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Lecture Notes on Physical Biochemistry - Spring 2005 | CHEM 156, Study notes of Chemistry

Material Type: Notes; Class: Physical Biochemistry; Subject: Chemistry; University: University of California - Los Angeles; Term: Spring 2005;

Typology: Study notes

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Uploaded on 08/30/2009

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Chem 156
Spring 2005
Prof. Hubbell
Summary of Lecture 16
1. Proteins with non-uniform charge distribution: implications for function Highly non-
uniform charge distributions in some proteins have apparently evolved for specific
functional purposes. Superoxide dismutase is an example where a positive electrostatic
potential around the active site serves to bias the diffusion of the negative substrate
super oxide to enhance the catalytic rate.
2. Electrostatics of planar surfaces (membranes): the Gouy-Chapman model (see
handout of the same title for details).
The solution of the Poisson-Boltzmann equation for a planar surface is given in two
parts, the potential at x = 0 (the surface potential, Φo) and the distance dependence:
valid for T = 298K, water as a solvent, 1:1 electrolyte.
=Φ
σ
sinh
0
C
Z
o
6.1360514. 1
Φo is in volts, σ is the surface charge density in charges/A2 (with correct sign), C is the
molar concentration of electrolyte, and Z is the valence of the electrolyte (absolute
value). The inverse hyperbolic sine is defined as
]
lnsinh ++ x
()
[
1
21 =
xx
The distance dependence from the surface is
()
x
oex
κ
Φ=
Φ
3. The Hydrophobic effect (see handout of the same title for details). For the phase
partition equilibrium of a non-polar solute S between a non-polar solvent and water:
()
X=KRT
X
RT
np
w
o
np
o
wlnln =
µµ
or, if the solute is a liquid and the equilibrium is between pure liquid S and water
(
)
KRT
oo lnln =Χ=
µµ
RT
wnpw
When the Henry’s law SS is chosen,
(
)
o
np
o
w
µµ
, the “free energy of transfer” , is just the
difference in free energy of the solute S with the two solvents. This is the quantity
needed to understand the insolubility of non-polar solutes in water.
pf2

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Chem 156 Spring 2005 Prof. Hubbell

Summary of Lecture 16

  1. Proteins with non-uniform charge distribution: implications for function Highly non- uniform charge distributions in some proteins have apparently evolved for specific functional purposes. Superoxide dismutase is an example where a positive electrostatic potential around the active site serves to bias the diffusion of the negative substrate super oxide to enhance the catalytic rate.
  2. Electrostatics of planar surfaces (membranes): the Gouy-Chapman model (see handout of the same title for details). The solution of the Poisson-Boltzmann equation for a planar surface is given in two parts, the potential at x = 0 (the surface potential, Φo) and the distance dependence:

Φ =^0 sinh  σ^ valid for T = 298K, water as a solvent, 1:1 electrolyte.

Z  C

o

Φ o is in volts, σ is the surface charge density in charges/A^2 (with correct sign), C is the molar concentration of electrolyte, and Z is the valence of the electrolyte (absolute value). The inverse hyperbolic sine is defined as

sinh −^1 ( ) x =ln[ x + x^2 + 1 ]

The distance dependence from the surface is

Φ ( ) x =Φ o e −^ κ x

  1. The Hydrophobic effect (see handout of the same title for details). For the phase partition equilibrium of a non-polar solute S between a non-polar solvent and water:

X

= RT K

X

RT

np

o w np

o

μ w −μ =− ln − ln

or, if the solute is a liquid and the equilibrium is between pure liquid S and water

(μ ow^ −μ npo ) =− RT ln Χ w =− RT ln K

When the Henry’s law SS is chosen, ( μ ow − μ onp ), the “free energy of transfer” , is just the

difference in free energy of the solute S with the two solvents. This is the quantity needed to understand the insolubility of non-polar solutes in water.

  1. Experimental values of the free energy of transfer

Experimental values of K as a function of chain length for the homologous series of alkanes, alkanols and fatty acids show that

884 CH 2 2102 CH 3

o np

o

μ w −μ = N + N alkanes

CH CH OH

o np

o

μ w −μ = 821 N 2 + 2039 N 3 − 2051 N alkanols

CH CH COOH

o np

o

μ w −μ = 825 N 2 + 2043 N 3 − 4718 N fatty acids

Note that the free energy contributions from –CH 3 and –CH 2 - groups are positive and essentially constant. The contributions of the polar groups are negative, and favor the transfer to water.