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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