Q1: We can use the principles of phase equilibrium to learn about the stability of protein sin biological systems. We consider the phase equilibrium of the protein lysozyme ()between its

native phase, n, and its denatured phase, d, where unfolding occurs (native denatured), as shown in the Figure below. The following thermochemical data are available. The heat capacity for each state is independent of temperature and is given by: \text { Denatured: } c_{P, d}=16.5 \frac{\mathrm{kJ}}{\mathrm{mol} . \mathrm{K}} \text { Native: } c_{P, n}=7.4 \frac{\mathrm{kJ}}{\text { mol. } K} At 28°C, the enthalpy and entropy differences between native and denatured states are given by: h_{d}-h_{n}=255 \frac{k J}{m o l} s_{d}-s_{n}=628 \frac{J}{\text { mol. } K} a) At 28°C, which phase is stable? Justify your answer. b) Develop a mathematical equation that can predict change in Gibbs free energy in terms of temperature. [5 (correct formulas) + 10 (correct procedure and calculations) + 5 (correct answer, units and explanation if required) = 20 points] c) Using MS Excel or a suitable tool, plot the difference in Gibbs energy from the native state to the denatured state (ga – gn) versus temperature, T. For this objective, you can use a temperature range of 130 – 390 K. [5 (correct formulas)+ 15 (correct procedure and calculations) + 15 (appropriate use of Excel or similar tool) + 5 (correct answer, units and explanation if required) = 40 points] i. The heat-denaturation temperature, which is given as the temperature above which the native protein is no longer thermodynamically stable. [10 points] ii. The cold-denaturation temperature, which is given as the temperature below which the native protein is no longer thermodynamically stable. [10 points] iii. Mention the temperature range(s) where the native state of the protein is stable. [10 points]