WEBVTT
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Welcome back to this lecture on Biochemical
Engineering. If you remember what we did in
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the last class, was essentially we are looking
at multiple receptors. And, what happens when
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the, in the species, in the complex, in the
ligand binds to multiple receptors. We talked
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about the fact that these multiple receptors
essentially bind independently to the ligand
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and the binding of one, does not influence
other. And then, we said, is it possible to
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determine separately, whether these multiple
receptors receptor 1 bound the amounts that
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receptor 1 binds to the ligand, the amount
that receptor 2 binds to the ligand. We said,
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well, it is not always easy to figure out
because what we do is, we label the ligands
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essentially.
So, the ligand, if there is a one kind of
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ligand, it will either bind to receptor 1
or bind to receptor 2. And, there is, you
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know we are just looking at the fluorescence
or the some kind of labeling. And, it is not
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possible to figure out in general, whether
receptor 1 binds or it is binds receptor 1
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or it binds to receptor 2. In some extreme
cases, obviously it is possible. And, what
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are those cases? So, the case that we talked
about was, when a receptor of a particular
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kind has a much higher binding infinity than
the receptor of the other kind.
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And, If you remember, so we, this was what
we look doing, if I will, if I go to the screen
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now. So, a receptor 1 and receptor 2 bind.
And, this steady state concentration of the
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complex is the amount of receptor 1 is binding
to the ligand plus the amount of receptor
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2 is binding to the ligand. So, the first
term you find here which is N C, C L naught
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N R T 1 over C L naught plus K D 1; that is
the amount of receptor ligands bind to receptor
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1. And, the second term that you see here,
C L naught N R T 2 over C L naught plus K
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D 2; that is the receptor amount of ligands
binding to receptor 2. So, the fact is that,
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each of this receptors act independently and
bind independently and there is no influence
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of the binding of 1 receptor to the binding
of the other. But, it is possible as I said
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to distinguish between the two values, if
the K D, that is the dissociation rate constant
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of receptor of the biding of receptor 1 is
much much higher or much much lower than the
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receptor binding rate constants of the second
one.
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Now, if you remember, so, we would went through
the calculations and I am not going to repeat
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all of that. And, when we did the calculations,
we took the case when the amount of complex
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that was there at the beginning was zero and
we did all the calculations.
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And, what I want to attract your attention
to is this factor this biphasic nature of
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the curve. So, what you see over here is N
C over N R T and this is time. So, you have
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a single receptor and the case of a single
receptor, the curve is monophasic; whereas
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in the case of two receptors, the curve is
biphasic.
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Now, as you can see over here, these are data
here and I am going to look at that. So, look
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at the K D 1. The K D 1 is 1 micro molar and
the K D 2 is 100 micro molar. So, there is
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the two orders of magnitude difference between
the dissociation rate constants of 1 and 2.
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So, if K D 1 is hundred times, so, here if
K D 1 is much much smaller than K D 2 as the
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case is, then K 1 K minus 1, let us say over
K 1, that is the backward rate constant over
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the forward rate constant is much smaller
than K minus 2 over K D 2; which means, affinity
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of receptor 1 is much much higher than affinity
of receptor 2. So, this is what we concluded;
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that the affinity of receptor 1 is much much
higher than receptor 2, if this is the case.
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And, these are cases, when we can really distinguish
between the two kinds of receptors because
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as I just showed in the biphasic response
that I showed that we had. So, what does this
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lead to, if K D 1 is much much smaller than
K D 2? What this leads to is that in the exponential
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graph that you have, the slopes of the exponential
graph all points of time will be very different.
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And, what you see is that, there is clear
biphasic nature; or in other words, the slope
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increase in a certain way. And, then they
decrease suddenly, say if I can go back to
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that screen quickly, see, you see that this
is the slope, say some kind of an exponential.
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But, this exponential slope of this curve,
whether it is exponential or something close
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to linear does not matter; is very very different
from the slope of the next curve.
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So, which means what? That, if these two receptors
are there and the K D 1 of 1 is much much
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smaller than the K D of the other; or in other
words, receptor 1 has a much higher affinity
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to than receptor 2. Then, this implies that
receptor whenever you let these ligands loose
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on the system, so, whenever you let these
ligands loose on the system, so receptor 1
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will capture these ligands, so that ligands
are around. So, receptor 1 will immediately
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come and bind with the ligand. So, as the
result, the initial amount of binding that
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will occur will essentially be because that
of receptor 1.
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So, once we understand than then we realize
that, yes the two slopes are very different
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from each other. So, initially you have a
much higher slope and then the slopes decrease
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with time. So, what this implies is that for
the case, when the slope is higher or in the
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initial time period, so as soon as you let
the ligands loose on the receptors, the receptor
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1 is binding with the ligand. So, this will
essentially imply that the receptor, the slope
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first slope that is the higher slope that
corresponds to the binding of receptor 1 to
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the ligand. It is not that receptor 2 does
not bind at all, but the fact of the matter
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is that much of the binding that occurs, maximum
most of the binding that occurs is because
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of receptor 1. And, if you come to think of
it, you know, if you come to think of these
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two as comparative binding that is receptor
1 is binding comparatively to receptor 2,
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then what you realize is that because the
rate constant or the affinity of receptor
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1 is hundred times that of receptor 2. So,
the binding should always be in the ratio,
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more or less on the ratio 1 is to 100 or 1
is to 99. So, if hundred ligands bind, 1 percentage,
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only 1 percentage of them is binding to receptor
2 and 99 percentage of them is binding to
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receptor 1.
So, how does this help us? How does this knowledge
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help us? Now, I gave you a problem at the
end of the last class and I will go back to
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that problem. This helps us in identifying
the rate constant of the system because just
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as I said that when we label the ligand, there
is no way for us to label the receptors, we
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are only labeling the ligand. So, when we
label the ligands what happens is that, we
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can only figure out that what is the total
amount of complex that has been found. And,
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there is no way for us to separate out the
receptor 1 from receptor 2 or the constant
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rate constants of receptor 1 from receptor
2. But, what we essentially want to measure
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are the rate constants of receptor 1 and receptor
2. So, how do we go about it?
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So, what we do? We choose the ligand, we purposely
choose the ligand. So, think of that you are
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doing an experiment and how do you go about
it. So, you can only label your ligands; you
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cannot label your receptors. So, what do you
do? You purposely choose a ligand such that,
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it binds both to receptor 1 and receptor 2.
But, the binding rate constant or the K D
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at the dissociation rate constant, the binding
affinity of the ligand for receptor 1 is much
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much higher than the binding affinity for
the ligand, for of the ligand for receptor
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2. So, that is the highlight. That is the
point that you need to exploit.
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So, once you have understood that what you
do is, you essentially figure out. So, if
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1 is hundred times, the other and if you let
the reaction happen for a certain period of
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time, initially what will happen is, if the
receptor 1’s affinity is higher, much much
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higher than receptor 2 what will happen is,
all of it will go. The ligands will, almost
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of the ligands will actually go and bind with
receptor 1. So, that is essentially that is
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going to happen. Then, what happens? As time
progresses, receptor 1 are going to be more
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or less saturated. And, then is, when you
see that bend in that curve that, just that
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I showed you just now.
So, receptor 1 is more or less saturated.
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As a result, what happens is, now receptor
2 starts to bind to the ligand. And, this
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is the mechanism or the hypothesis…, say
that is that is what we are going to explain
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as we try and solve the problems.
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So, let us go to the problem. So, this is
the problem that I gave you. And, what I asked
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you to do is that N C equals C L naught N
R T 1 over C L naught plus K D 1 plus C L
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naught N R T 2 over C L naught plus K D 2.
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So, let us write this. So, and C equals C
L naught N R T 1 over C L naught plus K D
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1 plus C L naught N R T 2 C L naught over
K D 2. So, one of the ways of tackling this
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is essentially, just what I said that, you
know there it is very hard to be able to figure
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out. So, you have, just have a one curve,
come to the think of it you just have a one
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curve. So, when you had… let us go back
to this. And, when you had a single receptor,
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this was your scatchard plot.
So, when you had a single receptor, you had
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two unknowns; one was N R T and the other
one was K D. And, you could draw a scatchard
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plot, which was linear. And, from the slope,
you figured out N R T. K D, sorry. And, from
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the intercept, you could figure out N R T.
So, two rate constants, two constants you
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needed to figure out. You had one plot. And,
from the slope, you figured one and for an
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intercept you figured another. But, here,
what happens is the the problem that I gave
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you. Sorry.
So, the problem that I gave you here, the
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problem with this is that you have one, two,
three, and four. So, four unknowns and1 plot.
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So, how do you go? So, this can utmost give
you two unknowns. So, how do you figure out
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the other two unknowns?
10:35.850 --> 10:42.850
So, what we do is, we obtain a scatchard plot.
Now, typically a scatchard plot would be linear.
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So, if you do N C over C L naught, over N
C. Then for a particular value C L naught,
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C L naught equal to say 1 micro molar or10
micro molar, something like that. This is
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what you get. This is your plot. But, in this
case what will happen is that, given that
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K D 1 is much much lesser than K D 2.
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So, your plot is going to be something like,
let us keep this formula next to each other.
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So, your plot is going to be something like
this, and this. Why is that going to be the
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case? The reason is that N C equals N C 1
plus N C 2. That is what my total. So, it
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is complex form from receptor 1 and the complex
form from receptor 2. That is what is going
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to be.
So, when that is the case, then when you have
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a limited amount of, a small amount of complex
that is formed, what will happen is, much
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of it will be from receptor 1. Is it clear.
Why is that going to be from receptor 1 because
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receptor 1’s affinity or affinity of receptor
1 is much much greater than that of receptor
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2. So, that is what will happen. So, you will
have, most of it will be formed from receptor
12:24.920 --> 12:29.510
1. Now, once receptor 1 is more or less saturated,
most of the receptors are bound to the ligand,
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then only receptor 2 starts to form. So, then
what happens is, you can exploit this as receptor
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1. You can call this as almost as receptor
1. And, this would be primarily receptor 2
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plus, may be receptor 1 also.
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So, what we do now? So, we have our equation
N C equals N C 1 plus N C 2, which equals
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C L naught N R T 1 over C L naught plus K
D 1 plus C L naught N R T 2 over C L naught
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plus K D 2.
Now, for the first part of the plot, so, if
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my scatchard plot is something like this,
N C over C L naught over N C, this N C that
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you get over here is essentially N C 1; this
value. So, I can straight away from the. So,
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if I take this this part of the plot over
here, I can straight away say that the slope
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of this equals the slope of the first part
equals minus 1 over K D 1 and intercept equals
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N R T over K D 1. I can straight away say
this. So, what happens is, essentially I use
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the first part of the slope to evaluate both
my K D 1 and N R T 1. And, the second part
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I can calculate, I can assume that both receptor
1 and receptor 2 are present and I can go
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ead and calculate everything I want. Or,
second part you, I can make an even more simplifying
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assumption and say that this is equivalent
to N C 2. In which case, the slope of this
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will give me minus K D1 over minus 1 over
minus K D 2 and the intercept will give N
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R T 2 over K D 2.
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So, that is the possibility. So, I can do
something like this. So, if that is the case,
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then slope, this is N C over C L naught over
N C. Then, slope equals minus 1 over K D 1
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intercept equals N R T 1 over K D 1 and slope
here, can be minus 1 over K D 2 and intercept
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equals N R T 2 over K D 2. That is the possibility,
or here you can assume both in receptor 1
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and receptor 2 to be acting; in which case,
you have to do a little more complicated calculation.
15:04.660 --> 15:09.149
But, this is something you can definitely
take. So, that is a, that is the way we solve
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this. Otherwise, there is no other way of
looking at it.
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Now, let us look at some other things here,
some other variations of the problem over
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here is, you can see the… what we are talking
over here is, now, so we start to talk about
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here is interconverting receptor subpopulation.
15:24.410 --> 15:30.850
So, something like, where the receptors convert
within, from two into each other, two different
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forms. So, essentially what happens is that,
in the previous part of the lecture what we
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talked about our receptor is that, work completely,
independently of each other, when they bind
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to the ligand independently. What we are going
to talk about now, there are cases where the
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receptors bind, but they do not bind independently.
They bind; there is a independence between
15:51.899 --> 15:57.860
these receptors. Or in other words, the receptors
populations are sub populations, which are
15:57.860 --> 16:01.519
interconverting to each other. And, what we
mean when we say, we are, they are interconverting
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to each other.
Essentially it is a system of logical change;
16:04.529 --> 16:09.339
you know this is conformational change that
is happening in the receptor. So, it is not
16:09.339 --> 16:14.600
the chemical composition of the receptors
as such that are changing, but the conformational
16:14.600 --> 16:17.449
of the receptor, the conformational changes
occurs in the receptor. And, what does the
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conformational change in tail; it essentially
means that the rate constant, the dissociation
16:21.589 --> 16:25.949
and association constants. Or in other words,
the binding rate constants would change for
16:25.949 --> 16:29.120
these cases. So, the slope, let us look at
there.
16:29.120 --> 16:33.540
So, interconverting receptor sub population;
so, if you look at the screen what happens
16:33.540 --> 16:39.550
is that, often the receptor undergoes a conformational
change, just as I said, after binding to the
16:39.550 --> 16:44.199
ligand. And, so, when we talking of two different
receptor populations, how does this differ
16:44.199 --> 16:48.610
from what we had done before is that, what
we had done before were two completely different
16:48.610 --> 16:52.649
receptors, which have affinity for the same
ligand. Here, it is not that case. It is not
16:52.649 --> 16:57.699
two completely different receptors, but the
same receptor which is undergoing conformational
16:57.699 --> 17:02.550
change. And, as a result, the dissociation
and association constants are differing here.
17:02.550 --> 17:09.550
So, the change, however does not affect subsequent
ligand receptor binding. So, and, but there
17:10.029 --> 17:13.620
is a conformational change that occurs. And,
in the most general case, the receptor is
17:13.620 --> 17:18.990
present as an interconverting sub population.
That is what we are going to talk about in
17:18.990 --> 17:25.299
moral, today. And, so the most general case
of receptors is present as an interconverting
17:25.299 --> 17:30.979
subpopulation. And, these subpopulations have
differences in rates of dissociation and association
17:30.979 --> 17:35.200
between receptors and ligand. So, just as
I said there, so because of the conformational
17:35.200 --> 17:40.769
change, what results; because of the conformational
change is, essentially the fact that these
17:40.769 --> 17:47.259
receptors have different kinetic constants.
So, now, let us look at the model that we
17:47.259 --> 17:51.629
have over here; the interconverting receptor
subpopulation model. So, as you can see over
17:51.629 --> 17:58.009
here, this is the most general form of the
model R 1 plus l. So, R 1 is my receptor population
17:58.009 --> 18:04.479
1; R 2, as I said is same, chemically the
same receptor but just a conformation, R 1
18:04.479 --> 18:11.479
with the conformational change. So, R 1 plus
L forms the complex C 1. And, R 1 also forms
18:12.570 --> 18:19.139
the complex, changes to R 2 and forms the
complex C 2. See, this picture may be slightly
18:19.139 --> 18:24.059
confusing. So, what I will do is I am going
to write this differently.
18:24.059 --> 18:31.059
So, R 1 plus L K 1 1 K minus 11 C 1; so, R
1 is essentially reacting with the ligand
18:38.899 --> 18:45.899
to form the complex C 1. And, R 1 is going
through a conformational change to form R
18:55.220 --> 19:02.220
2. And then, you have R 2 plus L. So, again
it reacts in the similar way as R 1. K 2 2
19:11.979 --> 19:17.049
and K minus 2 are its rate constants to form
C 2. And, these are the complexes; this is
19:17.049 --> 19:24.049
the complex, and this is the complex. And,
it turns out, this is the complex themselves
19:24.609 --> 19:31.609
can undergo conformational change to rotate
between each other. So, this is C 2 and the
19:32.759 --> 19:38.769
rate constants for this are K 1 2 and K minus
1 2.
19:38.769 --> 19:45.769
So, this is the set of reactions; interconverting
receptor, interconverting receptor sub populations.
19:53.479 --> 20:00.229
So, what is happening is that the receptor
1 is binding with ligand to form a complex
20:00.229 --> 20:04.200
and the rate constants, forward and backward
rate constants for that are K 1 1 and K minus
20:04.200 --> 20:11.200
1. Receptor 1 is undergoing a conformational
change to form R 2. And, the weight constants
20:11.559 --> 20:18.559
for that are K 2 1 K minus 2 1. And, R 2 reacts
with the ligand as it is, to form a complex
20:19.609 --> 20:25.259
C 2. And, C 2 and C 1 undergo conformational
change again.
20:25.259 --> 20:29.759
So, if you look over here, go back to the
screen and look over here, you find that this
20:29.759 --> 20:34.080
is the way we have represented it. This is
slightly confusing way of looking at it. And
20:34.080 --> 20:38.019
therefore, I drew it, but you can write this
cyclical way. So, remember the only thing
20:38.019 --> 20:42.599
I want to point out here is that, it is R
1. When I draw these, these this this line,
20:42.599 --> 20:46.710
for example, the one the vertical line on
the left; it is only the conversion between
20:46.710 --> 20:50.269
R 1 and R 2. And, L is not involved in this
reaction.
20:50.269 --> 20:57.269
Now, a special case of this is what you see
over here; where the receptor is themselves
20:57.539 --> 21:04.539
do not change conformation, but it is the
complex that changes conformation. So, R 1
21:05.359 --> 21:10.619
reacts with the ligand to form complex C 1.
And, it is C 1 that changes conformation to
21:10.619 --> 21:16.779
form complex C 2. And, the rate constants
are K 1 1 and K minus 1 and K 1 2 and K minus
21:16.779 --> 21:22.309
2. So, essentially what we have done is, if
you take this cyclical model appear, if you
21:22.309 --> 21:27.070
cut out these two parts, these two lines over
here and these two lines over here, you get
21:27.070 --> 21:32.820
R 1 plus L gives C 1, and C 1 gives C 2. So,
you get this triangle over here and that is
21:32.820 --> 21:36.340
the special case.
So, what we will do now in the in the remaining
21:36.340 --> 21:42.580
part of this lecture is, essentially look
at and model these. So, what we will do now
21:42.580 --> 21:47.379
is, to start with this. We will look at the
special case B first; the reason being that,
21:47.379 --> 21:49.190
it is easier to model.
21:49.190 --> 21:55.470
So, we will come back and look at A after
that. But, essentially we will look at B first.
21:55.470 --> 22:00.979
So, in B, let us look at this. So, R 1 plus
L, K 1 1 is the forward rate constant, K minus
22:00.979 --> 22:06.009
1 is the backward rate constant. And, C 1
and C 2 is the complex, which undergo conformational
22:06.009 --> 22:09.259
change through the rate constants K 1 2 and
K minus 1 2.
22:09.259 --> 22:16.259
So, what happens when there is steady state?
When there is steady state, you can relate
22:17.369 --> 22:23.179
the N C 2 and N C 1; that is the rate constant
1 and rate constant 2. Let me work out this
22:23.179 --> 22:24.639
for you.
22:24.639 --> 22:31.639
So, how this happens? Fine. C 1, K 1 2 K minus
1 2 and C 2. So, what is, let us write all
22:41.340 --> 22:48.340
these. So, d N R 1 d t, what is that equal,
to that equals K minus 1 1 N C 1 minus K 1
22:53.119 --> 23:00.119
1 N R 1 C L naught. Now, what would be my
d N C 1, d N C 1 d t that equals… that is
23:09.509 --> 23:16.509
a little longer one; so, K 1 1 N R 1 C L naught
minus K minus 1 1 N C 1 minus K 1 2 N C 1
23:30.279 --> 23:37.279
plus K minus 1 2 N C 2. And, what would be
my d N C 2 d t? That, simply equals K 1 2
23:42.590 --> 23:49.590
N C 1 minus K minus 2 1 N C 2. So, at steady
state, each of these would be zero.
23:58.450 --> 24:02.629
So, if that is zero, then from this equation,
the last equation let us call this equation
24:02.629 --> 24:09.629
a, b and c. Then, from equation c from equation
c, what we find is K 1 2 N C 1 at steady state
24:29.639 --> 24:35.399
equals K minus 1 2 N C 2. Let me go through
this again. So, these are my set of equations
24:35.399 --> 24:40.099
that I wrote this are the, these are the model
equations that dynamic model. Let us forget
24:40.099 --> 24:44.179
this steady state for a minute.
So, this is what I have and steady state model.
24:44.179 --> 24:51.179
So, N R 1 is being formed because of, so the
N R 1 is being formed because of this reaction.
24:51.529 --> 24:58.309
And, it is been destroyed or removed because
of the forward reaction. N C 1 is formed from
24:58.309 --> 25:03.229
this reaction as well as this reaction. And,
it also removes because of this reaction as
25:03.229 --> 25:07.309
well as these reactions.
So, four terms here and whereas, C 2 is formed
25:07.309 --> 25:12.320
in the forward direction and destroyed or
removed in the backward direction. At steady
25:12.320 --> 25:16.349
state, all these three have to be zero. So,
what we do, we first equate the simplest one;
25:16.349 --> 25:23.349
the second one to be equal to zero. So, this
is this is what we get. So, what we immediately
25:24.299 --> 25:31.299
get is that N C 2 equals K 1 2 over K minus
1 2 times N C 1. Or in other words, if K D
25:40.489 --> 25:47.489
1 2 is written as K minus 1 2 over K 1 2,
then this is written as N C 1 over K D K D
25:52.869 --> 25:57.440
1. So, that is what I get.
25:57.440 --> 26:04.440
Now, what is my N C? N C would be N C 1 plus
N C 2; which means that N C 1 plus N C 1 over
26:11.849 --> 26:18.849
K D 1. That equals, N C 1 plus 1 over K D
1. Fine. Now, what is my N R T? N R T is the
26:23.330 --> 26:30.330
total amount of receptors, which is N R 1
plus N C 1 plus N C 2. Or in other words,
26:37.739 --> 26:44.059
receptors that are completely free and receptors
that are in complex 1 and receptors that are
26:44.059 --> 26:49.539
as complex 2. So, this I can write now as
N R 1 plus. I have already covered calculated
26:49.539 --> 26:56.539
what my N C 1 and N C 2 together is which
is N C 1 plus 1 over K D 1.
26:56.649 --> 27:03.649
So, now if I go back to my equation over here,
if I go back to my equation over here, so
27:06.009 --> 27:10.389
this has to be zero at steady state, also
this has to be zero. So, what we can do is
27:10.389 --> 27:15.950
if I equate this to zero, I get a relationship
between N R 1 and N C 1 straight away.
27:15.950 --> 27:22.950
Let us do that. So, equating A to zero, so
A is this equation out here. So, A is this
27:27.769 --> 27:34.769
equation. I am equating this to zero; which
means that K minus 1 N C 1 equals K 1 1 N
27:36.440 --> 27:43.440
R 1 C L naught. So, N C 1 equals K 1 1 over
K minus 1 N R 1 C L naught. fine. So, this
27:53.859 --> 28:00.859
is what I get. So, N R 1 equals K minus K
minus 1 1 over K 1 1 times 1 over C L naught
28:12.549 --> 28:19.549
times N C 1 over C L naught. So, this I can
write as K D 1 1, previously I had written
28:20.589 --> 28:27.489
K D 1 2. So, I can use this for K D 1 1, so
C L naught. fine. So, this is one. So, N R
28:27.489 --> 28:34.489
1 equals K D 1 N R 1 equals, just let me just
put it over here like this yeah. So, N R 1
28:38.159 --> 28:45.159
equals K D 1 N C 1 over C L naught and N C
2 equals N C 1 over K D 1.
28:45.440 --> 28:52.440
Now, what was my constraint equation? My constraint
equation was that N R T equals simply equals
28:52.450 --> 28:59.450
N R 1 plus N C 1 plus N C 2 or I had written
this as N R 1 plus N C 1 times 1 over K D
29:06.529 --> 29:11.589
1. Now N R 1, I can write now from the from
the previous equation, I can write is this
29:11.589 --> 29:18.589
N R 1 N C 1 over C L naught plus N C 1 over
1 over K D 1. K D 1 2. Sorry. This is K D
29:24.479 --> 29:31.479
1 2.
So, now I can write this as N C 1 K D 1 1
29:32.289 --> 29:39.289
over C L naught plus 1 over K D 1 1 1 2. So,
this is what I get. So, this is my relationship
29:44.649 --> 29:51.649
between N C 1 and N R T. So, N R T times C
L naught if I write, then it will be N C 1
29:58.159 --> 30:05.159
K D 1 1 plus C L naught times 1 over K D 1
2. So, this is the relationship I have between
30:11.809 --> 30:12.659
these.
30:12.659 --> 30:19.659
So, N C equals N C equals N C 1 plus N C 2;
which I have written it previously if you
30:25.119 --> 30:32.119
remember, as N C 1 over 1 over K D 1 2. Now
N C 1, I can now write as N R T over times
30:37.359 --> 30:44.359
C L naught divided by K D 1 1 plus C L naught
1 plus K D 1 2 times, this factor over here;
30:52.339 --> 30:59.339
which is 1 over K D 1 2.
So, this is my relationship that I have between
31:03.659 --> 31:10.659
these numbers. So, I can, one of the things
I can do is try and convert this into a slightly
31:15.200 --> 31:22.200
more handle able form; which is K D 1 2 plus
1. And, divide this by K D 1 1 K D 1 2 plus
31:35.899 --> 31:42.899
C L naught 1 plus K D 1 2. I can write it
like this or I can make it even simpler.
31:47.989 --> 31:54.989
And, write this as N R T C L naught divided
by K D 1 1 K D 1 2 plus C L naught, divide
32:07.239 --> 32:14.239
this by 1 plus K D 1 2. So, this is my final
relationship between N C and N R T. So, if
32:16.549 --> 32:20.899
I want to draw my scatchard plot, which will
be N C over C L naught; that would simply
32:20.899 --> 32:27.899
be equal to N R T divided by C L naught plus
K D 1 1 K D 1 2 divided by 1 plus K D 1 2.
32:34.619 --> 32:36.809
So, I can draw a plot like this.
32:36.809 --> 32:43.320
So, if I now go back to the screen, this is
exactly I, what we derived; N C equals N C
32:43.320 --> 32:49.299
over C L naught equals N R T divided by C
L naught over K D 1 1 times K D 1 2 divided
32:49.299 --> 32:56.299
by 1 plus K D 1 2. So, this is the plot that
I want to obtain. So, why did I, why did I
32:57.830 --> 33:04.149
do it this way, why did I club club it this
way because if I am to draw a plot of N C,
33:04.149 --> 33:11.149
so, what do I do? I can write this as K D
apparent. My apparent rate constant I can
33:12.859 --> 33:16.739
write. So, if you look at this, it is even
has the units of rate constant correctly.
33:16.739 --> 33:22.769
So, I can write this as K D apparent, which
is K D 1 1 times K D 1 2 divided by K D 1
33:22.769 --> 33:23.309
2.
33:23.309 --> 33:30.309
So, what will happen is, if I can now write
my plot as N C L, N C over C L naught equals
33:30.729 --> 33:37.729
N R T over C L naught plus K D apparent. So,
what I can do is, if I take a, if I can either,
33:47.229 --> 33:53.759
I can either do it this way, which is C L
naught over N C equals C L naught over N R
33:53.759 --> 34:00.759
T plus K D apparent over N R T; where K D
apparent is this number, which is K D 1 1
34:10.710 --> 34:17.710
times K D 1 2 divided by K D 1 2 plus 1. Fine.
This is my number. So, what I can do over
34:20.560 --> 34:27.560
here is, if I draw a plot of C L naught over
N C versus C L naught, what do I get? I just
34:30.190 --> 34:36.579
get a straight line something like this. I
just get a straight line something like this.
34:36.579 --> 34:43.579
And, the slope of this straight line is 1
over N R T; whereas the intercept is K D apparent
34:49.980 --> 34:54.200
over N R T.
So, why can, what I can straight away do is,
34:54.200 --> 34:58.579
from the slope I can calculate my N R T and
from the intercept, I can calculate my K D
34:58.579 --> 35:04.430
apparent. So, what was the whole idea of this
entire process was to be able to we had many
35:04.430 --> 35:09.269
of constraints out there; K D 1 1, K D 1 2
and C L naught. And, C L naught is not a constant,
35:09.269 --> 35:12.390
but it is something that we can vary, but
N R T.
35:12.390 --> 35:17.920
So, what was the idea? The idea was to club
these three constants that we had into two
35:17.920 --> 35:21.490
because two is something very measurable.
And, write the equation in such a way that,
35:21.490 --> 35:26.010
we can plot it very easily. So, what we manage
to do is, write the equation in a way where
35:26.010 --> 35:32.920
it is a straight line. And, we could just
get the N R T from the intercept and K D;
35:32.920 --> 35:36.760
just get the N R T from the slope and the
K D apparent from the intercept.
35:36.760 --> 35:41.539
So, that was what we were trying, we are trying
to do. So, this is the special case. The case,
35:41.539 --> 35:47.519
where the receptors themselves they are not
interconverting, but the complexes are interconverting.
35:47.519 --> 35:52.740
So, what about the more complex case now?
And, as I just said that this set of reactions
35:52.740 --> 35:57.980
could be replaced by this set of reactions
are actually written by these set of reactions.
35:57.980 --> 36:03.250
And, what we do, you know, how can we look
at this and model this now? This is slightly
36:03.250 --> 36:06.430
more complicated one the process that we are
going to do. And, I probably, I am not going
36:06.430 --> 36:10.710
to have the time to work out the whole thing,
but we will start to work out. And, the process
36:10.710 --> 36:14.950
that we are going to have to do is exactly
the same process. So, this is, this is my
36:14.950 --> 36:20.569
plot, this is my, these are my sets of reactions.
And, all we have to do we have to write balances
36:20.569 --> 36:26.549
for each of these R 1, R 2, C 1, C 2. And,
let us try and do that, to start with.
36:26.549 --> 36:31.960
So, R 1 the balance from R for R 1, so it
just for a second, I will keep this out here,
36:31.960 --> 36:37.240
so that you can write these set of reactions
down. So, R 1 plus L K 1 a forward reaction
36:37.240 --> 36:44.210
K 1 1 backward reaction K minus 1 give C 1;
R 1 forward reaction K 2 1 and K minus 2 1
36:44.210 --> 36:51.019
gives R 2; R 2 plus L forward reaction K 2
2 backward K minus 2 2 give C 2 and C 1 forward
36:51.019 --> 36:56.809
reaction K 1 2 backward reaction K minus 1
2 gives C 2.
36:56.809 --> 37:03.809
So, let us try writing the balances for R
1 now, first. So, R 1 is being formed by this
37:04.539 --> 37:11.539
reaction, K minus 1 1 N C 1 and is being removed
or being depleted by this reaction N R 1 times
37:17.630 --> 37:23.490
C L naught. And, for another set of reactions
and R for that R 1 is being formed from K
37:23.490 --> 37:30.490
minus 2 1 N R 2 and being depleted from K
minus K 2 1 N R 1. Similarly, N R 2 d t is
37:40.019 --> 37:47.019
being formed for from K 2 K minus 2 two times
N C 2 and depleted from K 2 2 N R 2 times
37:59.569 --> 38:06.569
C L naught. And, there is other inter converting
reaction which is, it is being formed from
38:07.309 --> 38:14.309
K 2 1 N R 1 and depleted from K minus 2 1
N R 2. And, what about C 1 and C 2? That also,
38:19.279 --> 38:21.130
you have two sets of reactions.
38:21.130 --> 38:26.559
So, this is, this is their set of reactions
for N R 1 and N R 2. And, if you look at C
38:26.559 --> 38:33.559
1 and C 2, so d N C 1 d t equals K 1 1 N R
1 C L naught minus K minus 1 1 N C 1. And,
38:47.200 --> 38:54.200
then you have K minus 1 2 N C 2 minus K 1
2 N C 1. And, similarly we can write the balance
39:03.119 --> 39:10.119
for C 2, which is K 2 2 N R 2 C L naught minus
K minus 2 2 N C 2 plus K 1 2 N C 1 minus K
39:31.059 --> 39:38.059
minus 1 2 N C 2. So, these are my four balance
equations that I, that I write. And, what
39:40.410 --> 39:46.079
do we do with them? So, at steady state, I
have to equate all of these to be equal to
39:46.079 --> 39:49.549
zero d t.
39:49.549 --> 39:56.549
So, let us look at the total amount of complex
that we have. So, this is my set of equations
39:56.680 --> 40:01.430
for N C 1. So what, let us look at the total
amount of complex that we have, which is N
40:01.430 --> 40:08.029
C 1 plus N C 2. What is that number, if I
add all the terms that I get over here? I
40:08.029 --> 40:15.029
get K 1 1 N R 1 C L naught minus K minus 1
1 N C 1 plus K 2 2 N R 2 C L naught minus
40:22.329 --> 40:29.329
K minus 2 2 N C 2. And, for N R 1 and N R
2, if I add the two equations N R 1 plus N
40:34.460 --> 40:40.450
R 2, just this is not. yeah basically- basically,
this are adding these two equations. So, this
40:40.450 --> 40:47.450
is at steady state. This is and basically
this is d by d t of this at steady state this
40:47.869 --> 40:54.869
equals zero. Similarly, d by d t of N R 1
plus N R 2 equals K minus 1 minus 1 1 N C
41:02.150 --> 41:09.150
1 minus K 1 1 N R 1 C L naught plus K minus
2 2 N C 2 N R 2, sorry N C 2, minus K 2 2
41:21.250 --> 41:28.250
N R 2 C L naught. So, this is what we get.
So, and this equals zero at steady state.
41:29.569 --> 41:34.089
So, what we have to do essentially is that,
it is a little more cumbersome process. So,
41:34.089 --> 41:41.089
what we have to do essentially is that, we
have to solve for
N R 1, N R 2. And, let us say N C 2 in terms
41:52.970 --> 41:59.970
of N C 1, in terms of N C 1; let us say something
like that. So, when we do that, then what
42:08.509 --> 42:14.680
we can do from here, what we can get from
here is that, we can replace everything in
42:14.680 --> 42:20.299
terms of N C 1. Now, what was my constraint
equation? If you look at these two sets of
42:20.299 --> 42:23.450
reactions that I have written over here, this
is, let us go one by one.
42:23.450 --> 42:29.529
So, this is N R 1 and N R 2, and this is N
C 1 and N C 2. If you add these, all these
42:29.529 --> 42:34.279
four sets of reactions, what do I get? If
I add all these four, what I get is d d t.
42:34.279 --> 42:41.279
If you look at these equations of N R 1 plus
N R 2 plus N C 1 plus N C 2, the summation
42:44.180 --> 42:51.160
is d d t of that the summation is zero; which
means that N R 1 plus N R 2. This implies
42:51.160 --> 42:58.160
that, N R 1 plus N R 2 plus N C 1 plus N C
2, they all, there sum come out to be a constant,
42:59.829 --> 43:03.630
which is N R T; that is the total amount of
receptors present. Or, in other words, the
43:03.630 --> 43:08.900
total amount of receptors that is present
is divided into the receptor 1, that is free
43:08.900 --> 43:13.849
receptor 2, that is free receptor 1 complex
that has been formed and receptor 2 complex
43:13.849 --> 43:19.769
that has been formed. So, we go through these
calculations and I am skipping the details
43:19.769 --> 43:22.910
of the calculations because we do not really
have a lot of time.
43:22.910 --> 43:27.000
So, once we go through these calculations,
if you look at the screen, what you find over
43:27.000 --> 43:31.450
here is that this constraint equation that
I wrote. So, what you need to do? You have
43:31.450 --> 43:35.539
to keep substituting everything. So, you decide
that you want to substitute everything in
43:35.539 --> 43:41.759
terms of say, N C 1. So, you convert N R T
1 as; write it as in terms of N C 1. Substitute
43:41.759 --> 43:47.250
here, convert N R T 2 and write it in terms
of N C 2 and substitute here and N C 1 sorry
43:47.250 --> 43:52.380
and substitute here and write N C 2, also
in terms of N C 1 and substitute over here.
43:52.380 --> 43:57.190
So, once you do all that, what you will find
is just like in the previous case, just like
43:57.190 --> 44:01.630
in case B, as we did case B we did explicitly.
And, here also, you will find just like in
44:01.630 --> 44:08.630
case of B that N C could be written as C L
naught N R T over C L naught plus K D apparent.
44:10.079 --> 44:16.109
So, this is, this is what you can write. So,
N C over C L naught is N R T over C L naught
44:16.109 --> 44:22.359
plus K D apparent, where K D apparent is slightly
more complicated over here. It is K D 1 1
44:22.359 --> 44:29.349
times 1 plus K D 1 over K D 1 2 1 over 1 plus
K D 1 2, which could also be written as K
44:29.349 --> 44:36.349
D 2 2 times 1 plus K D 2 1 over 1 plus K D
1 2. And, these, all these numbers are written
44:36.420 --> 44:43.140
over here. So, K D 1 1 is the dissociation
rate constant for the first case here; that
44:43.140 --> 44:49.700
is K D minus 1 1 over K D 1 1, K D 2 1 is
K D minus 2 1 over K D 2 1; or in other words,
44:49.700 --> 44:54.789
the dissociation rate constraint over here,
between the two receptor subpopulations. So,
44:54.789 --> 45:00.220
this is the way the receptors the changing
conformation K D 1 2 equals K D minus 1 2
45:00.220 --> 45:05.890
over K D 1 2, which is how the complexes are
changing conformation. And, that is the dissociation
45:05.890 --> 45:12.890
rate constants for that, and K D 2 2 equals
minus K D 2 2 over K minus K 2 2 over K 2
45:13.039 --> 45:17.900
2 k. So, this is how the receptor, the second
receptor is binding to the complex. So, this
45:17.900 --> 45:21.309
this is the thing.
Now, what happens is, what you had been able
45:21.309 --> 45:25.470
to do or we had been able to do through these
processes. We had these one, two, three, four
45:25.470 --> 45:31.099
constants and N R T, of course the fifth constant.
So, we have been able to reduce these five
45:31.099 --> 45:36.210
constants into two; or in other words, K D
apparent, if you look at it involves all these
45:36.210 --> 45:39.170
constants. So, what we had been able to do
in the process is, we had been able to reduce
45:39.170 --> 45:44.049
these four constants into a single constant
K D apparent. And, why did we do that? The
45:44.049 --> 45:49.569
reason is again as the same reason that we
had before, which is that we, it is very hard
45:49.569 --> 45:54.680
to be to be able to handle all of these. So,
what? So, what we essentially have is just
45:54.680 --> 45:56.190
one equation over here.
45:56.190 --> 46:03.190
So, N C equals N C 1 plus N C 2 equals C L
naught N R T over C L naught plus K D apparent.
46:07.190 --> 46:14.190
So, this is my equation. So, the advantage
is N C over C L naught equals N R T over C
46:16.470 --> 46:23.470
L naught plus K D apparent. Fine. So, C L
naught over N C equals C L naught over N R
46:28.950 --> 46:35.950
T plus K apparent over N R T. So, what do
you do? You perform? take different values
46:37.779 --> 46:44.529
of C L naught, label the ligands and you perform
your experiments for each of these sets and
46:44.529 --> 46:49.869
then you find out that, what what you essentially,
you know what is the ratio of C L naught over
46:49.869 --> 46:55.200
N C for each of these sets and and measure
it in terms of C L naught.
46:55.200 --> 47:01.880
So, essentially this is your, if you want
to plot it over here, this is you plot. So,
47:01.880 --> 47:08.880
the C L naught over N C and this is C L naught.
So, just as I said that, this is going to
47:09.769 --> 47:16.769
be a straight line over here and the slope
is going to be equal to 1 over N R T and the
47:19.289 --> 47:26.289
intercept is going to be K D apparent over
N R T.
47:28.019 --> 47:35.019
So, this is, this is essentially what we get
and K D apparent over here, equals K D 1 1
47:36.869 --> 47:43.869
over 1 plus slightly complicated formula 1
plus K D 1 2 divided by 1 plus K D 2 1 right
47:49.049 --> 47:56.049
here and divided by K D 1 2. So, this is my
formula. So, what we essentially do? We conduct
47:56.890 --> 48:00.519
experiments with taking different amounts
of ligands and figure out, how much is my
48:00.519 --> 48:04.749
total fluorescence that I get; that is a total
amount of complexes that had been formed.
48:04.749 --> 48:10.079
And, we measure that and plot it against the
amount of ligands that are there.
48:10.079 --> 48:15.720
So, in case two, the case one, this is case
one. And, case two, we did separately, but
48:15.720 --> 48:19.799
if we have been able to do case one, which
is a lot more complicated case. And, case
48:19.799 --> 48:24.829
two, in case two would fall out, would be
a natural fall out of case one and it would
48:24.829 --> 48:31.829
come as 1 over it come. And, when the case,
for the case that 1 over K D 2 1 is zero or
48:32.940 --> 48:37.730
this case that is 1 over K D 2 1 is zero.
In other words, K D 2 1 is much much greater
48:37.730 --> 48:44.730
than K D minus 2 1 that is or in this case,
this case, this step is not there. So, if
48:45.259 --> 48:50.410
you go to case A, case B, you will see that
what is case B? Case B is when the receptor
48:50.410 --> 48:53.769
populations, they are themselves, they are
not interconverting, but it is the complexes
48:53.769 --> 48:58.549
that are interconverting. And, when does that
happen? When, there is no connection, no direct
48:58.549 --> 49:02.890
inter-conversion between the receptors.
So, if you look here, so what we are doing
49:02.890 --> 49:07.009
here is that, there is this reaction between
the receptor inter-conversion and that reaction
49:07.009 --> 49:13.880
has to be absent. So, I think, we more or
less looked at different cases over here.
49:13.880 --> 49:19.369
And, what you see on the screen right now
is a summary of the in different processes
49:19.369 --> 49:22.690
that we looked at. So, if you remember, so
this is I will, I will run you through very
49:22.690 --> 49:29.410
quickly. So, we started with the single receptor
case. And, let me go through this very quickly
49:29.410 --> 49:34.400
and try and summarize what we have done in
the last few cases. So, remember, this was
49:34.400 --> 49:39.160
the very important plots that we looked at
just a second.
49:39.160 --> 49:43.970
So, this is, this is where we started with.
This is the single receptor ligand binding
49:43.970 --> 49:47.480
case, where there is a receptor binding binding
to a single receptor, binding to a single
49:47.480 --> 49:52.839
ligand and we assume that, there is no receptor
depletion out here.
49:52.839 --> 49:59.299
So, we got straight and a linear scatchard
plots and stuffs like that. And then, we started
49:59.299 --> 50:03.619
to look at. So, this was the simplest case
possible. And, for this case, the scatchard
50:03.619 --> 50:07.380
plot is linear, is a straight line. The scatchard
plot is essentially the important plot that
50:07.380 --> 50:13.099
we will keep looking at here. And, N C over
C L naught versus C L naught and we said that
50:13.099 --> 50:16.970
this is linear, but then we said that, there
could be different deviations from this. And,
50:16.970 --> 50:20.670
what we had been studying in the last few
lectures are these deviations from the scatchard
50:20.670 --> 50:24.809
plot.
So, the scatchard plot is linear, if the receptor,
50:24.809 --> 50:30.119
the if the ligand is present is is in excess
and the receptor binds in the simple bio molecular
50:30.119 --> 50:33.569
kinetics.
Now, when the receptor there, they deviate
50:33.569 --> 50:38.309
from these, you have what is known as positive
co-operativity and negative cooperativity
50:38.309 --> 50:41.910
co-operativity. And, that is what we looked
at. So, you look at the scatchard plot over
50:41.910 --> 50:45.529
here and there is negative co-operativity
and positive co-operativity and let me.
50:45.529 --> 50:50.319
So, these are the different cases that we
looked at different deviations. So, first
50:50.319 --> 50:54.059
one we looked at is ligand depletion. So,
C L was no longer C L naught. It became a
50:54.059 --> 50:59.450
quadratic equation. We solved it and we showed
how to do that. Next was multiple receptors
50:59.450 --> 51:02.470
binding and what was the difference between
multiple, what is the point about multiple
51:02.470 --> 51:04.910
receptors they were binding independently
of each other.
51:04.910 --> 51:08.819
So, two different receptors binding to the
same ligand, but binding independently of
51:08.819 --> 51:13.730
each other. And then, we looked at cases where
multiple receptors, they are not actually
51:13.730 --> 51:17.799
multiple receptors, but the receptors were
changing conformations. As a result of which
51:17.799 --> 51:21.599
the rate constants were changing. And, they
are essentially behaving like multiple receptors.
51:21.599 --> 51:27.240
So, here is a summary of everything that we
did in the last couple of lectures, last few
51:27.240 --> 51:32.160
lectures. So, single receptor, when we have
a single receptor, the scatchard plot is linear
51:32.160 --> 51:36.309
and the dissociation kinetics is single exponential
as it is written on the table, in the table
51:36.309 --> 51:41.730
out here. When you have a single receptor,
but ligand is depleting, then the scatchard
51:41.730 --> 51:46.099
plot is non-linear and the dissociation kinetics
is still single exponential.
51:46.099 --> 51:50.789
Now, it is an interesting case of, next is
the interesting case of two different receptor
51:50.789 --> 51:57.339
populations. And, the scatchard plot is non-linear
and the dissociation kinetics is double exponential.
51:57.339 --> 52:00.930
And, we discussed this in the beginning of
this class that, if the dissociation constant
52:00.930 --> 52:05.029
of one is very different, is very large as
compared to the dissociation constant of the
52:05.029 --> 52:10.599
other, we are in business and we can separate
out the rate constant, the dissociation and
52:10.599 --> 52:14.869
the association rate constants of the two
cases. Otherwise, it is very hard. Next, we
52:14.869 --> 52:21.869
looked at the case where receptor ligand inter-conversion
and the scatchard plot is still linear as
52:22.480 --> 52:27.900
as we just showed. And, the dissociation kinetics
is two or more exponential. And, this inter-conversion
52:27.900 --> 52:32.349
is what kind of inter-conversion? The receptor
itself, receptor subpopulation itself changes
52:32.349 --> 52:36.180
conformation becomes another receptor.
So, because both these are chemically the
52:36.180 --> 52:40.890
same. They both bind to the same ligand, but
the dissociation rate constants are different
52:40.890 --> 52:45.940
and they form different complexes, which can
also interchange; that is interconvert through
52:45.940 --> 52:49.960
change of conformation. And then, there are
other cases which we did not really study,
52:49.960 --> 52:53.720
but these cases I just want to go through
quickly. One is inter-conversion of ligand
52:53.720 --> 52:58.160
to a non-dissociable form. And, and in this
case, the scatchard plot is linear and the
52:58.160 --> 53:02.509
dissociation kinetics is double exponential
as you see over here. So, inter-conversion
53:02.509 --> 53:07.539
of ligand to a non-dissociable form and then
then there is true co-operativity in which
53:07.539 --> 53:13.309
case that, scatchard plot is non-linear and
the dissociation kinetics is double exponential.
53:13.309 --> 53:20.309
So, let me go and show you this over here,
what we had before here. So, this multivalent
53:22.069 --> 53:25.869
ligand and these cases where receptor aggregations;
so, these are other cases that can happen
53:25.869 --> 53:29.740
that is, one ligand is bind to two receptors.
That is a straight forward case, but a little
53:29.740 --> 53:36.549
more complex cases, the case of receptor aggregation
which is that, two receptors R plus R together
53:36.549 --> 53:41.630
form an aggregate. And then, this binds step
by step with a ligand. So, essentially you
53:41.630 --> 53:46.930
form L R R L, which is two receptors binding
to two ligands, but the fine bond bind as
53:46.930 --> 53:51.109
an aggregate; which means that two receptors
bind to one ligand first. And then, this ligand
53:51.109 --> 53:54.309
with the two receptors binds to the second
ligand to form LRRL.
53:54.309 --> 53:58.480
So, again the kinetics here is going to be
little different. The reason this is going
53:58.480 --> 54:03.130
to be little different is that the kinetics
is going to be little different is because
54:03.130 --> 54:08.180
the dissociation, association rate constants
are varied over here. And, this is the case
54:08.180 --> 54:12.490
of co-operativity and I, if you remember at
the beginning of this course of this chapter,
54:12.490 --> 54:16.640
I gave you the example of haemoglobin molecules,
four hemoglobin molecules binding to oxygen
54:16.640 --> 54:20.930
step by step and this is the very parallel
example of that. So, two receptors binding
54:20.930 --> 54:25.660
to one ligand first and then the two receptor
one ligand complex binding to the second ligand
54:25.660 --> 54:30.440
and forming two receptor, two ligand complex.
So, there is the co-operativity out here and
54:30.440 --> 54:35.299
there is positive co-operativity. This is
similar to the haemoglobin oxygen binding.
54:35.299 --> 54:39.700
And then, what will happen is the rate constants
will be different at each of these steps and
54:39.700 --> 54:43.289
it, they would be different. So, two receptor
binding to two ligand is different from the
54:43.289 --> 54:48.430
rate constant. The simple case of one receptor
binding to rate constants would be different
54:48.430 --> 54:52.799
from one receptor to binding to one ligand.
So, these are the different kinds of receptor
54:52.799 --> 54:56.980
ligand bindings that we talked of the kinetics
of this. Now, where does these kinetics come
54:56.980 --> 55:01.440
into play, what is the, what are the real
physical processes where these kinetics are
55:01.440 --> 55:05.819
very important. And, I had talked a little
bit about this. And, the one particular case
55:05.819 --> 55:09.470
that we are going to look at and we are going
to look at with respect to particular disease,
55:09.470 --> 55:13.200
which is familial hypercholesterolemia in
the following lecture, is receptor-mediated
55:13.200 --> 55:16.989
endocytosis.
So, receptor-mediated endocytosis is the very
55:16.989 --> 55:22.630
important process. And, it is very important
for several biological and physiological processes.
55:22.630 --> 55:26.730
And, this is where the kinetics come in. And,
the disease that we are going to talk about
55:26.730 --> 55:30.529
familial hypercholesterolemia is a genetic
disorder. And, we are going to look at why
55:30.529 --> 55:35.150
this genetic disorder happens. It is because
of some kinetic disabilities that is, some
55:35.150 --> 55:40.269
kinetic rate constants is lower than what
they should have been and or some other reasons.
55:40.269 --> 55:45.519
You know the complexes not being formed properly
or not enough ligands or not enough receptors,
55:45.519 --> 55:48.480
what is the reason?
So, this is a very practical and interesting
55:48.480 --> 55:54.720
example and application of this theoretical
study of receptor ligand binding. So, we did
55:54.720 --> 55:58.690
a purely theoretical study till here of receptor
ligand binding, but we are going to apply
55:58.690 --> 56:03.239
it in the following lecture to different diseases
and then name. Main thing that we are going
56:03.239 --> 56:08.869
to focus on is the process call receptor-mediated
endocytosis. With that, I will stop to today
56:08.869 --> 56:12.999
and we will talk about receptor-mediated endocytosis
in the following lecture. Thank you.