BIOLOGICAL PHYSICS MAJOR OPTION 2015 - **MOCK PAPER**
Rubric: Answer two questions only. The approximate number of marks allocated to each
part of a question is indicated in the right-hand margin where appropriate. The paper
contains 3 sides including this one and is accompanied by a book giving values of
constants and containing mathematical formulae which you may quote without proof.
You should use a separate Answer Book for each question.
1
A first approach to modelling protein production is through an ODE for the protein
concentration p:
dp
= α − γp.
dt
(a) This kind of modeling is making three underlying assumptions:
(1)
1.The concentrations of the reactants evolve continuously and differentiably so
that description by differential equations is justified.
2.Reactions occur instantaneously and depend only upon the present state of the
system.
3.Reactants are distributed homogeneously in space.
Consider these assumptions in turn, and in light of what you know about
timescales of different processes explain if/why they may be valid for protein
expression in a bacteria cell. Can you give other examples of cell processes where
these might not hold?
Answer. Considerations on: (1) we can model as a continuous number if the
numbers are large, i.e. we don’t care about individual reaction events. As a rule of
thumb, this will be invalid for N < 100. (2) the ‘rate’ has no memory of the
previous states of the system. (3) if diffusion times are much shorter than reaction
times, then this should be valid. Otherwise meed to make models with explicit
space, and they will be partial differential equations.
We know that there are one or two orders of magnitude difference between
each of: diffusion time for small molecules; diffusion times of TF to promoter
sites; RNAP transcription. This makes assumption (3) valid for much transcription
activity. (1,2) not necessarily well satisfied. We have discussed that protein
expression is seen to happen in bursts, which is a violation of (2), most likely
related to residence times (an example of system memory) of repressors or RNAP.
(b) Consider the following expression for the probability of RNAP to be bound to
a promoter region:
[9]
2
c
c0
exp −β∆
,
1 + cc0 exp −β∆
pbound =
(2)
where c/c0 is the concentration of RNAP (relative to some standard value, c0 ) and
∆ is the change in energy when the RNAP binds.
Explain how this expression is obtained, and why is it acceptable to consider
the equilibrium RNAP occupancy as proportional to the average amount of
transcription of that gene.
Answer. Bookwork.
L ligands. Prob that 1 ligand is bound to receptor:
X
weight when receptor occupied = e−βb ×
e−β(L−1)sol ,
solution
where the summation is the sum over all ways of arranging the L − 1 ligands in
solution. Imagine Ω ‘lattice sites’ in solution. Then
X
Ω!
.
e−β(L−1)sol =
(L
−
1)![Ω
−
(L
−
1)]!
solution
The partition function is
X
Z(L, Ω) =
e−βLsol + e−βb
P
solution
e−β(L−1)sol .
solution
The sum in the second term has already been evaluated. The first term is
X
Ω!
e−βLsol = e−βLsol
.
L!(Ω − L)!
solution
Bringing both together,
Z(L, Ω) = e−βLsol
Ω!
Ω!
+ e−βb e−β(L−1)sol
.
L!(Ω − L)!
(L − 1)![Ω − (L − 1)]!
If we simplify considering
Ω!
' ΩL ,
L!(Ω − L)!
which is ok provided Ω >> L, then we can write the probability of being bound as:
Ω
e−βb (L−1)!
e−β(L−1)sol
L−1
pbound =
ΩL −βL sol
e
L!
Ω
+ e−βb (L−1)!
e−β(L−1)sol
L−1
Now defining ∆ = b − sol , we can simplify to:
pbound =
(L/Ω)e−β∆
,
1 + (L/Ω)e−β∆
.
[5]
3
which we can write in terms of a concentration c:
pbound =
(c/c0 )e−β∆
,
1 + (c/c0 )e−β∆
where c0 is a reference state of full occupation. For example, if we assume our
molecules to be of volume 1 nm3 , then c0 = 0.6M. It is always assumed that once
RNAP is bound, it will start and complete transcription. We are also assuming that
that binding/unbinding is a very fast variable. Then the equilibrium probability
gives us the fraction of time that the RNAP is active, and hence it appears as a
produce with the ‘unregulated’ transcription rate as the overall rate of production
of mRNA.
(c) If molecule numbers in the cell are small, we can only model concentration
probability distributions. A Master equation corresponding to this mRNA
production process is:
dP(m, t)
= αm P(m − 1, t) + βm (m + 1)P(m + 1, t) − βm mP(m, t) + αm P(m, t) ,
dt
where P(m, t) is the probability of finding m mRNA molecules at time t. Explain
the meaning of the terms in this expression.
Answer. The changes in state m are the 4 terms corresponding to gain (from
loss at state at m+1 and synthesis m-1) and loss (by synthesis or degradation).
[2]
(d) Show that from this Master equation, using a discrete Laplace transform
P
m
F(z, t) = ∞
m=0 z P(m, t) you can obtain the much simpler partial differential
equation:
∂F(z, t)
∂F(z, t)
= αm (z − 1)F(z, t) − βm (z − 1)
.
∂t
∂z
(3)
[2]
Answer. Should be a simple substitution and check.
(e) We are often interested in the mean and variance of the steady state
distribution, which can be calculated for this case. First explain why
F(z = 1, t) = 1. Then verify that
∂F(z, t) hm(t)i =
.
∂z z=1
Similarly, obtain an expression for hm2 (t)i − hm(t)i from F. Finally, considering
the steady state of equation 3, obtain expressions for the mean and the variance. Is
this consistent with a Poisson process?
Answer. F(z = 1, t) = 1 is the normalisation of P.
The second is a check.
(TURN OVER
[9]
4
Then can spot that:
∞
∂2 F(z, t) X
m−2
= hm2 (t)i − hm(t)i.
m(m
−
1)z
P(m,
t)
=
z=1
∂z2
m=0
The steady state ∂F/∂t = 0 solution is:
#
αm
F (z) = exp
(z − 1) .
βm
"
ss
Using the expressions above, at steady state:
hm(t)i = αm /βm
and hm2 (t)i − hm(t)i = (αm /βm )2 so variance σ2 = αm /βm . Mean = Variance,
ok with Poisson.
(f) What sources of noise is the model above trying to capture? What other
sources of noise do you know of, which would not lead to Poisson distributions?
Answer. Approach above is trying to catch purely the stochastic fluctuations of
number of molecules, which we called ‘intrinsic biochemical noise’. Other
sources of noise, for example variations of the number of active molecules
(ribosomes), or variations caused by external factors, not accounted for here. One
example seen in lectures is that protein expression occurs in bursts, which is
against the Poisson probability of reaction assumed here.
[3]
2
In the context of cell electrophysiology explain the significance of the Nernst
equation given by
!
ci
kB T
ln
.
νNernst = −
q
co
[2]
The Nernst equation gives the electrical potential, νNernst , that arises across the
cell membrane when a permeant ion species reaches equilibrium. This potential is
related to the ratio of concentrations of ions as follows
!
kB T
ci
ln
.
νNernst = −
q
co
where kB T is the thermal energy, q the ionic charge, co the concentration of ions outside
the cell and ci the concentration of ions inside the cell.
The table above lists the physiological ion concentrations of frog muscle cells at T
= 20 C which typically have a membrane potential of -90 mV. Calculate the Nernst
potential for each ion type and comment on your results. Are these cells in Donnan
Equilibrium?
◦
[4]
5
Donnan equilibrium concerns systems which contain multiple ionic species
diffusing across a semi-permeable membrane. In thermal equilibrium the Nernst
potentials of all diffusing ionic species are the same and equal to the membrane
potential; this situation is referred to as Donnan equilibrium. These cells are not in
Donnan Equilibrium which is consistent with what we know about living cells.
Describe how ion pumps maintain the membrane potential of axons at V0 = −60
mV, a voltage considerable more negative that the value that would be obtained in
thermal equilibrium.
Calculations using physiological ionic concentration differences and resting
membrane potentials reveal that cells are not in Donnan equilibrium e.g. in the case of
sodium the Nernst potential has the opposite sign to that of the membrane potential (as
for the cells in the example above) indicating that a high relative concentration of sodium
outside the cell is maintained against an unfavourable chemical gradient and electrical
potential. Ionic species are maintained out of the equilibrium by ion pumps which
actively pump ions against electrochemical gradients. The membrane potential is largely
established by the imbalances in the ion concentrations of sodium and potassium. These
are maintained by the sodium-potassium pump which carries out coupled transport of
sodium and potassium ions. It pumps out three Na+ ions from the cell and then pumps
two K+ ions into the cell to complete one cycle resulting in a net transport of one unit of
charge. This process is driven by ATP hydrolysis which releases approximately 20 kB T
worth of free energy. The free energy cost for pumping a sodium ion out is given by
−e(∆V − νNernstNa ) and for pumping potassium ion in −e(∆V − νNernstK ) where ∆V is the
membrane potential. As sodium is far from Nernst equilibrium the bulk of the energy is
spent pumping the three sodium ions. The free energy required for the entire process is
roughly 15-18 kB T for biological cells which suggests the pump runs with high
efficiency.
Outline the derivation of the Cable Equation that governs the evolution of the
(TURN OVER
[4]
6
membrane potential V and the ionic current j through the membrane in the axon of a
nerve cell,
!
2
∂V
∂2 V
=
j+C
,
∂x2
κa
∂t
where a is the radius of the axon, κ is the ionic conductance within the axon, and C is the
membrane capacitance per unit area.
Setting up the problem:
A patch of membrane dA can be modeled as an RC electric circuit (one segment of
circuit in Fig), where R represents the resistance of ion channels (all species) to the
passive ion permeation through the membrane, C is the capacitance of the membrane and
ν0 is the Nernst potential (from chord conductance formula). The axon can be modelled
as a chain of these modules each representing a cylindrical slice of membrane.
Neighbouring elements are connected through axial resistors dR x which represent the
resistive effects of the fluid inside the axon to the flow of ions. In comparison we assume
that there is little resistance to flow in the extracellular fluid such that the entire lower
wire is at a common potential. The currents IR (x) and I x (x) represent the net charge of all
ions leaving the axon interior (axoplasm) and the total current flowing to the right in
axoplasm.
Derivation:
The axial resistance dR x of a cylinder of fluid to axial current flow is given by its
dx
length divided by cross sectional area, dR x = (κπa
2 ) where κ is the fluids electrical
conductivity and a the radius of the axion. Ohm’s law gives
dV =
dxI x
πa2 κ
(4)
The radial current is the sum of charge flowing through the membrane : j2πadx (flux x
surface area axial segment) and a capacitive current : 2πadxC dV
where C is the
dt
membrane capacitance per unit area. Applying Kirchoff’s law at marked upper current
junction gives
dV
dI x = 2πa( j + C )dx
(5)
dt
Combining eqns (4) and (5) gives the cable equation
!
d2 V
2
dV
=
j+C
(6)
dx2
κa
dt
Let v be the membrane potential relative to the resting potential V0 . Show that for
Ohmic conductance of the channels in the membrane, such that the current through the
membrane varies as j = gv, the cable equation has the form
λ2
∂2 v
∂v
−τ =v
2
∂x
∂t
[4]
7
and find expressions for the time scale τ and length scale λ. In an idealised infinitely
long axon a small ion flux je is continuously injected locally at x = 0. Find the
steady-state solution to the cable equation in this case.
[5]
Answer: Assuming an Ohmic conductance and letting v(x, t) be the difference
between the interior potential and the resting potential V(x, t) − V0 gives j = vg.
Substituting into eqn (6) gives the linear cable equation
(λaxon )2
d2 v
dv
−τ =v
2
dx
dt
(7)
p
where λaxon = aκ/2g and τ = C/g are the axon’s space and time constants respectively.
The second part involves solving the linear cable equation for the steady state case
= 0 thus
where ∂v
∂t
∂2 v
λ2 2 = v,
∂x
subject to the boundary conditions v = 0 at x = ±∞ and v = v0 at x = 0 where v0 is the
voltage about the resting membrane voltage that the injected ion flux maintains. Define
X = x/λ which gives dX/dx = 1/λ and thus λ2 = (dx/dX)2 and 8 becomes
∂2 v
− v = 0,
∂X 2
which for the above boundary conditions has solutions v = v0 exp(− | x | /λ).
For an instantaneous pulse of ion flux injected at point x = 0 and t = 0, the
response is given by
!
x2 τ
−1/2
v∝t
exp(−t/τ) exp − 2 .
4λ t
Sketch both this and the steady ion flux injection solutions as a function of x. Comment
on this model in the context of action potential propagation.
[3]
A constant current injected at point x = 0 sets up a membrane perturbation which
decays exponentially in space as shown in Figure 1 but is constant over time. A current
impulse injected at point x = 0 and t = 0 also decays over a distance comparable to λaxon
and delocalises over a time scale comparable to the axon time constant τ as shown in
Figure 2. Neither of these solutions are travelling waves and indeed the the linear cable
equation does not permit travelling wave solutions and as such can’t describe action
potential propagation. To obtain travelling wave solutions rather than assuming Ohmic
conductance through the ion channels we must have voltage dependent ion channel
conductances.
Describe an experimental technique which has provided information about the
behaviour of individual sodium and potassium channels in the membrane.
(TURN OVER
[4]
8
1
v/v0
0.8
0.6
0.4
0.2
0
ï5
ï4
ï3
ï2
ï1
0
x/h
1
2
3
4
5
Figure 1: Linear Cable equation solution for constant current injection at x = 0.
In the patch clamp technique a small patch of membrane containing only a single
(or a few) voltage-gated ion channel is electrically isolated from the rest of the cell using
a patch electrode. Isolation is achieved by sealing a glass micropipette (1 um diameter)
onto the cell surface with suction, the pipette is filled with a bath solution matched to the
extracellular fluid of the cell of interest and an electrode is inserted into the pipette to
conduct electric current to an amplifier (pA currents). The current entering the cell
through these channels is recorded by a measurement device connected to the patch
electrode. In the voltage clamp mode a current is injected into the cell to maintain the
membrane potential at a command value. This technique represents a major
advancement in the investigation of single ion channel behaviour and allowed
measurements of conductances of individual ion channels. To measure channel
conductances channels were kept open by the application of chemicals, the current
through the channel was recorded as a function of command voltage. The slope of the
resultant graph gives the channel conductance. These measurements proved the
existence of ion specific ion channels e.g. the sodium channel which has a conductance
for sodium that is nearly 10 times that of other similar cations and a potassium channel
which admits potassium 50 times more readily that sodium. Patch clamp recordings also
confirmed the voltage gating hypothesis which was predicted by the first measurements
on action potentials by Hodgkin and Huxley. A depolarizing voltage step was applied to
the cell and the resultant current recorded on the patch clamp apparatus. The current
traces revealed individual sodium channels opening independently in response to the
depolarizing stimulus.
Explain qualitatively how the observed response of the channels to changes in the
local membrane potential leads to the propagation of an action potential along an axon.
[4]
9
0.35
0.3
v (arbitrary units)
t=o
0.25
0.2
0.15
0.1
t = 2o
0.05
t = 3o
0
ï5
ï4
ï3
ï2
ï1
0
axon
x/h
1
2
3
4
5
Figure 2: Linear Cable equation solution for a current impulse at x = 0 and t = 0.
Channels are observed to be ion specific and sodium channels are shown to be
voltage gated. Above-threshold depolarizing stimuli applied to an axon results in
traveling wave of excitation whose peak is independent of the strength of the initial
stimulus. During this nerve impulse the membrane potential temporarily reverses sign; it
goes from a resting potential which is comparable to the Potassium Nernst potential to a
value which is comparable to the Sodium Nernst potential. The action potential moves
down the axon at a constant speed with the peak potential independent of distance. After
the passage of the action potential the membrane potential overshoots becoming slightly
more negative than the resting potential.
Let’s consider what happens when the inside of the cell is made slightly less
negative corresponding to a depolarizing stimulus. Here we’ll look at the flow of
Potassium and Sodium ions only. The total charge that flows across the membrane can
be calculated
∆Q
= −(IK + + INa+ )
∆t
The Ohmic hypothesis gives us that I = g(∆V − νNernst ) therefore
∆Q
K+
Na+
= gK + (νNernst
− ∆V) + gNA+ (νNernst
− ∆V)
∆t
In the steady state there’s no charge flowing across the membrane and the membrane
potential is given by
Na+
K+
gK + νNernst
+ gNa+ νNernst
∆V =
gK + + gNa+
which shows the membrane species with the highest conductance sets the membrane
potential approximately equal to it’s Nernst potential. The propagation of the action
potential can thus be explained as follows:
(TURN OVER
10
1) Initially sodium conductance is much smaller than potassium conductance and
the membrane potential is close to the Nernst value for potassium.
2) An above threshold depolarizing stimulus causes the voltage-gated sodium
channels to open which dramatically increases sodium conductance above potassium
conductance (assumed to be fixed) and brings the membrane potential close to the Nernst
value for Sodium explaining the polarization reversal seen in action potentials.
3) This sharp increase in in the membrane potential of a patch raise the membrane
potential of the nearby patch above threshold leading to its depolarization which in turn
depolarizes the next patch and so forth giving rise to an action potential.
4) To fully account for the propagation of action potentials we must include
inactivation of sodium channels -it’s assumed sodium channels can go into an inactive
state and do so at a constant rate.
Write brief notes on two of the following topics:
(a) Physical models of transport mechanisms in biological cells (including both
passive and active mechanisms);
(b) Physical models to describe the lac operon;
(c) Networks of chemical reactions in a biological cell.
Answers do not require long prose, and can include equations, diagrams, etc.
Remember to define the terms in the subjects, first. Try not to just “brain dump”
everything you know; keep a logic in your exposition.
For example on (a). Can address first Passive versus Active transport.
- Passive transport = diffusion (no energy source required), active requires energy
source. (1)
- Diffusion has its limits (time taken for diffusion particle to travel a distance
grows with the square of the distance), still dominant form of transport at the sub-micron
level. (1)
- Active transport is carried out by molecular motors which are powered by ATP
(e.g. Kinesin) or ion gradients (e.g. flagellum). (1)
3
Then give some facts on models of transport.
- (Passive) : Diffusion can be modelled by using diffusion equation (explain
quantities and general behaviour predicted). (2)
- (Active) : Description of Rectification of Brownian motion (Brownian Ratchets)
- need for an energy source to do useful work. (2)
- Effects of introducing a load, mention Smoluchowski eqn. (2)
- Diagram of free energy landscape to support explanations. (2)
- Gibbs free energy - Chemical reactions - Cyclic ATP phosphorylation - need for
irreversible step. (2)
- Mechano-chemical coupling - proteins can change shape on binding other
molecules, need for asymmetrical binding potential. (1)
- Two headed Kinesin : coordination of motion between heads to produce
[15]
[15]
[15]
11
processive motion, tight coupling, powerstroke. (1)