so welcome back today we will going to start our lecture number thirty we are discussing
fluorescence anisotropy and we are done till the perrin equation so what is perrin equation
let me write it once again so perrin equation is written as r zero by r s s equal to one
plus tau f divided by tau or so where this r zero value is initial anisotropy r s s my
steady state anisotropy and this tau f is my fluorescence lifetime and tau over is orientational
relaxation time
so from here what you can see is that without if you have some idea about these value of
r zero right and this r s s can be measured very easily from this fluoremeter where you
have this polariser and analyser right normal fluoremeter does not have this fluorophore
polariser analyser but if you can install a polariser analyser so that their excitation
light is polarised light and the emission you are going to analyse which is the perpendicular
direction and which is in the intensity and what is the intensity of the perpendicular
direction and what is the intensity of the parallel direction
then you will be able to calculate this r steady state easily right and if you have
this knowledge of this fluorescence lifetime then you will be able to calculate this tau
over now if you recall that our stoke einstein relationship then this tau over can be written
as tau over equal to eta v by r t right this is my stoke einstein relationship where this
eta is equal to the viscosity of the medium v is molecular volume and r is my universal
gas constant and t is my absolute temperature
now if you plug in this expression for this tau or in the perrin equation what we will
going to see is the following one over r s s equal to one over r zero plus tau f v by
into r divided by r zero into v into temperature by viscosity right it means that now if i
plot or this is this is the measurable quantity this is your measurable quantity these is
the property at a particular temperature obviously you know the viscosity so then t by eta is
known quantity and i can measure this r steady state for a different value of t by eta against
the temperature obviously viscosity will change so t by eta will change and for a different
values of this t by eta i can measure this steady state anisotropy and then i can plot
it right so let us plot that same so if i plot what
you will going to see is one over r steady state versus t by eta we will going
to see a straight line and intercept will be equal to one over r zero so from here you
will get some idea of this r zero right and then slope will be equal to tau f r by r zero
v now if tau f the fluorescence lifetime is known r obviously is a constant and r zero
if you have fair idea of this r zero then you can calculate v that is the volume and
this is well used equation in application of the fluorescence anisotropy so from with
this you can actually determine the size of lets lets say you are working with a protein
and this protein is getting denatured once the protein is getting denature the size will
increase and if you have one fluorophore already attached to the protein then by measuring
the fluorescence anisotropy that means the fluorescence this r s s value by measuring
this r s s value what will be able to do you will be able to see that at this particular
condition what is the volume of the protein or that means the size of the protein and
so on right now one of the important thing here is that
what will be the value of r zero right thats thats the biggest question right so that should
be measured otherwise i cannot say anything about this and we have seen that limiting
value of r zero you remember our first equation i have written r equal to i parallel minus
i perpendicular divided by i parallel plus twice i perpendicular right now i am not talking
about i am not including this g because considering the g equal to one g is that one then just
a factor will be there right so considering the g value is equal to one and i am now doing
all those thing now in case of r time dependence then this
is nothing but r t is equal to i parallel t minus i perpendicular t and so on so that
means at time t equal to t equal to zero that is the my initial anisotropy r zero and that
is whatever i have written here as r zero so in this case i have to determine what will
be the value from that particular equation when i said that this i parallel minus i perpendicular
divided by i parallel plus twice i perpendicular obviously at time t equal to zero there was
no emission from this perpendicular component and there is only emission from this parallel
component makes it r zero equal to one but whether this is true and we are i also said
that when this r value will change as time goes because the intensity of the perpendicular
component will increase and parallel component will decrease because of the reorientations
of the molecules or transition dipoles when at at the excited state so as time goes
there will be the the intensity of the parallel and perpendicular components will be equal
and in that case you will see that r will going to be zero ultimately at a certain after
a certain time now let us see this things right in more detailed
way so let us now consider that you have excited the molecules along z axis so z axis is your
polarization for the excitation light ok so let me write here so polarization of the
excitation light is along z lets i consider it over here so if i consider that only those
molecules thats i i said the only the molecule whose transition moments are along the z axis
they will be preferentially excited thats what i said earlier again i am telling the
same thing so those molecule will be excited and if the direction of the absorption transition
moment and emission transition moments are same like they are co linear then the emission
will be also along z axis that is my condition now i am imposing here the direction of the
absorption transition moment and emission transition moment they are co linear then
the emission will be also along the z axis take a note on that ok
now let me tell you that now another molecule has been rotated an angle theta right and
now i am looking at this molecule so now this bold arrow is my molecule which has been rotated
at an angle theta from the direction of the excitation and now i am looking at this molecule
so here you see the rotation of the molecule makes a component of this i parallel s cosine
theta as you know and but the molecule could can rotate like if this is my x z axis the
molecule can rotate from here to this side as well as from here to this side in all the
cases the angle is theta this side angle is theta this side angle is theta that means
i need to worry about all these different angles for a fixed value of theta that means
if i if i assign this angle as pi then the value of pi should be zero t two pi thats
what i have written over here so now here you see for this perpendicular component
this should be sin theta sin phi correct so the intensity those are the electric field
so those are the intensity i parallel will be equal to cos square theta and i perpendicular
will be sin square theta sin square phi but here also please note that we should take
all values of sin square phi that means i should take the sin square phi average right
so i should take the sin square phi average so let me find out that
so sin square phi average is equal to integration zero to two pi sin square phi d phi divided
by integration zero to two pi d phi so and i already got this i parallel equal to like
this i perpendicular is like this so if i can calculate this if i can calculate this
i can simply replace this sin square phi with this this i can do as i said because that
all values between all values of phi between zero to two pi will have the equal
probability so now i can evaluate this integration zero to two pi sin square phi d phi right
that part so if you integrate this then you will
get phi by two minus one over four sin square phi limit i will put zero to two pi divided
by this will going to this is simple this is just two pi so then this will going to
be pi by two pi this is equal to half so as i got this sin square phi average is this
half so i will simply write i parallel is equal to cosine square theta and i perpendicular
equal to half sin square theta you see that this is now half sin square theta
now r is equal to i parallel minus i perpendicular divided i parallel plus twice i perpendicular
which will going to give me cos square theta minus half sin square theta divided by cosine
square theta plus plus two half and two this will cancels sin square theta so this is this
is equal to three and then i will just simplify this and what you will get is cosine square
theta minus one divided by two now as you can see here that for the value of theta if
the theta equal to zero right the theta value could be anything right so lets let me put
for theta equal to zero that means it has not oriented because you have excited along
the z axis and just immediately after excitation you are looking at the r value right so theta
equal to zero so for theta equal to zero this is nothing but three minus one divided by
two sorry this is equal to one for theta equal to zero the value of r equal
to one so here i go the initial anisotropy is one right and now as the molecule will
rotate the theta will change theta will increase and it will increase and increase and ultimately
at some value of theta then isotropy will be zero so when the cos square theta equal
to one by three as you can see from this equation that when cos square ok let me write this
thing this is important so when for theta equal to zero r i got equal to one and when
theta equal to fifty four point seven five degree it means cos square theta equal to
one third what you get you get r equal to zero right that means the molecule is not
necessarily has to orient itself ninety degree to make the anisotropy equal to zero only
fifty four point seven degree orientations will make the anisotropy equal to zero
now whatever i said here right is the ideal situation that means that your your excitation
light is polarised along z axis only the molecules which whose transition moment absorption transition
moment is oriented along z axis only those will going to be excited like that now why
and now this molecules will reorient themself and the theta value will change i consider
that the direction of the transition absorption transition and emission transitions are co
linear so the emission will also come along that line right as i have excited and then
i got this equation however it is not true that once you use a
vertically polarised light or polarised light electric filed is oscillating along the z
axis that will only excite the molecules which are oriented along the z direction obviously
there will be some component of even the electric field is like this and the molecules like
this this the the component which is along this direction the electric the component
of the electric field which is the along this direction that will also be used to excite
the molecule little bit it is true that the those molecules which are oriented just perpendicular
to the direction of the electric field of the excitation light those will not excited
at all thats true but there is a possibility that the molecules which are oriented at
certain angle from the co linear geometry of compared to the excitation polarization
there will be some possibility of the excitation and that is known as the excitation photo
selection of the fluorophore let me discuss these things over here
so see this is the excitation photo selection of fluorophore
this is what i said the molecules which are oriented like this way they will be excited
thats sure but the molecules which are oriented like this way will also some chance for the
excitation right thats what i said and that that is proportional to the cos square theta
if the molecule is oriented at an angle theta from the polarisation of the excitation light
right so that will be proportional to cosine square theta ok so let me write this over
here so as you use ok let me write it like this way so the excited state population will
be distributed symmetrically around z axis in this particular
case ok so that distribution right i can write this
as f theta d theta that is possible to show that the is equal to cosine square theta sin
theta d theta right so and this anisotropy is given by i have already seen three cosine
square theta minus one by two right thats what i already got so in this a case what
i will do let let us continue so in this case that cosine square theta right not here in
this case that cosine square theta in my earlier equation is that the value of theta is not
a single value because now the theta value is different and that is my distribution over
here what i said this this this is distributed so in a molecule so an each molecules is make
a different angle theta and for a different different theta value all the five values
are possible right so this is a like a complicated situation
and in this case i should not write the equation as cos square theta but i should write the
average cos square theta so the then i have to take i have to take the average cos square
theta right so for this average cos square theta what i will write sorry average cos
square theta equal to cos square theta f theta because this is my distribution f theta d
theta is my distribution and that integration from zero to pi by two divided by zero to
pi by two f theta d theta and if you evaluate this this is integration zero to pi by two
cosine to the power four theta sin theta d theta divided by zero to pi by two cos square
theta sin theta d theta so if you evaluate you will see that these
value is three by five right so all this theta what is the distribution of theta is because
of the excitation photo selection and this distribution is as this distribution is because
of the excitation photo selection this is my initial value of the cos square theta average
so now if put this cos square theta average in my original equation lets see that r equal
to three cos square theta but now this is average because i have so many things minus
one divided by two so then this is equal to for the initial value three into five minus
one by two this is nothing but zero point four so r zero what i will get is a zero point
four and this is true for when the absorption and emission transition moments are co linear
with each other if they are not then this value will not like that if not if the absorption
and emission transition moments will make an angle beta
in between them right then this r zero equation right now i am writing
r zero i am not writing r right now so r zero will can be written as two by five to three
cosine square beta divided by say minus one by two right so when this beta equal to zero
that means they are co linear right immediately what you will going to see is that r zero
equal to zero point five if beta equal to fifty four point seven five then what will
immediately get that r zero equal to zero if beta equal to ninety degree that means
if this absorption and emissions are perpendicular to each other right then for example for example
let me give you example that will then it will be much clear to you
lets say i have this molecule our famous example many times i use this molecule enthroning
right and i told you this absorption for the s zero s one is along this direction and s
zero s two is along this direction i consider that emission is also the same but if you
now excite from s zero s two and but you will get the emission from the s zero s one from
s one to s zero because of the so you are exciting along this direction so here this
direction is your excitation because you are exciting from s zero to s two but emission
is along this direction is different right so they are perpendicular to each other
now if you excite emission is from s one to s zero now if you excite from s zero to s
one and then s one to s zero then they are co linear to each other right then they are
collinear so in that case that is the value beta value is equal to zero and in this case
the beta value is equal to ninety degree so for beta value equal to ninety degree what
you will get is this r zero value equal to minus zero point two
so what you can see is that the the value of r zero or initial anisotropy or fundamental
anisotropy right it depends it depends on many factors right many factors means basically
this is what is the angle between the absorption transition moment and the emission transition
moment right so for a different different molecules this angle will be different so
the value will be different so this has to be measured right and we can measure it
by time resolve anisotropy decay as i told you if you have a very very high time resolution
like a ten to second time resolution using the fluorescence up conversion system then
by putting this polariser in the ten to second up conversion system either at the parallel
or in the perpendicular position compared to the direction of the excitation or the
polarisation of the excitation is lesser been you will be able to find out the value of
r at very very early time with a very a high time resolution and that will be going to
be your r zero value right ok so this all about this fluorescence anisotropy
and as we have little more time then i think that we should continue our discussion on
little more atmospheric of this fluorescence anisotropy which is the associated anisotropy
decay associated anisotropy decay so right now what i said is that the decay our anisotropy
decay of a fluorophore is present in a particular media right in a homogeneous media now i am
thinking of that if i have two def two fluorophores right but present together in a solution but
their environment are different their the environment of two different types of fluorophores
are different but they will going to give me give the contribution to the anisotropy
decay then what will be the effect so lets take this as that i have i of the
i th component is defined as alpha i so this my lifetime part right t by tau f i and my
r t part r i t is equal to r zero i into e to the power minus t by tau or i right so
i can write the fractional intensity of the i the component if it is only two then first
and second component at any time t just i write this as f i t equal to alpha i e to
the power minus t by tau f i divided by sum over i alpha i e to the power minus t by tau
f i that means the time dependent anisotropy that r t r t should be given by sum over i
because if it has more weightage the early time then the weightage of the r r r i
t will be there because it is now more weightage and in the later time
so i can simply write r t equal to sum over i f i t r i t so you see this is a complicated
thing right and as expected the anisotropy will also be complicated usually anisotropy
looks like that is started from some value r zero and exponentially decay but in this
case if you see this anisotropy decay it will look something like this but using such kind
of formula you can analyse it without any problem so if i now plot r t as a function
of t over here then it could looks like this like this or it could look like this so it
doesnt look is anisotropy but this is because of this two different fluorophore two or more
different fluorophores present whose environments are quite different one quick example is that
the tryptophan right so if you have tryptophan in a in a protein
right when tryptophan is present in the protein the r t value right or this tau over value
is very large because the protein is very large but if it is now free right then the
tau over value is really small and the lifetime is also quite different so for such kind of
complicated cases what you will going to see such kind of associated anisotropy decay and
you need to really careful you have to be really careful for the analysis of such kind
of anisotropy decay ok so we will finish here and
thank you for your kind attention