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Chem 340 - Notes 13 - Biochemical thermodynamics (Engel Ch. 10)
Biological systems are not at equilibrium, they exist as open systems with transport of
matter, charge, heat and work in and out of systems. To keep concentrations near
constant, biological reactions must be regulated to control rates and turn on and off
The cells produce work, electrical or mechanical, but are not heat engines and have
higher efficiencies for conversion of energy (chemical) to work
Plants can convert photon energy (sun light) into compounds of high free energy
6CO2 + 6H2O C6H12O6 + 6O2 Go’ = 2868 kJ/mol
This is endergonic, G o’ (+) not spontaneous at T = 298 K, also Ho’ = 2813 kJ/mole
and So’ = (G o’ - H o’)/T = -182 J/molK -- assembling low Gibbs energy to highGfo’
Animals do the reverse steps, use carbohydrates, proteins, fats to do work and break
down to low Go’ species, anerobic glycolysis, glucose to pyruvic acid to lactic acid, 1st
C6H12O6 2 CH3COCOOH + 2H2O Go’ = -217 kJ/mol exergonic, Go’(-)
Under aerobic conditions the oxidation of pyruvic acid continues to form small
molecules, so overall the opposite reaction to photosynthesis occurs (respiration)
C6H12O6 + 6O2 6CO2 + 6H2O Go’ = - 2868 kJ/mol
This conversion to work can occur at high efficiency (~60%)
Engel fig 10.1, fig 10.2
These use a coupled reaction
ATP + H2O ADP + Pi for energy transfer
So instead of capturing heat, like heat engine, to do work, biological systems capture
free energy in ATP and couple that to processes to do work, electrical, osmotic, etc.
It is not sufficient for the energy to be just in the cell, it would just turn to heat, the
endergonic reaction must couple to an exergonic one by a common intermediate
Example: glucose + fructose form sucrose, but Go’ = 23kJ/mol (endergonic)
Couple to ATP + H2O ADP + Pi where Go’ = -30.5 kJ/mol
or glucose + fructose + ATP sucrose + ADP + Pi Go’ = -7.5 kJ/mol
Two step process: a) ATP + glucose ADP + G-1-P Go’ = -9.6 kJ/mol
b) G-1-P + fructose sucrose + Pi Go’ = 2.1 kJ/mol
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this is all under standard conditions, in cell likely different, more or less exergonic
e.g. if Q < K, then ATP, fructose, glucose > ADP, Pi, sucrose, then G more negative
recall G’ = RTln(Q/K) - ratio of concentrations compared to equilibrium at issue
note: G-1-P (glucose -1-phosphate) in both (a&b) cancels, reaction independent conc.
ATP must be formed by reaction along reverse path way: ADP + Pi ATP +H2O
But this is endergonic, Go’ = 30.5 kJ/mol, so change source of phosphate
e.g. couple to 1,3-bisphosphoglycerate + ADP 3-phosphoglycerate + ATP
in using Mg+2 and phosphoglycerate kinase (an enzyme), Go’ = -18.8 kJ/mol
1,3-bisphosphoglycerate from phosphorylate 3-phosphate glycerate: Go’ = 49.3 kJ/mol
RCOO- + H+ + HPO4-2 RCOPO3
-2 + H2O R = -CH(OH)CH2OPO3-2
this is endergonic, get energy from redox couple: NAD+ + glyceraldehyde-3-phosphate:
RCHO + NAD+ + H2O RCOO- + NADH +2H+ Go’ = -43.1 kJ/mol
All together these form coupled reactions, with the 1,3-bisphosphoglycerate common
RCHO + NAD+ + ADP + HPO4-2 RCOO- + NADH +H+ + ATP
Go’ = (-43.1 + 49.3 -18.8) kJ/mol = -12.6 kJ/mol
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Formulas Engel p251
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Phosphate transfer potential, is negative of hydrolysis of phosphates: Go’hyd
Larger value will be the phosphate donor in a coupled reaction
Membranes and ion transport
Ions in and outside of cell membrane, if nothing else would equilibrate, but
macromolecules can also be charged and cannot pass membrane, so leads to potential
As previously discussed : = out - in = RT ln (cout/cin) since ion std state same out/in
Passive transport, if cin > cout, then flow inout, reduces , = G (-)
Opposite direction takes work: w = zF(out – in) (Nernst) = -(RT/zF) ln (cout/cin)
Na-K pump, couples transport of Na+ and
K+ to ATP + H2O ADP + Pi
Na+ higher outside, K+ higher inside,
moving Na+ out and K+ in both endergonic
Facilitated by pump whose energy from a
coupled reaction with ATP
Subunit get phosphorylated in presence
of Na+, causes structural change resulting
in transport 3Na+ from interior to exterior.
This form can bind K+ on exterior, if bind 2
K+ cause hydrolysis of PO4 group,
conformational change and 2K+
transported to interior.
Overall reaction: 3Na+in + 2K+
out + ATP +H2O ADP + Pi + 3Na+out +2K+
in
Reversible work: G = 3Nain-out + 2K
out-in = RT[3 ln(cNaout/c
Nain) +2 ln(cK
in/cK
out)] + F
Go’hyd (kJ/mol)
Compound Go’hyd (kJ/mol) Phosphate transfer potential
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ATP, ADP, Pi all are in complex acid-salt equilibrium and have many states at pH7
Basically the terminal phosphate group is protonated around pH7 (pK values)
Reaction: Kobs = (cADP/co) (cPi/c
o) /(cATP/co) ~ (cADP +cHADP )(cHPO4 +cH2PO4)/(cATP +cHADP)
Relate to acid dissociation equilibria: KADP = cADP cH+ /cHADP and others similar
Kobs = cADP (1+ cH+/KADP ) cHPO4 (1+ cH+/KPi ) / cATP (1+ cH+/KATP )
Inverse of values in parentheses is mole fraction of that component
xATP-4 = cATP-4/cATP = (1+ cH+/KATP )-1 xADP-3 = cADP-3/cADP = (1+ cH+/KADP )-1
if define K1 = cATP-4 cHPO4-2 cH+/cATP-4 then Kobs = (K1 /cH+)( xATP-4/xADP-3xPi)
Goobs = -RT ln Kobs = -RT[(K1 /cH+)( xATP-4/xADP-3xPi)]
Similarly Hoobs = [(Go
obs /T)/(1/T)] = RT2 [(ln Kobs)/T]
and Soobs = [(Go
obs)/T] = R [(T ln Kobs)/T]
with all the H+ exchange reactions contributing as pH allows
adding Mg+2 or Ca+2 affects equilibria of PO4
-n containing species, due to complexation
multiple exchange equilibria affect overall thermodynamics and exergonic character
Reaction Name pK(I=0.2) Go H
o S
o
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Calorimetry
Earlier we discussed differential scanning calorimetry (DSC) as a way to determine
heats for transitions in molecules. One good application would be for protein folding
studies, where the enthalpy required to unfold the protein is determined by differentially
heating it and a reference solution. The actual instrument output is the heat capacity as
temperature is increased, which can be integrated to give the enthalpy
Hcal = ∫CpdT
This measured change includes all components of the sample, so it needs to be
corrected for solvent (by reference) and for heat capacity of the system as temperature
varies. For protein (or nucleic acid) folding experiments, we are interested in the excess
heat capacity due to the phase transition of the macromolecule. This is usually obtained
by correcting for the baseline in the scan, which can be simple, if the folded and
unfolded states have the same heat capacity and a simple transition (e.g. BPTI mutants,
left). But for complex proteins that have substates, the transitions can have multiple
peaks and baselines (IgG forms, right).
It is even possible to study multiple proteins, e.g. human plasma (left) and recreate the
thermogram by weighting for individual protein contributions (lower left), and attempts
are being made to use these patterns to diagnose for cancer and other diseases (right)
C. Johnson, Arch. Biochem.Biophys. 2013, 531, 100-109
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Isothermal titration calorimetry
Many protein and other biochemical systems
function by interacting with various ligands or
substrates. This binding can be exothermic
or endothermic and its measurement can
give information on thermodynamics of these
processes as well as their mechanism. In
ITC a sample is titrated by successive
addition to a protein (or other) solution of
small amounts of solution containing the
ligand or other species that will bind the
protein. The heat released (or absorbed) is
measured by comparing the sample to a reference while maintaining both at a constant
temperature. These data need to be corrected for dilution of the sample and for any
heat developed by salvation of the ligand, via a separate experiment where ligand is
added to just buffer or everything but the protein (blank).
Assume the macromolecule (e.g. protein) has one binding site, the fraction bound sites:
fM = [M٠L]/([M]+[M·L]) fM = [L]K/(1+[L]K) using K = [M٠L]/[M][L] (divide out [M])
since [L] and [M] can be hard to determine let: [L]tot = [L] + [M·L] and [M]tot = [M] + [M·L]
combine to get also [L]tot = [L] + fM[M]tot so use this with fM expression:
fM2 - fM(1/[M]totK + [L]tot/[M]tot + 1) + [L]tot/[M]tot = 0
solve this quadratic form to get fM as function to total concentrations and K only
In ITC each drop added will evolve heat, as add more less of each drop will be bound
until the macromolecule becomes saturated. So less heat each step, but the sum of the
progression of heats evolved will be Q: depends on fM, V0 - the initial volume and Horxn
Q = fM([M]totV0)Horxn
If have a series of j injections,
heat per injection is Qj – Qj-1
these correspond to the
peaks in the ITC data plots
Example - drug binding assay
using ITC. The relative
binding affinity is (A) 0.9 mM
& (B) 4 M, KD dissoc.
const.~1/bind so the
candidate taken for further
pharma development was (B)
Note vertical scale (B) much
higher enthalpy per injection than (A) saturate fast, almost 1:1 at equilibrium
H
1:1 bind
H
1:1 bind
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ITC gives K and H, by a best fit of curve encompassing the heat per injection
(integrate the peaks) then determine G = -RTlnK and S = (G -H)/T from the data
ITC data for protein – lipid vesicle binding – from Ning Ge, Ge Zhang, UIC
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Titrations can go either way, company literature promotion:
Above examples are looking directly at heat involved in (DSC) phase change of
macromolecule or (ITC) binding of macromolecule to other species (ligand or lipid).
That yields H and if we then look at this as q, heat, we can get S = qrev/T, and putting
them together gives us G and from that the equilibrium constant: K = exp(-G/RT)
However, since K is a function of concentration and there are many ways to get at
concentrations, one can take an alternate way provided the molecules cooperate. The
conceptually easy way to start is weighing pure samples, but that does not tell what are
the equilibrium concentrations. One could get them by titrating if reagent is pH or simple
reaction sensitive (volumetric). Also can do some sort of reaction to capture and
separate one reagent (gravimetric).
In biochemistry this does not work so well, as the macromolcules and their ligand
associated forms would have similar chemistries. Also, biochemical samples tend to not
be pure, so weighing does not always yield a good total concentration. Electrochemistry
can be used (Nernst relation) to get concentrations for species that undergo some sort
of electron transfer (redox). This is sensitive, and can be used for macromolecules,
particularly metalloproteins, whose active sites are often redox coupled. The way people
often get concentration and concentration changes is with Spectroscopy, using light to
probe the sample and select out the component of interest due to the differences in its
absorbance or fluorescence or other characteristics. This also affects Kinetics and both
will be major components of our course (Chem 344) next semester.
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DNA duplex formation
DNA is a polymer of phosphor ribose units with a sequence of (four) bases attached
To form the double strand the sequences need to be complementary, A-T and G-C, this
interaction is vital for replication and PCR and many biochemical processes. The
equilibria, and thus G, can be monitored by the change in uv absorbance for ds form
Analysis of these data show that the thermodynamics canot be represented by just
summing up the A-T and G-C G values, but is dependent on what comes next, in that
it is affected by the base stacking (London – van der Waals forces)
G = Go(initiate) + Go(near neighbor) , Go(initiate)(+) unfavorable
For ( ss) 5’-ATAGCA-3’ + 5’-TGCTAT-3’ (ds) 5’-ATAGCA-3’/3’-TATCGT-5’
Go = Go(initiate)+Go(AT/TA)
+Go(TA/AT)+Go(AG/TC)+Go(GC/CG)+Go(CA/GT)
From table:
Go = 8.1-3.7-2.4-5.4-9.3-6.0 =
-18.7 kJ/mol
K310 = e-G/RT = 1416
Similarly Ho and So
can be determined from tables
in same way
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Ion effect on Protein-nucleic acid interaction
Nucleic acids (and some proteins) are multipley charged (each PO2- group has
associated an ion, eg Na+), if ligand L interacts with n phosphate groups n Na+
originally on phosphates will be released
P+L PL + nNa+
For this K = [PL][Na+]n/[P][L] and Kobs = [PL]/[P][L]
So Kobs = K[Na+]n --> so observed equilibrium strongly depend on Na+ concentration
Slope of log Kobs vs [Na+] will give n