Chapter 3, Problem 1P

Interpretation Introduction

Interpretation:

The temperature at which the unfolding of lysozyme become favorable should be determined.

Concept Introduction:

Entropy changes can be determined by using enthalpy changes and standard free energy changes values. The Gibbs energy change can be defined as the difference between the enthalpy of the system and product of the temperature and entropy of the system. The value of standard Gibbs energy change also helps in predicting the spontaneity of a reaction, that whether the reaction favorable or not.

Explanation of Solution

If ${\text{ΔG}}^{\text{o}}\text{=}\text{+ve}$ then unfolding of lysozyme become favorable.

The folding starts when ${\text{ΔG}}^{\text{o}}\text{=0}$

Thus,

  $\begin{array}{l}\text{ΔH° =}\text{TΔS°}\\ \text{T}\text{=}\frac{\text{ΔH°}}{\text{ΔS°}}\\ \text{T}\text{=}\frac{\text{-280}}{\text{-0}\text{.79}}\text{=}\text{354}\text{.4 }\text{K}\end{array}$

So, above $\text{354}\text{.4 }\text{K}$ unfolding of lysozyme become favorable.

Interpretation Introduction

Interpretation:

The temperature at which the ratio of unfolded protein to protein becomes 1:5 should be determined.

Concept Introduction:

Entropy changes can be determined by using enthalpy changes and standard free energy changes values. The Gibbs energy change can be defined as the difference between the enthalpy of the system and product of the temperature and entropy of the system. The value of standard Gibbs energy change also helps in predicting the spontaneity of a reaction, that whether the reaction favorable or not.

Explanation of Solution

  $\text{Keq}{\text{.=e}}^{\text{-(ΔG/KT)}}$

We will now substitute the value of Gibbs energy change in the above equation. But, we will also reverse the sign as protein is going from an unfolded state to a folded state. So, the equation becomes as

  $\text{Keq}{\text{.=e}}^{\text{-(-ΔG/KT)}}$

  $\text{Keq}{\text{.=e}}^{\text{-((TΔS-}\Delta \text{H)/KT)}}$

Now, we will calculate T as

  $\text{T=ΔH/}\left(\text{ln}\left({\text{K}}_{\text{eq}}\right)\text{K+ΔS}\right)$

Changing the value of $\text{ΔS}$ and $\text{ΔH}$ to Kcal

  $\text{ΔS= -790 J/mol K= -}\text{.188 kCal/mol}\text{.K}$

  $\text{ΔH= -280 kJ/mol = -66}\text{.9 kCal/mol}\text{.}$

  $\text{K = 0}\text{.00198 kcal /K}\text{.mol}$

  ${\text{K}}_{\text{eq}}\text{= }\frac{\text{folded}}{\text{unfolded}}\text{ = }\frac{\text{5}}{\text{1}}\text{ =5}$

  $\text{T=ΔH/}\left(\text{ln}\left({\text{K}}_{\text{eq}}\right)\text{K+ΔS}\right)$

  $\text{T= }\frac{\text{-66}\text{.9}}{\left[\text{ln}\left(\text{5}\right)\times \text{0}\text{.00198-0}\text{.188}\right]}$

  $\text{T = -66}\text{.9 }\left(\text{1}\text{.609×00198 - }\text{.188}\right)$

  $\text{T = -66}\text{.9 }\left(\text{.00318 - }\text{.188}\right)$

  $\text{T = -66}\text{.9 }\left(\text{-}\text{.184}\right)$

  $\text{T= 12}\text{.31 K}$

Thus, the temperature at which the ratio of unfolded protein to protein becomes 1:5 is 12.31 K.