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 Incoming links: Coherent feed-forward loop

Coherent feed-forward loop Created 2026-10-20 Updated 2026-10-25

The direct path and the indirect path of a coherent feed-forward loop have matching overall signs. The all-activating case with an AND rule can delay activation while allowing rapid removal of the production signal when the direct input disappears.

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Past exam of the biological physics course of the University of Oxford / 2021 / 2 / a / Solution 2026-10-25

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Use the logic approximation for gene regulation, zero basal production and initially vanishing X, Y and Z. Let

K = K_{x y} = K_{x z} = K_{yz} = K_{xx} /2

Y_{\infty} = Z_{\infty} = β_{y} / α

, and assume the activated Q level exceeds the X activation threshold. The circuit combines a coherent feed-forward loop with negative autoregulation. While X is below its self-repression threshold it obeys

\dot{X} = β_{x} - α X

, hence X first crosses

K

t_{X} = - \frac{1}{α} ln (1 - \frac{α K}{β _{x}}) .

(1)

Y production begins then. While X remains above

K

Y = Y_{\infty} [1 - e^{- α (t - t_{X})}]

. The AND gate for Z first becomes active when Y reaches

K

, so

T_{m i n} = t_{X} - \frac{1}{α} ln (1 - \frac{α K}{β _{y}}) ≃ - \frac{1}{α} ln (1 - \frac{α K}{β _{y}}) .

(2)

The approximation drops the short X build-up time because

β_{x} ≫ β_{y}

. Strictly the input must last longer than this onset time to produce a nonzero amount of Z. This requires

β_{y} / α > K

: at equality the threshold is approached only asymptotically, and below it Z never switches on. No numerical delay follows without the threshold-to-production ratio.

In the strong negative autoregulation limit, X is held close to

K_{xx} = 2 K

; a smooth repression law regularizes the rapid switching of the ideal Boolean rule. After the input disappears at

t_{0}

, production of X stops and

X ≃ 2 K e^{- α (t - t_{0})}

. It takes

(ln 2) / α

for X to fall below

K

. Until then Y remains at

Y_{\infty} > K

and the Z AND gate stays on, so Z stays at

Z_{\infty}

. After that Z decays by dilution, taking another

(ln 2) / α

to halve. Therefore

t_{Z, 1/2} - t_{0} ≃ \frac{2 ln 2}{α} .

(3)

More generally replace the first

ln 2

ln [X (t_{0}) / K]

. This distinguishes disappearance of the external ligand from disappearance of the persistent X transcription factor.

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Past exam of the biological physics course of the University of Oxford / 2022 / 2 / a / Solution 2026-10-25

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Use the logic approximation for gene regulation: production is either

β

or zero, and every species has linear removal rate

α

. Put

C = β / α

and assume zero basal expression, inactive X before time zero, and all required thresholds strictly between

0

and

C

. The input is

u (t) = 1

for

0 < t < T

and zero otherwise. The common ordinary differential equation is

\overset{c}{˙} = β g - α c

, with Boolean production gate

g

. Both networks therefore have

Y (t) = C (1 - e^{- α t}) (0 < t < T), Y (t) = C e^{- α (t - T)} (t > T),

(1)

where the latter uses the stipulated attained steady state. Define the threshold-crossing delay

d (K) = - α^{- 1} ln (1 - K / C), t_{r} = d (K_{Y Z}), t_{1} = d (K_{YY 1}) .

(2)

Because

K_{YY 1} > K_{Y Z}

t_{1} > t_{r}

. In both networks,

Y_{1}

stays zero until

t_{1}

and then follows

Y_{1} = C [1 - e^{- α (t - t_{1})}]

until the OFF response. Write

K_{A}

for the activation threshold of

Y_{1}

on Z in diagram (a), and

K_{I}

for the level of

Y_{1}

that blocks Y's repression in diagram (b); these thresholds were not supplied numerically.

Diagram (a). All three inputs X, Y and

Y_{1}

must permit Z production. Since Y has already crossed its Z threshold before

Y_{1}

starts, the final permission arrives at

t_{a} = t_{1} + d (K_{A}) .

(3)

Thus

Z = 0

until

t_{a}

and

Z = C [1 - e^{- α (t - t_{a})}]

subsequently. Solving

Z = C /2

gives

t_{Z, 1/2} = d (K_{YY 1}) + d (K_{A}) + \frac{ln 2}{α} .

(4)

For the frequently adopted but extra convention that both required thresholds are

C /2

, this is

3 ln 2/ α

. At the OFF step, X's direct AND input disappears, so

Z = C e^{- α (t - T)}

immediately. Y also decays immediately.

Y_{1}

remains at

C

while Y remains above

K_{YY 1}

; it starts decaying at

T + α^{- 1} ln (C / K_{YY 1})

. This is a coherent feed-forward loop persistence response: activation is delayed, while removal of the direct required input switches production off promptly.

Diagram (b). The production gate is

g_{Z} = u [\neg (Y > K_{Y Z}) \lor (Y_{1} > K_{I})]

. Z initially rises, Y switches Z production off at

t_{r}

, and

Y_{1}

restores production at

t_{b} = t_{1} + d (K_{I})

. The continuous solution during the ON phase is

Z (t) = ⎩ ⎨ ⎧ C (1 - e^{- α t}), C (1 - e^{- α t_{r}}) e^{- α (t - t_{r})}, C + [Z (t_{b}) - C] e^{- α (t - t_{b})}, 0 < t < t_{r}, t_{r} < t < t_{b}, t_{b} < t < T .

(5)

The derivative changes from positive to negative at

t_{r}

, and from negative to positive at

t_{b}

. Consequently the first temporary minimum following the initial pulse occurs at

t_{m i n} = d (K_{YY 1}) + d (K_{I}) .

(6)

Its value is

Z (t_{b}) = K_{Y Z} e^{- α (t_{b} - t_{r})}

. At the OFF step Z again decays immediately because X is required for production, with Y and

Y_{1}

behaving as above. If a required threshold is unattainable, these late switching events do not occur; equal production and removal rates alone do not fix threshold values.

Figure 1.
Logic approximation for gene regulation: delayed activation and a transient repression window
. **Details**.

The plotted thresholds are explicitly illustrative, with

K_{Y Z} / C = 0.25

K_{YY 1} / C = 0.55

and

K_{A} / C = K_{I} / C = 0.45

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Past exam of the systems biology course of the University of Cambridge / 2024 / Paper 3 / 3 / d / Solution 2026-10-20

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Add the activating link from

Y

Z

and multiply the two Hill activation functions for the target:

\frac{d Y}{d t} = a_{Y} h (X; K_{X Y}, n_{X Y}) - Y, \frac{d Z}{d t} = a_{Z} h (X; K_{XZ}, n_{XZ}) h (Y; K_{Y Z}, n_{Y Z}) - Z .

(1)

The extra Hill constant

K_{Y Z} > 0

has units of

Y

concentration and

n_{Y Z} > 0

is its Hill coefficient. Both inputs are required for substantial target production: if either is zero, the production term for

Z

vanishes. This all-activating circuit is a coherent feed-forward loop. A product of graded input functions implements the specified AND rule; it is a phenomenological rule rather than a complete molecular promoter model.

Retain the parameter values of parts (b) and (c), and set

K_{Y Z} = 1

n_{Y Z} = 4

. Continue the same Python script with the altered right-hand side:

def and_rhs(state, X):
    Y, Z = state
    return np.array([2*hill(X, 1, 2) - Y,
                     3*hill(X, 2, 4)*hill(Y, 1, 4) - Z])

times = np.linspace(0, 10, 1001)
plt.figure(figsize=(7, 4))
for X in [1, 3]:
    trajectory = integrate(lambda s: and_rhs(s, X), [0, 0], times)
    plt.plot(times, trajectory[:, 0], "--", label=f"Y, X={X}")
    plt.plot(times, trajectory[:, 1], label=f"Z, X={X}")
plt.xlabel("Time")
plt.ylabel("Concentration")
plt.legend()
plt.tight_layout()
plt.show()

X = 1

Y^{*} = 1

and

Z^{*} = (3/17) (1/2) = 3/34 \approx 0.08824

. At

X = 3

Y^{*} = 1.8

and

Z^{*} \approx 2.2873

. In both simulations

Y

first accumulates and opens the additional gate;

Z

therefore initially grows more slowly than under direct activation alone. The equilibrium concentration of

Y

is unchanged by the added outgoing interaction.

Figure 1.
AND feed-forward dynamics and the effect of activation thresholds
. The left panel compares fixed inputs X=1 and X=3 from the zero state. The right panel compares the full feed-forward equilibrium response with direct activation alone and with individual Hill constants increased. **Details**.

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Past exam of the systems biology course of the University of Cambridge / 2024 / Paper 3 / 3 / e / Solution 2026-10-20

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The quasi-steady-state approximation for

Y

sets

Y = a_{Y} h (X; K_{X Y}, n_{X Y})

. Substituting into the

Z

equation gives

\frac{d Z}{d t} = a_{Z} h (X; K_{XZ}, n_{XZ}) h (a_{Y} h (X; K_{X Y}, n_{X Y}); K_{Y Z}, n_{Y Z}) - Z .

(1)

Its equilibrium concentration is therefore

Z_{AND}^{*} (X) = a_{Z} \frac{X ^{n_{XZ}}}{K _{XZ}^{n_{XZ}} + X ^{n_{XZ}}} \frac{( a _{Y} X ^{n_{X Y}} / ( K _{X Y}^{n_{X Y}} + X ^{n_{X Y}} ) ) ^{n_{Y Z}}}{K _{Y Z}^{n_{Y Z}} + ( a _{Y} X ^{n_{X Y}} / ( K _{X Y}^{n_{X Y}} + X ^{n_{X Y}} ) ) ^{n_{Y Z}}} .

(2)

For constant input this is also the exact equilibrium of the full two-variable model: the fast-

Y

assumption is needed to replace the transient dynamics, not to calculate a fixed point.

Relative to the direct-activation result, $Z_{AND}^{*} = Z_{direct}^{*} h (Y^{*}; K_{Y Z}, n_{Y Z}) \leq Z_{direct}^{*}$ . The extra activation gate suppresses the response at low input and can sharpen its rise. The high-input limit is

a_{Z} h (a_{Y}; K_{Y Z}, n_{Y Z})

, which need not reach

a_{Z}

because

Y^{*}

saturates at

a_{Y}

For the parameters in part (d), the following Python continuation plots the expression and changes each Hill constant separately:

def z_equilibrium(X, Kxy=1, Kxz=2, Kyz=1):
    Yeq = 2 * hill(X, Kxy, 2)
    return 3 * hill(X, Kxz, 4) * hill(Yeq, Kyz, 4)

X_grid = np.linspace(0, 5, 501)
plt.figure(figsize=(7, 4))
plt.plot(X_grid, 3*hill(X_grid, 2, 4), ":", label="Direct activation")
for parameters, label in [((1, 2, 1), "Baseline"),
                          ((2, 2, 1), "Kxy=2"),
                          ((1, 3, 1), "Kxz=3"),
                          ((1, 2, 2), "Kyz=2")]:
    plt.plot(X_grid, z_equilibrium(X_grid, *parameters), label=label)
plt.xlabel("X concentration")
plt.ylabel("Equilibrium Z")
plt.legend()
plt.tight_layout()
plt.show()

Increasing

K_{X Y}

lowers

Y^{*}

at fixed finite

X

, delaying opening of the indirect gate. Increasing

K_{XZ}

delays opening of the direct gate. Neither changes the limiting response as

X \to \infty

, provided the constants remain finite. Increasing

K_{Y Z}

makes a given

Y

less effective and also lowers the high-input plateau. These changes are visible in the preceding figure. At small

X

, the response is proportional to

X^{n_{XZ} + n_{X Y} n_{Y Z}}

, while both gates are far below saturation, explaining the increased low-input suppression.

For a persistence detector, retain finite

Y

dynamics and use, for example,

a_{Y} = a_{Z} = 1, K_{X Y} = K_{XZ} = 1, n_{X Y} = n_{XZ} = 4, K_{Y Z} = 0.7, n_{Y Z} = 8,

(3)

with unit loss rates,

X_{off} = 0

X_{on} = 2

, and initially

Y = Z = 0

. On switching the input on,

Y_{\infty} = 16/17

and

Y (t) = \frac{16}{17} (1 - e^{- t}) .

(4)

The time to reach the half-activation threshold of the indirect gate is

t_{gate} = - lo g (1 - \frac{0.7}{16/17}) = lo g (160/41) \approx 1.362 .

(5)

An isolated input pulse of duration

0.25

ends well before this threshold; a duration-four pulse persists long enough to open the gate. The large but finite Hill coefficient makes this a graded filter, not a perfectly sharp cutoff. Once

X

vanishes, its direct gate closes immediately and stops production of

Z

, which then decays with time constant one. The protein output itself does not drop discontinuously. This delayed-on, prompt-production-off mechanism is the characteristic persistence detector interpretation of an AND-type coherent feed-forward loop.^[ref]

This final Python continuation simulates both pulse durations. Each discontinuity is placed on the time grid and the input is held constant within each integration step:

def persistence_rhs(state, X):
    Y, Z = state
    return np.array([hill(X, 1, 4) - Y,
                     hill(X, 1, 4)*hill(Y, 0.7, 8) - Z])

def pulse_trajectory(duration, dt=0.005):
    times = np.arange(round(8/dt) + 1) * dt
    out = np.zeros((len(times), 2))
    for j in range(len(times)-1):
        X = 2.0 if 1 <= times[j] + dt/2 < 1 + duration else 0.0
        # Each step is integrated with its constant input.
        out[j+1] = integrate(lambda s: persistence_rhs(s, X),
                             out[j], [0, dt])[-1]
    return times, out

fig, axes = plt.subplots(1, 2, figsize=(10, 4))
for duration in [0.25, 4]:
    times, out = pulse_trajectory(duration)
    axes[0].plot(times, out[:, 0], label=f"Y, duration {duration}")
    axes[1].plot(times, out[:, 1], label=f"Z, duration {duration}")
    print(duration, "maximum output", out[:, 1].max())
axes[0].axhline(0.7, color="black", linestyle=":")
for ax in axes:
    ax.set_xlabel("Time")
    ax.set_ylabel("Concentration")
    ax.legend()
plt.tight_layout()
plt.show()

The chosen example produces peak

Z

of about

1.72 \times 1 0^{- 6}

for the short pulse and

0.766

for the long pulse. These are original illustrative simulations, not fitted experimental measurements.

Figure 1.
A coherent feed-forward loop distinguishes short and persistent input pulses
. Both inputs switch from zero to X=2 at time one. One ends after 0.25 time units and the other after four. The intermediate Y crosses its activation threshold only for the sustained input, producing a much larger target response. **Details**.

Finally, the strict instantaneous-

Y

reduction used for the equilibrium calculation has no separate intermediate accumulation delay. Its remaining one-variable dynamics can still attenuate short pulses through ordinary output relaxation, but it cannot demonstrate the additional delay caused by finite

Y

buildup. The pulse test therefore uses the full two-variable model.

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Persistence detector Created 2026-10-20 Updated 2026-10-25

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A persistence detector responds strongly to a sustained input and weakly to short isolated inputs. In an activating coherent feed-forward loop, an intermediate protein accumulates toward a threshold while the target also requires the direct input. Eliminating that intermediate by an instantaneous quasi-steady-state approximation removes its accumulation delay.

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