Consider the protein lysozyme. Protein molecules have native (folded) conformations (label the folded state n ), and denatured (un-folded) conformations (label the unfolded state d ). We can consider the two conformations n and d to be separate phases, and use the principles of thermodynamic equilibrium to analyze protein folding. Assume that the heat capacities for each phase of lysozyme are independent of temperature, and are given by: CP,d=15.1kJ/molKCP,n=6.0kJ/molK At 25C, the enthalpy and entropy differences between native and denatured states are given by: HdHn=242kJ/molSdSn=0.619kJ/molk (a) At 25C, which conformation is stable and why? (b) Find algebraic expressions for [HdHn](T) and [SdSn](T). (That is, the difference as a function of temperature.) (Hint: you can use a convenient path approach.) (c) Plot [GdGn](T) over a temperature range of 200 to 375K. From your plot, estimate the colddenaturation and heat-denaturation temperatures of lysozyme. ("Cold denaturation" refers to the loss of protein stability at low temperature)