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.

Question

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Answer 1

Question

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.

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 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.

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]

T = -66.9 (1.609×00198 - .188)

T = -66.9 (.00318 - .188)

T = -66.9 (-.184)

T= 12.31 K

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

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Answer 2

Question

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.

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 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.

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]

T = -66.9 (1.609×00198 - .188)

T = -66.9 (.00318 - .188)

T = -66.9 (-.184)

T= 12.31 K

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

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Answer 3

Question

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.

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 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.

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]

T = -66.9 (1.609×00198 - .188)

T = -66.9 (.00318 - .188)

T = -66.9 (-.184)

T= 12.31 K

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

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Answer 4

Question

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.

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 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.

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]

T = -66.9 (1.609×00198 - .188)

T = -66.9 (.00318 - .188)

T = -66.9 (-.184)

T= 12.31 K

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

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

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.

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 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.

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]

T = -66.9 (1.609×00198 - .188)

T = -66.9 (.00318 - .188)

T = -66.9 (-.184)

T= 12.31 K

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

Answer 6

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.

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 K unfolding of lysozyme become favorable.

Answer 7

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.

Answer 8

Expert Solution

Explanation of Solution

If ΔGo = +ve then unfolding of lysozyme become favorable.

The folding starts when ΔGo =0

Thus,

ΔH° = TΔS°T = ΔH°ΔS°T = -280-0.79 = 354.4 K

So, above 354.4 K unfolding of lysozyme become favorable.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

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.

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]

T = -66.9 (1.609×00198 - .188)

T = -66.9 (.00318 - .188)

T = -66.9 (-.184)

T= 12.31 K

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

Answer 13

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.

Answer 14

Expert Solution

Explanation of Solution

Keq.=e-(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

Keq.=e-(-ΔG/KT)

Keq.=e-((TΔS-ΔH)/KT)

Now, we will calculate T as

T=ΔH/(ln(K eq)K+ΔS)

Changing the value of ΔS and ΔH to Kcal

ΔS= -790 J/mol K= -.188 kCal/mol.K

ΔH= -280 kJ/mol = -66.9 kCal/mol.

K = 0.00198 kcal /K.mol

Keq= foldedunfolded = 51 =5

T=ΔH/(ln(K eq)K+ΔS)

T= -66.9[ln(5)×0.00198-0.188]