ink to Ointment
Biochemistry
.
Biochemistry
|
September 20, 2000 |
Lecture 10: Biochemical Energetics 2
Lecture 10 (PDF) This document is available as a PDF image.
Assigned reading in Campbell: Chapter 11.6-11.10.
(Omit section 11.11)
Key Terms:
|
Liposomes
Metabolism
Anabolism |
Catabolism
Oxidation-reduction reactions
Coupling |
|
Take the Review Quiz on Lecture
10 concepts.
Protein
Folding-Unfolding Transition: a Shockwave movie of an "aa
structure" that allows you to set the reaction temperature and view
the results on a transition curve graph.
|
|
Please note: The shaded sections of these revised lecture notes will
not be covered this term.
10.13.00
|
11.6 Hydrophobic Interactions: A Case Study in Thermodynamics
The descriptions in Campbell are a good introduction to the concepts involved
in describing the formation of liposomes and the folding of proteins into their
native conformation. The following sections add to those concepts some experimental
results that use for their interpretation the relationships between
DG,
DH,
DS, and the equilibrium
constant.
|
A. Measurement of protein folding/unfolding equilibria
For the reversible protein unfolding reaction,
N <=> U with Keq = [U]/[N]
the observed "transition curve" between native (N) and unfolded
(U) states is a steep function of temperature, reflecting the cooperativity
of the stabilizing forces. The examples below show the unfolding transitions
of Protein G plotted as the fraction unfolded vs. the temperature. Curves
for the wild type and two different mutant proteins are shown.
Experimental data are analyzed using the following approach:
The van't Hoff equation relates the equilibrium constant to temperature
and allows us to extract DH and DS
for the transition and to calculate protein stability at any temperature:
DG = -RT lnKeq = DH
- TDS
rearrange:
lnKeq = -DH/RT + DS/R
A plot of lnKeq vs 1/T yields a straight line with slope
= -DH/R.
Three steps in using the van't Hoff equation, starting from a transition
curve:
- Determine the spectroscopic signal (e.g. A280nm
or fluoresence) for 100% native protein (N) and 100% unfolded protein
(U). Measure the signal as a function of temperature (T) to produce
a transition curve. Determine the fraction of unfolded (or native)
protein as a function of temperature.
- Calculate K as a function of T:
K = Fu / Fn Note that K = 1 and DG
= 0 at Tm.
Tm is the midpoint of the transition curve where [N] =
[U]; it also called the "melting temperature".
- Plot lnK vs 1/T (K-1) a "van't Hoff
plot"
- Determine DH of unfolding: = -R*slope
- Determine DS of unfolding: = DH/Tm
- Calculate the stability of the protein at any other temperature
using: DG = DH
- TDS
B. Example of Protein G unfolding (and some typical calculations)
These calculations are based on the curves shown above.
- For the unfolding reaction of wild type Protein G (Thr 53):
- DH = 50.4
kcal/mol from a van't Hoff plot (not shown).
- We calculate DS
at Tm:
- DS = DH/Tm
= 50,400/342 = 147.4 cal/mol/K.
- For the Ala 53 mutant Protein G:
- DH = 43.7
kcal/mol.
- DS = 132.4
cal/mol/K.
Now compare the wild type protein to the Ala 53 mutant at the same
temperature.
- At 69°C (the Tm for wild type):
DGwt = 0
DGmutant = DH
- TDS = 43,700 - 342*132.4 = -1.58 kcal/mol.
Thus, DGwt - DGmutant
= DDG = 0 -
(-1.58) = 1.58 kcal/mol.
Conclusion: Unfolding of wild type protein is less favorable
than the Ala 53 mutant by 1.58 kcal/mol.
- At 27°C:
DGwt = 50,400 - 300*147.4 =
6.2 kcal/mol.
DGmutant = 43,700 - 300*132.4
= 4.0 kcal/mol.
DDG = 6.2 -
4.0 = 2.2 kcal/mol.
- The equilibrium constant at 27°C for wild type protein unfolding
is calculated from:
- DG = -RTlnKeq
Keq = exp(-6.2/0.6) (RT = 0.6 kcal/mol/K
at 300K)
- Keq = 3.25*10-5
- The fraction of wild type protein in the unfolded state at this
temperature is:
- Fu = Keq/(1 + Keq)
- Fu = 3.3*10-5
There are two remarkable and general features of these protein stability
results:
- The DG for Protein G folding is
modest. -- only -6.2 kcal/mol at room temperature. This is also the
case for other proteins where the DG values
range from about -5 to -15 kcal/mol.
- The modest -DG for protein folding
in general, represents the small difference between a large favorable
DH and an almost-as-large unfavorable TDS
contribution.
C. The hydrophobic effect can be measured as a DG
of transfer.
During protein folding, the transition from the countless unfolded states
to a single native state is accompanied by the burial of solvated nonpolar
sidechains (and polar peptide units) into the nonsolvated core of the
protein. The reduction in solvent-accessible area of these groups favors
the folded state. An empirical correlation that appears to have predictive
value also provides insight into the energetics of the hydrophobic effect.
The correlation and further description are on a separate page, Lecture
10 (Supplement). This material is not "assigned reading" in
the course. |
D. Factors that stabilize/destabilize proteins
- Temperature: Cold and heat denaturation.
- pH: Attractive and repulsive electrostatic interactions.
- Detergents: SDS (sodium dodecyl sulfate) solubilizes surface and core
residues.

- Salts affect water structure.
- sulfate < phosphate < acetate < Chloride < bromide <
Iodide < ClO4 < SCN-
- ammonium , Cs+, K+, Na+, > Li+
> Mg2+ > Ca2+ > Ba2+.
- Urea (8M) and guanidinium ion (3M)
For example, proteins are most stable in ammonium sulfate and least stable
in barium thiocyanate
11.7 The Nature of Metabolism
11.8 The Nature of Oxidation and Reduction
11.9 Coenzymes in Biological Important Oxidation-Reduction Reactions
These three sections cover topics that were mostly covered in introductory
biology. Review the concepts in the Key Terms list. Knowledge of the
structures of nucleotides and coenzymes are not required at this point in
Biochemistry I.
11.10 Coupling of Production and Use of Energy
The first step in the glycolytic pathway is the phosphorylation of glucose to
make glucose-6-phosphate:
Glucose + ATP <=> ADP + G-6-P
The reaction is catalyzed by the enzyme, hexokinase.
Use the values in Table 11.1 (Campbell, p. 411) to calculate the following for
the hexokinase reaction:
- DG°'
- Keq
- Will the above values be different in the cell, where the substrate and
product concentrations are in the mM range?
Compare your answers to those provided in lecture.
Return to Home Page.
9.18.00
(back to links)
March 21, 2001
| all
rights reserved copyright 1999 |
contact
us for
more
information |
from
our mind to yours   |
 |