Molecular binding and free energy

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Review of the molecule trappers and introduction to the free energyExamples of DNA deposition onto a solid surface and Osmotic flow

The free energy and the entropy

Molecular binding and free energy

Jie Yan

September 5, 2006

Jie Yan Molecular binding and free energy

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Contents

I Molecular binding problem and examples.

I The free energy and the entropy.


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Contents

I Molecular binding problem and examples.

I The free energy and the entropy.


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Time slots

I Tutorial starts from week 3 (28AUG-1SEP).

I A class tutorial every week, on Friday 9 am - 10 am.

I A small group tutorial every even week (4,6,8,10), on Mon(12.00-12.50pm) or on Wed (5.00-5.50pm). The small grouptutorial classroom is S13-04-02.

I Group A(30%): week 4, 8,12, Mon; group B(70%): week6,10, 12, Wed.

I For those who cannot come at the two slots, please come todiscuss with me at Phys E-lab from 2:30 pm to 3:30 pm onWed of week 4 and 8.


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Molecular trapper


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Molecular trapper

I Now let’s consider that the particle is bound to the center bya short range interaction, so Es(x) = −ε when−d/2 < x < d/2, and Es(x) = 0 when x < −d/2 or x > d/2.

I Z =∫ −d/2−L/2 dx1 +

∫ d/2−d/2 dxe

− −εkBT +

∫ L2

d/2 dx1 = L− d + deε

kBT .

I ρ(x) = 1

L−d(1−eε

kBT )e− Es (x)

kBT .

I Ptrap =∫ d/2−d/2 dxρ(x) = de

εkBT

L−d(1−eε

kBT )= γe

εkBT

1−γ(1−eε

kBT ), where

γ = d/L.


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Molecular trapper


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Molecular trapper


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Introduction to the free energy

I We have re-understood our two-trapper problem as follows:

Pi = nieεi

kBT

n2eε2

kBT +n1eε1

kBT, where −εi is the trapping energy of the

particle in region i , and ni is the number of rooms in region i .

I By ni = e ln(ni ) = ekBT ln(ni )

kBT , We can re-write the above formula

as: Pi = e(εi +kBT ln(ni ))

kBT

e(ε2+kBT ln(n1))

kBT +e(ε1+kBT ln(n2))

kBT

= e−FikBT

e−F1kBT +e

−F2kBT

, where

Fi = −εi − kBT ln(ni ) is called the ”free energy” of the i-thtrapper.


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Components in the free energy

I We have seen that two sources contributing to the ”freeenergy” of each trapper: the trapping energy −ε, and howmany room the trapper contains.

I In biological binding problem, the solution is always a trapper,ε = 0, and n = V /ν, where V is the solution volume and ν isthe volume of the molecule. Usually n is a huge number. Thesolution free energy is thus F = −kBT ln(V /ν).

I Reducing the solution volume leads to a proportionalreduction of the room number, or a proportional increase inthe molecule concentration c ≈ (the volume fraction of themolecules). So the solution free energy can be written in termof the concentration c as: F = kBT ln(c).


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Properties of the free energy

I A molecule tends to bind to a trapper to minimize the freeenergy.

I We can extend our argument to the multiple-trapper problem:

Pi = e−FikBTPN

j=1 e−FikBT

, where Fi = −εi − ln(ni ) is the free energy of

the ith trapper.

I The ratio: piPj

is determined by the difference between the free

energies of the molecule binding the two trappers:

piPj

= e−(Fi−Fj )

kBT .


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Examples: DNA deposit onto a surface

I The conformation of a DNA can be considered to bedetermined by a 3-d random walk with a step size ofb = 100nm. For large N of the walking steps, the random coilsize of the DNA is R =

√Nb. A lambda DNA has a mass

m ≈ 5× 10−11ug .

I A lambda DNA has a contour length L = 16 microns, soN = 160. The random coil size is thusR ≈

√160× 100nm = 1.3 micron. Its volume is about

ν = R3 = 2.2× 10−12cm3 = 2.2× 10−9ul(1ul = 10−3ml = 10−3cm3).


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I Assuming we have a DNA solution with a concentration1ng/ul = 2× 107DNA/ul = 4.4× 10−2ul/ul , its volumefraction is thus c = 4.4× 10−2. The solution free energy for alambda DNA is then Fs = kBT ln(c) = −3.12kBT .

I Assuming the surface is positively charged and it trapped aDNA base pair with −1kBT trapping energy, then−εsur ≈ −50, 000kBT since a lambda DNA has about 50, 000base pairs. Assuming we want to image a surface area of100µm2, then the room number is nsur ≈ 60. The surface freeenergy for a lambda DNA is thenFsur = −εsur − kBT ln(nsur ) ≈ −50, 000kBT .


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I When DNA is short (< 100nm ≈ 300bp), it is rigid enough soits size can be approximated by its contour length. In average,1bp DNA has a mass 10−12ng .

I By assuming each DNA occupies a volume ∼ 100nm3, a DNAsolution with a mass concentration 1ng/ul has a volumefraction c = 3. The solution free energy for the DNA is thenFs = kBT ln(c) ≈ 1kBT .

I Assuming the surface trappes a DNA base pair with −1kBTtrapping energy, then −εsur ≈ −300kBT . For a surface areaof 100µm2, the room number is nsur ≈ 10, 000. The surfacefree energy for the 100bp DNA is then Fsur ≈ −310kBT .


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The free energy

I Previously we defined the free energy of a molecule that canbind to a trapper to be Fi = −εi − kBT ln(ni ), where −εi isthe potential energy (or the interaction energy) between themolecule and the trapper, and ni is the number of rooms ornumber of ways of hosting the molecule.

I Degree of disorder: ni represents the ”degree of disorder” ofthe one particle system. The more ways of hosting themolecule, the higher the degree of disorder of the system.


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The entropy

I Boltzmann defined a quantity to measure the degree ofdisorder of a system: S = kB ln Ω, where Ω is the totalnumber of states of a system. In our simple one particletrapper system, Ω = ni , and Si = kB ln(ni ). S is called theentropy. By defining the entropy, the free energy becomesFi = −εi − TSi .

I A particle tends to bind to minimize its free energy, by eitherreducing the interaction energy (increasing εi ), or byincreasing its entropy Si .


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Osmotic flow

Osmotic flow: the chamber is separated by a semipermeable membrane allowing only water molecule passing through. The pistons can freely sliding. At the beginning, a clump of sugar is placed at the right chamber. a) At small load, the pistones will move to the right to minimize the free energy.b) Bigger load moves the pistones to the left, free energy is increased.The example is taken from Nelson’s book


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Molecular binding and free energy