= Solution
The <quasi-steady-state approximation> for $Y$ sets $Y=a_Yh(X;K_{XY},n_{XY})$. Substituting into the $Z$ equation gives
$$
\frac{dZ}{dt}=a_Zh(X;K_{XZ},n_{XZ})
h\!\left(a_Yh(X;K_{XY},n_{XY});K_{YZ},n_{YZ}\right)-Z.
$$
Its <equilibrium concentration> is therefore
$$
\boxed{Z^*_{\rm AND}(X)=a_Z\frac{X^{n_{XZ}}}{K_{XZ}^{n_{XZ}}+X^{n_{XZ}}}
\frac{\left(a_Y X^{n_{XY}}/(K_{XY}^{n_{XY}}+X^{n_{XY}})\right)^{n_{YZ}}}
{K_{YZ}^{n_{YZ}}+\left(a_Y X^{n_{XY}}/(K_{XY}^{n_{XY}}+X^{n_{XY}})\right)^{n_{YZ}}}}.
$$
For constant input this is also the \b[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, \b[$\boxed{Z^*_{\rm AND}=Z^*_{\rm direct}\,h(Y^*;K_{YZ},n_{YZ})\leq Z^*_{\rm direct}}$]. The extra activation gate suppresses the response at low input and can sharpen its rise. The high-input limit is $a_Zh(a_Y;K_{YZ},n_{YZ})$, 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 $K_{XY}$ 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 $K_{YZ}$ 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 $X^{n_{XZ}+n_{XY}n_{YZ}}$, while both gates are far below saturation, explaining the increased low-input suppression.
For a \b[<persistence detector>], retain finite $Y$ dynamics and use, for example,
$$
\boxed{a_Y=a_Z=1,\quad K_{XY}=K_{XZ}=1,\quad
n_{XY}=n_{XZ}=4,\quad K_{YZ}=0.7,\quad n_{YZ}=8},
$$
with unit loss rates, $X_{\rm off}=0$, $X_{\rm on}=2$, and initially $Y=Z=0$. On switching the input on, $Y_\infty=16/17$ and
$$
Y(t)=\frac{16}{17}(1-e^{-t}).
$$
The time to reach the half-activation threshold of the indirect gate is
$$
\boxed{t_{\rm gate}=-\log\!\left(1-\frac{0.7}{16/17}\right)
=\log(160/41)\approx1.362}.
$$
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>.<Structure and function of the feed-forward loop network motif by Shmoolik Mangan and Uri Alon>[p. 11981, Equations for Gene Regulation Reactions; pp. 11983–11985, coherent FFL kinetics and Coherent FFLs as Persistence Detectors]{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 $1.72\times10^{-6}$ for the short pulse and $0.766$ for the long pulse. These are original illustrative simulations, not fitted experimental measurements.
\Image[/past-exam-of-the-systems-biology-course-of-the-university-of-cambridge/2024/paper-3-persistence.png]
{title=A coherent feed-forward loop distinguishes short and persistent input pulses}
{description=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.}
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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