Download 2 Applications I

Applications I 1
2
Applications I
Diverse Applications of the Canonical Distribution
The canonical distribution was presented as a central tool of statistical mechanics.
We now use this distribution for some practical applications in order to
consolidate before we develop additional theory and techniques.
Ion Channels
Membranes are an integral part of all living cells. They act as gatekeepers
allowing some molecules to pass while others are rebuffed. Ion channels are pores in
membranes that may be open or closed to specific ions. Consider an ion channel that can
be in only one of two states, open or closed, with respective energies o and c. Applying
the canonical distribution (Eq.16 of chapter 1), the probability po of the channel being
open can be written as
1
(1)
po 
1  exp    
where   c  o . Assuming this is positive, a graph of po verses   takes the shape
shown in the figure. Note that only the difference in energies is relevant to the result.
1
1.
Derive Eq.(1) for the two-state ion channel.
0.5
Barometric Pressure
0
0


5
5
Barometiric altimeters use atmospheric pressure to determine altitude A relation
between altitude and pressure is built into the altimeter’s scale. We can derive this
famous relation using the canonical distribution. Make the approximation that
temperature is constant from the ground to a height y above ground. Consider an
atmospheric molecule of mass m at altitude y with potential energy mgy . Let py and p0
represent the respective probabilities of finding the molecule at elevations y and 0. We are
interested in the ratio p y / p0 and no calculation of Z needs to be considered. Probabilities
are proportional to densities and for ideal gases these are proportional to the
corresponding pressures Py and P0:
p y Py

p0 P0
Applications I 2
2.
Under the assumptions of this section, use the canonical distribution (do
not evaluate Z) to derive the barometric equation,
  Mgy 
Py  P0 exp 

 RT 
where M is the mass of a mole of gas, M  N Am .
3.
Consider a 290 K atmosphere composed of nitrogen molecules with a
molecular weight M of 28. Take the sea level pressure to be 1 ATM.
Calculate the pressure at 1.609 km (1 mile) above. [ans 0.83 ATM]
Useful Manipulations
Applications often require manipulations of factorials. In particular, dividing the
(binomial coefficient) expressions,
N
CL 
N!
by
N  L ! L!
N
CL1 
N!
N  L  1! L  1!
is readily seen to give N  L  1 / L . When N  L an approximation is warranted,
N  L 1 N
 .
L
L
4.
Repeat the treatment above to show that when N  L ,
N
CL
N

N
C L1 L
These manipulations will be useful in the next two problems.
(2)
Ligand-Receptor
Ligands are any atoms, molecules, or ions that can bind with some receptor. For
example, oxygen binding to hemoglobin or metal ions binding to an electrode. Picture
local space to be divided into N cells. Let L ligands be randomly distributed in these cells.
The figure shows the two kinds of energy states, free and bound. The dots represent
ligands. The respective numbers of unbound and bound microstates is




N!
N!

 and 

 N  L ! L! 
 N  L  1! L  1! 
N boxes and L dots
Each possible arrangement
has energy L
N boxes and L –1 dots
Each possible arrangement
has energy (L-1) + B

Applications I 3
An outline of the bound probability, pB, calculation is then
pB 
number of bound microstate s  exp  L  1   B 
Z
with
Z  number of bound microstate s  exp  L  1  B  number of free microstate s  exp  L
5.
(a) Construct the bound state probability following the outline above. Use
the approximations of Eq.(2) and show that the result can be reduced to
the form
1
pB 
N
1    exp   
L
with   B   (usually negative).
(b) Introduce ligand concentration c  L /V and “standard concentration”
c0  N / V (usually 1 M) into the ratio (N/L) above.
Gene Expression
Gene expression begins when a protein, RNA polymerase, binds to a specific site
(the promoter) on the long DNA molecule (the gene). The gene can be modeled as a long
chain of N boxes each of which can bind one of L molecules of RNA polymerase. Only
the promoter site, however, will result in gene expression.
L
RNA polymerase
N sites
6.
promoter site
The probability of binding the promoter site and therefore causing gene
expression is found in much the same way as the ligand-receptor analysis.
Show that
1
probabilit y of gene expression 
N
1    exp   
L
with    B   in obvious notation.