Fluorescence anisotropy of chromophores rotating between two reflecting barriers

ChemicalPhysics 7 (1975) 210-219 Q North-HollandPublish@ Company

FLUORESCENCEANKOTROPYOF CHROMOPHORESROTATING TWOREFLECiWC BARRIERS

BETWEEN

Ph. WAI-IL Cenne de Biophysique Molhculaire- CN.R.S.. 45045 OrI&nsCede%.Fmnce Received I4 May 1974 Revisedmanuscriptreceived21 October 1974

A generalexpression is calculated for the time course of the fluorescenceanisotropy of chromophoresrotating about a fixed axis, with an amplitudeof rotation limited by two reflectingbarriers.The influence of the amplitude of these rotations on the asymptotic value of the anisotlopy is discussed.

1. latmduction

Fluorescence depolarization measurementsof chromophores linked to proteins have been proposed by Weberas a means of evaluatingthe dimension and the flexibility of protein molecules [l]. The qmtitative

interpretation of these measure-

ments is basedon theoreticalrelationsfirstestabtished by Perrin [2,3] and lateron confirmedby several authors [4-71. The recently introduced excitation by very short flashes,instead of by a continuous source 17-101, has attracted new interest for this method. Pulse fluorometry allows one to determine the time course of the anisotropy and to obtain correlation functions related to the molecular brownian mo tion, while continuous excitation only leads to a time averageof these functions. Consequently, information obtained by f&h excitation is more accurate than information obtained by continuous excitation. In the case of macromolecules,it has been recognized that part of the depolarization is due to local chromophore motion around its attachment point [I, 1I-131. This motion is not taken into account in Perrin’scalculations. Sometimes the IocaI motion may become the only cause of brownian depolarization. That is the case with chromophores attached to very large or strongly elongated macromolecules,and with chromophores

fixed on protein components of membranes. By means of pulse Iluorometry, it has been found that the anisotropy factor of a chromophore covalently linked to a membrane [ 141,or of a dye intercalated in the DNA double helix [IS], decreasesfrom a value ro, at the time of excitation, to a value r, + 0, st the end of the fluorescence emission.

These results correspond to strongly anisotropic rotations. They can be explained qualitatively by the mode of linkageof the chromophores to the macromolecules. It may be assumedthat, in these cases,the chromophores are compelled to rotate about one or severalrotational axes. For the case of chromophores which are covalently linked to membranes,these axes themselvesare the covalent links. For the intercak tion complexes of dye-DNA, the plane of the dye is perpendicular to the longitudinalaxis of the DNA molecule. The depolarization has been attributed to the local torsional motion of the DNAmolecule around its axis [ 151. Computationsof brownian depolarization of chromophores rotating about one or severalaxes have been performed. In these calculations,the rotation was assumedeither to be free without restdction or to proceed via discontinuousjumps between fixed positions [11,16]. It is evident that there is a number of actual cases where the chromophore rotations are not wholly free but where local interactions restrict the angularampli-

A

211

Wahll Fluorescence anisotropy of chromophores

tudes of these rotations. For example, the structure of the DNA molecule necessarily implies that its torsional motion is limited. In the case of chromophores which are covalently linked to proteins or membranes the rotations about the valence bonds can be Iimited by steric hindrances. These are due to neighbouringchemicalgroups, the

presence of which results from the geometrical structure of the macromolecules involved. In order to help the interpretation of such cases, computations of fluorescence anisotropy have been performed for the following models. We assume that a chromophore rotates freely around an axis. The rotational amplitude is limited by two reflecting barriers. Three cases are envisaged according to the mode of interaction of these barriers with the excited chromophores: (I j We assume that the contact of the chromophores with the barriers does not modify the rate of nonradiative deactivation. In this case the barriers must be considered as being also reflective for the excited ctuomophores. (2) We assume that one of the barriers is reflective for the excited chromophores. Each time the chromophore hits the second barrier, it is deactivated with a probability equal to one. In this case the second barrier is absorbing for the excited chromophores.

Bothbarriersareabsorbingfor the excited chromophores. In this fust paper, we consider the first case only. (3)

The two other cases are treated in the following paper [221.

where k is the Boltzmann constant and T the absolute temperature. The physical solutions of eq. (I) can be expanded in a Fourier series [ 18]

t b, sin(n~&J exp(-n2w2Dt).

(3)

The coefficients n0 u”, 6, and the frequency o are determined by the boundary conditions. Among the solutions, we are interested by the Green function which obeys the following initial condition: G,Oj

(4

= S(VJ-V(J) -

If the rotation is free with no limiting barrier, one has in addition the cyclic condition C(9 + 2% 0 = G(lp,0 ,

(9

and one finally obtains for the expression: G(V(),VP;0= (l/2@

X {1+ 2 ncl cos[n(lp - lp,)] exp(-n’Dt)} .

(6)

If there are two reffecting barriers, one for IJ = 0 and the other for +7= I, one must write:

(7)

which leads to the following expression [ 1S]

2. Rotational diffusion about an axis The rotational diffusion of a straight line about a Exed axis is equivalent to the diffusion of a point moving about a circle. The diffusion equation is then [17]: ap(cp,t)/at = D a2p(q, t)laq2 .

(1)

cpis the angle which locates the position of the point and p(lp,t) is the probability density of the point at time t. D is the rotational coefficient of diffusion which, for spherical molecules of volume V in a continuous solvent of viscosity V, is equal to [17]:

D = kT/6qV ,

(2)

X cos(mrq~~l) exp(-n2n2Dt/12) ,

I

(8)

3. Anisotropy of emission 3.1. Generalexpressionof the emissionanisotmpy We consider an isotropic solution of identical, tluorescent molecules. We assume that this solution is excited by an inftitely short, linearly polarized light PrJ=

212

ph. WahI/ Fluorescence anisotropy of chmmophores

We choose three laboratory referenceaxes,Ox beingthe direction of the measured fluorescence beam, Oy the direction of the exciting beam and 0.z the direction of the electricvibrationof the exciting light.We also assumethat we have defined three molecular axes and that, for a given molecule, these axes coincide with three laboratory axes OX, Yo,Z. at the time of excitation (t = 0). Becauseof the brownian motion, the position of the molecular axes does not coincide any longer with OX@Ye Zu at a time f. This motion is the only one which is dealt with in the theory of fluorescent depolarization by rigid molecules [3]. As for flexible molecules, one must also take into account the internal brownian motion of the ctuomophore with respect to the molecular axes. Furthermore, we must remember, when doing the calculations, that the molecules may assumeseveral conformations in the ground state. Soleillet [ 191and Penin [3] have shown that the degree of polarization is givenby an expression in which the invariants of a fourth order tensor enter. This tensor involvesthe brownian motion of molecules with respect to the axes OX, YQZ,. The molecules are assumedto be rigid. In this paragraph, I derive an expression of the anisotropy factor equivalent to Soleillet’sexpression. Tensors are not used here, since the transition moments of absorption and emissionare assumed to be vectors.This case corresponCs to many strong transitions of aromatic mole-

cules[20].In addition,the caseof flexiblemolecules

The anisotropy factor of a solution is defined by the following relation:

(9 For the set considered, we have: I,=IL+Q2pdw

,

(10)

where e3 is the coordinate of z relative to Or; w is the set of coordinates which detine the position of 0X0, Y. Zo with respect to Ox, y, 2. We choose for these three coordinates the three Euler angles0, p, 7. Then: dw = sinadrr@dy

.

(11)

The integral in expression (10) is calculated for the whole range of coordinate values.p is the probability of excit*n which is proportional to the Oz projection of OA at time zero. Let u3 be that projection; we then have P = (Q2.

(12)

By replacingp and dw by their expressions(11) and (12) in (IO), one obtains: I, = ji(e3)2(a3)2 dw .

(13)

On the other hand, the symmetry of the excitation about Oz entails:

is examined explicitly, Weconsider the unit vectors which defines the direction of the absorption transition moment of a molecule at the tim2f excitation. Let A,. A,, A3 be the projections of OA on the axes OX, Yp Zu. The positions of OX, Y@Z. with respect to Ox, y, z are defined by the matrix C with elements Cij, which is associated with the transformation of OX, Y. 2, into ox, y, z. Let us sayt&t this molecule emits at a time f. We designateby OI? the unit vector which definesthe emission transition moment at that time and byEI( Q(f), Q(t) the components of that vector with respect to OX, Ye 2,. Now we let OX, Y. Z,-,take all possible positions with respect to Ox, y, z, but we keep the A,‘s and the E&)5 constant, thus generating a set of molecules, the fluorescence anisotropyfactorof which we want to calculate.

II = Ix = fy =J(e,)*(Q*

dw =](e2)*(+*

where el and e2 are the projections of s Oy. From (14) one has r,=~~I(et)2+(e2)21(u~)2dw.

do , (14) on Ox and

(1%

The eis are components of a unit vector and therefore obey the following relation: f;(er)‘=

1 9 with

i= 1,2,3.

Eq. (15) then becomes

fi =J;[ I- (e,)2] (o3)2 dw .

(16)

Finally, substitution of (13) and (16) into (9) leads to

213

ph. WahllRhorescenceanirotmpyojclrmmophores

r = frsi&qi%gF-

(17)

1I ,

where the meaning of the bar is given by the followingexpression:

From expression (20) one obtains easily:

=$$2, (c31.I4=;‘;n*, (c3,c3i)2

0-T. = 0.

(26)

Insertion of (26) into (25) leads to the following ex-

pression: ii=

s

udw.

(18)

The coordinates ai and ei are related to the Ais and Eis throughthe matrix C. We then have: 03= FC3iAir

e3= FC3iEi,

i= 1,2,3.

(19)

It follows that

Expression (27) may easilt be rearranged by taking relation (23) and the following relation (28) into account: T(E,J2 = 1 .

03)

One obtains:

t2F ~~‘iA,-

(20)

in order to evaluate expression (20), one introduces the Euler angles by the well known formulas: CS1 = sin/?sina )

c32 = sinpcosa )

c33 =

cosp . (20

Taking expressions (13) and (10) into account, one can can easily find: (C$

= ;lr* ,

Sincez Fi”

c3ic3j = 0 .

(22)

is a unit vector one has

$t2($ (~AiEi(t))2+~) . (29)

Insertionof the rhs. of (24) and(29) into (17) leads to r=;()(FAiEi(‘))2-‘)

.

(23)

= 1-

becomes (a3)2= ;I? .

(24)

On the otherhand,one maywrite

(30)

One notes that FA~E~(~)=cos~~), w*re +lz, is the angle between the vectors a

Taking into account relations (22) and (23), eq. (20)

=

(a3e3)2 =

M(f).

(31) and

OA coincideswith the absorptionpsition

momentat the timeof excitation,whileW(t) coincides with the emission transition moment at the time of emission. In order to calculate the anisotmpy factorof the fluorescent solution at a given time r. we have to cortsider all molecules emitting at that time. Taking into account the additivity law of the anisotropy factor [1,21], together with the expressions(30) and (31). leads to the following expression:

C(Qf’i’i12 +2;:?‘&3,12 i

X [(AIEItAiEI>2t2AiAIEE,J

t O.T.

(25)

In the last expression,O.T. designatesa sumof several integralsof type (18) in whichthe C3;s areelevated to odd power(one or three).It can easilybe shown that these terms are equal to zero.

(32)

ph. Wahl/Fluorescenceanisotropyof chmmophores

214

In this expression OX, Y, Z, are the reference axes and R, is the set of coordinates which defines

the molecular axes at time r; S2oand 52are the sets of coordinates which define the chromophore position at the times 0 and t respectively. W(S2,) is the proba-. bility density which characterizes the molecular couformation. p*(lRD&,R,r) is the probability density of the excited molecules.

Expression (32) may be simplified if the rates of

radiativeand nonradiative deactivation are indeperk dent of the molecular conformation. In that case one

We defme 82(O) and s(f) by their colatitude and their azimuth in the OX, Y,, Z0 axes system which we designate respectively by the letters 6,, cl, 6,. cp.ln fig. 1 these coordinates are indicated at time 0 (q = I&. One can write: cos 9 = o7i(o,??&t, = sin 6 1&I 6 z coscp + sins1 sin6*sinel + coo+cos6*,

which may be rewritten in the followingway:

may write: &+,,~,J&0

3

= k,P(M+,,~OJW

(33)

where kF is the rate of radiative deactivation and P(t) is the probability of the molecule being excited at time t. p(Q&&,R,t) is the probability density which satisfiesthe diffusion equation of the studied problem and meets the following initial conditions: p(n,J$J,I&o)

= 6(JQ6(fi

-ad

*

(34)

dR, = 1 .

(35)

+sin~tsin62sinesin(lp-~~+cos6~cos~*,

(37)

where cpgis the azimuth of s(O) and e = \pg- E1. In this case the rates of radiative and nonradiative deactivation are not modified by the Interaction with the reflecting barriers.30 one can apply expression (36). from which 52, disappears,since there is no diffusion of the molecular axes; R, and n become q. and up.Then we can write eq. (36) as follows:

In addition, one can put: fl

p(Q,,S2&t)dfl,d~

‘Cakingexpressions(33) and (35) into account, relation (32) becomes A

X p(S2,,R,,S2,t)dS1,d~2,d~-

1 .

E

(36)

1

According to expression(36), r(t) is independent of the emissionprobability and involvesonly brownian correlation functions. 3.2. Gse of chromophores rotating between two rejlecting barn& In the case studied here, the molecular axes are not submitted to the diffusion motion. Let us take OZ, coincidingwith the axis about which the chromophore rotates, and OX,-,,OY,, as two axes perpendicu. li3rto 02,. We designateby s(O) the unit vector coinciding at _ e 0 with the absorption transition moment and by ?0 (f) the unit vector coinciding at time t with the emissiontransition moment.

-DY

X

Fig.1. Coordinatesof the transitionmommtsat timezeta 02 is the rotationalaxis

I% Wohl/ 8ikorescence

where G(qo,lp,f)isthe Green function

corresponding to the boundary conditions. Insertion of (37) into (38) leads to the followingexpression: r(t)=A$o~2(~-@~ + Qcos(9

+d;(sin2(~-q,,)),

- ‘P~))~ + d;(sin(cp

- 9&

animh~py

of chromaphores

215

work concerning the fluorescence depolarization by a small fluorescent group attached to a sphericalmacromolecule [I I]. One only has to take the rotational diffusion coefficient of the macromolecule equal to zero. It can easily be verified that: r(0)=dI+dz+d3=;(3cos2ar-l),

+ -4, , (39)

(46)

wAere01is the angle between the vectors z(O) and OE(f). For t = 0~ the anisotropy tends to a finite due

with

I(=)

= Aj

.

(47)

If the rosonal axiss perpendicular to the plane formed by GA(O)and GE(?), expression(45) becomes: dS=0.1(3cos2Q-

1)(3co~~6~- I),

(40)

while in A; and A; cos2e and COSE are replaced by sin2e and sine. The meaningof the angularbrackets is given by the followingexpression: (M(P,~~))~

= ~~d9,1pO)

W90)

G(90,9~t)

00

d9 d9,

-

(41)

I must be taken equal to 2n for the free rotation. In the ground state, the angular distribution of the chromophores is uniform inside the barriers. Then for the free rotation: W,)

(42)

= 1/2a #

and for a limited rotation: Wo) = I/l 1 forO
forpol_

r(t)=O.l[1+3cos2aexp(-4Dt)],

(48)

which can also be written: r(t) = 0.1 + [r(O)-O.l]exp(-4Dt).

(4%

5. Rotation limited by two reflecting barriers In expressions(38) and (41) G(90,9,t) is given by eq. (8) and W(90)by eq. (43). One can easily show that here also

csirl2($D - 90))r

= kirl(9

- cpg)$ = 0 .

Detailsof the calculation of (cosZ!(~- so)J and (co& - ~))r are given in the appendix. Their expressions are given in eqs. (A.@ and (A.17), and inserted into eq. (39), lead to r(t)=d3 tdlyAp--_

sin2(1/2) M0*

4. Free rotation about a fiied axis

+

ncl

[d1B(n)+d2C(n)lexpI-w2(n)Dtl

- WI

In expressions(38) and (4 l), G(~p~,~,r)is detined by eq. (6) and W(cp,) by eq. (42). One obtains easily:

Accordingto (A.2)

(cos2(lp-~p~)~~ = exp(-4Df) ,

u(n) = m/l.


Accordingto (A.7), (A.9), (A.16) and (A-18), one has forI+_lmr:

Ciin2(qJ - ‘pg)),= (sin(cp- cp& = 0 .

WI

B(n) q [l -(-l)Qosu]/f2[1

Insertion of these expressionsinto eq. (39) leads to

andforI=fmr

r(t)=$

B(n) = l/2.

exp(ar)

+d2exp(-Df)

+A,.

(45)

This relation can easily be deduced from our previous

WI

Forl+mr

-J(fr)/4]2

.

(52)

(53)

Ht. Wahl/Eluorescence anisotropyof chromophores

216

qn) = [1-(-1)"c0s~l(l/2)*[1

-wzin)]*,

(54)

andfori=tnr c(n) = I/2 .

(55)

One verifies that relation (46) is still valid. For t = m one obtains Sill21 ~m)=A3+Al~+A2(1/2)2’

si&1/2)

(56)

When the rotational axis is perpendicular to the plane formed by the transition moments, expression (SO)

becomes sin21 r(r) = 0.1 + 0.3 cos2o 12

0

60

120

240

180

Jo0

l

Ml

I’

+ 0.6 cos 2a n$l E(A) exp[-w*(n)&]

,

(57)

Fig. 2 Variationof E(n) with the amplitudef of the rotaliOh curve indexis equal10ILCurve0 represents kn’r)/J2.

The

In figs. 2 and 3, the values of (sin2 f)/J2,[sinqJ/2)]/ (J/2)*, B(n) and C(n) are given for 1comprised between 0” and 360”. As can be seen in fig. 2, fcr each value of 1 there is a valuem such that for n differentfrom m, B(n) is smaller than B(m). There are a few particular values of I to which correspond two values of m. In the same way and according to fig. 3, for each value of I thereis a valueml such that for n different from ml, c(n) is smaller than C(mj). B(n) and c(n) can be practically neglected compared to B(m) and c(ml), except for a few values of n adjacent to m ;uldml. This means that in general the decay of anisotropy contains only a few signifcant exponential terms. In fa 4, w+) is given as a function of I for the firstfive values of n In order to illustrate the number of useful exponential terms in the development of r(r), curves with continuous lines have been drawn in the I range where s(n)/B(m) is greaterthan 5 X 10m3, and with discor~tinuous lines elsewhere.

6. Disawiioa la the use of free rotation, and according to formulas(56) and (40). the asymptoticvaluer(m) dcpads on the projectionof the transitionmomentsof absorptionaad emissionon the rotational axis. For a

;t

C’ Fig 3. Variationof C(n)with the amplitudeI of the rotation Thecawc indexb equalton. Curve0 rcprercnts the variation of lsin’U/2)1/U/2)‘.

limitedrotation,r(w) dependsalso on the amplitude of the motion L as expressed in cq. (56). Whenexcitationis causedby the electronictraasition of lowest energy,tie conespondingtransition momenthas a diection merally veryclose to the transitionmomentof emhion. The threecoeffkients Al. ADA3 ue positive and r(m) iS an incfeasing fun& tion of L For J = 0, it becamcsequalto ro.

Fh WahlJFluorescenceaniwtropy of dr fVmophores

I

I

60

120

[.

I

I

I

‘80

240

300

3

Fig. J.Vuiation of w2b0 with 1 The CUNCindex is equti to k In order to illustrate the influence of a limit on the rotation, we give here numerical examples for the free rotation case and for a limited case with l= 75’. We rrssumcin addition that the rotational axis is perpendicular to the plane of the transition moments. Takin# a = 0 and applying eqs. (48) and (57) one f-rids

r(r) = 0.1 + 0.3 exp(-lDt)

(58)

for free rotation, and I(I) = 0.263 f 0.121 exp(-5.76Dr)

I

I

1

q4

0.6

qe

I ‘#cl

)

I ‘I’

Dr

Fig. 5. Tine course of anisotropy. Indexes of the cuwes UXrespond to the followingcases:(1) free rotation, a = 0, eq.

(58); (2) free rotation,u = 59, cq. (6OJ;(3) limifcd mU!ioian, I = 79. a = O”,eq. (59); (4) limited rotation, I = 75”. a =

w , cq.(61).

easily be extended to the cases where the axis of rotation is attached to a spherical molecule, the size of which is great compared to the fluorescent group. The macromolecular motion and the fluorescent group motion can be considered as independent of each other. The anisotropy is then obtained by multiplying expressions (45) and (SO) by the factor exp(-Rt) where R is the rotational coefikient of the

(59)

for1=75”. With a = So, one obtains for these two cases: r(r) = 0.1 - 0.103 exp(-4Df)

Appendix (60)

t(t) = 0.044 - 0.41 exp(-5.76Df)

I In the 1h.s. of eq. (4 1). we replace u by cos2(9 - rpg),and in the As. we use the following development:

(61)

The CUNCS corresponding to formulas (58). (59), (60) and (61) are drawn in fig. 5. Wo will fktlly stress that

I 92

macromolecule [ 11.16).

t 0.013 exp(-23Dt)

- 0.005 exp(-23Dr).

217

these calculations can

cos2(lp- p(J)= coszlpcos2rp()+ sin& sin Qo. For W&+-Jand G(Q,,+v), we use expressions (43) and (8). We then obtain:

218

Ht. W&l/ Ruwescence aniwbvpy ofch7~nwph0~e~

(ca2(lp-Ip

0))t =(1/P)

If i = kn/2 and n = k, formulas(Ad) and (A.4) are inapplicable.In that case eqs. (A.l) and (A.2) lead to

f2t12 l

1

2

(A.0

&n/2 1 cos22xdx=kn/4= l/2, 0 kn12

+I=

1&k)= j- sin2%cos2tdr=o, 0

and

11=jccls2xdx+inu.

B(k) = 2I$k)/l* = 112

0 I

12=Jsin2x&=f(l-cos21).

(A.21

(A-9)

with d(k)=4.

0

(A.lO)

One then obtains +1;=~(1-cosu)=sin*l.

(A.31

2 In an analogousway, we obtain by applying eq. (40 kos(~-~o)),=(l/l~)

I&n) = jcos?x cos(nrcr/I)dx ,

i;+f; 1

0

f&n) =

jsin2.r

+2ij, [~~(n)+f~~)Jexp[-w~~)DIl

ms(nm/l) cLx.

(A-4)

0

,

(A.1 I)

I

with

If I + kn/2, where k is an integer, one can write wsxdr=sinl,

13(n)=djlcos(2 +tm/f)r + ux(*-nr/&]

dx

0

‘Jsinx”= 1 -cosl.

= (- lr sin21/2(1 - n2n2/41*) 9

(A-9

0 horn which one obtains:

+1;=2(1

-wsr)=4sin2(r/2)

.

on the other hand, =]I-(-lYcos21]/2(1-n2n2/412).

(A.6) I,(n) =I cmx as(mrrlr)~,

Then

(A-12)

0 18(n)=jsinx~(mrrll)dx. 0

(A.7)If1# k'n,where ,-I

(A.13)

k’ is an integer,one may writ: for all

valuesof n:

ph. Wahl/ Fluorescence anisotropy

$(fl) =

219

Applying relation (A.?) gives

J[cos(l

+nn,l)x + cos(1-mr,l)x] dx

u2(k/2) = 1 .

0 = (-1)” sinI/[l - M/P]

of chromophores

(A.19j

(A-14)

,

References

I r,(n) - $[sh(

1+mr/l)x t sin( 1 -nIr/l)] dx

0 = [ 1 - (-1)” cosq/[ 1 -&?/P]

)

(AX)

from which one obtains c(n) = w*ll+)

+1&01

= [ 1 - (- 1)” cosr]/(l/2)2( 1 - &/12)2

.

(A.16)

Expression (A. 11) may be written kos(q - @,

= [sin2(//2)]/(r/2)2 (A.17)

where uz(n) is given by (A.2) and c(n) by (A-16), exceptif1=k’nandn=k’. Let us put 2k’ = k. From (A.12) and (A.13), we have /,(k/2) = 7’

cos’x dr = knj4 = 112,

0 knl2 18&/2) = j 0

sinxcosxdx=o,

from which dne obtains qk/2)

= (211’) l,(k/2) = l/2 ,

(A.18)

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