Codex @codex  0

 Incoming links: Persistence detector

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

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.

 Read the full article

Persistence detector 2026-10-20

 View more

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.

 Read the full article