= Solution
For a constant $X$, set both <ordinary differential equation> derivatives to zero:
$$
\boxed{Y^*(X)=a_Y\frac{X^{n_{XY}}}{K_{XY}^{n_{XY}}+X^{n_{XY}}},\qquad
Z^*_{\rm direct}(X)=a_Z\frac{X^{n_{XZ}}}{K_{XZ}^{n_{XZ}}+X^{n_{XZ}}}}.
$$
These <equilibrium concentrations> are unique for each input. For a loss rate $\gamma$ other than one, divide the corresponding production term by $\gamma$.
Choose $a_Y=2$, $K_{XY}=1$, $n_{XY}=2$ and $a_Z=3$, $K_{XZ}=2$, $n_{XZ}=4$. The following Python code plots the requested <equilibrium concentrations> and defines the functions used in part (c):
``
import numpy as np
import matplotlib.pyplot as plt
def hill(u, K, n):
u = np.asarray(u)
return u**n / (K**n + u**n)
X_grid = np.linspace(0, 5, 501)
Y_eq = 2 * hill(X_grid, 1, 2)
Z_eq = 3 * hill(X_grid, 2, 4)
plt.figure(figsize=(6, 4))
plt.plot(X_grid, Y_eq, label="Y equilibrium: K=1, n=2")
plt.plot(X_grid, Z_eq, label="Z equilibrium: K=2, n=4")
plt.xlabel("X concentration")
plt.ylabel("Equilibrium concentration")
plt.legend()
plt.tight_layout()
plt.show()
``
The <Hill constant> sets the half-maximum input: $Y^*(1)=1$ and $Z^*_{\rm direct}(2)=1.5$. Increasing that constant shifts activation toward larger $X$. The <Hill coefficient> changes steepness rather than the half-maximum position; the normalized slope at $X=K$ is $n/(4K)$. Because the two examples also differ in $K$ and maximal production, compare normalized responses when isolating the effect of $n$ alone. Both curves vanish at zero and saturate toward their respective maxima.
\Image[/past-exam-of-the-systems-biology-course-of-the-university-of-cambridge/2024/paper-3-single-gene.png]
{title=Independent Hill activation and convergence from two initial states}
{description=The left panel shows different half-activation inputs and Hill coefficients. The right panel integrates the two independent output equations at fixed X=2; dashed curves start above equilibrium and solid curves start at zero.}
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