Welcome back to this lecture on Biochemical Engineering. If you remember what we did in
the last class, was essentially we are looking at multiple receptors. And, what happens when
the, in the species, in the complex, in the ligand binds to multiple receptors. We talked
about the fact that these multiple receptors essentially bind independently to the ligand
and the binding of one, does not influence other. And then, we said, is it possible to
determine separately, whether these multiple receptors receptor 1 bound the amounts that
receptor 1 binds to the ligand, the amount that receptor 2 binds to the ligand. We said,
well, it is not always easy to figure out because what we do is, we label the ligands
essentially. So, the ligand, if there is a one kind of
ligand, it will either bind to receptor 1 or bind to receptor 2. And, there is, you
know we are just looking at the fluorescence or the some kind of labeling. And, it is not
possible to figure out in general, whether receptor 1 binds or it is binds receptor 1
or it binds to receptor 2. In some extreme cases, obviously it is possible. And, what
are those cases? So, the case that we talked about was, when a receptor of a particular
kind has a much higher binding infinity than the receptor of the other kind.
And, If you remember, so we, this was what we look doing, if I will, if I go to the screen
now. So, a receptor 1 and receptor 2 bind. And, this steady state concentration of the
complex is the amount of receptor 1 is binding to the ligand plus the amount of receptor
2 is binding to the ligand. So, the first term you find here which is N C, C L naught
N R T 1 over C L naught plus K D 1; that is the amount of receptor ligands bind to receptor
1. And, the second term that you see here, C L naught N R T 2 over C L naught plus K
D 2; that is the receptor amount of ligands binding to receptor 2. So, the fact is that,
each of this receptors act independently and bind independently and there is no influence
of the binding of 1 receptor to the binding of the other. But, it is possible as I said
to distinguish between the two values, if the K D, that is the dissociation rate constant
of receptor of the biding of receptor 1 is much much higher or much much lower than the
receptor binding rate constants of the second one.
Now, if you remember, so, we would went through the calculations and I am not going to repeat
all of that. And, when we did the calculations, we took the case when the amount of complex
that was there at the beginning was zero and we did all the calculations.
And, what I want to attract your attention to is this factor this biphasic nature of
the curve. So, what you see over here is N C over N R T and this is time. So, you have
a single receptor and the case of a single receptor, the curve is monophasic; whereas
in the case of two receptors, the curve is biphasic.
Now, as you can see over here, these are data here and I am going to look at that. So, look
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
the two orders of magnitude difference between the dissociation rate constants of 1 and 2.
So, if K D 1 is hundred times, so, here if K D 1 is much much smaller than K D 2 as the
case is, then K 1 K minus 1, let us say over K 1, that is the backward rate constant over
the forward rate constant is much smaller than K minus 2 over K D 2; which means, affinity
of receptor 1 is much much higher than affinity of receptor 2. So, this is what we concluded;
that the affinity of receptor 1 is much much higher than receptor 2, if this is the case.
And, these are cases, when we can really distinguish between the two kinds of receptors because
as I just showed in the biphasic response that I showed that we had. So, what does this
lead to, if K D 1 is much much smaller than K D 2? What this leads to is that in the exponential
graph that you have, the slopes of the exponential graph all points of time will be very different.
And, what you see is that, there is clear biphasic nature; or in other words, the slope
increase in a certain way. And, then they decrease suddenly, say if I can go back to
that screen quickly, see, you see that this is the slope, say some kind of an exponential.
But, this exponential slope of this curve, whether it is exponential or something close
to linear does not matter; is very very different from the slope of the next curve.
So, which means what? That, if these two receptors are there and the K D 1 of 1 is much much
smaller than the K D of the other; or in other words, receptor 1 has a much higher affinity
to than receptor 2. Then, this implies that receptor whenever you let these ligands loose
on the system, so, whenever you let these ligands loose on the system, so receptor 1
will capture these ligands, so that ligands are around. So, receptor 1 will immediately
come and bind with the ligand. So, as the result, the initial amount of binding that
will occur will essentially be because that of receptor 1.
So, once we understand than then we realize that, yes the two slopes are very different
from each other. So, initially you have a much higher slope and then the slopes decrease
with time. So, what this implies is that for the case, when the slope is higher or in the
initial time period, so as soon as you let the ligands loose on the receptors, the receptor
1 is binding with the ligand. So, this will essentially imply that the receptor, the slope
first slope that is the higher slope that corresponds to the binding of receptor 1 to
the ligand. It is not that receptor 2 does not bind at all, but the fact of the matter
is that much of the binding that occurs, maximum most of the binding that occurs is because
of receptor 1. And, if you come to think of it, you know, if you come to think of these
two as comparative binding that is receptor 1 is binding comparatively to receptor 2,
then what you realize is that because the rate constant or the affinity of receptor
1 is hundred times that of receptor 2. So, the binding should always be in the ratio,
more or less on the ratio 1 is to 100 or 1 is to 99. So, if hundred ligands bind, 1 percentage,
only 1 percentage of them is binding to receptor 2 and 99 percentage of them is binding to
receptor 1. So, how does this help us? How does this knowledge
help us? Now, I gave you a problem at the end of the last class and I will go back to
that problem. This helps us in identifying the rate constant of the system because just
as I said that when we label the ligand, there is no way for us to label the receptors, we
are only labeling the ligand. So, when we label the ligands what happens is that, we
can only figure out that what is the total amount of complex that has been found. And,
there is no way for us to separate out the receptor 1 from receptor 2 or the constant
rate constants of receptor 1 from receptor 2. But, what we essentially want to measure
are the rate constants of receptor 1 and receptor 2. So, how do we go about it?
So, what we do? We choose the ligand, we purposely choose the ligand. So, think of that you are
doing an experiment and how do you go about it. So, you can only label your ligands; you
cannot label your receptors. So, what do you do? You purposely choose a ligand such that,
it binds both to receptor 1 and receptor 2. But, the binding rate constant or the K D
at the dissociation rate constant, the binding affinity of the ligand for receptor 1 is much
much higher than the binding affinity for the ligand, for of the ligand for receptor
2. So, that is the highlight. That is the point that you need to exploit.
So, once you have understood that what you do is, you essentially figure out. So, if
1 is hundred times, the other and if you let the reaction happen for a certain period of
time, initially what will happen is, if the receptor 1’s affinity is higher, much much
higher than receptor 2 what will happen is, all of it will go. The ligands will, almost
of the ligands will actually go and bind with receptor 1. So, that is essentially that is
going to happen. Then, what happens? As time progresses, receptor 1 are going to be more
or less saturated. And, then is, when you see that bend in that curve that, just that
I showed you just now. So, receptor 1 is more or less saturated.
As a result, what happens is, now receptor 2 starts to bind to the ligand. And, this
is the mechanism or the hypothesis…, say that is that is what we are going to explain
as we try and solve the problems.
So, let us go to the problem. So, this is the problem that I gave you. And, what I asked
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
naught N R T 2 over C L naught plus K D 2.
So, let us write this. So, and C equals C L naught N R T 1 over C L naught plus K D
1 plus C L naught N R T 2 C L naught over K D 2. So, one of the ways of tackling this
is essentially, just what I said that, you know there it is very hard to be able to figure
out. So, you have, just have a one curve, come to the think of it you just have a one
curve. So, when you had… let us go back to this. And, when you had a single receptor,
this was your scatchard plot. So, when you had a single receptor, you had
two unknowns; one was N R T and the other one was K D. And, you could draw a scatchard
plot, which was linear. And, from the slope, you figured out N R T. K D, sorry. And, from
the intercept, you could figure out N R T. So, two rate constants, two constants you
needed to figure out. You had one plot. And, from the slope, you figured one and for an
intercept you figured another. But, here, what happens is the the problem that I gave
you. Sorry. So, the problem that I gave you here, the
problem with this is that you have one, two, three, and four. So, four unknowns and1 plot.
So, how do you go? So, this can utmost give you two unknowns. So, how do you figure out
the other two unknowns?
So, what we do is, we obtain a scatchard plot. Now, typically a scatchard plot would be linear.
So, if you do N C over C L naught, over N C. Then for a particular value C L naught,
C L naught equal to say 1 micro molar or10 micro molar, something like that. This is
what you get. This is your plot. But, in this case what will happen is that, given that
K D 1 is much much lesser than K D 2.
So, your plot is going to be something like, let us keep this formula next to each other.
So, your plot is going to be something like this, and this. Why is that going to be the
case? The reason is that N C equals N C 1 plus N C 2. That is what my total. So, it
is complex form from receptor 1 and the complex form from receptor 2. That is what is going
to be. So, when that is the case, then when you have
a limited amount of, a small amount of complex that is formed, what will happen is, much
of it will be from receptor 1. Is it clear. Why is that going to be from receptor 1 because
receptor 1’s affinity or affinity of receptor 1 is much much greater than that of receptor
2. So, that is what will happen. So, you will have, most of it will be formed from receptor
1. Now, once receptor 1 is more or less saturated, most of the receptors are bound to the ligand,
then only receptor 2 starts to form. So, then what happens is, you can exploit this as receptor
1. You can call this as almost as receptor 1. And, this would be primarily receptor 2
plus, may be receptor 1 also.
So, what we do now? So, we have our equation N C equals N C 1 plus N C 2, which equals
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
plus K D 2. Now, for the first part of the plot, so, if
my scatchard plot is something like this, N C over C L naught over N C, this N C that
you get over here is essentially N C 1; this value. So, I can straight away from the. So,
if I take this this part of the plot over here, I can straight away say that the slope
of this equals the slope of the first part equals minus 1 over K D 1 and intercept equals
N R T over K D 1. I can straight away say this. So, what happens is, essentially I use
the first part of the slope to evaluate both my K D 1 and N R T 1. And, the second part
I can calculate, I can assume that both receptor 1 and receptor 2 are present and I can go
ead and calculate everything I want. Or, second part you, I can make an even more simplifying
assumption and say that this is equivalent to N C 2. In which case, the slope of this
will give me minus K D1 over minus 1 over minus K D 2 and the intercept will give N
R T 2 over K D 2.
So, that is the possibility. So, I can do something like this. So, if that is the case,
then slope, this is N C over C L naught over N C. Then, slope equals minus 1 over K D 1
intercept equals N R T 1 over K D 1 and slope here, can be minus 1 over K D 2 and intercept
equals N R T 2 over K D 2. That is the possibility, or here you can assume both in receptor 1
and receptor 2 to be acting; in which case, you have to do a little more complicated calculation.
But, this is something you can definitely take. So, that is a, that is the way we solve
this. Otherwise, there is no other way of looking at it.
Now, let us look at some other things here, some other variations of the problem over
here is, you can see the… what we are talking over here is, now, so we start to talk about
here is interconverting receptor subpopulation.
So, something like, where the receptors convert within, from two into each other, two different
forms. So, essentially what happens is that, in the previous part of the lecture what we
talked about our receptor is that, work completely, independently of each other, when they bind
to the ligand independently. What we are going to talk about now, there are cases where the
receptors bind, but they do not bind independently. They bind; there is a independence between
these receptors. Or in other words, the receptors populations are sub populations, which are
interconverting to each other. And, what we mean when we say, we are, they are interconverting
to each other. Essentially it is a system of logical change;
you know this is conformational change that is happening in the receptor. So, it is not
the chemical composition of the receptors as such that are changing, but the conformational
of the receptor, the conformational changes occurs in the receptor. And, what does the
conformational change in tail; it essentially means that the rate constant, the dissociation
and association constants. Or in other words, the binding rate constants would change for
these cases. So, the slope, let us look at there.
So, interconverting receptor sub population; so, if you look at the screen what happens
is that, often the receptor undergoes a conformational change, just as I said, after binding to the
ligand. And, so, when we talking of two different receptor populations, how does this differ
from what we had done before is that, what we had done before were two completely different
receptors, which have affinity for the same ligand. Here, it is not that case. It is not
two completely different receptors, but the same receptor which is undergoing conformational
change. And, as a result, the dissociation and association constants are differing here.
So, the change, however does not affect subsequent ligand receptor binding. So, and, but there
is a conformational change that occurs. And, in the most general case, the receptor is
present as an interconverting sub population. That is what we are going to talk about in
moral, today. And, so the most general case of receptors is present as an interconverting
subpopulation. And, these subpopulations have differences in rates of dissociation and association
between receptors and ligand. So, just as I said there, so because of the conformational
change, what results; because of the conformational change is, essentially the fact that these
receptors have different kinetic constants. So, now, let us look at the model that we
have over here; the interconverting receptor subpopulation model. So, as you can see over
here, this is the most general form of the model R 1 plus l. So, R 1 is my receptor population
1; R 2, as I said is same, chemically the same receptor but just a conformation, R 1
with the conformational change. So, R 1 plus L forms the complex C 1. And, R 1 also forms
the complex, changes to R 2 and forms the complex C 2. See, this picture may be slightly
confusing. So, what I will do is I am going to write this differently.
So, R 1 plus L K 1 1 K minus 11 C 1; so, R 1 is essentially reacting with the ligand
to form the complex C 1. And, R 1 is going through a conformational change to form R
2. And then, you have R 2 plus L. So, again it reacts in the similar way as R 1. K 2 2
and K minus 2 are its rate constants to form C 2. And, these are the complexes; this is
the complex, and this is the complex. And, it turns out, this is the complex themselves
can undergo conformational change to rotate between each other. So, this is C 2 and the
rate constants for this are K 1 2 and K minus 1 2.
So, this is the set of reactions; interconverting receptor, interconverting receptor sub populations.
So, what is happening is that the receptor 1 is binding with ligand to form a complex
and the rate constants, forward and backward rate constants for that are K 1 1 and K minus
1. Receptor 1 is undergoing a conformational change to form R 2. And, the weight constants
for that are K 2 1 K minus 2 1. And, R 2 reacts with the ligand as it is, to form a complex
C 2. And, C 2 and C 1 undergo conformational change again.
So, if you look over here, go back to the screen and look over here, you find that this
is the way we have represented it. This is slightly confusing way of looking at it. And
therefore, I drew it, but you can write this cyclical way. So, remember the only thing
I want to point out here is that, it is R 1. When I draw these, these this this line,
for example, the one the vertical line on the left; it is only the conversion between
R 1 and R 2. And, L is not involved in this reaction.
Now, a special case of this is what you see over here; where the receptor is themselves
do not change conformation, but it is the complex that changes conformation. So, R 1
reacts with the ligand to form complex C 1. And, it is C 1 that changes conformation to
form complex C 2. And, the rate constants are K 1 1 and K minus 1 and K 1 2 and K minus
2. So, essentially what we have done is, if you take this cyclical model appear, if you
cut out these two parts, these two lines over here and these two lines over here, you get
R 1 plus L gives C 1, and C 1 gives C 2. So, you get this triangle over here and that is
the special case. So, what we will do now in the in the remaining
part of this lecture is, essentially look at and model these. So, what we will do now
is, to start with this. We will look at the special case B first; the reason being that,
it is easier to model.
So, we will come back and look at A after that. But, essentially we will look at B first.
So, in B, let us look at this. So, R 1 plus L, K 1 1 is the forward rate constant, K minus
1 is the backward rate constant. And, C 1 and C 2 is the complex, which undergo conformational
change through the rate constants K 1 2 and K minus 1 2.
So, what happens when there is steady state? When there is steady state, you can relate
the N C 2 and N C 1; that is the rate constant 1 and rate constant 2. Let me work out this
for you.
So, how this happens? Fine. C 1, K 1 2 K minus 1 2 and C 2. So, what is, let us write all
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
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
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
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
N C 1 minus K minus 2 1 N C 2. So, at steady state, each of these would be zero.
So, if that is zero, then from this equation, the last equation let us call this equation
a, b and c. Then, from equation c from equation c, what we find is K 1 2 N C 1 at steady state
equals K minus 1 2 N C 2. Let me go through this again. So, these are my set of equations
that I wrote this are the, these are the model equations that dynamic model. Let us forget
this steady state for a minute. So, this is what I have and steady state model.
So, N R 1 is being formed because of, so the N R 1 is being formed because of this reaction.
And, it is been destroyed or removed because of the forward reaction. N C 1 is formed from
this reaction as well as this reaction. And, it also removes because of this reaction as
well as these reactions. So, four terms here and whereas, C 2 is formed
in the forward direction and destroyed or removed in the backward direction. At steady
state, all these three have to be zero. So, what we do, we first equate the simplest one;
the second one to be equal to zero. So, this is this is what we get. So, what we immediately
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
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
1. So, that is what I get.
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
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
total amount of receptors, which is N R 1 plus N C 1 plus N C 2. Or in other words,
receptors that are completely free and receptors that are in complex 1 and receptors that are
as complex 2. So, this I can write now as N R 1 plus. I have already covered calculated
what my N C 1 and N C 2 together is which is N C 1 plus 1 over K D 1.
So, now if I go back to my equation over here, if I go back to my equation over here, so
this has to be zero at steady state, also this has to be zero. So, what we can do is
if I equate this to zero, I get a relationship between N R 1 and N C 1 straight away.
Let us do that. So, equating A to zero, so A is this equation out here. So, A is this
equation. I am equating this to zero; which means that K minus 1 N C 1 equals K 1 1 N
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
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
times N C 1 over C L naught. So, this I can write as K D 1 1, previously I had written
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
1 equals K D 1 N R 1 equals, just let me just put it over here like this yeah. So, N R 1
equals K D 1 N C 1 over C L naught and N C 2 equals N C 1 over K D 1.
Now, what was my constraint equation? My constraint equation was that N R T equals simply equals
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
1. Now N R 1, I can write now from the from the previous equation, I can write is this
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
1 2. So, now I can write this as N C 1 K D 1 1
over C L naught plus 1 over K D 1 1 1 2. So, this is what I get. So, this is my relationship
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
K D 1 1 plus C L naught times 1 over K D 1 2. So, this is the relationship I have between
these.
So, N C equals N C equals N C 1 plus N C 2; which I have written it previously if you
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
C L naught divided by K D 1 1 plus C L naught 1 plus K D 1 2 times, this factor over here;
which is 1 over K D 1 2. So, this is my relationship that I have between
these numbers. So, I can, one of the things I can do is try and convert this into a slightly
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
C L naught 1 plus K D 1 2. I can write it like this or I can make it even simpler.
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
this by 1 plus K D 1 2. So, this is my final relationship between N C and N R T. So, if
I want to draw my scatchard plot, which will be N C over C L naught; that would simply
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.
So, I can draw a plot like this.
So, if I now go back to the screen, this is exactly I, what we derived; N C equals N C
over C L naught equals N R T divided by C L naught over K D 1 1 times K D 1 2 divided
by 1 plus K D 1 2. So, this is the plot that I want to obtain. So, why did I, why did I
do it this way, why did I club club it this way because if I am to draw a plot of N C,
so, what do I do? I can write this as K D apparent. My apparent rate constant I can
write. So, if you look at this, it is even has the units of rate constant correctly.
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
2.
So, what will happen is, if I can now write my plot as N C L, N C over C L naught equals
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,
I can either do it this way, which is C L naught over N C equals C L naught over N R
T plus K D apparent over N R T; where K D apparent is this number, which is K D 1 1
times K D 1 2 divided by K D 1 2 plus 1. Fine. This is my number. So, what I can do over
here is, if I draw a plot of C L naught over N C versus C L naught, what do I get? I just
get a straight line something like this. I just get a straight line something like this.
And, the slope of this straight line is 1 over N R T; whereas the intercept is K D apparent
over N R T. So, why can, what I can straight away do is,
from the slope I can calculate my N R T and from the intercept, I can calculate my K D
apparent. So, what was the whole idea of this entire process was to be able to we had many
of constraints out there; K D 1 1, K D 1 2 and C L naught. And, C L naught is not a constant,
but it is something that we can vary, but N R T.
So, what was the idea? The idea was to club these three constants that we had into two
because two is something very measurable. And, write the equation in such a way that,
we can plot it very easily. So, what we manage to do is, write the equation in a way where
it is a straight line. And, we could just get the N R T from the intercept and K D;
just get the N R T from the slope and the K D apparent from the intercept.
So, that was what we were trying, we are trying to do. So, this is the special case. The case,
where the receptors themselves they are not interconverting, but the complexes are interconverting.
So, what about the more complex case now? And, as I just said that this set of reactions
could be replaced by this set of reactions are actually written by these set of reactions.
And, what we do, you know, how can we look at this and model this now? This is slightly
more complicated one the process that we are going to do. And, I probably, I am not going
to have the time to work out the whole thing, but we will start to work out. And, the process
that we are going to have to do is exactly the same process. So, this is, this is my
plot, this is my, these are my sets of reactions. And, all we have to do we have to write balances
for each of these R 1, R 2, C 1, C 2. And, let us try and do that, to start with.
So, R 1 the balance from R for R 1, so it just for a second, I will keep this out here,
so that you can write these set of reactions down. So, R 1 plus L K 1 a forward reaction
K 1 1 backward reaction K minus 1 give C 1; R 1 forward reaction K 2 1 and K minus 2 1
gives R 2; R 2 plus L forward reaction K 2 2 backward K minus 2 2 give C 2 and C 1 forward
reaction K 1 2 backward reaction K minus 1 2 gives C 2.
So, let us try writing the balances for R 1 now, first. So, R 1 is being formed by this
reaction, K minus 1 1 N C 1 and is being removed or being depleted by this reaction N R 1 times
C L naught. And, for another set of reactions and R for that R 1 is being formed from K
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
being formed for from K 2 K minus 2 two times N C 2 and depleted from K 2 2 N R 2 times
C L naught. And, there is other inter converting reaction which is, it is being formed from
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,
you have two sets of reactions.
So, this is, this is their set of reactions for N R 1 and N R 2. And, if you look at C
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,
then you have K minus 1 2 N C 2 minus K 1 2 N C 1. And, similarly we can write the balance
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
minus 1 2 N C 2. So, these are my four balance equations that I, that I write. And, what
do we do with them? So, at steady state, I have to equate all of these to be equal to
zero d t.
So, let us look at the total amount of complex that we have. So, this is my set of equations
for N C 1. So what, let us look at the total amount of complex that we have, which is N
C 1 plus N C 2. What is that number, if I add all the terms that I get over here? I
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
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
R 2, just this is not. yeah basically- basically, this are adding these two equations. So, this
is at steady state. This is and basically this is d by d t of this at steady state this
equals zero. Similarly, d by d t of N R 1 plus N R 2 equals K minus 1 minus 1 1 N C
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
N R 2 C L naught. So, this is what we get. So, and this equals zero at steady state.
So, what we have to do essentially is that, it is a little more cumbersome process. So,
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
of N C 1, in terms of N C 1; let us say something like that. So, when we do that, then what
we can do from here, what we can get from here is that, we can replace everything in
terms of N C 1. Now, what was my constraint equation? If you look at these two sets of
reactions that I have written over here, this is, let us go one by one.
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
four sets of reactions, what do I get? If I add all these four, what I get is d d t.
If you look at these equations of N R 1 plus N R 2 plus N C 1 plus N C 2, the summation
is d d t of that the summation is zero; which means that N R 1 plus N R 2. This implies
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,
which is N R T; that is the total amount of receptors present. Or, in other words, the
total amount of receptors that is present is divided into the receptor 1, that is free
receptor 2, that is free receptor 1 complex that has been formed and receptor 2 complex
that has been formed. So, we go through these calculations and I am skipping the details
of the calculations because we do not really have a lot of time.
So, once we go through these calculations, if you look at the screen, what you find over
here is that this constraint equation that I wrote. So, what you need to do? You have
to keep substituting everything. So, you decide that you want to substitute everything in
terms of say, N C 1. So, you convert N R T 1 as; write it as in terms of N C 1. Substitute
here, convert N R T 2 and write it in terms of N C 2 and substitute here and N C 1 sorry
and substitute here and write N C 2, also in terms of N C 1 and substitute over here.
So, once you do all that, what you will find is just like in the previous case, just like
in case B, as we did case B we did explicitly. And, here also, you will find just like in
case of B that N C could be written as C L naught N R T over C L naught plus K D apparent.
So, this is, this is what you can write. So, N C over C L naught is N R T over C L naught
plus K D apparent, where K D apparent is slightly more complicated over here. It is K D 1 1
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
D 2 2 times 1 plus K D 2 1 over 1 plus K D 1 2. And, these, all these numbers are written
over here. So, K D 1 1 is the dissociation rate constant for the first case here; that
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,
the dissociation rate constraint over here, between the two receptor subpopulations. So,
this is the way the receptors the changing conformation K D 1 2 equals K D minus 1 2
over K D 1 2, which is how the complexes are changing conformation. And, that is the dissociation
rate constants for that, and K D 2 2 equals minus K D 2 2 over K minus K 2 2 over K 2
2 k. So, this is how the receptor, the second receptor is binding to the complex. So, this
this is the thing. Now, what happens is, what you had been able
to do or we had been able to do through these processes. We had these one, two, three, four
constants and N R T, of course the fifth constant. So, we have been able to reduce these five
constants into two; or in other words, K D apparent, if you look at it involves all these
constants. So, what we had been able to do in the process is, we had been able to reduce
these four constants into a single constant K D apparent. And, why did we do that? The
reason is again as the same reason that we had before, which is that we, it is very hard
to be to be able to handle all of these. So, what? So, what we essentially have is just
one equation over here.
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.
So, this is my equation. So, the advantage is N C over C L naught equals N R T over C
L naught plus K D apparent. Fine. So, C L naught over N C equals C L naught over N R
T plus K apparent over N R T. So, what do you do? You perform? take different values
of C L naught, label the ligands and you perform your experiments for each of these sets and
then you find out that, what what you essentially, you know what is the ratio of C L naught over
N C for each of these sets and and measure it in terms of C L naught.
So, essentially this is your, if you want to plot it over here, this is you plot. So,
the C L naught over N C and this is C L naught. So, just as I said that, this is going to
be a straight line over here and the slope is going to be equal to 1 over N R T and the
intercept is going to be K D apparent over N R T.
So, this is, this is essentially what we get and K D apparent over here, equals K D 1 1
over 1 plus slightly complicated formula 1 plus K D 1 2 divided by 1 plus K D 2 1 right
here and divided by K D 1 2. So, this is my formula. So, what we essentially do? We conduct
experiments with taking different amounts of ligands and figure out, how much is my
total fluorescence that I get; that is a total amount of complexes that had been formed.
And, we measure that and plot it against the amount of ligands that are there.
So, in case two, the case one, this is case one. And, case two, we did separately, but
if we have been able to do case one, which is a lot more complicated case. And, case
two, in case two would fall out, would be a natural fall out of case one and it would
come as 1 over it come. And, when the case, for the case that 1 over K D 2 1 is zero or
this case that is 1 over K D 2 1 is zero. In other words, K D 2 1 is much much greater
than K D minus 2 1 that is or in this case, this case, this step is not there. So, if
you go to case A, case B, you will see that what is case B? Case B is when the receptor
populations, they are themselves, they are not interconverting, but it is the complexes
that are interconverting. And, when does that happen? When, there is no connection, no direct
inter-conversion between the receptors. So, if you look here, so what we are doing
here is that, there is this reaction between the receptor inter-conversion and that reaction
has to be absent. So, I think, we more or less looked at different cases over here.
And, what you see on the screen right now is a summary of the in different processes
that we looked at. So, if you remember, so this is I will, I will run you through very
quickly. So, we started with the single receptor case. And, let me go through this very quickly
and try and summarize what we have done in the last few cases. So, remember, this was
the very important plots that we looked at just a second.
So, this is, this is where we started with. This is the single receptor ligand binding
case, where there is a receptor binding binding to a single receptor, binding to a single
ligand and we assume that, there is no receptor depletion out here.
So, we got straight and a linear scatchard plots and stuffs like that. And then, we started
to look at. So, this was the simplest case possible. And, for this case, the scatchard
plot is linear, is a straight line. The scatchard plot is essentially the important plot that
we will keep looking at here. And, N C over C L naught versus C L naught and we said that
this is linear, but then we said that, there could be different deviations from this. And,
what we had been studying in the last few lectures are these deviations from the scatchard
plot. So, the scatchard plot is linear, if the receptor,
the if the ligand is present is is in excess and the receptor binds in the simple bio molecular
kinetics. Now, when the receptor there, they deviate
from these, you have what is known as positive co-operativity and negative cooperativity
co-operativity. And, that is what we looked at. So, you look at the scatchard plot over
here and there is negative co-operativity and positive co-operativity and let me.
So, these are the different cases that we looked at different deviations. So, first
one we looked at is ligand depletion. So, C L was no longer C L naught. It became a
quadratic equation. We solved it and we showed how to do that. Next was multiple receptors
binding and what was the difference between multiple, what is the point about multiple
receptors they were binding independently of each other.
So, two different receptors binding to the same ligand, but binding independently of
each other. And then, we looked at cases where multiple receptors, they are not actually
multiple receptors, but the receptors were changing conformations. As a result of which
the rate constants were changing. And, they are essentially behaving like multiple receptors.
So, here is a summary of everything that we did in the last couple of lectures, last few
lectures. So, single receptor, when we have a single receptor, the scatchard plot is linear
and the dissociation kinetics is single exponential as it is written on the table, in the table
out here. When you have a single receptor, but ligand is depleting, then the scatchard
plot is non-linear and the dissociation kinetics is still single exponential.
Now, it is an interesting case of, next is the interesting case of two different receptor
populations. And, the scatchard plot is non-linear and the dissociation kinetics is double exponential.
And, we discussed this in the beginning of this class that, if the dissociation constant
of one is very different, is very large as compared to the dissociation constant of the
other, we are in business and we can separate out the rate constant, the dissociation and
the association rate constants of the two cases. Otherwise, it is very hard. Next, we
looked at the case where receptor ligand inter-conversion and the scatchard plot is still linear as
as we just showed. And, the dissociation kinetics is two or more exponential. And, this inter-conversion
is what kind of inter-conversion? The receptor itself, receptor subpopulation itself changes
conformation becomes another receptor. So, because both these are chemically the
same. They both bind to the same ligand, but the dissociation rate constants are different
and they form different complexes, which can also interchange; that is interconvert through
change of conformation. And then, there are other cases which we did not really study,
but these cases I just want to go through quickly. One is inter-conversion of ligand
to a non-dissociable form. And, and in this case, the scatchard plot is linear and the
dissociation kinetics is double exponential as you see over here. So, inter-conversion
of ligand to a non-dissociable form and then then there is true co-operativity in which
case that, scatchard plot is non-linear and the dissociation kinetics is double exponential.
So, let me go and show you this over here, what we had before here. So, this multivalent
ligand and these cases where receptor aggregations; so, these are other cases that can happen
that is, one ligand is bind to two receptors. That is a straight forward case, but a little
more complex cases, the case of receptor aggregation which is that, two receptors R plus R together
form an aggregate. And then, this binds step by step with a ligand. So, essentially you
form L R R L, which is two receptors binding to two ligands, but the fine bond bind as
an aggregate; which means that two receptors bind to one ligand first. And then, this ligand
with the two receptors binds to the second ligand to form LRRL.
So, again the kinetics here is going to be little different. The reason this is going
to be little different is that the kinetics is going to be little different is because
the dissociation, association rate constants are varied over here. And, this is the case
of co-operativity and I, if you remember at the beginning of this course of this chapter,
I gave you the example of haemoglobin molecules, four hemoglobin molecules binding to oxygen
step by step and this is the very parallel example of that. So, two receptors binding
to one ligand first and then the two receptor one ligand complex binding to the second ligand
and forming two receptor, two ligand complex. So, there is the co-operativity out here and
there is positive co-operativity. This is similar to the haemoglobin oxygen binding.
And then, what will happen is the rate constants will be different at each of these steps and
it, they would be different. So, two receptor binding to two ligand is different from the
rate constant. The simple case of one receptor binding to rate constants would be different
from one receptor to binding to one ligand. So, these are the different kinds of receptor
ligand bindings that we talked of the kinetics of this. Now, where does these kinetics come
into play, what is the, what are the real physical processes where these kinetics are
very important. And, I had talked a little bit about this. And, the one particular case
that we are going to look at and we are going to look at with respect to particular disease,
which is familial hypercholesterolemia in the following lecture, is receptor-mediated
endocytosis. So, receptor-mediated endocytosis is the very
important process. And, it is very important for several biological and physiological processes.
And, this is where the kinetics come in. And, the disease that we are going to talk about
familial hypercholesterolemia is a genetic disorder. And, we are going to look at why
this genetic disorder happens. It is because of some kinetic disabilities that is, some
kinetic rate constants is lower than what they should have been and or some other reasons.
You know the complexes not being formed properly or not enough ligands or not enough receptors,
what is the reason? So, this is a very practical and interesting
example and application of this theoretical study of receptor ligand binding. So, we did
a purely theoretical study till here of receptor ligand binding, but we are going to apply
it in the following lecture to different diseases and then name. Main thing that we are going
to focus on is the process call receptor-mediated endocytosis. With that, I will stop to today
and we will talk about receptor-mediated endocytosis in the following lecture. Thank you.