Biochemistry I
Lecture 9
February 1, 2013
Lecture 9: Protein Stability.
Assigned reading in Horton - 4.9, 10. Nelson -1.3 and 4.4
Key Terms:


OLI Quiz: 3


Thermal Denaturation
Native state
TM
Ho, So Determination.
Thermodynamics
N→U
Description
ΔHo HBOND = HU-Hbond - HN-Hbond
The enthalpy change due to H-bonds
is positive; heat is required to break the
H-bonds in the folded protein.
ΔHoVDW = HU-VDW - HN-VDW
The enthalpy change for vdw is
positive; heat is required to break the
vdw interaction in the folded protein
ΔSoCONF = SU-CONF - SN-CONF
The entropy change of the chain is
positive, and very large. The unfolded
state has more disorder.
ΔSoSOLV = SU-SOLV - SN-SOLV
The entropy change of the solvent is
negative, due to the exposure of the
hydrophobic groups during unfolding.
Stabilizes
Native
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.
Feature of Tertiary Structure Driven
by Thermodynamic Factors
Hydropho
-bic Effect
1. Hydrophobic residues in core.
2. Well packed core.
3. Hydrogen bonded secondary structures.
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
van der
Waals
H-bonds
Stabilizes
Unfolded
Biochemistry I
Lecture 9
February 3, 2012
Thermal Denaturation of Proteins: The relative energy of the native and unfolded state can be
changed with temperature: 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
Temp (K)
G 0  H o  TS o
G 0   RT ln K EQ
K EQ 
[U ] EQ
[ N ]EQ
Obtaining H0: The temperature dependence of the equilibrium
constant can be used to determine Ho. Equating the two
expressions for Go:
G o  H o  TS o   RT ln K EQ
ln K EQ 
This is the equation of a straight line, if lnKEQ is
plotted versus 1/T.
6
Slope = -Ho /R or Ho = -slope  R
This plot is referred to as a van't Hoff Plot.
4
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".
0
 H o 1 S o

R T
R
y = -24188x + 72.202
2
lnKeq
2
-2
-4
-6
-8
-10
0.0027
0.0029
0.0031
1/T
0.0033
Biochemistry I
Lecture 9
February 3, 2012
Obtaining So: 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 (fN = fU) , therefore:
0 = Ho - TMSo,
So = Ho/TM
giving:
Example Question 1: Given Ho and So, predict the fraction unfolded at any given temperature.
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 the production of lysine.
The reaction is normally run at 290 K (~ room temperature). You
know that the rate of the reaction, and therefore the Lysine
production, will increase at higher temperatures. Consequently,
you increase the reaction temperature to 310K (37o 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).
[N ]
[ N ]  [U ]
fU 
[ N ] /[ N ]
[ N ] [U ]

[N ] [N ]

fN 

fN 
1
1  K EQ
[U ]
[ N ]  [U ]
[U]/[N]
[N] [U ]

[N] [ N ]
K EQ
fU 
1  K EQ
i. Calculate Go at the required temperature, Go=Ho-TSo
Go = 300 – (310)x(1.0) = -10 kJ/mol
ii. Calculate KEQ at the required temperature:
K EQ  e  G
o
/ RT
= e+10/2.57 = e4 = 48.9
iii. Calculate fU using KEQ: fu = 48.9/(1+48.9) = 0.98.
CH3
Example Question2: The denaturation curves for both wild-type and mutants
of a protein were measured to obtain Ho and So. Explain the effects on both
Ho and So.
H
N
OH
O
Threonine(Thr)
Wild-Type
210.0 kJ/mol
616 J/mol-deg
H
So
o
Mutant
206 kJ/mol
611 J/mol-deg
Denatured (Unfolded)
Native (Folded)
Wild-type
O
Thr 53
H=+210 kJ/mol
CH3
O
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
3
H
N
S
O
Methionine(Met)
CH3