Chapter 4, Problem 4.3IA

(a)

Interpretation Introduction

Interpretation: The form of the equation for Gibbs free energy of unfolding for the given helical polypeptide has to be justified.

Concept introduction: The Gibbs free energy is a thermodynamic quantity which allows the maximum amount of reversible work that can be performed.  It is given by the product of the temperature and entropy subtracted from the enthalpy of the system.

Answer to Problem 4.3IA

The form of the equation for Gibbs free energy of unfolding for the given helical polypeptide is justified below.

Explanation of Solution

The Gibbs free energy is calculated by the formula,

  $\Delta G=\Delta H-T\Delta S$                                                                                             (1)

Where,

• $\Delta G$ is the Gibbs free energy.
• $\Delta H$ is the molar enthalpy.
• $\Delta S$ is the molar entropy.
• $T$ is the temperature.

The polypeptide has $N$ amino acid residues and $N-4$ hydrogen bonds are formed to form an $\alpha$ helix.

Let ${\Delta }_{\text{hb}}H$ be the molar dissociation enthalpy of $1$ hydrogen bond.

Thus, the molar dissociation enthalpy of $N-4$ hydrogen bonds is $\left(N-4\right){\Delta }_{\text{hb}}H$.

Since, $N-2$ residues in the polypeptide have restricted motion.

Thus, the molar entropy for $N-2$ residues is $\left(N-2\right){\Delta }_{\text{hb}}S$.

Substitute the value of molar dissociation enthalpy and molar entropy in equation (1).

    ${\Delta }_{\text{unfold}}G=\left(N-4\right){\Delta }_{\text{hb}}H-\left(N-2\right)T{\Delta }_{\text{hb}}S$

(b)

Interpretation Introduction

Interpretation: The melting temperature $T_{\text{m}}$ may be written as $T_m=\frac{\left(N-4\right){\Delta }_{\text{hb}}H}{\left(N-2\right){\Delta }_{\text{hb}}S}$ is to be shown.

Concept introduction: The Gibbs free energy at phase transition is zero.  Hence, the value of enthalpy becomes equal to the product of the temperature and molar entropy.

Answer to Problem 4.3IA

The melting temperature $T_{\text{m}}$ may be written as $T_m=\frac{\left(N-4\right){\Delta }_{\text{hb}}H}{\left(N-2\right){\Delta }_{\text{hb}}S}$.

Explanation of Solution

The Gibbs free energy for the unfolding of the given protein is given as,

    ${\Delta }_{\text{unfold}}G=\left(N-4\right){\Delta }_{\text{hb}}H-\left(N-2\right)T{\Delta }_{\text{hb}}S$                                                        (2)

Where,

• ${\Delta }_{\text{unfold}}G$ is the Gibbs free energy of unfolding.
• ${\Delta }_{\text{hb}}H$ is the dissociation enthalpy of hydrogen bonds present.
• ${\Delta }_{\text{hb}}S$ is the molar entropy of the hydrogen bonds.
• $T_{\text{m}}$ is the melting temperature of the polypeptide at which it undergoes phase. transition.

At phase transition, Gibbs free energy for the unfolding $\left({\Delta }_{\text{unfold}}G\right)$ is $0$.

Substitute the value of ${\Delta }_{\text{unfold}}G$ in equation (2).

    $0=\left(N-4\right){\Delta }_{\text{hb}}H-\left(N-2\right)T_{\text{m}}{\Delta }_{\text{hb}}S$

Hence,

  $\begin{array}{c}\left(N-4\right){\Delta }_{\text{hb}}H=\left(N-2\right)T_{\text{m}}{\Delta }_{\text{hb}}S\\ T_{\text{m}}=\frac{\left(N-4\right){\Delta }_{\text{hb}}H}{\left(N-2\right)T{\Delta }_{\text{hb}}S}\end{array}$

Hence, the melting temperature $T_{\text{m}}$ may be written as $T_m=\frac{\left(N-4\right){\Delta }_{\text{hb}}H}{\left(N-2\right){\Delta }_{\text{hb}}S}$ is justified.

(c)

Interpretation Introduction

Interpretation: The graph of $T_{\text{m}}/\left({\Delta }_{\text{hb}}{H}_{\text{m}}/{\Delta }_{\text{hb}}{S}_{\text{m}}\right)$ has to be plotted for $5\le N\le 20$.  The value of $N$ for which $T_{\text{m}}$ changes by less than $1\%$ when $N$ increases by $1$ has to be determined.

Concept introduction: The Gibbs free energy is a thermodynamic quantity which allows the maximum amount of reversible work that can be performed and is given by the product of the temperature and entropy subtracted from the enthalpy of the system.

Answer to Problem 4.3IA

The graphs of $T_{\text{m}}/\left({\Delta }_{\text{hb}}{H}_{\text{m}}/{\Delta }_{\text{hb}}{S}_{\text{m}}\right)$ for $5\le N\le 20$ is plotted below.  The value of $N$ for which $T_{\text{m}}$ changes by less than $1\%$ when $N$ increases by $1$ is greater than $\underset{\_}{16}$

Explanation of Solution

The melting temperature is given by the equation,

  $T_{\text{m}}=\frac{\left(N-4\right){\Delta }_{\text{hb}}{H}_{\text{m}}}{\left(N-2\right){\Delta }_{\text{hb}}{S}_{\text{m}}}$                                                                                       (3)

Equation (3) follows the equation of a line, $y=mx+c$.

Where

• $T_{\text{m}}$ lies on the x axis.
• $\left(\frac{{\Delta }_{\text{hb}}{H}_{\text{m}}}{{\Delta }_{\text{hb}}{S}_{\text{m}}}\right)$ lies on the y axis.

Thus, the graph will be linear.

Slope of equation (3) is,

  $m=\frac{\left(N-4\right)}{\left(N-2\right)}$

For $N=5$,

    $\begin{array}{c}m=\frac{\left(5-4\right)}{\left(5-2\right)}\\ =0.33\end{array}$

For $N=20$,

    $\begin{array}{c}m=\frac{\left(20-4\right)}{\left(20-2\right)}\\ =0.88\end{array}$

Hence the graph is represented as

The value of $T_{\text{m}}$ changes by less than $1\%$ when $N$ increases by $1$.

  $T_{\text{m,f}}-T_{\text{m,i}}<0.01$

Let,

  $T_{\text{m,f}}<1.01T_{\text{m,i}}$

Or,

  $\frac{T_{\text{m,f}}}{T_{\text{m,i}}}<1.01$                                                                                                      (4)

Therefore,

    $T_{\text{m,f}}-T_{\text{m,i}}<0.01$

From equation (3),

  $T_{\text{m,i}}=\frac{\left(N-4\right){\Delta }_{\text{hb}}{H}_{\text{m}}}{\left(N-2\right){\Delta }_{\text{hb}}{S}_{\text{m}}}$                                                                                   (5)

When $T_{\text{m,i}}$ changes by less than $1\%$, $N$ increases by $1$.

Hence,

  $\begin{array}{c}{T}_{\text{m,f}}=\frac{\left(N-4+1\right){\Delta }_{\text{hb}}H}{\left(N-2+1\right){\Delta }_{\text{hb}}S}\\ =\frac{\left(N-3\right){\Delta }_{\text{hb}}H}{\left(N-1\right){\Delta }_{\text{hb}}S}\end{array}$                                                                                  (6)

Divide equation (6) by (5).

  $\frac{T_{\text{m,f}}}{T_{\text{m,i}}}=\frac{\frac{\left(N-3\right){\Delta }_{\text{hb}}H}{\left(N-1\right){\Delta }_{\text{hb}}S}}{\frac{\left(N-4\right){\Delta }_{\text{hb}}{H}_{\text{m}}}{\left(N-2\right){\Delta }_{\text{hb}}{S}_{\text{m}}}}$

Substitute equation (4) in above equation.

    . $\begin{array}{c}\frac{\left(N-3\right){\Delta }_{\text{hb}}{H}_{\text{m}}}{\left(N-1\right){\Delta }_{\text{hb}}{S}_{\text{m}}}\times \frac{\left(N-2\right){\Delta }_{\text{hb}}{H}_{\text{m}}}{\left(N-4\right){\Delta }_{\text{hb}}{S}_{\text{m}}}<1.01\\ \left(N-3\right)\times \left(N-2\right)<1.01\times \left(N-1\right)\left(N-4\right)\\ 0.01N^2-0.05N-1.96>0\end{array}$

Hence,

  $N>\underset{\_}{16}$

Hence, the value of $N$ for which $T_{\text{m}}$ changes by less than $1\%$ when $N$ increases by $1$ is $\underset{\_}{>16}$.