titration produces ultrasensitivity in this kind of setting. The precise relationship required among $K_d$, $K_w$, and $W_{\text{total}}$ for the curve to behave this sharply was left as the opening problem for the next lecture, rather than derived here.

Exercises¶

These are the two questions the lecture itself raised and left open, rather than problems from a separate problem set.

For the simple binding equilibrium $F_b = S/(K_d+S)$, work out $F_b$ exactly in the case $E_{\text{total}} = K_d$ (with, as usual, $S_{\text{total}}$ comparable to $K_d$ so the answer is not trivial). The lecture only guessed at the value ("around a third or a fifth") without computing it.
For the molecular titration scheme — $X$ binding the promoter with dissociation constant $K_d$, and binding a decoy protein $W$ with dissociation constant $K_w$ — determine what relationship between $K_d$, $K_w$, and $W_{\text{total}}$ is needed to get a genuinely ultrasensitive response (expression staying near zero until $X_{\text{total}}$ exceeds $W_{\text{total}}$, then rising sharply), as opposed to a gradual one.

Sources¶

Transcript of the lecture: recordings/lly1u2aghiq.md.
Binding equilibrium, units of $K_d$, and the four limits of $F_b$: 00:00–35:40.
Michaelis–Menten kinetics, $K_m$, and the thermodynamic limitation of the one-way scheme: 35:40–48:00.
Competitive vs. catalytic-step inhibitors and reading mechanism off $V_{\max}$/$K_m$: 48:00–57:28.
Separation of time scales, the dilution/degradation rate $\alpha$, and the symmetry of relaxation: 57:28–1:10:17.
Ultrasensitivity via cooperative (Hill) binding and via molecular titration: 1:10:17–1:17:34.
This is a transcript-only source: whatever the lecturer drew on the board (the actual curves for $F_b$, the Michaelis–Menten plot, the Hill functions, the titration curve, and the time-course of $X$, $X^*$, and $Y$) was not captured. The shapes reconstructed here — the two figures above, and the qualitative descriptions elsewhere — follow only what was said about them out loud.
Referred to but not supplied: Uri Alon's textbook (called "Uri's book" in the lecture, on the separation-of-time-scales argument for gene expression and on typical measured Hill coefficients in real promoters), and unpublished conversation with Nick Buchler on the mechanism of molecular titration. Homework on Michaelis–Menten kinetics and on inhibitor mechanisms was mentioned as upcoming but was not part of this task's inputs.

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