WEBVTT
00:17.020 --> 00:21.200
welcome to the lecture number twenty four
the last few lectures what we were discussing
00:21.200 --> 00:29.290
is the fluorescence quenching and we have
seen basically few things let me list it out
00:29.290 --> 00:44.910
dynamic quenching
i also have seen static quenching
00:44.910 --> 01:21.330
i have seen combined static and dynamic quenching
i also have seen that apparent static quenching
01:21.330 --> 01:43.101
this is because of the quencher is near to
the fluorophore right and now what i will
01:43.101 --> 02:01.709
i will try to tell you is the another type
of quenching which is for the fractional quenching
02:01.709 --> 02:09.310
lets say i have two such kind of fluorophore
in a molecule and the proteins are one of
02:09.310 --> 02:13.530
the typical example for such kind of fractional
quenching
02:13.530 --> 02:23.410
so lets say i have this protein over here
i have just drawn my protein like this way
02:23.410 --> 02:34.500
looks good i have one fluorophore which is
present here and one fluorophore which is
02:34.500 --> 02:42.740
present over here what i can comment about
the position of this two fluorophore this
02:42.740 --> 03:09.850
fluorophore is easily reached by the quencher
but this is not this is present in buried
03:09.850 --> 03:27.430
location ok so this is
accessible and this is not accessible if so
03:27.430 --> 03:38.770
now if i add quencher so quencher will come
and interact with this fluorophore but quencher
03:38.770 --> 03:47.030
cannot come and interact with this fluorophore
so the fluorescence intensity due to these
03:47.030 --> 03:59.500
fluorophore will decrease this fluorophore
will decrease fluorescence
03:59.500 --> 04:11.860
intensity of this fluorophore will decrease
but fluorescence intensity of this fluorophore
04:11.860 --> 04:38.300
will be unaffected right fluorescence intensity
of these fluorophore will be unaffected clear
04:38.300 --> 04:49.590
now let me write this as site a because this
is accessible so this is site a and lets me
04:49.590 --> 04:57.550
write this as my site b because this is in
the buried without addition of quencher when
04:57.550 --> 05:06.719
there is no quencher then the total fluorescence
intensity i zero can be written as i zero
05:06.719 --> 05:18.620
for a position of the a and plus i zero of
b i can simply write that one now as i argued
05:18.620 --> 05:24.599
that when you will add the quencher the quencher
will quench the fluorescence of a but it will
05:24.599 --> 05:31.000
not going to quench the fluorescence of b
that i argued like that way so the fluorescence
05:31.000 --> 05:38.580
intensity in presence of quencher which is
typically denoted by i i can write this i
05:38.580 --> 05:52.819
as one over one plus k d right this transformer
quenching constant for this so let me write
05:52.819 --> 06:05.699
k d a into q it initial intensity i zero a
right remember i zero by i equal to one plus
06:05.699 --> 06:15.039
k d a into q
so from here i have written plus i zero b
06:15.039 --> 06:21.339
because the i zero b fluorescence unaffected
because quencher cannot reach to that site
06:21.339 --> 06:26.719
b site two quench the fluorescence of this
fluorophore which is present here which is
06:26.719 --> 06:36.270
marked as b so i zero b will be i zero b even
after you will add the quencher molecule clear
06:36.270 --> 06:49.069
so what i can do i can take the ratio of i
zero by i as usual i zero by i is equal to
06:49.069 --> 07:16.039
ok i zero a plus i zero b into one plus k
d a into q divided by i zero a plus i zero
07:16.039 --> 07:29.800
b into multiply multiply it by one plus k
d a into q now what i intended to do i intend
07:29.800 --> 07:38.139
to plot i zero by i versus q obviously this
is a complicated right think and if you plot
07:38.139 --> 07:56.710
this i zero by i versus q let me plot it somewhere
here so this is my i zero by i versus q what
07:56.710 --> 08:05.399
you will see that the curve will look something
like this the curve will look something like
08:05.399 --> 08:12.879
this
it means that this is not a straight line
08:12.879 --> 08:20.759
obviously it is not so in this case my question
is how i will going to calculate what is the
08:20.759 --> 08:27.229
value of k d a right
let us move forward and see how can we do
08:27.229 --> 08:42.460
that lets write this i equal to i zero a by
one plus k d a into q plus i zero b thats
08:42.460 --> 08:54.630
what i was writing and let me write this these
quantity i zero minus i like so what is i
08:54.630 --> 09:05.870
zero write here i zero equal to i zero
a plus i zero b right so i zero minus i i
09:05.870 --> 09:21.620
can write it as i zero a plus i zero b and
then minus i minus i zero a divided by one
09:21.620 --> 09:30.320
plus k d a i dont have any k d b because that
fluorophore present in that b site but it
09:30.320 --> 09:52.000
site is not accessible to the quencher into
q minus i zero b so this is equal to i zero
09:52.000 --> 10:14.300
a to one minus one over one plus k d a into
q ok this is equal to i zero a to one plus
10:14.300 --> 10:34.350
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
10:34.350 --> 10:50.870
a into q divided by one plus k d a into q
so this one i write as delta i i can also
10:50.870 --> 11:07.410
write this as a delta i so one over delta
i is equal to one plus k d a into q divided
11:07.410 --> 11:31.100
by i zero a k d a into q
now i multiply both side by i zero by i zero
11:31.100 --> 11:49.140
i will get i zero by delta i is equal to i
zero one plus k d a into q divided by i zero
11:49.140 --> 12:04.810
a k d a into q so if you just simplify this
and proceed what you will get is i zero a
12:04.810 --> 12:18.290
plus rearranging after rearranging i zero
b divided by i zero a k d a into q plus i
12:18.290 --> 12:43.650
zero a plus i zero b k d a into q divided
by i zero a k d a into q lets assume that
12:43.650 --> 12:56.600
f a this fraction is equal to i zero a divided
by i zero a plus i zero b then this equation
12:56.600 --> 13:15.450
will change to so i zero by delta i this equation
will be equal to one over f a k d a q plus
13:15.450 --> 13:31.750
one over f a remember that f a is equal to
i zero a divided by i zero a plus i zero b
13:31.750 --> 13:40.420
now you see we got this equation now if i
plot instead of plotting i zero by i now i
13:40.420 --> 13:53.950
am plotting i zero by delta i is a function
of one over q if i plot i will get a straight
13:53.950 --> 14:14.350
line right where the slope will be equal to
one over f a k d a and the intercept will
14:14.350 --> 14:24.230
be equal to one over f a so like this i will
be able to calculate what is the value of
14:24.230 --> 14:39.510
k d a ok now the question is that as i said
that in this particular case this position
14:39.510 --> 14:48.560
is completely buried position is it is not
accessible at all right now if both are partially
14:48.560 --> 15:05.790
accessible so in this case my equation will
be anyway so i zero is equal to i zero a plus
15:05.790 --> 15:18.230
i zero b but in this case i will be equal
to i zero a divided by one plus k d a into
15:18.230 --> 15:29.730
q plus i zero b divided by one plus k d b
it is for the site b so i have written k d
15:29.730 --> 15:33.870
b
in this case what i am telling in this case
15:33.870 --> 15:40.610
both the sides are accessible but not at same
extent right so the stun harmer quenching
15:40.610 --> 15:45.529
constant may be defined for the different
side although the same molecule here my condition
15:45.529 --> 15:49.220
is the quencher is the same molecule it is
also interacting with the site a it is also
15:49.220 --> 15:52.850
interacting with the site b the fluorophore
which is present in site a and site b they
15:52.850 --> 15:57.940
are also be same but their accessibility is
little bit different right and thats why all
15:57.940 --> 16:03.880
the differences are coming the first case
what i showed is that the one is fully accessible
16:03.880 --> 16:10.630
another is zero accessibility there is completely
inaccessible to that quencher but here i said
16:10.630 --> 16:21.310
that is somewhat accessible ok so then this
will be equal to q so now if i just write
16:21.310 --> 16:28.270
this ratio i by i zero what will happen i
will simply get you can just see later on
16:28.270 --> 16:46.370
this will be equal to i zero a plus i zero
b into one over one plus k d a into q plus
16:46.370 --> 16:59.690
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
16:59.690 --> 17:11.809
will come over here into q so this i will
write as f a that is thats what i have defined
17:11.809 --> 17:23.010
like this way here you see f a so f b also
this will be small a anyway it doesnt matter
17:23.010 --> 17:37.720
f a f b same thing so i zero b divided by
i zero a plus i zero b is defined like this
17:37.720 --> 17:47.860
way
so this is f a to one by one plus k d a to
17:47.860 --> 18:06.290
q plus f b to one by one plus k d b into q
so for having more than one site more than
18:06.290 --> 18:19.120
two sites so i will just simply write the
sum is f i divided by one plus k d i into
18:19.120 --> 18:30.170
q so this will going to be little complicated
so and the one of the limiting thing is little
18:30.170 --> 18:39.770
easier but let us proceed with this let me
show you the emission spectra of that what
18:39.770 --> 18:45.590
I showed you earlier that protein having two
fluorophore this is the same diagram let me
18:45.590 --> 18:59.980
draw over here thats what i that that one
right the same thing now i have one over here
18:59.980 --> 19:08.000
as i said i have one over here and these protein
having these two fluorophore present in two
19:08.000 --> 19:19.160
different location and consider that for this
fluorophore mu e is greater than mu g and
19:19.160 --> 19:34.720
this whole protein is present inside water
buffer right so outside there is a buffer
19:34.720 --> 19:42.690
so this fluorophore is accessible to buffer
more compare to this fluorophore is that its
19:42.690 --> 19:47.400
a double moment is more than grossly double
moment you will remember the solver to provision
19:47.400 --> 19:55.200
of a fluorescence so the fluorescence of this
one will be red shifted compared to this one
19:55.200 --> 20:15.780
let me write that red
shifted fluorescence this will be blue shifted
20:15.780 --> 20:24.680
fluorescence so if you take this fluorescence
spectra of this what you will going to see
20:24.680 --> 20:37.590
is fluorescence spectra emission spectra right
intensity versus wavelength
20:37.590 --> 20:42.590
see this something over here probably you
will going to see such kind of thing looks
20:42.590 --> 20:50.420
like there is something over here right because
one is little red shifted other is little
20:50.420 --> 21:05.980
blue shifted so in this case this part of
fluorescence is given by this fluorophore
21:05.980 --> 21:10.500
and this part of the fluorescence because
this is a red shift higher longer wavelength
21:10.500 --> 21:19.510
lower energy this part of the fluorescence
is given by this fluorophore as i said that
21:19.510 --> 21:26.500
one of them is accessible more accessible
compare to the other that means these fluorescence
21:26.500 --> 21:34.740
intensity that that these these i by i zero
thats what i just have calculated this is
21:34.740 --> 21:40.380
a function of wavelength
if i calculate at this wave length hm at this
21:40.380 --> 21:49.080
wave length then this i by i zero value and
if i calculated this wave length then the
21:49.080 --> 21:54.120
i by i zero value will be different suddenly
it will be because when i will calculate the
21:54.120 --> 21:58.800
i by i zero for this wavelength then there
will be lot of change in the fluorescence
21:58.800 --> 22:04.429
intensity when i will calculate the i by i
zero at this wavelength then there will be
22:04.429 --> 22:16.410
less change in the fluorescence so here is
the less change is a more change in the fluorescence
22:16.410 --> 22:28.929
intensity so these are the function of lambda
ok so this function of lambda then that summation
22:28.929 --> 22:36.910
what i have written this f i that also has
to be function of lambda because at this particular
22:36.910 --> 22:45.450
wavelength the contribution of this one and
this one is something different than the at
22:45.450 --> 22:51.460
this wavelength because the emission spectrum
shifted for the location a and location b
22:51.460 --> 23:00.970
you remember here is this is my location a
and this is my location b so emission spectrum
23:00.970 --> 23:05.880
is shifted if the emission spectra is same
exactly same for the location a and location
23:05.880 --> 23:13.390
b then i want to have such kind of wavelength
dependency and this f i is also the wavelength
23:13.390 --> 23:17.460
independent
but in this case as there is a change so i
23:17.460 --> 23:25.380
have this f i is a function it depends on
the wavelength so then this divided by one
23:25.380 --> 23:50.820
plus k d i lambda into q so i can write this
k d i lambda right this k d i lambda whether
23:50.820 --> 24:01.140
it should be depends on the wavelength or
not this is my question whether this k d i
24:01.140 --> 24:09.130
i means for the i th one it could be a it
can be b if you have multiple then c d and
24:09.130 --> 24:21.580
so on and so forth so whether for a particular
site either its k d a or k d b is does
24:21.580 --> 24:31.930
it depends on the wave length obviously no
so i should write this only k d i this k d
24:31.930 --> 24:40.460
i lambda right it does not depend on the wavelength
because its a stung harmer quenching constant
24:40.460 --> 24:44.520
so it it will not going to depends on the
wavelength it is the interaction between the
24:44.520 --> 24:52.650
fluorophore and the quencher right so the
where that molecule will emit i mean it doesnt
24:52.650 --> 25:01.110
matter for it
ok so ultimately what will going to have this
25:01.110 --> 25:23.180
i lambda by i zero lambda is equal to sum
over i f i lambda by one plus k d i into q
25:23.180 --> 25:43.170
like if i have three such different sites
then i will i can simply write i lambda by
25:43.170 --> 25:56.450
i zero lambda equal to f one that is for my
first site divided by one plus k d one to
25:56.450 --> 26:29.549
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
26:29.549 --> 26:43.010
denotes my excitation sorry emission spectra
right
26:43.010 --> 26:53.370
emission spectrum this i lambda has the component
of three different species in this case you
26:53.370 --> 27:07.120
see this is for the hm f one f two and f three
right so this i lambda if i multiply it by
27:07.120 --> 27:19.799
f i lambda that will going to give me intensity
for the i th site at this particular lambda
27:19.799 --> 27:26.520
i will simply get that one right so obviously
this i lambda is a measured quantity right
27:26.520 --> 27:34.190
i will get this one this is my i lambda it
has the component of b and it has component
27:34.190 --> 27:42.929
of a so i measured it together and now this
f i is that fractional fluorescence intensity
27:42.929 --> 27:53.720
right for this i th component and i will get
this i i lambda right so if i analyse lets
27:53.720 --> 28:11.740
say for example if i analyse these thing in
a global fitting method then what you will
28:11.740 --> 28:19.040
get you will get the global fitting method
means that same equation will be applied for
28:19.040 --> 28:31.169
all different wavelengths note it down
these contribution f one f two f three they
28:31.169 --> 28:38.410
are wavelength dependent i by i zero wavelength
dependent but k d two or k d one or k d three
28:38.410 --> 28:45.260
they are not wavelength dependent so for each
wavelength if you plot i by i zero and then
28:45.260 --> 28:53.750
just feet this one right so you will get the
unique value of k d one k d two k d three
28:53.750 --> 29:14.870
so you will get unique values of k d one k
d two k d three and also all the f i lambdas
29:14.870 --> 29:24.110
at each wavelength it f i lambdas at ok
f i lambda means at each wavelength ok and
29:24.110 --> 29:36.110
what i know another information is f i sum
of our i f i is equal to one that i know that
29:36.110 --> 29:45.410
means what you will going to get is all these
a f i lambda right you will get all these
29:45.410 --> 29:56.950
f i lambda and this is your measured spectrum
so if you multiply this measured one with
29:56.950 --> 30:07.100
all these f i lambda you will get the spectrum
for the i th species and thats what we called
30:07.100 --> 30:20.049
this as quenching resolved
30:20.049 --> 30:36.290
emission
spectra ok so let us stop here and we will
30:36.290 --> 30:39.150
continue on the next day
thank you very much