Assignment 4
Biophys 4322/5322
Tyler Shendruk
February 28, 2012
1
Problem 1) Phillips 7.3
1.1
Part a)
Re-consider dimoglobin utilizing the canonical ensemble (meaning rederive Eq. 31 from Phillips Chapter 7).
For a canonical ensemble
p (Ei ) =
e−Ei /kB T
Z
(1)
P
with the partition function is Z = states e−Ei /kB T .
For our case there are N molecules of O2 and they can reside at any of the
Ω sites. What states can we have? Answer:
1. No O2 bound to the dimoglobin
2. An O2 bound to site A but not site B
3. An O2 bound to site B but not site A
4. An O2 bound each site A and B.
We can think of this in a couple of different ways. Perhaps the easiest way is to
break the total partition function Z into the sum of the partition function due
to each of the above cases:
X
Z=
e−Ei /kB T
all states
= Z1 + Z2 + Z3 + Z4
X
X
X
e−Ei /kB T +
e−Ei /kB T +
=
e−Ei /kB T +
case 1
|
case 2
{z
Ei = 0
∀i
}
|
case 3
{z
Ei = A
∀i
}
|
X
e−Ei /kB T .
case 4
{z
Ei = B
∀i
}
|
{z
Ei = A + B + J
}
∀i
If you wanted to be fancy you could write something like E = σA A + σB B +
σA σB J where σi [0, 1]. More importantly though each of these partition
functions has degeneracy (Phillips calls it multiplicity sometimes). Let’s look
at each:
1
1. For Z1 : There are
Ω!
N ! (Ω − N )!
ways of placing N O2 molecules on Ω sites.
2. For Z2 : There are
Ω!
(N − 1)! (Ω + 1 − N )!
ways of placing one O2 at site A and (N − 1) O2 molecules on Ω sites.
3. Dido for Z3 .
4. For Z4 : There are
Ω!
(N − 2)! (Ω + 2 − N )!
ways of placing one O2 at site A and one at B and the rest on the Ω sites.
Therefore, we can write the partition function as
Z = Z1 + Z2 + Z3 + Z4
X
X
X
X
=
e−0/kB T +
e−A /kB T +
e−B /kB T +
e−(A +B +J)/kB T
case 1
case 2
case 3
case 4
Ω!
Ω!
e0 +
e−A /kB T
=
N ! (Ω − N )!
(N − 1)! (Ω + 1 − N )!
Ω!
Ω!
+
e−B /kB T +
e−(A +B +J)/kB T
(N − 1)! (Ω + 1 − N )!
(N − 2)! (Ω + 2 − N )!
Ω! e−A /kB T + e−B /kB T
Ω!e−(A +B +J)/kB T
Ω!
+
+
.
(2)
=
N ! (Ω − N )!
(N − 1)! (Ω + 1 − N )!
(N − 2)! (Ω + 2 − N )!
Up to this point we’ve been totally general but now we recognize that if we want
to re-create Phillips’ Eq. 7.31 we should let:
= A = B
Ω!
ΩN 1→
≈ ΩN .
(Ω − N )!
(3a)
(3b)
Putting these into the partition function, we get
Z = Z1 + (Z2 + Z3 ) + Z4
Ω! e−A /kB T + e−B /kB T
Ω!
Ω!e−(A +B +J)/kB T
=
+
+
N ! (Ω − N )!
(N − 1)! (Ω + 1 − N )!
(N − 2)! (Ω + 2 − N )!
ΩN
ΩN −1 −/kB T
ΩN −2 −(2+J)/kB T
≈
+2
e
+
e
.
(4)
N!
(N − 1)!
(N − 2)!
2
What we are looking for is the average number bound so we have to add up
the average number just bound to A, just bound to B, and bound to both A
and B i.e.
hN i = hNA i + hNB i + hNAB i
=
=
Z2
Z3
Z4
(Z2 + Z3 ) + 2Z4
+
+2
=
Z
Z
Z
Z
ΩN −1 −/kB T
ΩN −2 −(2+J)/kB T
2 (N −1)! e
+ 2 (N −2)! e
ΩN
N!
N −1
Ω
−/kB T +
+ 2 (N
−1)! e
ΩN −2 −(2+J)/kB T
(N −2)! e
.
To move on from here, let’s multiply the numerator and the denominator by
N !/ΩN to get
N !/ΩN
hN i = 2
N !/ΩN
N
ΩN −2 −(2+J)/kB T
ΩN −1 −/kB T
+ (N
(N −1)! e
−2)! e
ΩN −1 −/kB T
ΩN −2 −(2+J)/kB T
ΩN
+ (N
N ! + 2 (N −1)! e
−2)! e
2
N
−(2+J)/kB T
−/kB T
e
+
Ω
Ω2 e
=2
.
−/kB T + N 2 e−(2+J)/kB T
1+2 N
Ω e
Ω2
(5)
Either we can recognize N/Ω to be the concentration or compare Eq. (5) to Eq.
7.31 from Phillips and see that it is. So the average number bonded is
2
c
c
−/kB T
e−(2+J)/kB T
e
+
c0
c0
hN i = 2
.
(6)
2
c
c
−/k
T
−(2+J)/k
T
B
B
1 + 2 c0 e
+ c0 e
1.2
Part b)
Write expressions for the probabilities p0 , p1 and p2 .
We basically have this from Part a).:
p0 =
Z1
=
Z
ΩN
N!
+
ΩN
N!
ΩN −1 −/kB T
ΩN −2 −(2+J)/kB T
2 (N
e
+ (N
−1)!
−2)! e
1
=
1+2
c
c0
e−/kB T +
2
c
c0
(7)
e−(2+J)/kB T
N −1
Z2 + Z3
p1 =
=
Z
Ω
−/kB T
2 (N
−1)! e
N −1
ΩN
N!
N −2
Ω
−/kB T + Ω
−(2+J)/kB T
+ 2 (N
−1)! e
(N −2)! e
2 cc0 e−/kB T
=
2
1 + 2 cc0 e−/kB T + cc0 e−(2+J)/kB T
(8)
N −2
Z4
p2 =
=
Z
Ω
−(2+J)/kB T
2 (N
−2)! e
ΩN
N!
N −1
N −2
Ω
−/kB T + Ω
−(2+J)/kB T
+ 2 (N
−1)! e
(N −2)! e
2
2 cc0 e−(2+J)/kB T
.
=
2
1 + 2 cc0 e−/kB T + cc0 e−(2+J)/kB T
3
(9)
Now plot them.
2
2.1
Problem 2) Phillips 7.4
Part a)
Find the probabilities of different states of occupance in the MWC
model of dimoglobin. From Phillips pg. 273 Eq. 7.34 and 7.35 we have
Z = 1 + 2e−β(T −µ) + e−β(2T −2µ) + e−β 1 + 2e−β(R −µ) + e−β(2R −2µ)
= 1 + 2x + x2 + 1 + 2y + y 2 e−β
(10a)
2
(10b)
x + x2 + y + y 2 e−β
hN i =
Z
c
x=
e−β(T −µ0 )
(10c)
c0
c
y=
e−β(R −µ0 ) .
(10d)
c0
Phillips implicitly points out that this can be written Z = ZT + ZR with
ZT = 1 + 2x + x2
(10e)
ZR = 1 + 2y + y 2 e−β
(10f)
However, there is another way of looking at it, which I think is better for our
purposes:
Z=
Z0
|{z}
none bound
+
2Z1
|{z}
+
one bound
Z2 .
|{z}
(11)
two bound
A little bit of though allows us to identify the terms in ZT and ZR which make
up Z0 , Z1 and Z2 . When no interaction occurs there is always a corresponding
1 in the partition function; therefore
Z0 = 1 + e−β .
(12a)
If for no other reason than the presence of the “2”, we identify
Z1 = x + ye−β
(12b)
and the rest of the terms must belong to the last part of the partition function
Z2 = x2 + y 2 e−β .
(12c)
You can convince yourself of this in a more rigorous way if you’d like. The nice
thing about breaking the partition function up in this manner is that now the
probabilities are obvious.
1 + e−β
Z0
=
Z
Z
2 x + ye−β
Z1
p1 =
=
Z
Z
Z2
x2 + y 2 e−β
p2 =
=
.
Z
Z
p0 =
4
(13)
(14)
(15)
There’s really only one more qualm we can have about these solutions: if we are
to plot the probabilities we have two variables (x and y) both of which represent
a sort of non-dimensionalized concentration. From the definitions of x and y,
we can get rid of the y
c
x=
e−β(T −µ0 ) = yeβ(R −T )
c0
= yeβ∆
(16)
where we defined ∆ ≡ R − T since only the difference in binding energies
really matters. Now the probability distributions and the partition function are
Z = 1 + 2x + x2 + 1 + 2xe−β∆ + x2 e−2β∆ e−β
(17)
1 + e−β
Z
2x 1 + e−β(+∆)
p1 =
Z
x2 p2 =
1 + e−β(+2∆) .
Z
p0 =
(18)
(19)
(20)
Now just plot these as a function of x for some reasonable values of ∆.
3
Problem 3) Phillips 7.5
3.1
Part a)
Use the Hill equation to calculate the probability that hemoglobin will
be saturated with oxygen and the probability that it will be saturated
with carbon monoxide.
• For O2 :
n
pO2 =
([O2 ] /KO2 ) O2
n
n
1 + ([O2 ] /KO2 ) O2 + ([CO] /KCO ) CO
pCO =
([CO] /KCO ) CO
n
n
1 + ([O2 ] /KO2 ) O2 + ([CO] /KCO ) CO
(21)
• For CO:
n
(22)
The partial pressures are measures of the concentration so for O2 , according
to the question we have 21% which leads to
[O2 ] = 21% × 1atm = 0.21 × 760mmHG
= 160mmHg.
Plus the question gives the disassociation constant to be
KO2 = 26mmHg
5
and the Hill coefficient
nO2 = 3.0
.
Likewise, we are just told the partial pressure of CO is
[CO] = 2mmHg,
the Hill coefficient is
nCO = 1.4
and that the disassociation constant is simply related to KO2 as
KO2
240
= 0.11mmHg.
KCO =
Plug all these numbers into the probabilities (Eq. (21) and Eq. (22)) to find
3.2
pO2 ≈ 80%
(23)
pCO ≈ 20%
(24)
Part b)
Plot pO2 as a function of partial pressure of CO.
Just substitute everything into pO2 except [CO] to find
n
pO2 =
([O2 ] /KO2 ) O2
n
n
1 + ([O2 ] /KO2 ) O2 + ([CO] /KCO ) CO
3
=
=
=
(6.15)
3
1.4
1 + (6.15) + ([CO] /.11)
233.0
1.4
234.0 + 21.981 [CO]
1
1.004291 + 0.094320 [CO]
1.4 .
(25)
Plot this.
3.3
Part c)
What is the condition for equal probability of binding? For what
partial pressure of CO does this occur?
pO2 = pCO
n
nO2
([O2 ] /KO2 )
([CO] /KCO ) CO
nO2
nCO =
n
n
1 + ([O2 ] /KO2 )
1 + ([O2 ] /KO2 ) O2 + ([CO] /KCO ) CO
+ ([CO] /KCO )
n
n
([O2 ] /KO2 ) O2 = ([CO] /KCO ) CO
nO2 nCO
[O2 ]
KCO
= 1.
(26)
KO2
[CO]
6
We can rearrange it again to find
[CO] = KCO
[O2 ]
KO2
nO2 /nCO
,
(27)
which if we plug our values into gives
[CO] = 5.4mmHg.
4
(28)
Problem 4) Harden
4.1
Part a)
Derive an expression for the mole fraction xm of m-mers as a function
of m, x1 and α.
In case it’s not clear, we have particles that can bind together to form a
linear chain. The chain length is m.
• The (mole) fraction of free, single monomers (m = 1) is x1 .
• A dimer is m = 2 with fraction x2 .
• A trimer is m = 3 with fraction x3
The question is asking “What is the (mole) fraction that are some specific length
m?”
We could write the free energy (grand potential) or we can just grab the
answer from pg 242 of Phillips:
c
µi = E + kB T ln
.
(29)
c0
For this question there is no reason that the chemical potential should be any
different when adding the mth monomer than it was when adding the (m − 1)th .
So µi = µ ∀i. To be super clear, µ s the chemical potential of each monomer.
Also we can related the concentrations and the mole fractions c/c0 = xm /m
which means
x m
mµ = mm + kB T ln
m
mµ − mm
xm = m exp
kB T
mµ − mb − αkB T
= m exp
(30)
kB T
but the question asks for xm (x1 , α) and so far all that we have provided is
xm (µ, b , α). The key difference is x1 . Notice
µ − b − αkB T
x1 = xm=1 = exp
kB T
µ − b
= exp
exp (−α) ,
(31)
kB T
7
which (with a little rearrangement) we can see inside of xm .
mµ − mb − αkB T
xm = m exp
kB T
µ − b
= m exp m
exp (−α)
kB T
m
µ − b
= m exp
exp (−α) exp ([m − 1] α)
kB T
(m−1)α
= mxm
1 e
4.2
(32)
Part b)
Determine the critical aggregation concentation (CAC).
The CAC occurs when the mole fraction of monomers (m = 1) is equal to
the mole fraction of the monomers in any randomly chosen m. i.e.
xc = x1 =
xm
m
α(m−1)
= xm
1 e
α(m−1)
= xm
c e
x1−m
= e−α(1−m)
c
xc = e−α
4.3
(33)
Part c.i)
Show that above CAC the equilibrium distribution is
m
1
e−α
xm = m 1 − √
xB eα
(34)
where xB is the total monomer mole fraction in the solution.
Compare this to Eq. (32)
(m−1)α
xm = mxm
1 e
m
= m (x1 eα ) e−α .
Really all we have to demonstrate is
1− √
1
?
= x1 eα .
xB eα
(35)
To do this let’s consider this new fraction that was introduced:
xB = total monomer fraction
∞
∞
X
X
α(m−1)
=
xm =
mxm
1 e
m=1
= e−α
m=1
∞
X
m (x1 eα )
m
m=1
= xc
∞
X
m=1
8
m
m (x1 eα ) .
That series is similar to a geometric series:
n
X
kz k =
1 − (n + 1) z n + nz n+1
2
(1 − z)
k=1
.
(36)
Clearly there’s a problem here. This goes to n whereas our series goes to the
limit n → ∞. Let’s leave it as n for a second and then we’ll take the limit. So
the bulk mole fraction is
#
"
n=∞
α n
α n+1
X
1
−
(n
+
1)
(x
e
)
+
n
(x
e
)
1
1
m
xB = xc
m (x1 eα ) = xc
.
2
(1 − (x1 eα ))
m=1
Now we see why we had to be above the critical aggregation concentration.
n
If x1 xc = e−α then x1 eα < 1 and so limn→∞ (x1 eα ) → 0, which would
meaning
#
"
1−0+0
lim xB ≈ xc
2
n→∞
(1 − (x1 eα ))
xc
(37)
=
2.
(1 − (x1 eα ))
And this is all that we had to demonstrate. In case, you can’t see it by eye from
comparing Eq. (35) to Eq. (37), I’ll go a bit further (recall xc = e−α )
xB =
xc
2
(1 − x1 eα )
1
xB
=
2
xc
(1 − x1 eα ))
r
xB
1=
(1 − x1 eα )
xc
r
r
xB
xB
x1 eα
=
−1
xc
xc
1
x1 eα = 1 − p
xB /xc
1
=1− √
.
xB eα
4.4
Part c.ii)
What m = mmax gives the maximum in xm ?
Let’s use the form we just verified.
∂xm
=0
∂m
"
!m
#
∂
1
−α
=
m 1− p
e
∂m
xB /xc
!m
!m
!
1
1
1
0= 1− p
+m 1− p
ln 1 − p
xB /xc
xB /xc
xB /xc
9
(38)
So then
!mmax
1
1− p
xB /xc
= −mmax
!mmax
1
1
mmax = ln 1 − p
xB /xc
r
xc
≈
xB
4.5
!
ln 1 − p
xB /xc
1− p
xB /xc
!
1
1 = −mmax ln 1 − p
1
xB /xc
!
(39)
(40)
Part d)
Calculate the mean and standard deviation.
The mean is defined as
hmi =
∞
X
m=1
mxm =
∞
X
α(m−1)
m2 xm
= e−α
1 e
m=1
∞
X
m
m2 (x1 eα ) .
(41)
m=1
Again we have a series like the geometric series. This one has the form
n
2
X
1 + z − (n + 1) z n + 2n2 + 2n − 1 z n+1 − n2 z n+2
2 k
k z =
.
3
(1 − z)
k=1
(42)
Obviously let’s keep assuming that we’re above the critical aggregation concentration so that just like with the last one like this, we can say
hmi = xc
∞
X
m
m2 (x1 eα )
m=1
≈ xc
1 + x1 eα
(1 − x1 eα )
(43)
3
To find the standard deviation we need
∞
∞
X
2 X
m
m =
m 2 xm = xc
m3 (x1 eα ) .
m=1
(44)
m=1
The same sort of series with the same sort of less than 1 condition, resulting in
∞
X
2
m
m = xc
m3 (x1 eα )
m=1
≈ xc
1 + 4x1 eα + (x1 eα )
(1 − x1 eα )
4
2
2
σm
= m2 − hmi .
10
2
(45)
(46)