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The MOLECULES
of LIFE
Physical and Chemical Principles
Solutions Manual
Prepared by James Fraser and Samuel Leachman
Chapter 10
Chemical Potential and
the Drive to Equilibrium
Problems
True/False and Multiple Choice
1. Molecules move spontaneously from regions of low
chemical potential to regions of high concentration.
True/False
2. The difference in chemical potential for a region
with 500 mM of molecule B and a region with 1 M of
molecule B is equal to:
a. kBT ln(0.5)
b. 0
c. 1
d. kBT ln(500)
e. kBT ln(1)
3. The proton concentration in pure water at standard
state (298 K) is:
a.
b.
c.
d.
e.
equal to 14
always less than –7
the square root of the ion product
107
81 kJ•mol–1
7. During protein folding, the entropy of water:
a. increases
b. decreases
c. is equal to the protein entropy change
d. is zero
Fill in the Blank
8. A region with a high chemical potential for molecule A
has a _______ concentration of molecule A than a region
with low chemical potential.
Answer: greater/higher
9. To make a buffer, add a weak acid to its conjugate
_______.
Answer: base
10. _____, ______, and ______ sidechains generally have
a pKa less than 7.0. _______, ________, and ______
sidechains have a pKa greater than 9.
4. The melting temperature (TM) is the temperature at
which 100% of the protein molecules are unfolded.
True/False
5. Which of the following must be independent of
temperature when properly applying the van’t Hoff
equation?
a. Keq
b. ∆So
c. 1/T
d. pH
6. The pKa of a protein sidechain depends only on
the chemical identity of the sidechain, not on the
surrounding environment.
True/False
Answer: His, Asp, Glu; and Tyr, Lys, Arg
11. The integral of the melting curve of heat capacity versus
temperature yields the _____________ change of protein
unfolding.
Answer: enthalpy change
12. The _______________ change for a reaction determines
the direction of spontaneous change.
Answer: free energy or ∆G
Quantitative/Essay
13. Two regions of an ideal dilute solution have a difference
in concentration of potassium ions (K+). At 293 K, what
is the difference in chemical potential between region
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
2
Chapter 10: Chemical Potential and the Drive to Equilibrium
1, with a concentration of 0.5 M K+, and region 2, which
has a concentration of 2 mM?
Answer:
C2
∆µ = RT ln
C1
= 8.31 J•K–1•mol–1 × 293 K × ln
18. The arginine-tRNA synthetase enzyme catalyzes the
reaction that charges a tRNA with the amino acid
arginine:
0.002 M
0.5 M
ATP + arginine + tRNA
= –13.5 kJ•mol–1
Answer:
0.2 M
x
0.2 M
x
C2
C1
0.2 M
x
= 2.05
= 7.79
15. A cell with an internal calcium ion (Ca2+) concentration
of 20 µM is placed in media with a Ca2+ concentration
of 70 mM. What is the difference in chemical potential
for Ca2+ ions between the inside and outside of the cell
at 310 K?
Answer:
∆µ = µin – µout = RT ln
Cin
Cout
∆µ = 8.31 J•K–1•mol–1 × 310 K × ln
20 × 10–6 M
70 × 10–3 M
∆µ = 21.0 kJ•mol–1
16. At equilibrium, in a test tube, the concentration of GDP
is 1 M, of GTP is 20 µM, and of Pi is 1 M. What is the
equilibrium constant of the reaction, GTP GDP + Pi?
Answer:
[1][1]
[20 × 10–6]
= 50000
17. The reaction, A + 2B C, has an equilibrium constant of
2000. During a reaction, the concentration of A is 0.01
M, of B is 0.2 M, and of C is 0.5 M.
a.
b.
Answer:
[500 × 10–6] [500 × 10–6] [RtRNA*]
[2 × 10–6] [500 × 10–3] [10 × 10–6]
[RtRNA*] = 4.52 × 10–5 = 45.2 µM
19. The pKa of a weak acid is 5. What is the pH when
the concentration of the acid form is 0.5 M and the
concentration of the conjugate base form is 0.05 M?
Answer:
pH = pKa + log10([base]/[acid]) = 5 + log10(0.05/0.5) =
5 + log10(0.1) = 4
x = 25.7 mM
K=
At equilibrium, following an in vitro reaction, the
concentration of ATP is 2 µM, of arginine is 500 mM, and
of arginyl-tRNA is 10 µM. The concentrations of AMP
and of PPi are 500 µM. What is the concentration of
arginyl-tRNA?
K = 1.13 =
5000 J•mol–1 = 8.31 J•K–1•mol–1 × 293 K × ln
ln
AMP + PPi + arginyl-tRNA
The value of the equilibrium constant is 1.13.
14. The difference in chemical potential for a particular
molecule between two regions of an ideal dilute
solution is 5 kJ•mol–1. The region with the higher
chemical potential has a concentration of 200 mM.
What is the concentration of the molecule in the other
region at 293 K?
∆µ = RT ln
Answer:
a. Q = [C]/([A][B]2) = 0.5/(0.01 × 0.22) = 1250
b. The reaction will proceed towards the right
since Q < Keq.
What is the reaction quotient (Q)?
In what direction will the reaction proceed?
20. The pH of a 0.15 M propionic acid/0.1 M sodium
propionate buffer is 4.71. What is the pKa of propionic
acid?
Answer:
pKa = pH – log([base]/[acid]) = 4.71 – log(0.1/0.15) =
4.71 + 0.18 = 4.89
21. Consider a protein with a surface-exposed histidine
residue in a pH 4 solution. What is the fraction of protein
molecules in which this histidine residue is charged?
(Assume that the pKa is 6.0.)
Answer:
[His]/[His+] = 10(pH – pKa) = 10(4 – 6) = 10–2 = 0.01
Fraction[His] = [His]/[His+] / ([His]/[His+] + 1) = 0.9901
Therefore, 99% of the sidechains will be charged.
22. For a protein with a surface-exposed aspartic acid,
at what pH will this residue be neutral in 75% of the
protein molecules? (Assume that the pKa is 4.0.)
Answer:
[Asp–]/[Asp] = 0.25/0.75 = 1/3
pH = pKa + log(1/3)
pH = 4 – 0.48
pH = 3.52
23. A histidine is involved in an interaction with a glutamic
acid that stabilizes the charged form of the histidine,
such that the value of ∆Go for deprotonation is 15
kJ•mol–1 at pH 7.0 and 293 K (calculated using the
biochemical standard state). What is the pKa of this
histidine?
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
Answer:
[His]/[His+] = e(–∆G/RT) = e(–15,000/(293 × 8.31)) = 0.002118
pKa = pH – log([His]/[His+]) = 7 – log10(0.002118) = 9.67
24. At the TM of a protein (55°C), the value of ∆Hounfolding is
15 kJ•mol–1. What is the value of ∆Sounfolding?
Answer:
∆Sounfolding = ∆Hounfolding/TM = 15 kJ•mol–1/328 K =
45.7 J•mol–1•K–1
25. A protein has a ∆Hunfolding value of 140 kJ•mol–1 at 25°C.
The value of ∆CP is 7.5 kJ•K–1•mol–1. The value
of ∆Hounfolding at the TM is 230 kJ•mol–1. What is the
value of TM?
Answer:
∆CP (298 K – TM) = ∆Hounfolding @ 25 – ∆Hounfolding @ TM
298 K – TM = (∆Hounfolding @ 25 – ∆Hounfolding @ TM)/∆CP
TM = –[(∆Hounfolding @ 25 – ∆Hounfolding @ TM)/∆CP – 298 K]
TM = 310 K or 37°C
26. A lysine sidechain has four torsion angles, each of
which can take on three different values (60°, –60°, and
180°). Each unique combination of angles is called a
rotamer. For example, a lysine residue where the first,
second, third, and fourth torsion angles are all 60° is
one unique rotamer, whereas a residue where the first,
second, and third torsion angles are 60° and the fourth
torsion angle is 180° is a second rotamer.
In contrast, a serine sidechain has only one torsion
angle, which can take on three different values.
Assume that all possible dihedral angles are allowed
at each angle for residues at this surface-exposed
position.
a. What is the difference in molar entropy between a
protein with a surface-exposed lysine and an otherwise
identical protein with a serine mutation at that position?
b. Why might the simplification that each lysine
torsion angle is able to adopt any of the three staggered
positions, independent of the conformation at other
torsion angles, lead to an overestimate of the number of
low-energy conformations that lysine can adopt?
Answer:
a.
WK = 34
WS = 31
∆S = R ln
WK
WS
= 8.31 J–1•K–1•mol–1 × ln
34
27. In the hydrophobic core of a folded protein, there are
three alanine and five phenylalanine residues that are
buried, and do not interact with water. Assume:
• In solution, waters can take on seven energetically
equal states.
• Two waters are ordered around each alanine in the
unfolded state.
• Six waters are ordered around each phenylalanine
in the unfolded state.
• In the unfolded state, waters are ordered around
alanine or phenylalanine residues and can take on only
two energetically equal states.
What is the difference in the entropy of the water due
to the burying of these residues as this protein folds?
Answer:
Total water molecules = 2 × number of Ala +
6 × number of Phe
=2×3+6×5
= 36 waters
Sfolded = R ln736 = 8.31 J•K–1•mol–1 × 36 × ln7 =
582 J•K–1•mol–1
Sunfolded = R ln236 = 8.31 J•K–1•mol–1 × 36 × ln2 =
207 J•K–1•mol–1
∆So = Sfolded – Sunfolded = 375 J•K–1•mol–1
28. Why do proteins denature at cold temperatures?
Answer:
While the physical mechanism behind cold denaturation
is not yet understood, the phenomenon can be
predicted from the curvature of protein stability curves.
The constant curvature, arising from the difference in
heat capacity between the folded and unfolded states,
means that because ∆Go = 0 at the TM, it must also
equal zero at some other point.
29. How do hydrophobic interactions provide favorable
entropy for protein folding?
Answer:
Since water molecules cannot hydrogen bond with
hydrophobic groups on the protein, the rotation of
water around these groups is restricted. When the
hydrophobic groups collapse into the core of the
protein, water no longer surrounds them. Thus, the
water that previously was restricted around the
hydrophobic groups is released to the bulk solvent and
free to move between many configurations (increasing
the entropy of the system).
31
∆S = 27.4 J–1•K–1•mol–1
b. Some rotamers might be disallowed because
they sterically clash with other residues, so not all
independent conformations of torsion angles might be
possible.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
4
Chapter 10: Chemical Potential and the Drive to Equilibrium
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science