Born
Lets now think about a somewhat different, but related problem that is associated with
the name Born.
Using a name (usually the name of the guy who first discovered or solved the problem) to
describe a particular type of problem is a common in science. This nomenclature provides
us with a shorthand code to talk about certain kinds of problems.
If I say Born, you should from now on immediately associate this with the concept of
electrostatic self-energy of a charged species. If I say Coulomb, you should think about
the electrostatic work associated with moving charges relative to one another etc.
But now to the topic of the class: The Born energy is the energy it takes to establish or
maintain a charge on an object (i.e. the self-energy of that charge)
The concept that a charge by its mere existence should possess energy seems pretty nonintuitive to many people, so here is a way to think about it.
Lets start with a conducting, but neutral sphere. It will not cost us any energy to deposit a
miniscule charge on its surface. But if we now add another very small charge of the same
sign, the two charges will now repel each other. We have to expend work (as defined by
Coulomb law) to bring these two charges together. As we add more and more charges we
have to expend more and more work for each fractional charge we add to those that are
already on the surface. This is, because we are working against a higher charge on the
sphere.
We can think of this as integration over some parameter  that is 0 for the uncharged
sphere and 1 for the fully charged sphere.
s ( ) 
Gel 

q
4 0 Da

1
0
s ( )  q  d 
q
q2

q

d


 0 4 Da
4 0 Da
0
1
1 1
q2

2 4 0 Da
 0   d  
1
This looks very similar to Coulomb doesn’t it? The only difference to Coulomb
is the factor of 1/2.
So we can immediately get to calculating numbers because we remember that we can pull
together all the constants for the conversion of meter into angstrom etc. into one constant.
So
1 q2
o
GBorn
  int.  333kcal/mol
2 Da

Where a is the radius of the sphere we are charging.
So how much energy does it cost to charge up a mole of 2 Å spheres with one charge on
each of the spheres.
1
1
GBorn  
 333kcal/mol 1kcal/mol
2 78.54  2
How much in a membrane or a protein interior


1 1
GBorn  
 333kcal/mol  27kcal/mol
2 3 2
Wait, if we can calculate the energy of charging a sphere in a medium we can actually
use a little thermodynamic cycle to calculate the energy for transferring an ion from one
dielectric (e.g. water) into a medium with another dielectric (e.g. a membrane).
Our energy for the transfer of an ion from a medium with dielectric constant D1 to a
second medium with dielectric constant D2 is given by
G


0
transfer(D1 D2 )
333 q 2 333 q 2 333 q 2



1/D2 1/D1
2aD1
2aD2
2a
So transferring an ion with a single charge and a radius of 2Å from water to a protein
interior (in this case lets say the protein has a dielectric of 3) requires the following free
energy.
333
0
Gtransfer(D

1/3 1/78.54  26.7kcal/mol
1 D2 )
2 2
That’s why you hardly ever see a charge buried in a non-polar solvent.
These numbers are amazingly good. In particular if you include the fact that the effective
ion size is a bit larger than the VDW radius of the ion itself, you get almost perfect
numbers. And because there is no entropy loss etc involved (we are not moving anything,
We are not working against atmospheric pressure etc.) This is a true delta G
Using the concept of Born energy to understand the charge distribution in a protein
active site.
Now its time to put some of those equations to work and see if we can use them to help
us understand how actual proteins work, if we can make sense of some experimentally
measured data.
Lets have a look together at the active site of my favorite protein, the photoreceptor
protein PYP. PYP uses a para-hydroxycinnamic acid, which is buried in the protein’s
interior, as a chromophore to sense blue light. In order for this to work properly the
chromophore has to be negatively charged. The protonated chromophore absorbs light in
the wrong part of the spectrum and undergoes the light-driven chromophore
isomerization reaction, that ultimately leads to receptor activation, with minimal
efficiency.
Charged groups inside protein interiors are not as uncommon as the high Born energy of
solvation might make us think. After all, charged groups are often required to carry out
key functions in protein active sites. The proteins most certainly have to expend the
energy to burry those charges during the folding process, but this is possible, as long as
there are enough favorable interactions elsewhere in the structure that make up for the
energetic penalty of burying a charge. So while burying a charge is not impossible, those
charges that we find in the interior of a protein are there by random chance, but most
likely serve a critical function that made it worthwhile for the protein to expend all that
energy for burying that charge.
Now lets look at the PYP active site. In it we find the chromophore located right next to a
glutamic acid. And as a matter of fact the chromophore actually forms a hydrogen bond
to the glutamic acid.
What should puzzle you is that the chromophore is deprotonated, even though in water it
has a much lower nominal pKa than the glutamic acid. The question you should be asking
yourself is, “Why does that proton not simply jump over to the chromophore!” All that
would be needed is a change in the hydrogen atom position of just 0.5 Å or so.
A recent ultra-high resolution crystal structure (Getzoff, Gutwin & Genick, Nature
Structural Biol. 2003) gave a hint. The structure shows that the chromophore adopts a
charge state that is a hybrid between the phenolic and the quinonic form.
In other words the charge is distributed evenly over the entire 12 atoms when it is on the
chromophore. On the other hand the charge would be distributed over only three atoms
(those of the carboxylic group) if it were localized on the glutamic acid.
Think: chromophore = big sphere -> lower Born energy of carrying a charge
Glutamate = little sphere -> higher Born energy of carrying a charge
Could the difference in the Born energy quantitatively explain that when glu and the
chromophore are buried inside the protein we find the charge on the chromophore, even
though in solution the chromophore’s pKa is 4.5 pH units higher than that of the glutamic
acid.
To find out, lets set up a thermodynamic box.
where, because we are dealing with a closed thermodynamic box,
G 0 Born Glu proteinwater  G 0 protontransfer H 20  G 0 Born Chr  water protein  G 0 protontransfer protein  0
so
G 0 protontransfer protein  (G 0 BornGlu proteinwater  G 0 protontransfer H 20  G 0 BornChr  water protein)

Out of these energies we can calculate the two Born energies for transferring the charged
chromophore and the charged glutamic acid from water to the protein interior. We can
further calculate the energy for transferring the proton from the chromophore to the
glutamic acid when the two are in water. We can do the latter because we can either
measure the pKa of the chromophore and glutamic in a simple, old-fashioned titration
experiments, or, easier yet, can look them up in a table.
Here is how we get the energy for the transfer reaction from the pKa’s:
The reaction is H-Chr +Glu -  Chr - +H-Glu
We can determine the energy for this reaction from the equilibrium constant.
G 0  RT ln

[Chr ][HGlu]
[HChr][Glu ]
but where can we get the concentrations for the reactants. For this we use the HendersonHasslebalch equation.
pH  pK a  log
[A ]
[HA]
so
[A ]
 10 pH pKa
[HA]
See how our equilibrium constant is the product of two such terms.

[Chr ]
 10 pH  pKaChr
[HChr]
and
[HGlu]
1
pKa  pH
 pH  pKa glu  10 glu

[Glu ] 10
so
[Chr ][HGlu]
pKa  pH
pKa  pKa
 10 pH  pKaChr 10 glu
 10 glu Chr

[HChr][Glu ]
 104.5

note that this energy is independent of pH and independent of concentration
(dimensionless K) substituting this result into our equation for delta G0 we get
G 0  RT ln104.5
G 0  6.2kcal
Now we have to calculate the energies of transferring the charges inside the protein. This
will be done with the Born equation.

Lets now see if we can find a reasonable approximation to calculate the solvation energy
of the chromophore and the glutamic acid using Born.
1) Lets assume that the difference in the energy of solvation is purely determined by
Born. This seems reasonable, because the volume and Van der Waals interaction
surfaces stay exactly the same if we have a charge or if we have no charge.
2) From basic physics we know that the energy of distributing charges over a sphere
is dependent on the size of the sphere’s surface. So we should be able to
approximate the energy of burying a molecular group of arbitrary shape by the
energy of burying a sphere with the same surface area and the same charge.
In the case of the chromophore we have the charge distributed over the surface of 12
atoms each with a radius of ~1.6 Å. Here we will model the chromophore simply as a
sphere that has a surface area that is 12 times larger than a sphere of 1.6 Å. This is
obviously an approximation, because the chromophore is not spherical, but neither is the
glutamic acid (which we will model as a sphere with a surface 3 times larger than a single
atom) and we can assume that the two approximations roughly cancel each other out. I.e.
the relative energy of charging the two surfaces will depend predominantly on the surface
area, while the deviation from the spherical shape is similar in both cases, cancels and
can therefore we neglected.
How does the surface area of a sphere change if we change its radius?
surface  Asphere  4 r 2
rsphere 
ra

rb

Asphere
4
Aa
4 
Ab
4
Aa
Ab
So spheres with surface areas 12 times and 3 times larger than that of an sphere with 1.6Å
radius have radii of:
surface of chromophore  12 1.6Å  5.54 Å
and
surface of carboxylate group  3 1.6Å  2.77Å
What is now the Born energy of burying singly charged spheres of these diameters?

G 0 Born(water protein) 
333  1
1 
 



2a 
Dprotein Dwater 
G 0 Born( proteinwater)(Glu ) 
G 0 Born(water protein)Chr  

333  1
1 
 


 19.2kcal
2aglu 
Dwater Dprotein 
333
2aChr
 1
1 
 


D
 9.6kcal
 protein Dwater 
So we have to expend 9.6 kcal/mol less to burry the negatively charged chromophore
than we need to expend on burying the glu.
Lets plug all those numbers into the equation for our thermodynamic box.
G 0 protontransferin protein  (G 0 Born Glu proteinwater  G 0 protontransferH 20  G 0 Born Chr  water protein)
G 0 protontransferin protein   (19.2  9.6  6.2) kcal/mol
G 0 protontransferin protein  3.4 kcal/mol

In other words transferring the charge from the chromophore to the glutamic acid is an
“up-hill” reaction that would cost 3.4 kcal/mol and therefore the charged chromophore is
highly favored.
In other words, simple considerations of Born energies explain why the proton does not
simply hop from the glutamic acid to the chromophore, the way an initial consideration
of pKa’s from aqueous solutions would predict. Another way of looking at it is that the
Born considerations tell us that the pKa of Glutamic acid and the chromophore are very
different inside the protein than they are in solution (Glu’s pKa goes up relative to chr’s
by more than 4.5 pH units if we go from water to the protein interior).