Transcript Diapositiva 1

Dottorato in Scienze Biomediche (DRSBM)
Analysis and interpretation of
single channel records
Mauro Toselli
University of Pavia, Italy
http://www-3.unipv.it/tslmra22/
What kind of information is
contained in a single channel current
record?
The information in a single channel record is contained in:
 the amplitudes of the openings,
closed
open
1 pA
 the durations of the open and shut periods,
C1
C2
C3
C4
C5
C6
C7 C8
C9
closed
open
O1
O2
O3
O4
O5
O6
O7
A 8 O9
O10
20 ms
 and in the order in which the openings and shuttings occur. For instance:

 C2 

 O
C1 


or

 O 

 C2
C1 


Various aspects of single channel behavior can be predicted, on the basis of a
postulated kinetic mechanism for the channel.
This will allow the kinetic mechanism to be tested, by comparing the predicted behaviour
with that observed in experiments.
How does a ionic channel as a single molecule “behave?”
This kind of question is mostly approached experimentally by observing the collective
behavior of a very large number of identical molecules.
1st Example
The decay of an endplate current (EPC) at the neuromuscular junction provides an
example of a measurement of the rate of a macroscopic process, i.e. a measurement from a
large numbers of molecules.
0
Why a decay?
PSC (nA)
-1
The decay of the EPC occurs by a sudden reduction
to zero of the ligand concentration.
-2
-3
The new equilibrium condition will be approached
with a single exponential time course
-4
-5
0.00
0.02
0.04
Time (s)
0.06
Endplate current evoked by nerve
stimulation at -130 mV. The observed
current is shown by the filled circles
0
The continuous line is a fitted single
exponential curve with a time constant tM.
PSC (nA)
-1
-2
-3
-4
-5
(Modified from Colquhoun & Sheridan (1981).
Single exponential
fitting of EPC
decay with time
constant tM=7.8
ms
0.00
0.02
0.04
Time (s)
This feature can be represented by the following
reaction scheme:
k12

 S
O
k
0.06
1
21
2
The rate of this process can be expressed as an observed time constant, tM (M stays for
Macroscopic). It can be demonstrated (see Appendix A) that these values are given by:
tM = 1/(k12 + k21)
But this is a lucky situation because if the concentration of neurotransmitter is zero, the
channels will not open any more, so that k21→0, and therefore tM≈1/k12 ……
Generally this is not the case and tM depends on both rate constants k12 and k21, as shown in
the next example.
2nd Example
Here the voltage-gated current activation provides another example of a measurement of
the rate of a macroscopic process, i.e. a measurement from a large numbers of molecules.
The activation of the V-G current occurs by a
depolarizing voltage step DV.
DV
1400
The new equilibrium condition will be approached
with a single exponential time course.
1200
Current (pA)
1000
Single exponential
fitting of current
activation with
time constant
tM=19 ms
800
600
400
200
This feature can be rapresented by the following
reaction scheme:
k12

 O
S
k
0
0.0
0.1
0.2
Time (s)
0.3
0.4
1
21
2
Also the rate of this process can be expressed as an observed time constant, tM. It can be
demonstrated (see Appendix A) that these values are given by:
tM = 1/(k12 + k21)
In conclusion
Generally, the value of the macroscopic time constant tM depends on both of the reaction
transition rates.
Therefore the only measurement of tM does not allow to obtain separate values for k12 and k21 (*)
This is because we are observing the average behavior of many molecules.
We shall see that if we could directly observe single macromolecules functioning, we could
derive more information about their mechanisms — about how they work — than we can
from the macroscopic, ensemble-averaged experiments.
SINGLE MOLECULES IN ACTION
Ionic channels are membrane proteins that undergo a conformational change from a “shut"
to an "open" state, and it is experimentally possible to observe these two conformations
directly at the single-molecule level.
shut
open
The question we want to wrestle with here is:
What kind of behavior should we expect for a protein undergoing conformational
changes between two such states:
k12

 O
S
k
1
21
2
where O represents the open channel and S the shut.

 O
S
1
2
Suppose that the channel is in the shut conformation, but at some moment (call it zero
time) the channel opens, i.e., enters state 1.
t0
shut (2)
Open (1)
Dtopen
We ask:
•
•
?
How long does it stay there (state 1) before changing to shut (state 2) again?
What is the "lifetime" of the open state?
In order to get a good idea of the channel's open-state lifetime, we will have to do this
measurement many times because the single-molecule behavior is intrinsically statistical
i.e., channel lifetime is a so called random variable
The light-bulb analogy
We have bought a lot of 1000 bulbs – for each of them it will be measured the
“lifetime” (i.e, the time it stays on before burning out).
Sensible questions:
• What is the average lifetime of the bulb?
• How much variation around the average does this lot of bulbs give?
By contrast, the question: “what is the lifetime of the light bulb?" is not so meaningful (*),
because the answer is intrinsically statistical:
What it is really important to know is the statistical distribution of lifetimes.
i.e., what fraction of the bulbs have lifetimes between 0 and 100 hr, between 100-200 hr, etc.
Lifetime prob.
0-100
0.001
100-200
0.017
200-300
0.04
300-400
0.18
400-500
0.21
500-600
0.27
600-700
0.18
700-800
0.08
800-900
0.02
900-1000
0.002
0.3
Prob. of bulb lifetime
Lifetime
0.2
0.1
0.0
0
200
400
Time (hr)
600
800
1000
Prob. Of bulb lifetime
0.3
This statistical distribution answers for example the
question “what fraction of lifetimes falls in the range
500-600 hr?“
0.2
This fraction (about 0.27) is actually a formal
probability -- the probability of finding a lifetime, t,
in the range of 500-600 hr:
0.1
0.0
0
200
400
600
800
1000
Prob{t  (500,600)}, or Prob {t  (500, 500+Dt)},
where Dt=100 hr.
Bulb lifetime (hr)
Starting from this formal probability we will define the
probability density function (or pdf)
Probability density
The pdf, written f(t), is a function of t, and is defined in terms of the formal probability above:
f (t) 
Lifetime prob. density
(hr-1)
0-100
0.001
1E-5
100-200
0.017
1.7E-4
200-300
0.04
4E-4
300-400
0.18
0.0018
400-500
0.21
0.0021
500-600
0.27
0.0027
600-700
0.18
0.0018
700-800
0.08
8E-4
800-900
0.02
2E-4
900-1000
0.002
2E-5
In the example
above: the
probability that
the bulb burns in
the next 1 hr if it
has alredy been
on for 500 hours
is: Prob {t 
(500,600)}/100 =
0.27/100 =
0.0027 hr-1.
0.0030
0.0025
-1
Lifetime
prob.
Probability density (hr)
Lifetime
Prob t  (t, t  Dt )
Dt
0.0020
0.0015
0.0010
0.0005
0.0000
0
200
400
600
800
1000
Lifetime
The pdf is NOT a probability: it is not dimentionless: it has units of time-1 !!
It is rather a probability per unit time, otherwise known as a rate constant.
Cumulative distribution
Knowing f(t), it is also possible to answer this other question:
what's the probability that this light bulb I just bought
will burn out sometime before 500 hours?
Probt  500   f(t)dt
0
-1
500
Probability density (hr )
The answer is:
0.0030
0.0025
0.0020
0.0015
0.0010
0.0005
0.0000
0
200
400
600
Lifetime
800
1000
Or, what’s the probability that a bulb will last at least 500 hours?
Probt  500   f(t)dt
500
0.0030
-1

Probability density (hr )
That is:
0.0025
0.0020
0.0015
0.0010
0.0005
0.0000
0
200
400
600
Lifetime
800
1000
This introduces another important statistical function: the cumulative distribution function (CDF)
Cumulative Probability
Probability
Lifetime
Lifetime prob.
Lifetime
∑ Lifetime prob.
0-100
0.001
0-100
0.001
100-200
0.017
100-200
0.018
200-300
0.04
200-300
0.058
300-400
0.18
300-400
0.238
400-500
0.21
400-500
0.448
500-600
0.27
500-600
0.718
600-700
0.18
600-700
0.898
700-800
0.08
700-800
0.978
800-900
0.02
800-900
0.998
900-1000
0.002
900-1000
1
∑ lifetime prob.
∑ lifetime prob.
0.2
0.1
0.0
0
200
400
600
Bulb lifetime (hr)
800
1000
Cumulative Prob. of bulb lifetime
Prob. of bulb lifetime
0.3
1.0
0.8
0.6
0.4
0.2
0.0
0
200
400
600
Bulb lifetime (hr)
800
1000
Random data
Such data, like bulb lifetime, are called random data or stochastic
This means that any result is random
Indeed, we have seen that the bulb lifetime can assume many values: each observation
of the phenomenon, the lifetime of every single light bulb, will be unique.
In order to analyze random data, specific mathematical tools, like those shown in the
example of light bulbs lifetime analysis, are needed.
In Appendix B the basic mathematical tools for analyzing random data are briefly described.
Back to single channel molecules
What works for light bulbs also works for single macromolecules such as ionic channels.
If we could know what the probability density function f(t) is, then we could find out
many things about the random behavior of these macromolecules.
We will have to derive an expression for the probability density function of lifetimes
in the open state and in the closed state
How do we do this?
We must first make up a mental picture - a model - of what we mean by the statement: "the
channel opens at t=0 and then at some random time later, it closes."
This is the heart of the problem: to think up a model of randomness that applies to this
particular situation.
Open-state lifetimes for a single channel
Let’s make up a model of what we mean by the statement: "the channel opens at t=0 and
then at some random time later, it closes”
Suppose a channel protein can undergo two conformational states: open (1) and shut (2)

 O
S 
2
1
Suppose a channel enters the open state at t=0.
QUESTION: what is the probability that the channel will still be open at some time later t,
i.e., that it has not yet closed at time t? Let us call this probability p1(t).
t0=0
shut (2)
Open (1)
p1(t)=?
t
QUESTION
What is the probability that the channel has not yet closed at some time later t?
t0=0
shut (2)
Open (1)
p1(t)=?
t
To answer that question
We have to consider two "axioms:"
A. The probability that the channel closes in a given small time interval Dt is simply
proportional to that time interval. (*)
B. The probability that the channel closes in a given time-interval is totally independent
of how much time has previously passed. (This is the assumption that the channel does
not "remember" anything about its past history. (**)
Now we translate each axiom into a mathematical expression:
Axiom A: [Prob (channel closes between t and t+Dt open at t)] = aDt
where a is the proportionality factor (units: time-1)
Axiom B: a ≠ a(t), i.e., a is a constant, independent of time
(*)
The channel obviously must either close or not close during Dt. Hence,
Prob (channel does not close between t and t+Dt open at t)] = 1 - aDt
So that the probabilities for these two alternatives must add to unity.
We have previously defined p1(t)=Prob(channel stays open throughout the time from 0 to t)
The following question is:
What is the probability that the channel stays open throughout the time from 0 to t + Dt , i.e.,
what is p1(t+Dt)?
For the channel to still be open at time t+Dt, two things must happen:
1) the channel must be open throughout 0 and t (its probability is p1(t) and,
2) the channel does not close during the little time interval Dt (its probability is 1 - aDt).
Moreover, by Axiom B, these two events are totally independent.
(**)
Because of this independence, we can multiply the individual probabilities to get the joint
probability of the two events:
p1(t+Dt)= p1(t)(1-aDt)
p1 (t  Δt)  p1 (t)
 p1 (t)  α
Δt
i.e., by arrangement,
By definition of derivative, in the limit of Dt → 0, the left side of
is simply dp1/dt, therefore: dp1/dt=-ap1
By integration we obtain: p1(t)=exp(-at)
p1(t)=[Prob(channel stays open throughout the time from 0 to t)]
=Prob(open lifetime <t)
p1 (t  Δt)  p1 (t)
 p1 (t)  α
Δt
Then, Prob(open lifetime ≥t)=1-p1(t)=1- exp(-at)
This defines the cumulative distribution function (CDF) of open channel lifetimes F1(t).
1.0
F1 (t)  1  exp(α  t)
0.6
1-p1(t)
p1(t)
0.8
0.4
p1 (t)  exp(α  t)
0.2
0.0
0
1
2
3
Time
4
5
What about the probability density function (pdf) of channel opening f1(t)?
We know from statistics that the pdf is the first derivative of the CDF
f1(t) = dF1(t)/dt
d
f1 (t)  (1  e αt )  α  exp( α  t)
dt
2.5
-1
pdf for channel opening (time )
then:
a
2.0
1.5
1.0
f1 (t)  α  exp(α  t)
0.5
0.0
0
1
2
3
4
5
Time
b


S 
O
a
2
1
Similar procedures for pdf for channel closing yield:
f2(t) = bexp(-bt) (t≥0)
(*)
The pdf of channel opening f1(t) = ae-at and the pdf of channel closing f2(t) = be-bt give the
probability density of the time that the channel is in the open and closed states respectively
(i.e, the dwell time distributions of the channel once it enters the open or closed states)
Suppose now we observe
such a single channel activity
Measuring the time lengths of each opening segment (ton) and each closing segment (tcn) in
the trace (the trace of course must be much longer than that in the example) and plotting a
histogram for opening and one for closing, one usually observes the following results:
pdf of the channel
open lifeime;
interpolation curve
f1(t) = ae-at
= t11e-t/t
pdf of the channel
closed lifeime;
interpolation curve
f2(t) = be-bt
= t21e-t/t
1

Note that:

 f (t)dt   α  exp(αt)dt  1
1
0
2
0

and
f
0

2
(t)dt   β  exp(βt)dt  1
0
pdf of the channel
open lifeime;
interpolation curve
f1(t) = ae-at
= t11e-t/t
pdf of the channel
closed lifeime;
interpolation curve
f2(t) = be-bt
= t21e-t/t
1
2
Since both open lifetime and closed lifetime histograms are well fitted by
single exponential functions, we can suppose the following kinetic model:
b

 O
S
a
2
1
In this simple two state scheme: t1=1/a and t2=1/b
The general rule that we can extrapolate from the previous considerations is that
the mean lifetime of each state (time constant of the dwell times histogram)
equals the inverse of the sum of the rate constants that run away from that state
This rule has a general validity and holds for any n-state kinetic scheme, as we shall see next
Relationship between single-channel events and whole-cell EPC
0
Endplate current evoked by nerve stimulation at -130 mV.
The observed current is shown by the filled circles
Single exponential
fitting of EPC
decay with time
constant tM=7.8
ms
-2
-3
-4
-5
0.00
0.02
0.04
Time (s)
1
τ M   τ O  7.8ms
α
40
N. of events
0.06
30
The lower part (B) shows
the sum of these five
channels with a smooth
exponential curve
superimposed on it. The
curve has a time constant
of 7.8 ms.
20
10
0
0
10
20
30
40
Time (ms)
50
60
open
They have exponentially
distributed lifetimes with
a mean of 7.8 ms.
a


O  S
b
Single exponential
fitting of open
time distribution
with mean open
time to=7.8 ms
50
shut
Simulation of the decay A
of a single channel EPC.
The upper part (A) shows
five individual channels.
0
B
N. of open channels
PSC (nA)
-1
0.02
0.04
Time (s)
0.06
0
1
2
3
4
5
Single exponential
fitting with time
constant t=7.8 ms
Relationship between single-channel events and whole-cell VGC
DV
Voltage-gated current evoked by membrane depolarization.
The observed current is shown by the filled circles
1400
1200
DV
Simulation of the
A
activation of a VG single
channel current. The
upper part (A) shows five
individual channels.
Single exponential
fitting of current
activation with
time constant
tM=19 ms
800
600
400
200
0
0.0
0.1
0.2
Time (s)
0.3
0.4
τ τ
1
τM 
 O C  19ms
α  β τO  τC
60
Open time distr.
tO=44 ms
50
40
80
N. of events
N. of events
100
30
20
60
40
20
10
0
0
0
50
100
150
200
Time (ms)
250
300
350
0
open
shut
They have exponentially
distributed dwell times.
b


S 
O
a
The lower part (B) shows
Closed time the
distr.sum of these five
tC=34 ms
channels with a smooth
exponential curve
superimposed on it. The
curve has a time constant
40
80
120
160
200
of ~19
ms.
Time (ms)
0
B
N. of open channels
Current (pA)
1000
5
4
3
2
1
0
0.1
0.2
Time (s)
0.3
0.4
Some more realistic mechanisms:
kinetic schemes with more than two states
So far the only concrete examples that we have considered were simple because they had only
two states, and because these two states could be distinguished on an experimental record.
We must now consider what happens when there are more than two states, and, in particular,
when not all of the states can be distinguished from one another by looking at the experimental
record.
•
In general, it is usually observed, for virtually every type of channel, that the p.d.f.
required to fit the distribution of shut times (non conducting states) has more than one
exponential component.
Closed time distribution
•
The number of components in this distribution provides a (minimum) estimate of the
300
number of non conducting states. Interpolation with the sum of
Number of closings
•
350
250
two exponentials
200
Likewise the number of components
in the open
time distribution indicates the

1=2.3 ms
150 (conducting) tstates.
(minimum) number of open
100
t2=25 ms
50
We shall consider two simple mechanisms, each of which has two closed
0
states and one open state. 10 20 30 40 50 60
Closing duration (ms)
A channel block mechanism
An example with two closed states and one open state is provided by the case where a
channel, once open, can be subsequently plugged by an antagonist molecule in solution.
b'
This can be written:
 O 

 B
S


a
k
k B . x B
3
1
B
2
open
shut channel
channel
blocked
It has three
states, and the lifetime
in each of
these states is expected
to be exponentially
unblocked
channel
distributed with, according to the rule given before, the mean lifetimes for states open (1),
blocked (2) and shut (3) being respectively
m1 = 1/(a + k+BxB)
m2 = 1/ kB
m3 = 1/b,
where xB (M) is the free concentration of the blocking molecule.
The blocking molecule binding constant is KB= k-B/ kB
The four rate constants have the following units:
k+B = M-1s-1
k-B =˙ s-1
a =˙ s-1
b’ = s-1
(*)
How does look like a macroscopic current in the presence of a blocking agent?
tMs
tMf
Time course of relaxation of an evoked
Time course of relaxation
of an endplate
endplate current
recordedcurrent
in the evoked
by nerve stimulation
at -130
in the blocking
absece ofagent
blocking
presence
of mV
a channel
agents.
(gallamine 5 mM).
The observed current relaxation can be interpolated by a
single exponential function
tMf=1.4 ms, tMs=28.1 ms.
(from Colquhoun & Sheridan (1981).
Here the time course of relaxation is described by the sum of two exponentials with
different time constants (tMs and tMf for the slower and faster values respectively).
This situation could be described by a mechanism with three states of the type
k B x B
b



 B
S
O



a
k
'
3
.
1
tMs
B
2
where the macroscopic time constants
and tMf are indirectly - and in a very complex way related to all of the four transition rates in the underlying mechanism (not shown here).
We shall see that from single channel current analysis we obtain the four transition
rates in a easier and much more direct way than by macroscopic curren analysis.
b'
 O 

 B
S


a
k
k B . x B
3
B
1
2
The channel block mechanism has one conducting state (O)

it predicts a single exponential open time p.d.f.
The open time p.d.f. will therefore be
fo(t) = to1exp(t/to)
where to is the mean open time given.
350
Open times distribution
Number of openings
300
250
Single exponential fit
with time constant to
200
150
100
50
0
5
10
15
20
Dwell time (ms)
25
30
Linear dependence of the open lifetime on blocker concentration
b'
k B x B



 B
S
O



a
k
.
3
B
1
Obtained from the analysis
of open time distributions
2
k-B
a
k+BxB
xB
to
1/to=a+k+B·xB
k+B=(1/to-a)/xB
ms-1
ms-1
ms-1
ms-1
mM
ms
ms-1
M-1·ms-1
0.5
0.1
1
0.480
50
0.675
1.481
0.0096
0.235
25
0.81
1.235
0.0094
0.095
10
0.913
1.100
0.0095
0.048
5
0.954
1.053
0.0096
-
0
1
1.000
-
-1
1/to= a+k+B·xB (ms )
b’
Plot of 1/to vs xB
1.6
k+B
1.4
1.2
a
0.8
0
10
20
30
xB (mM)
40
50
b'
 O 

 B
S


a
k
k B . x B
3
B
1
2
The channel block mechanism has two not conducting states (S and B)

it predicts a double exponential shut time p.d.f
The shut time p.d.f. should have the double-exponential form
fc(t) = asts1exp(t/ts) + aftf1exp(t/tf) ,
where as and af are the relative areas under the p.d.f. accounted for by the slow and fast
components respectively, and ts and tf are the observed ‘time constants’, or ‘means’, of
the two components. (*)
350 Closed times distribution
Number of closings
300
250
Double exponential fit
with time constants tf
and ts
200
150
100
50
0
10
20
30
Dwell time (ms)
40
50
60
Bursts of channel openings
Whenever the closed time distribution has at least two time constants, and one time constant is
much shorter than the other,
tf << ts
1) the openings, will appear to be grouped together in bursts of openings
2) the experimental record will contain short closed periods
3) and some much longer ones.
gap
between
bursts
gap
between
bursts
gaps
within
bursts
closed
0.4 pA
open
20 ms
burst
burst
burst
Now we must enquire how the two mean closed times tf and ts are related to the underlying
reaction transition rates.
k . x
b'
B
B



S
O
B



a
k
3
B
1
2
Schematic illustrations of
transitions between various
states (top)
blocked
shut
blocked
shut
and
the corresponding observed
single channel currents
(bottom).
•
No direct transition is possible between the two sorts of not conducting states.
•
Every closing of the channel must, therefore, consist of
1) either a single sojourn in the shut state (state 3),
2) or a single sojourn in the blocked state (state 2).

Therefore the two time constants tf and ts of the not conducting states (S and B) in this
case are simply the mean lifetimes of these two states,
ts=m3=1/b
tf=m2=1/kB.
Relationship between bursts and macroscopic currents with channel block
tMs
tMf
Evoked endplate current
Simulation
of presence
EPC decay
recorded in the
of a when
blocking
agent
achannel
channel
blocker
is supposed
(gallamine
5 mM).
to
be present
so some of the
openings
(all butexponential
channels 2
The fitted double
and
aretime
interrupted
one or
curve3)has
constants by
tf=1.37
ms,
ts=28.1
ms.
more
blockages.
(from Colquhoun & Sheridan (1981).
tMs
tMf
The sum of all seven channels
is shown in the lower part; a
double exponential decay curve
is superimposed on it (the two
separate exponential
components are shown as
dashed lines).
An agonist activated channel
An agonist (A) binds to a receptor (R), following which an isomerization to the active state
(i.e. the open channel, R*) may occur.
This can be written:
k 1. x A
b
*





A R 
AR 
AR

a
k
3
1
2
1
It has three states, and the lifetime in each of these states is expected to be exponentially
distributed with, according to the rule given above, the mean lifetimes for states 1, 2 and
shut channel
shut channel
open
3 being respectively
channel
agonist
bound
agonist unbound
m1 = 1/a
m2 = 1/(b + k1)
m3 = 1/(k+1xA)
where xA (M) is the free concentration of the agonist.
The equilibrium constant for the initial binding step is: KA=k1/k+1
The four rate constants have the following units:
k+1=M-1 s-1
k-1=˙s-1
a=˙s-1
b=˙s-1
(*)
Open times for the agonist mechanism
k 1. x A
b
*




A  R 
AR
AR


a
k
1
3
2
1
By measuring the distribution of open lifetimes from the experimental record we can obtain
the mean open time that will provide an estimate of a directly (a=1/m1).
Example of open lifetimes distribution from a single channel record:
a means to calculate the rate constant a
Dwell Time Histogram
C ount (N )
2000
1500
m1=2.3 ms
1000
500
0
1 pA
10 ms
0
4
8
12
Dwell Time (ms)
(Level 1)
16
20
24
Shut times for the agonist mechanism
k 1. x A
b
*




A  R 
AR
AR


a
k
1
3
2
1
This scheme is a bit more complicated than that for channel block; the reason for this is that
the two shut states intercommunicate directly.
Furthermore, the two sorts of shut states (states 2 and 3) cannot be distinguished on the
record (see below).
Thus an observed shut period might consist of
1) a single sojourn in AR (state 2) followed by reopening of the channel,
but it might also consist of
2) a variable number of transitions from AR to R and back before another opening occurred
(both states are shut so these transitions would not be visible on the experimental record).
This diagram illustrates the actual
molecular transitions that underlies the
bursting behaviour in the scheme.
Notice the ‘invisible’ oscillations
between the two shut times that result
from agonist occupancies that fail to
produce opening of the channel.
Shut times for the agonist mechanism
k 1. x A
b
*




A  R 
AR
AR


a
k
1
3
2
1
Example of shut lifetimes distributions from a single channel record:
1 pA
10 ms
where:
m2 = 1/(b + k1)
m3 = 1/(k+1xA)
2000

k+1xA=1/m3
1000
0
0
10
20
30
40
50
Dwell Time (ms)
(Level 0)
60
70
Square Root Count (N)
Count (N)
3000
30
m2=0.8 ms
25
20
15
m3=29.1 ms
10
5
0
-0.5
0
0.5
1
Log Dwell Time (ms)
(Level 0)
1.5
2
Bursts with agonists:
How to calculate the two rate constants b and k-1
k 1. x A
b
*




A  R 
AR
AR


a
k
3
1
2
1
We have just seen that the mean length of the short gaps within a burst (m2) will be
approximately 1/(b+k1), and that m2 can be obtained from the pdf of the closed time
distribution.
But m2 alone does not allow us to estimate the separate values of b and k1.
We shall see below that the two rate constants (b and k1) are also related to the
mean number of openings per burst
Therefore, if we are able to measure the mean number of openings per burst, then
the channel opening rate constant b, and the agonist dissociation rate constant k1
can both be separately estimated.
The number of openings per burst for agonist
.x
b
k 1 A
*





AR
AR
Consider a channel in the intermediate state AR (state 2): A R 


a
k
1
3
Its next transition may be either
1) reopening (rate b)
2
1
β
β  k 1
AR*
π 21 
R
p23=1p21
AR
or
2) agonist dissociation (rate k1)
AR
The relative probability of reopening happening (regardless of how long it takes before it
happens), which we shall denote p21, is thus
π 21 
β
β  k 1
and the probability of the latter (agonist dissociation) happening is therefore
p23=1p21.
By definition a burst must contain at least one opening;
AR*
1
.x
b
k 1 A
*





A R 
AR
AR


a
k
3
1
2
1
P[one opening]=p21
AR
(r-1) re-openings
AR*
1
2
3
r
For the channel to re-open r1 times before
dissociation occurs, opening must occur r1 times
before agonist dissociation.
AR
AR*
1
2
3
r
The probability of r openings in a burst (r=1, 2, 3, ...,
∞) is thus:
AR
P[r openings and then shutting] =p21(r-1)·p23
R
.x
b
k 1 A
*





A R 
AR
AR


a
k
1
3
2
1
We have just seen that the probability of r openings in a burst (r=1, 2, 3, ..., ∞) is
P(r) πr211  π23
that can be written also as:
P(r) (1  p 23 )( r 1)  p 23
This is the geometric distribution of the number of openings per burst.
(*)
The mean number of openings per burst can be found from the general expression for the
mean, m, of a discrete distribution:
μ  r 1 r  P(r) r 1 r  (1  p 23 )( r 1)  p 23 

Therefore:
μ

1
p 23
In conclusion,

1
p 23
1
1

 1  b / k 1
b
1  p 21 1 
b  k 1
μ  1  b / k1
is the other relationship we need to calculate separately b and k-1.
Example of distribution of the number of openings within bursts
7
23
6
25
6
29
1 pA
10 ms
Distr. of the N. of openings in bursts
Where: m=1+b/k-1
20
Furthermore we have previously calculated
the mean shut lifetime within bursts:
m2 = 1/(b + k1)
Count (N)
15
m=16.7
10
Now, from the two relationships we can
obtain the rate constants b and k-1.
5
0
0
10
20
30
Events in Burst
40
50
60
ko
k i



 I



C
O
k
k
3
c
1
i
2
m1 = 1/(kC+k+i)
m3 = 1/ kO
m2 =1/k-i
For the V-G Na channel it is normally k+i>>k-i so that the transition O→I is almost
irreversible (I is the adsorbing state) and inactivation very strong.
m=(kC/ k+i)+1 (the mean number of openings per burst when flikering is between states C and O)
Alternative model
H. & H. model
Rapid activation
Slow inactivation
Slow activation
Rapid inactivation
+10
+10
-100
-100
0
10
20
30
40
50
0
10
20
30
40
50
H. & H. model
Alternative model
Slow activation
Rapid inactivation
Rapid activation
Slow inactivation
+10
+10
-100
-100
By latency analysis we can discriminate between the two models
(*)
45
100
40
35
30
l=2 ms
60
N. of Events
N. of Events
80
40
l=10 ms
25
20
15
10
20
5
0
0
0
2
4
6
8
10
12
14
16
18
20
0
10
First latency (ms)
C
400
20
O
100
20
30
40
50
First latency (ms)
I
C
100
20
O
800
I
That’s all folks!
N. of Events
xB=50 mM
to=0.676 ms
Open time (ms)
xB=25 mM
xB=10 mM
xB=5 mM
xB=0
Appendix A
First order kinetics
Let’s consider a 1st order reaction kinetics
an
1-n
n
bn
where n particles make transitions between the permissive and non permissive forms
with rate constants an and bn
If the initial value of the probability is known, subsequent values can be calculated by
solving the differential equation
dn
 α(1  n)  βn
dt
By solving the differential equation we obtain:
n(t )  n  (n  n0 )  exp[t  (a  b )]
where n0 is the value of n at t=0
and n∞ is the value of n at the steady state (equilibrium)
1/(anbn)=tn rapresents the time constant of the kinetic reaction (*)
At the steady state (equilibrium) it will be:
and therefore,
n 
αn
αn  βn
dn
 α(1  n)  βn  0
dt
Appendix B
Basic descriptive properties of random data
When a variable can assume many values, so that any result is random, than it is called a
random variable.
DEFINITION: a random variable X is a function that associates to every elementary event
a unique real number. As a matter of fact it is a variable whose numerical value is
determined by the result of a trial.
CONTINUOUS RANDOM VARIABLE: if it can assume any value within a specific
interval.
Examples: weight of students in a school; lifetime of a lot of light bulbs; lifetime of
radioactive particles.
DISCRETE RANDOM VARIABLE: if it assumes a finite or numerable amount of results.
Examples: number of parasites on a leave, number of calls received by a call centre, number
of heads following five throws of a coin.
Data representing a random physical phenomenon cannot be
described by an explicit mathematical relationship, because each
observation of the phenomenon will be unique.
Let consider the output voltage from a thermal noise generator recorded as a function of
time.
A single time history observed over
a finite time interval is defined as a
sample function or sample record.
Here are represented three sample
records of thermal noise generator
outputs.
Let consider the collection (ensemble) of time-history records (sample functions) defining. a
random process.
The mean value (first moment) of the
random process at some time tl can be
computed by taking the instantaneous
value of each sample function of the
ensemble at time tl, summing the values,
and dividing by the number of sample
functions.
That is, for the random process {x(t)},
where the symbol { } is used to denote an
ensemble of sample functions, the mean
value mx(tl) is given by:
In most cases, however. it is also possible to describe the properties of a stationary random
process by computing time averages over specific sample functions in the ensemble. For
example. consider the kth sample function of the random process illustrated in figure. The
mean value mx(k) of the kth sample function is given by:
Other statistical functions to describe the properties of random data are:
 Mean square value: furnishes the general intensity of any random data and is the
average of the square values of the time history of the random process x(t)
1
T  T
x 2  lim

T
0
x 2 (t )dt
 Variance: describes the fluctuating component of the data
1
  lim
T  T
2
x
 x (t )  m  dt
2
T
0
x
 Probability functions

Probability density function (pdf)

Probability distribution function or cumulative distribution function (cdf)
Probability and probability distributions
If X is a random variable, we are usually interested in the
probability that X takes on a value in a certain range.
Probability density function
A probability density function (or probability distribution function) is a function f
defined on an interval (a, b) and having the following properties.
(a) f(x) ≥ 0 for every x

(b)
 f(x)dx  1
i.e the area under the entire graph of f (x) = 1
-
Probability Associated with a Continuous Random Variable
A continuous random variable X is specified by a probability density function f. The
probability P(c ≤X≤ d) is specified by:
d
P(c ≤X≤ d) =  f(x)dx
0.0030
0.0025
c
That is, the probability that X takes on a
value in the interval [c; d] is the area
above this interval and under the graph of
the density function.
pdf (hr-1)
0.0020
0.0015
0.0010
0.0005
0.0000
0
200
c
400
600
d
800
1000
Lifetime (hr)
Mean or Expected Value of a probability distribution
If X is a continuous random variable with probability density function f defined on an
interval with (possibly infinite) endpoints a and b, then the mean or expected value of
X is
b
mx=E(X) =
 x  f(x)dx
a
E(X) is also called the average value of X. It is what we expect to get if we take the
average of many values of X obtained in experiments.
Variance and Standard Deviation of a probability distribution
Let X be a continuous random variable with density function f defined on the interval
(a, b), and let μ = E(X) be the mean of X. Then the variance of X is given by
b
Var(X) = E((X - μ)2) =
2
(x
μ)
 f(x)dx

a
The standard deviation of X is the square root of the variance,
(X) = (Var(X))
Exponential Density Function
An exponential density function is a function of the form
f(x) = ae-ax (a is a positive constant)
with domain [0 +∞). Its graph is shown in the figure.
a
f(x)  a  eax
Mean of an Exponential Distribution
If X has the exponential distribution function f(x) = ae-ax, then E(X) = 1/a.
Variance and Standard Deviation of an Exponential Distribution
If X has the exponential distribution function f(x) = ae ax, then
Var(X) = 1/a2
And
(X) = 1/a.
Probability of a geometric distribution
It’s the probability that the first occurrence of success requires k number of independent
trials, each with success probability p. If the probability of success on each trial is p, then
the probability that the kth trial (out of k trials) is the first success is:
P (k )  (1  p)( k 1)  p
for k = 1, 2, 3, ....
The mean of a geometrically distributed random variable X is 1/p and the variance is (1 −
p)/p2.
The above form of geometric distribution is used for modeling the number of trials until the
first success.
P(k)=(1-p)(k-1)*p
0.18
0.16
0.14
0.12
P(k)
Example. A die is thrown repeatedly until
the first time a "1“ appears. The
probability distribution of the number of
times it is thrown is supported on the
infinite set { 1, 2, 3, ... } and is a
geometric distribution with p = 1/6.
0.10
0.08
0.06
0.04
0.02
0.00
The mean value is k=6.
1
11
21
k
31
41
Normal Density Function
A normal density function is a function of the form
2

 xm  
 exp 0.5  
 
f(x)=

  2p

 

1
with domain (- , + ). The quantity μ is called the mean and can be any real number, while
 is called the standard deviation and can be any positive real number. The graph of a normal
density function is shown in the following figure.
Properties of a Normal Density Curve
(1) It is "bell-shaped" with the peak at x = μ.
(2) It is symmetric about the vertical line x = μ.
(3) It is concave down in the range μ- ≤x ≤μ + .
(4) It is concave up outside that range, with inflection
points at x = μ- and x = μ+.
Mean of a Normal Distribution
If X is normally distributed with parameters μ and , then
E(X) = μ.
Variance and Standard Deviation of a Normal Distribution
If X is normally distributed with parameters μ and , then
Var(X) = 2
and
(X) =  .
Cumulative distribution function
The cumulative distribution function F(x) for a continuous random variable X with
probability density function f is defined for every number x by
x
x

F(x) = P(X≤x) = f(y)dy

 x  f(x)dx

For each x, F(x) is the area under the density curve to the left of x.
pdf f(x)
0.0025
1.0
cdf F(x)
0.0030
0.8
0.0020
0.6
0.0015
Fx(a)
0.4
0.0010
0.2
0.0005
P(X≤a)
x
0.0000
0
200
400
600
a
800
1000
x
0.0
0
200
400
600
a
Proposition
If X is a continuous random variablr with pdf f(x) and cdf F(x), then at every x at
which the derivative F’(x) exists, F’(x) = f (x).
800
1000
Appendix C
Probabilities
Probabilities
t
t+Dt
ooen
Consider the behavior of six individual
ion channels. Imagine that these six
channels behave in a manner typical of a
much larger number of channels so that
the ratios given are good estimates of the
true or long-term average values of the
probabilities.
1
shut
2
3
4
5
6
Open at t
Shut at t+Dt
√
X
√
√
√
√
X
√
X
X
√
X
Only two channels out of the six are both open at t and shut at t + Dt, so
Prob(open at t and shut at t + Dt) = (number open at t and shut at t + Dt)/(total number) = 2/6.
However, if we ask the conditional question "What is the probability of a channel being closed
at t+Dt given that it was open at t?" we arrive at a different answer. In the above example only
four of the channels were open at time t. Of these four, two were closed at time t + Dt.
Therefore,
Prob(closed at t+Dt open at t) = 2/4 = 1/2.
We can formulate a general rule of probability for this example which states that for any
events A and B
In this case, A is "open at t "; B is "shut at t + Dt."
Prob(B A) = Prob(A and B)/Prob(A)
or
Prob(A and B) = Prob(A)Prob(B A)
(1)
In the above example,
Prob(A and B) is open at t and closed at t + Dt = 2/6
and
Prob(A) is open at t = 4/6.
Therefore,
Prob(B A) = (2/6) / (4/6) = 1/2
In general, if A and B are independent, then the probability of B does not depend on
whether or not A has occurred, therefore Prob(B A) can be written simply as Prob(B) and
equation (1) reduces to the simple multiplication rule of a combined probability
Prob(A and B) = Prob(A)Prob(B)
Appendix D
Bursts during block
The number of openings per burst for the channel block mechanism
b'
 O 

 B
S


a
k
3
k B . x B
1
B
2
The distribution of the number of openings per burst, and hence an expression for the
mean number, can be derived as follows. Consider an open channel (state 1 in the
scheme). Its next transition may be either blocking (going to state 2 with rate k+BxB), or
shutting (going to state 3 with rate a). The relative probability of the former happening
(regardless of how long it takes before it happens), which we shall denote p12, is thus
π12 
k B  x B
α  k B  x B
and the probability of the latter happening is therefore p13=1p12. Notice also that a
blocked channel (state 2) must unblock (to state 1) eventually, so p21=1. The probability of
blocking and then reopening is therefore p12p21=p12. A burst will contain r openings (and
r1 blockages) if an open channel blocks and unblocks r1 times (probability p12r1 ), and
then returns after the last opening to the long- lived shut state, state 3 (with probability
p13).
The probability of seeing r openings (i.e. r1 blockages) in a burst is thus
r -1
P(r) π12
 π13
(*)
The mean number of openings per burst can be found from the general expression for the
mean, m, of a discontinuous distribution, viz.
μ  r 1 r  P(r)

In the present example,
b'
k B



 B
S
O



a
k
3
1
B
2
the mean number of openings per burst is therefore
μ
1
k x
 1  B B
1  π12
α
The end