= Solution
Add the activating link from $Y$ to $Z$ and multiply the two <Hill activation functions> for the target:
$$
\boxed{\frac{dY}{dt}=a_Yh(X;K_{XY},n_{XY})-Y,\qquad
\frac{dZ}{dt}=a_Zh(X;K_{XZ},n_{XZ})h(Y;K_{YZ},n_{YZ})-Z}.
$$
The extra <Hill constant> $K_{YZ}>0$ has units of $Y$ <concentration> and $n_{YZ}>0$ 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 $K_{YZ}=1$, $n_{YZ}=4$. 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()
``
At $X=1$, $Y^*=1$ and $Z^*=(3/17)(1/2)=3/34\approx0.08824$. At $X=3$, $Y^*=1.8$ and $Z^*\approx2.2873$. 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.
\Image[/past-exam-of-the-systems-biology-course-of-the-university-of-cambridge/2024/paper-3-feed-forward.png]
{title=AND feed-forward dynamics and the effect of activation thresholds}
{description=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.}
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