UCL quadrangle, 20 March 2003 - University of California...

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UCL quadrangle, 20 March 2003

Transcript of UCL quadrangle, 20 March 2003 - University of California...

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UCL quadrangle, 20 March 2003

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http://www.ucl.ac.uk/Pharmacology/dc.html http://www.dcsite.org.uk

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Sources: Changeux, Unwin

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Channels opened by a low concentration of acetylcholine

Human recombinant muscle type nicotinic receptor (α2βεδ), expressed in HEK cell, Openings are downward deflections, and have amplitude of about 6 pA at –100 mV.

D. Colquhoun 2003

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AR AR*Rk+1 βk−1

αRate constants interpreted as probabilities

Prob(open channel shuts during ∆t) = α ∆t + o(∆t)

Prob(channel shuts between t and t + ∆t | open at t = 0) = α ∆t + o(∆t)

so the law of mass action rate constant is

α = Prob(channel shuts between t and t + ∆t | open at t = 0) / ∆tlim [∆t →0

]

+−+−

+−=

)()(

)(

32313231

23232121

13121312

qqqqqqqqqqqq

Q

In the general treatment, the rates for transition from state i to state j(qij say) are tabulated in the Q matrix. The diagonal elements are defined so that the sum of each row is zero.

The Q matrix

D. Colquhoun 2003

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Macroscopic (deterministic) behaviour

Define p(t) as a row vector (1 × k) that contains the occupancy of each state at time t :

p(t) = [p1 p2 p3 . . . . .pk ]

Qpp )(d

)(d ttt

=

When Q is constant, the solution is just like the simple scalar case

)exp()0()( tt Qpp =

This is all there is to say about macroscopic kinetics! D. Colquhoun 2003

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Partitioned matricesAny matrix can be divided up into rectangular subsections, each of which is itself a matrix. We shall often want to divide the k states in a mechanism into kA open states (set A) and kF shut states (set F) (k = kA + kF). By convention the open states are given the lowest numbers so they appear in the upper left hand part of the Q matrix.

−+β−βαα−

=

++

−−

xkxkkk

11

11

0)(

0Q

AA AF

FA FF

[ ]α−=AAQ

+β−=

++

−−

xkxkkk

11

11FF

)(Q

=

FFFA

AFAA

QQQQ

QThis can be written as

where the submatrices are

β=

0FAQ

For the del Castillo-Katz mechanism we have k = 3 states, divided into kA = 1 open state and kF = 2 shut states

[ ]0AF α=Q

D. Colquhoun 2003

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An example from muscle nicotinic receptors

−++β−β

+β−βα+α−

α+α−

=

++

−+−+

−−

−−

++

A1A1

1A211A21

2222

2*22

*2

1A*2A

*21

22000)(0

02)2(000)(00)(

xkxkkxkkxk

kkkk

xkxk

Q

FFFA

AA AF

R AR A2R

A2R*

k−1

2k+1

α2β2

k+2

2k−2

AR*

β1 α1

5 4 3

1 2

shut

open

Colquhoun & Sakmann, 1985

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What is PAA(t)?

For the purposes of macroscopic currents we are interested only in the state (open or shut) at time t, and it does not matter what happens between 0 and t, so we start with probabilities defined thus

0]at statein at stateProb[in )( == titjtpij

But for the purposes of single channels it does matter what happens between 0 and t, so to find the lifetime of the open state (length of time spent continuously in set A, the open states, we define

0]at statein at time statein and , to0ut A throughoset within nsProb[remai)( == titjttpijA

A, ∈ji

If the Apij are the elements of a matrix PAA(t) (value in ith row and jth column),then, when the transition rates are constant,

)exp()( AAttAA QP = Chapman-Kolmogorov equation D. Colquhoun 2003

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What is GAB? (1)Next we need to define a density that describes the probability of staying open(within the subset of states, A) for a time t, and then leaving A for a state outside A, in subset B say.

gij(t) = lim[Prob(stay in A from time 0 to t and leave A for state j∆t→0

between t and t + ∆t | in state i at t = 0)/ ∆t ] Bji ∈∈ A,

Probabilities in PAA(t) Probabilities from QAB ∆t

If the gij(t) are elements of a matrix GAB(t), this is given (see C&H82 p10 for details) by

GAB(t) = PAA(t) QAB = exp(QAAt) QAB

D. Colquhoun 2003

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What is G*AB(s) ?

The Laplace transform of these matrices is what is actually used in mostof the derivations

st

+=−α

1)]α[exp(LThe Laplace transform of an ordinary exponential is

1AAAA )()][exp(L −−=− QIQ stThe Laplace transform of a matrix exponential is

GAB(t) = exp(QAAt) QABSince

its transform is

G*AB(s) = (sI − QAA)−1 QAB

If we are interested only in the probabilities of particular transitions, regardless of how long they take to occur, simply set s = 0:

G*AB(0) = − QAA

−1 QAB GAB, which, for brevity, we write as D. Colquhoun 2003

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The distribution of the open timeCall the set of all shut states F (= B∪C). The matrix GAF(t) describes the time from the start of an opening, through any number of transitions within open states (set A), until eventual exit to any of the shut states (set F), so the pdf is

AAAAAo )()exp()( uQQφ −= ttfFAFAAo )exp( uQQφ t=FAFo )()( uGφ ttf =

or

(simple exponential pdf: one open state)f(t) = exp(−λt) λcompare

[Here, ϕo is a 1 x kA row vector that contains the probabilities that an opening starts in each of the kA open states, and uA is a kA x 1 unit vector.]

The mean open time can be found by using the Laplace transform

A1

AAo0

*

)(d

)(dµ uQφ -

sssf

−=

−=

=(compare µ = λ−1 in simple case)

AAA1

AAo* )()()( uQQIφ −−= −ssf (NB QAAuA+ QAFuF = 0 because rows sum to zero) D. Colquhoun 2003

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brief shut time in A2R

Open time (in A2R*)

R AR A2R

A2R*

k−1

2k+1

α2β2

k+2

2k−2

AR*

β1 α1

5 4 3

1 2

(1)From the ‘physiological’ point of view they are the effective channel openings, and they provide the connection between the single molecule behaviour and the macroscopic currents that flow in synapses.

(2) The fine structure of bursts provides information about the channel opening mechanism. It allows the separation of effects on binding and gating and the estimation of rate constants.

Channel openings usually occur in ‘bursts’

Bursts are interesting for two reasons

D. Colquhoun 2003

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The Q matrix, is again partitioned into A(open states), B (brief shut states) and C(long shut states)

CC

AC

BC

AA

BA

AB

BBCA CB

−++β−β

+β−βα+α−

α+α−

=

++

−+−+

−−

−−

++

A1A1

1A211A21

2222

2*

22*

2

1A*

2A*

21

22000)(0

02)2(000)(00)(

xkxkkxkkxk

kkkk

xkxk

Q

An example in which the burst contains two open states and twoshort-lived shut states

+−+−

−−

++

)2(2]A[])A[(

*22

*2

*2

*21

kkkk

αα

QAA =For example, the top left hand sub-matrix is

Bursts of openings are spent in states A∪B = E (‘burst states’)

CCCE

EE EC

−++β−β

+β−βα+α−

α+α−

=

++

−+−+

−−

−−

++

A1A1

1A211A21

2222

2*

22*

2

1A*

2A*

21

22000)(0

02)2(000)(00)(

xkxkkxkkxk

kkkk

xkxk

Q

D. Colquhoun 2003

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ABC

shutopen

Set

∑∞

=1rbφ Cu

3*BA

*AB )]()([ ss GG BC

*AB )( GG s )(*

AC sG

1*BA

*AB )]()([ −rss GG BC

*AB )([ GG s ])(*

AC sG+=)(* sf

burst length, r = 4 burst length, r = 1

The distribution of the burst length is found by convolving the distributions of the intervals that constitute the burst.

A burst will contain a random number of openings, the probability of getting r openings in a burst, regardless of their duration, being ABAAB

1BAABb )()()( uGGIGGφ −= −rrP

The distribution of the burst length (1)

openshort shutlong shut

D. Colquhoun 2003

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The distribution of the burst length (2)

We now have the Laplace transform of the pdf as

C*ACBC

*AB

1

1*BA

*ABb

* )]()([)]()([)( usssssfr

r GGGGG += ∑∞

=

−ϕ

C*ACBC

*AB

1*BA

*ABb )]()([)]()([ ussss GGGGGI +−= −ϕ

It is not obvious how to invert this, because s occurs in four different places, and in fact it is one of the trickier inversions. The method is described in appendix 1 of C&H 1982. The result is, however, simple and elegant,

time spent in the burst states (E) starting and ending in open state (A)

burst fails to continuec.f scalar case P(r) = π21

r−1 (1−π21)

ABAABAAAAEEb ))(()][exp()( uGGIQQφ −−= ttf

Note: exp(QEE) is a kE × kE matrix so the result will be a mixture of kE exponentialsNote: we need only the AA subsection of exp(QEE), i.e. the upper kA × kA part.

D. Colquhoun 2003

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The spectral expansion of QWhen Q is a k × k matrix, with k eigenvalues, we can also calculate, from the eigenvectors of Q, the k matrices, Am,

IA =∑=

k

mm

1

andnmnm ≠= ,0AA mrm AA =, but

Once the A matrices have been found they can be used to define any analytic function of a matrix, Q, as that function of the eigenvalues of Q (also OK with repeated

eigenvalues if Q is diagonalisable, i.e. process is reversible).

D. Colquhoun 2003

∑=

=k

mmmff

1)λ()( AQ Sylvester’s theorem

In particular, in our case, we find the ubiquitous matrix exponential as

∑=

=k

mmm

1)λexp()exp( AQ

This allows all the distributions to be expressed in scalar form as a linear combination of exponentials with exponents that are the eigenvalues of the matrix, exp(λmt).

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A first look at single channel data: empirical fits

(1) Idealise: make list of open and shut times

# ms pA1 3.045 02 0.064 2.433 0.035 04 …. …

Gly 10 µM

(5) Attempt to estimate physical rate constants retrospectively from

the distributions

(4) Build a plausiblemechanism

(2) Display data asdistributions: open times, shut times, burst lengths, correlations..

Note –resolution 30 µs

(3) Fit exponentials to DISTRIBUTIONS

Minimum number of shut and open states, time constants, connectivity…

D. Colquhoun 2003

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A better way: direct fit of a mechanism to data

# ms pA1 3.045 02 0.064 2.433 0.035 04 …. …

Gly 10 µM

Gly300 µM

# ms pA1 9.325 02 2.562 2.513 0.050 04 …. …

Gly 100 µM

# ms pA1 5.120 02 1.312 2.483 0.044 04 …. …

Gly 1000 µM

# ms pA1 2.215 02 5.327 2.473 0.539 04 …. …

(1) Idealise resolution = 30 µs

(2) Postulate a plausible mechanism

(5) Do these rates, with the mechanism, describe the data accurately?

(3) Choose rate constants to maximise likelihood

HJCFIT(with exact missed event method)

(4) Rate constantsα2 = 1250 ± 120β2 = 1250 ± 120k+2 = . . . . .

D. Colquhoun 2003

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open1 = 2 ms

shut1 = 10 ms

open2 = 5 ms

Maximum likelihood fitting to a sequence

Likelihood of this sequence = Prob[open1 = 2 ms]× Prob[shut1 = 10 ms given open1=2 ms]× Prob[open2 = 5 ms given (shut1 = 10 ms and open1=2 ms)]

These probabilities depend on the values for the rate constants in the mechanism and on the resolution of the observations, from which they are calculated, by the method of Hawkes, Jalali & Colquhoun (1990, 1992).The values of the rate constants are adjusted until the likelihood is maximised.

To infer, for example, whether a mutation affects the binding site, we need values for the underlying rate constants. This method fits what we want directly to the observations, so avoiding entirely the difficult step between fitting exponentials to distributions, and inference of values for the underlying rate constants. D. Colquhoun 2003

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The equation for the likelihood can be evaluated from the known resolution and the Q matrix, which specifies the mechanism, and the values of the rate constants in it. The parameters are not now the τ1, τ2, a1 , etc for arbitrary exponentials, but are the elements the Q matrix, i.e the rate constants in the mechanism.

φΑ is a (1 × kA) vector that gives the probability that an opening starts in each of the open states at equilibrium, and when the next factor in eq. 6.1 is included, φΑ

eGAF (to1) is a (1 × kF) vector that gives the probability that the next shut period starts in each of the shut states following an opening of length to1; and φΑ

eGAF (to1) eGFA (ts1) is again a (1 × kA) vector that gives the probability that the next opening starts in each of the open states following an opening of length t01 and a shutting of length ts1, and so on, up to the end of the data

φΑeGAF(to1) eGFA(ts1) eGAF (to2) eGFA(ts2) eGAF(to3) . . . uF = L

1× kA

1× kA kF × 1 1× 1

1× kF

1× kF

1× kA

D. Colquhoun 2003

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Testing the quality of the fit

apparent shut time

N o

f eve

nts/

frequ

ency

den

sity

√ D. Colquhoun 2003

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Several types of distribution at a range of concentrations100 µM

shut times

open times

100 µMN

of e

vent

s/fre

quen

cy d

ensi

ty

shut times

open times

D. Colquhoun 2003

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Mea

n ap

p. o

pen

time

(ms)

apparent open time apparent shut time

frequ

ency

den

sity

frequ

ency

den

sity

frequ

ency

den

sity

conditional open time (openings adjacent to apparent shut times shorter than 100 µs)

Testing whether the fitted rate constants describe the data accurately

Blue: observed mean open timeRed: HJC predictions from fit

Perfect resolution

HJC prediction from fit

All openings

D. Colquhoun 2003

adjacent shut time (ms)

HJCsim Figure 6sim39b

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The mechanism Wild type εL221F

R-AR

AR-R

AR-R*

A2R

R-AR*

R-R51700

1770

415

115 41200

A2R*

4750

1730

2.2×108 12600

1.45×109

AR-R

AR-R*

A2R

R-AR*

R-R70400

1070

115

153 50700

A2R*

4390

1390

2.2×108

3.31×109R-AR

3210

For the doubly-occupied receptor (mean of 5 patches in each case):total dissociation rate 14300 ± 1640 s−1 4600 ± 741 s−1

opening rate (β) 51700 ± 2300 s−1 70400 ± 2170 s−1

shutting rate (α) 1770 ± 113 s−1 1070 ± 119 s−1

Thus the main effect of the mutation is to slow agonist dissociation from theshut state, and so to prolong the burst of openings (channel activation)

D. Colquhoun 2003

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Come to our course (14th-18th June 2004 )

D. Colquhoun 2003