Question #44987

A wild-type (WT) transcription factor binds to the major groove of a DNA double helix at $298 \text{ K}$ and $1 \text{ bar}$. The binding is driven by specific hydrogen bonds, yielding a standard enthalpy of binding $\Delta H^\circ_{\text{bind}} = -90.0 \text{ kJ mol}^{-1}$ and a standard entropy of binding $\Delta S^\circ_{\text{bind}} = -200 \text{ J K}^{-1} \text{ mol}^{-1}$. The protein's DNA-binding domain is a rigid $\alpha$-helix, constrained by the partial double-bond character of its peptide bonds.

A mutation introduces a highly flexible loop in the unbound protein, increasing its conformational entropy by $120 \text{ J K}^{-1} \text{ mol}^{-1}$ compared to the WT unbound state. The enthalpy of the unbound state remains unchanged. To achieve the required 'Lock and Key' fit in the major groove, the mutant protein must adopt the exact same rigid bound conformation as the WT protein.

Calculate the standard free energy of binding ($\Delta G^\circ_{\text{bind}}$) for the mutant protein at $298 \text{ K}$.