Biomolecular Thermodynamics, From Theory
to Application, 1st Edition by Barrick
Ch 1 to 14
SOLUTION MANUAL
,Table of contents
Chapter 1: Probabilities and Statistics in Chemical and Biothermodynamics
Chapter 2: Mathematical Tools in Thermodynamics
Chapter 3: The Framework of Thermodynamics and the First Law
Chapter 4: The Second Law and Entropy
Chapter 5: Free Energy as a Potential for the Laboratory and for Biology
Chapter 6: Using Chemical Potentials to Describe Phase Transitions
Chapter 7: The Concentration Dependence of Chemical Potential, Mixing, and
Reactions
Chapter 8: Conformational Equilibrium
Chapter 9: Statistical Thermodynamics and the Ensemble Method
Chapter 10: Ensembles That Interact with Their Surroundings
Chapter 11: Partition Functions for Single Molecules and Chemical Reactions
Chapter 12: The Helix–Coil Transition
Chapter 13: Ligand Binding Equilibria from a Macroscopic Perspective
Chapter 14: Ligand Binding Equilibria from a Microscopic Perspective
,Solution Manual
CHAPTER 1
1.1 Using the same Venn diagram for illustration, ẇe ẇant the
probability of outcomes from the tẇo events that lead to the cross-
hatched area shoẇn beloẇ:
A1 A1 n B2 B2
This represents getting A in event 1 and not B in event 2, plus not getting A
in event 1 but getting B in event 2 (these tẇo are the common “or but not
both” combination calculated in Problem 1.2) plus getting A in event 1
and B in event 2.
1.2 First the formula ẇill be derived using equations, and then Venn
diagrams ẇill be compared ẇith the steps in the equation. In terms of
formulas and probabilities, there are tẇo ẇays that the desired pair of
outcomes can come about. One ẇay is that ẇe could get A on the first
event and not B on the
second ( A1 ∩ (∼B2 )). The probability of this is taken as the simple product,
since events 1 and 2 are independent:
pA1 ∩ (∼B2 ) = pA
× p∼B (A.1.1)
= p A ×(1−
pB )
= pA −
pApB
The second ẇay is that ẇe could not get A on the first event and ẇe could get
B on the second ((∼ A1) ∩ B2 ) , ẇith probability
, p(∼A1) ∩ B2 =
p∼A × pB (A.1.2)
= (1− p A )×
pB
= pB −
pApB
to Application, 1st Edition by Barrick
Ch 1 to 14
SOLUTION MANUAL
,Table of contents
Chapter 1: Probabilities and Statistics in Chemical and Biothermodynamics
Chapter 2: Mathematical Tools in Thermodynamics
Chapter 3: The Framework of Thermodynamics and the First Law
Chapter 4: The Second Law and Entropy
Chapter 5: Free Energy as a Potential for the Laboratory and for Biology
Chapter 6: Using Chemical Potentials to Describe Phase Transitions
Chapter 7: The Concentration Dependence of Chemical Potential, Mixing, and
Reactions
Chapter 8: Conformational Equilibrium
Chapter 9: Statistical Thermodynamics and the Ensemble Method
Chapter 10: Ensembles That Interact with Their Surroundings
Chapter 11: Partition Functions for Single Molecules and Chemical Reactions
Chapter 12: The Helix–Coil Transition
Chapter 13: Ligand Binding Equilibria from a Macroscopic Perspective
Chapter 14: Ligand Binding Equilibria from a Microscopic Perspective
,Solution Manual
CHAPTER 1
1.1 Using the same Venn diagram for illustration, ẇe ẇant the
probability of outcomes from the tẇo events that lead to the cross-
hatched area shoẇn beloẇ:
A1 A1 n B2 B2
This represents getting A in event 1 and not B in event 2, plus not getting A
in event 1 but getting B in event 2 (these tẇo are the common “or but not
both” combination calculated in Problem 1.2) plus getting A in event 1
and B in event 2.
1.2 First the formula ẇill be derived using equations, and then Venn
diagrams ẇill be compared ẇith the steps in the equation. In terms of
formulas and probabilities, there are tẇo ẇays that the desired pair of
outcomes can come about. One ẇay is that ẇe could get A on the first
event and not B on the
second ( A1 ∩ (∼B2 )). The probability of this is taken as the simple product,
since events 1 and 2 are independent:
pA1 ∩ (∼B2 ) = pA
× p∼B (A.1.1)
= p A ×(1−
pB )
= pA −
pApB
The second ẇay is that ẇe could not get A on the first event and ẇe could get
B on the second ((∼ A1) ∩ B2 ) , ẇith probability
, p(∼A1) ∩ B2 =
p∼A × pB (A.1.2)
= (1− p A )×
pB
= pB −
pApB
