### Analyze the free energy function. Is a pure state ever preferred? (Is the minimum ever at \(\phi=0\) or \(\phi=1\)). In case of large repulsive interactions between solvent and protein, at what concentration is the minimum of the free energy?

No, a pure state is never preferred. Complete separation is strongly penalized by entropy. This is reflected in the infinite derivative of the entropy term at \(\phi=0\) or 1. Even if the interactions between different molecules are strongly repulsive, there will we some residual mixing. The minimum free energy obeys

$$ kT(\log\phi + \log(1-\phi)) -(2\phi - 1)\frac{z(u_{sp} - u_{ss} - u_{pp})}{2} = 0 $$

At very small \(\phi\), this condition can be approximated as

$$ kT\log\phi + \frac{z(u_{sp} - u_{ss} - u_{pp})}{2} = 0 $$

with solution

$$ \phi = e^{-\frac{z(u_{sp} - u_{ss} - u_{pp})}{2kT}} $$

Hence repulsive interactions make the preferred mixing concentration exponentially small, but it is always different from 0.

### How does the free energy landscape change with temperature? What contributions to the free energy are affected?

The entropy part in the simple lattice model we considered is purely combinatorial. Entropy enters the free energy with a prefactor \(kT\). But all relevant quantities depend on the ratio of the free energy to \(kT\).

$$ e^{-\frac{G}{kT}} = e^{-\frac{U- kTS}{kT}} = e^{S}e^{-\frac{U}{kT}} $$

So its really the energy part that is sensitive to temperature, but the relative weight of the entropy part increases as temperature increases.