welcome to the lecture number twenty four the last few lectures what we were discussing
is the fluorescence quenching and we have seen basically few things let me list it out
dynamic quenching i also have seen static quenching
i have seen combined static and dynamic quenching i also have seen that apparent static quenching
this is because of the quencher is near to the fluorophore right and now what i will
i will try to tell you is the another type of quenching which is for the fractional quenching
lets say i have two such kind of fluorophore in a molecule and the proteins are one of
the typical example for such kind of fractional quenching
so lets say i have this protein over here i have just drawn my protein like this way
looks good i have one fluorophore which is present here and one fluorophore which is
present over here what i can comment about the position of this two fluorophore this
fluorophore is easily reached by the quencher but this is not this is present in buried
location ok so this is accessible and this is not accessible if so
now if i add quencher so quencher will come and interact with this fluorophore but quencher
cannot come and interact with this fluorophore so the fluorescence intensity due to these
fluorophore will decrease this fluorophore will decrease fluorescence
intensity of this fluorophore will decrease but fluorescence intensity of this fluorophore
will be unaffected right fluorescence intensity of these fluorophore will be unaffected clear
now let me write this as site a because this is accessible so this is site a and lets me
write this as my site b because this is in the buried without addition of quencher when
there is no quencher then the total fluorescence intensity i zero can be written as i zero
for a position of the a and plus i zero of b i can simply write that one now as i argued
that when you will add the quencher the quencher will quench the fluorescence of a but it will
not going to quench the fluorescence of b that i argued like that way so the fluorescence
intensity in presence of quencher which is typically denoted by i i can write this i
as one over one plus k d right this transformer quenching constant for this so let me write
k d a into q it initial intensity i zero a right remember i zero by i equal to one plus
k d a into q so from here i have written plus i zero b
because the i zero b fluorescence unaffected because quencher cannot reach to that site
b site two quench the fluorescence of this fluorophore which is present here which is
marked as b so i zero b will be i zero b even after you will add the quencher molecule clear
so what i can do i can take the ratio of i zero by i as usual i zero by i is equal to
ok i zero a plus i zero b into one plus k d a into q divided by i zero a plus i zero
b into multiply multiply it by one plus k d a into q now what i intended to do i intend
to plot i zero by i versus q obviously this is a complicated right think and if you plot
this i zero by i versus q let me plot it somewhere here so this is my i zero by i versus q what
you will see that the curve will look something like this the curve will look something like
this it means that this is not a straight line
obviously it is not so in this case my question is how i will going to calculate what is the
value of k d a right let us move forward and see how can we do
that lets write this i equal to i zero a by one plus k d a into q plus i zero b thats
what i was writing and let me write this these quantity i zero minus i like so what is i
zero write here i zero equal to i zero a plus i zero b right so i zero minus i i
can write it as i zero a plus i zero b and then minus i minus i zero a divided by one
plus k d a i dont have any k d b because that fluorophore present in that b site but it
site is not accessible to the quencher into q minus i zero b so this is equal to i zero
a to one minus one over one plus k d a into q ok this is equal to i zero a to one plus
k d a into q minus one divided by one plus k d a into q this is equal to i zero a k d
a into q divided by one plus k d a into q so this one i write as delta i i can also
write this as a delta i so one over delta i is equal to one plus k d a into q divided
by i zero a k d a into q now i multiply both side by i zero by i zero
i will get i zero by delta i is equal to i zero one plus k d a into q divided by i zero
a k d a into q so if you just simplify this and proceed what you will get is i zero a
plus rearranging after rearranging i zero b divided by i zero a k d a into q plus i
zero a plus i zero b k d a into q divided by i zero a k d a into q lets assume that
f a this fraction is equal to i zero a divided by i zero a plus i zero b then this equation
will change to so i zero by delta i this equation will be equal to one over f a k d a q plus
one over f a remember that f a is equal to i zero a divided by i zero a plus i zero b
now you see we got this equation now if i plot instead of plotting i zero by i now i
am plotting i zero by delta i is a function of one over q if i plot i will get a straight
line right where the slope will be equal to one over f a k d a and the intercept will
be equal to one over f a so like this i will be able to calculate what is the value of
k d a ok now the question is that as i said that in this particular case this position
is completely buried position is it is not accessible at all right now if both are partially
accessible so in this case my equation will be anyway so i zero is equal to i zero a plus
i zero b but in this case i will be equal to i zero a divided by one plus k d a into
q plus i zero b divided by one plus k d b it is for the site b so i have written k d
b in this case what i am telling in this case
both the sides are accessible but not at same extent right so the stun harmer quenching
constant may be defined for the different side although the same molecule here my condition
is the quencher is the same molecule it is also interacting with the site a it is also
interacting with the site b the fluorophore which is present in site a and site b they
are also be same but their accessibility is little bit different right and thats why all
the differences are coming the first case what i showed is that the one is fully accessible
another is zero accessibility there is completely inaccessible to that quencher but here i said
that is somewhat accessible ok so then this will be equal to q so now if i just write
this ratio i by i zero what will happen i will simply get you can just see later on
this will be equal to i zero a plus i zero b into one over one plus k d a into q plus
i zero b divided by i zero a plus i zero b to one over one plus k d b yeah that k d b
will come over here into q so this i will write as f a that is thats what i have defined
like this way here you see f a so f b also this will be small a anyway it doesnt matter
f a f b same thing so i zero b divided by i zero a plus i zero b is defined like this
way so this is f a to one by one plus k d a to
q plus f b to one by one plus k d b into q so for having more than one site more than
two sites so i will just simply write the sum is f i divided by one plus k d i into
q so this will going to be little complicated so and the one of the limiting thing is little
easier but let us proceed with this let me show you the emission spectra of that what
I showed you earlier that protein having two fluorophore this is the same diagram let me
draw over here thats what i that that one right the same thing now i have one over here
as i said i have one over here and these protein having these two fluorophore present in two
different location and consider that for this fluorophore mu e is greater than mu g and
this whole protein is present inside water buffer right so outside there is a buffer
so this fluorophore is accessible to buffer more compare to this fluorophore is that its
a double moment is more than grossly double moment you will remember the solver to provision
of a fluorescence so the fluorescence of this one will be red shifted compared to this one
let me write that red shifted fluorescence this will be blue shifted
fluorescence so if you take this fluorescence spectra of this what you will going to see
is fluorescence spectra emission spectra right intensity versus wavelength
see this something over here probably you will going to see such kind of thing looks
like there is something over here right because one is little red shifted other is little
blue shifted so in this case this part of fluorescence is given by this fluorophore
and this part of the fluorescence because this is a red shift higher longer wavelength
lower energy this part of the fluorescence is given by this fluorophore as i said that
one of them is accessible more accessible compare to the other that means these fluorescence
intensity that that these these i by i zero thats what i just have calculated this is
a function of wavelength if i calculate at this wave length hm at this
wave length then this i by i zero value and if i calculated this wave length then the
i by i zero value will be different suddenly it will be because when i will calculate the
i by i zero for this wavelength then there will be lot of change in the fluorescence
intensity when i will calculate the i by i zero at this wavelength then there will be
less change in the fluorescence so here is the less change is a more change in the fluorescence
intensity so these are the function of lambda ok so this function of lambda then that summation
what i have written this f i that also has to be function of lambda because at this particular
wavelength the contribution of this one and this one is something different than the at
this wavelength because the emission spectrum shifted for the location a and location b
you remember here is this is my location a and this is my location b so emission spectrum
is shifted if the emission spectra is same exactly same for the location a and location
b then i want to have such kind of wavelength dependency and this f i is also the wavelength
independent but in this case as there is a change so i
have this f i is a function it depends on the wavelength so then this divided by one
plus k d i lambda into q so i can write this k d i lambda right this k d i lambda whether
it should be depends on the wavelength or not this is my question whether this k d i
i means for the i th one it could be a it can be b if you have multiple then c d and
so on and so forth so whether for a particular site either its k d a or k d b is does
it depends on the wave length obviously no so i should write this only k d i this k d
i lambda right it does not depend on the wavelength because its a stung harmer quenching constant
so it it will not going to depends on the wavelength it is the interaction between the
fluorophore and the quencher right so the where that molecule will emit i mean it doesnt
matter for it ok so ultimately what will going to have this
i lambda by i zero lambda is equal to sum over i f i lambda by one plus k d i into q
like if i have three such different sites then i will i can simply write i lambda by
i zero lambda equal to f one that is for my first site divided by one plus k d one to
q plus f two one plus k d two to q plus f three one plus k d three into q now if i lambda
denotes my excitation sorry emission spectra right
emission spectrum this i lambda has the component of three different species in this case you
see this is for the hm f one f two and f three right so this i lambda if i multiply it by
f i lambda that will going to give me intensity for the i th site at this particular lambda
i will simply get that one right so obviously this i lambda is a measured quantity right
i will get this one this is my i lambda it has the component of b and it has component
of a so i measured it together and now this f i is that fractional fluorescence intensity
right for this i th component and i will get this i i lambda right so if i analyse lets
say for example if i analyse these thing in a global fitting method then what you will
get you will get the global fitting method means that same equation will be applied for
all different wavelengths note it down these contribution f one f two f three they
are wavelength dependent i by i zero wavelength dependent but k d two or k d one or k d three
they are not wavelength dependent so for each wavelength if you plot i by i zero and then
just feet this one right so you will get the unique value of k d one k d two k d three
so you will get unique values of k d one k d two k d three and also all the f i lambdas
at each wavelength it f i lambdas at ok f i lambda means at each wavelength ok and
what i know another information is f i sum of our i f i is equal to one that i know that
means what you will going to get is all these a f i lambda right you will get all these
f i lambda and this is your measured spectrum so if you multiply this measured one with
all these f i lambda you will get the spectrum for the i th species and thats what we called
this as quenching resolved
emission spectra ok so let us stop here and we will
continue on the next day thank you very much