Biochemistry I
Lecture 9
February 2, 2015
Lecture 9: Protein Stability.
Skills:
 Determine the ΔHo and ΔSo from thermal denaturation curves.
 Predict amount of folded protein, given ΔHo and ΔSo.
 Interpret changes in ΔHo and ΔSo due to mutations.
Thermodynamics
Description
F→U
Sign
Stabilizes
(F → U)
F
ΔH
o
HBOND
= HU-Hbond –HF-Hbond
U
The enthalpy change due to H-bonds is
positive; heat is required to break the Hbonds in the folded protein.
ΔHoVDW = HU-VDW – HF-VDW
The enthalpy change for vdw is positive;
heat is required to break the vdw interaction
in the folded protein
ΔSoCONF = SU-CONF – SF-CONF
The entropy change of the chain is positive,
and very large. The unfolded state has
more disorder.
ΔSoSOLV = SU-SOLV – SF-SOLV
The entropy change of the solvent is
negative, due to the exposure of the
hydrophobic groups during unfolding.
The features of folded globular proteins are a consequence of the thermodynamics forces that
stabilize the folded form of the protein, listed in order of significance, from left to right.
Conformational entropy has been left off of this chart, remember it is very destabilizing for the folded
state.
Hydrophobic
residues in core.
Well packed core.
Hydrophobic Effect
van der Waals
H-bonds
Interplay between 1o, 2o and 3o structure.
If α-helices and β-sheets are equally stable from the perspective of
vdw and H-bonds between mainchain atoms, why is one favored over
another?
1
Hydrogen bonded
secondary structures.
Biochemistry I
Lecture 9
February 2, 2015
Thermal Denaturation of Proteins: The relative energy of the native and unfolded state can be
changed with temperature. Unfolding occurs at high temperature due to the positive ΔSo:
G 0  H o  TS o
1
0.9
Fraction Unfolded
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
300
310
320
330
340
350
360
Tem p (K)
Go = -RTlnKEQ
Obtaining ΔHo and ΔSo:
Predict
fn, fu
Exp
Data
Ho , So
Calculate
Go= Ho - T So
Obtaining H0: The temperature dependence of the equilibrium constant can be used to determine
Ho. Equating the two expressions for Go, and performing a little algebra:
o
o
o
This is the equation of a straight G  H  TS   RT ln K EQ
ln K EQ 
line, if lnKEQ is plotted versus 1/T.
 H o 1 S o

R T
R
Slope = -Ho /R
Ho = -slope  R
This plot is referred to as a van't Hoff Plot.
Steps:
1. Vary temperature, record T & fraction native.
2. Convert temperature from C to K (C+273).
3. Calculate fraction unfolded, fU = (1-fN).
4. Calculate KEQ = fU/fN
5. Plot ln(Keq) versus 1/T (K).
6. Use the graphing wizard in Excel (Scatter plot).
7. Select plotted data with mouse.
8. Under Chart options, select "Add Trendline".
9. Use linear, under "Options" select "Display
equation on chart".
Obtaining S : Once the enthalpy is known, the
change in entropy can be calculated from the TM and
from Ho. At the melting temperature, TM, the energy
difference between the native and unfolded states
is zero.
o
2
6
4
y = -24188x + 72.202
2
0
q
e-2
K
ln
-4
-6
-8
-10
0.0027
0.0029
0.0031
1/T
0.0033
Biochemistry I
Lecture 9
February 2, 2015
Example Question 1: Given Ho and So, predict the fraction
[F ]
[U ]
unfolded at any given temperature.
fF 
fU 
[ F ]  [U ]
[ F ]  [U ]
You work for a company that uses an enzyme to make
the amino acid Lysine, an important amino acid in dinosaur
food at Jurassic park. Your supervisor tells you to increase
[ F ] /[ F ]
[U]/[F]


the production of lysine. The reaction is normally run at 290
[ F ] [U ]
[F] [U ]


K (~ room temperature). You know that the rate of the
[F ] [F ]
[F] [ F ]
reaction, and therefore the Lysine production, will increase at
K EQ
1
higher temperatures.
Consequently, you increase the
fF 
fU 
o
1  K EQ
1  K EQ
reaction temperature to 310K (37 C). Given the following
thermodynamic properties of the unfolding of the enzyme
used in the reaction: Ho= +300 kJ/mol, So= +1000 J/mol-K, have you just lost your job? (RT=2.57
kJ/mol at 310 K).
i. Calculate Go at the required temperature, Go=Ho-TSo
Go = 300 – (310)x(1.0) = -10
kJ/mol
ii. Calculateo KEQ at the required temperature:
K EQ  e  G
/ RT
= e+10/2.57 = e4 = 48.9
CH3
iii. Calculate fU using KEQ: fu = 48.9/(1+48.9) = 0.98.
H
Example Question2: The denaturation curves for both wild-type and
mutants of a protein were measured to obtain Ho and So.
Ho
So
Tm
Wild-Type
210.0 kJ/mol
616 J/mol-deg
341 K
N
OH
H
N
O
Mutant
206 kJ/mol
611 J/mol-deg
337 K
Threonine(Thr)
S
CH3
O
Methionine(Met)
1.0
Wild-type
Fu
0.5
Explain the effects on both H and S .
o
o
Mutant
0
Denatured (Unfolded)
Native (Folded)
Wild-type
O
Thr 53
H=+210 kJ/mol
CH3
O
337 341
Temperature
O
O
O
CH3
O
O H
H
O
Thr 44
H
S=+616
O
O
CH3
O
Altered Protein
Met53
S
CH3
O
O
H=+206 kJ/mol
O
S
CH3
O
O
H
S=+611
O
O
Thr44
O
CH3
O
 SoH2O
 SoH2O
 SoH2O
 SoCONF
(except 53)
 SoH2O (Thr53)
 So OBS = 616
3
 SoH2O
 SoCONF
(except 53)