WEBVTT
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Continuation of what we are doing on receptor
Ligand binding in the last lecture. So, what
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we looked at in the last lecture was a case
of receptor binding to a Ligand. So, I talked
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about the receptor being on the cell surface,
a part of it inside the cell surface cell
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you know in the cytoplasm a part of it outside
the cytoplasm the Ligand comes and swims in
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you know binds to the receptors. So, then
we looked at the kinetics of receptor Ligand
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binding and we looked at the simplest possible
kinetics which is the case of R plus R receptor
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binding to the Ligand directly R plus L giving
R l. So, and then we talked about free free
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receptors and total number of receptors and
free receptors in the complexes form.
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So, today what we will do is, we will look
at other kinds of kinetics. So, this is a
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simplified kinetics that may or may not occur
all the time. But in a realistic situation
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you can have other kinds of kinetics coming
in. Before we do that let us very quickly
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go through what we did in the last class because
I hurried a little bit towards the end of
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the lecture.
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So, this is a kinetics that we looked at L
plus R giving C complex and then the complex
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will then you know, that is that is what is
the required over here and K 1 and K minus
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1 being the forward and the backward grade
constant for the system and this is how we
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wrote the kinetic equation, the basic balance
equation for the complex D. So, d C dt of
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C equals a forward reaction K 1 times C R
times C L minus K minus 1 C C if you remember
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it.
Now, what we said that we do not want to express
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this in terms of concentration. We want to
express this in terms of numbers. So, N R
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being the number of receptors, N C being the
number of complexes so, free receptors is
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N R, N C being the number of complexes and
N R T is the total number of receptors which
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includes both the complex and the free receptors.
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So, then we said that how do we convert these
concentrations into numbers. So, Ligand is
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obviously, in a concentration because it is
in the liquid form, but we found out a way
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of converting this to numbers because the
total amount of Ligand C L naught equals the
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amount of Ligand that is present right now
plus the amount of Ligand that is reacted
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with the receptor to form the complex.
So, that is given. The concentration of that
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is given by n times N C over N A. N being
the number of cells per unit volume. So, if
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N C is the number of complexes per cell then
N N and N is the number of cells per unit
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volume then N times N C would be the number
of complexes per unit volume. That you divide
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by the Avogadro number to give the get the
molar concentration of the complex that is
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formed. And so, therefore, that you. So, the
C L that you have the concentration of the
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Ligand is initial concentration minus this
value or in other words initial concentration
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is current value plus whatever current free
Ligand plus whatever has formed complexes
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correct. So, then we can substitute it back.
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So, now, we can have this entire equation
almost as a, in the form of a number right.
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And then we said that we can go ahead and
solve it we can go ahead and solve this also
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as you as you can see over here this is a
second order equation in N C. And this could
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be solved is in partial fractions, but we
meant we wanted to wanted our life little
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simpler and we made a simplifying assumption
which is that C L naught is much, much larger
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than this. So, this could be taken as a constant
and it becomes a first order equation. So,
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we had this first order equation over here
and then N C naught, we assumed N C naught
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to be 0. So, we had exponential part and then
we figured out how to evaluate the half time
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and this is the plot and so on.
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So, I just want to do the last part, bit yes
I think this is where I was little fast. So,
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N C is the you know, so, if N C naught is
0 then you have the exponential variation
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and we figured out how to evaluate these constants.
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Now, the thing that I want to talk about is
this one the last thing I did which is a Scatchard
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plot. So, scatchard plot is essentially a
steady state diagram. So, for example, here
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when you are plotting this, this is this is
your basic expression the one in the box.
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Now, when you are plotting this over here
you are doing an unsteady state plot, but
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this scatchard plot is the steady state version
of it.
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So, steady state means this expression with
time going to infinity. So, the steady state
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would be N C would be N R T C L naught over
K D plus C L naught. And that is a very important
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number for us or a important expression for
us because it is easy to measure because you
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know, it is always harder to measure dynamic
dynamics of a process because we have to keep
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measuring at every time instant at every time
interval whereas, for steady state you can
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let it happen as far as long as you want and
then you can measure it. So, this so, this
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steady state is easier to measure and that
is why it is important in our calculations
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and you will see that this is what we used
mostly. So, N C A steady state would be N
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R T C L naught K D plus C L naught clear?
So, that is what we do when we plot in the
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do a scatchard plot.
So, this is what we start with. So, this is
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the N C max this is what we call N C max.
Why? Because that is the maximum as you can
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see over the from the plot over here. That
is the maximum value of N C that is attainable.
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So, the maximum value is attained at steady
state. So, we call this steady N C max and
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that is given as N N R T C L naught over K
D plus C L naught. This I can rearrange you
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know dividing the both sides by C L naught,
I can rearrange this as N C max over C L naught
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equals N R T over K D minus N C max over K
D. And this I can plot because easily because
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I can measure N C max. So, I can what I do
is I vary in my C L naught. That is the Ligand
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concentration I use different Ligand concentrations.
Let it happen for as long as it happens. Then
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take this steady state, measure my N C max.
How will I measure? Because I told you the
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other day that we will label these Ligands.
So, we label Ligands. We will limit fluorescence
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or something some kind of labeling if it is
a fluorescence some kind of labeling then,
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we can measure the complexes straight away.
So, at steady state we measure how much complexes
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have been formed and we plot this N C over
C L naught versus N C. We plot this and then
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we can get the rate constant. The rate constants
K D as well as N R T fine from the slope and
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the intercept. So, the slope will give you
minus 1 over K D. The intercept will give
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you N R T over K d. So, you can evaluate both
N R T and K D fine. This is known as the scatchard
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plot and it is a very important plot because
we will keep referring to this. What I want
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to mention to you is that scatchard plots
can come in different ways you know. I think
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I gave you the assignment also where you had
to plot the scatchard plot. So, scatchard
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plot in come can come in various ways. The
only thing that is common or unifying between
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these different ways is this scatchard plot
appears appears a is the fact that these are
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steady state plots steady state plots of complex
versus Ligand in some form.
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So, you can keep varying the form of this
scatchard plot, but it still remains a scatchard
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plot you know as long as the complex versus
the Ligand is in one axis it is you have the
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complex and the other axis you have the Ligand
fine. So, this is I think the problem that
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I gave you and I hope you did that. So, today
we start something. So, this was all these
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analysis that we did was for the case where
we assumed simple reaction kinetics R plus
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L giving a complex and we assumed that it
is a second order in the forward direction
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of first order in the backward direction.
Now, that may not be the case all the time.
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So, that is what we are going to look at.
So, as you see on the screen what this this
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is called deviations from bimolecular kinetics.
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So, what we had looked at till now is bimolecular
kinetics so, deviations from simple bimolecular
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kinetics. So, R plus L giving C is the biomolecular
kinetics and this is if I am doing a scatchard
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plot again this is a scatchard plot. So, N
C over C L naught versus N C. So, this is
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scatchard plot and this scatchard plot is
something like linear. Just as you we plotted
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it here right this is this is the linear scatchard
plot because N C over C L naught versus N
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c. So, this linear this line in the middle
that you see corresponds to the corresponds
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to the bimolecular kinetics. Now, this line
below is this curved line and the line above
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this corresponds to deviation from the bimolecular
kinetics. So, it does not have to be necessarily
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bimolecular and what we will do in today’s
lecture, we will study that why could it does
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it not have to be necessarily bimolecular?
If it is not bimolecular what is the options
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how does it react? And try and understand
this and this is known as cooperativity. Positive
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cooperativity and negative cooperativity and
in earlier course you know we had studied
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this if you remember that oxygen plus hemoglobin
you know.
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So, this was the reaction that we studied
oxygen plus hemoglobin is H b O 2 in one of
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the earlier cases and what we figured is that
1 can assume it to be 4 O 2 plus H b 4 because
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hemoglobin comes as a molecular for H b 4
O 8 K 1 K minus 1. So, this is very important
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physiological reaction. As you all know the
reaction between oxygen and hemoglobin and
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that is how oxygen is carried into our blood
through oxyhemoglobin and this is the formation
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of the oxyhemoglobin. So, this is what is
written standard. This is a variation on that.
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this is the variation on that. So, this is
a standard bimolecular form. This is bimolecular.
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standard bimolecular form.
So, this is assumed that one molecule of oxygen
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reacts with one molecule of hemoglobin that
that was the assumption. But then later it
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was found out that these hemoglobin molecules
remain in cluster of four. So, therefore,
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one molecule of oxygen cannot react with the
cluster of four. It has to be four molecules
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of oxygen reacting with the cluster of four.
So, the bimolecular form is no longer preserved
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over here. So, it is one molecule of haemoglobin
reacting with four molecules of oxygen essentially.
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But in reality, this is also a hypothesis.
In reality what happens is that, when you
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write the oxygen hemoglobin equation you find
that if I am to write it in a certain way
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say for example, S o 2 that is the saturation
of oxygen it turns out it goes as O 2 to the
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power N divided by this O 2 to the power N
something like this. So, where N equals 2
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point 1 or 2 point 3 or 2 point 3 4 something
like that which means that it is neither it
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is neither 4 molecules nor 1 molecule, but
some fractional molecule reacting with hemoglobin
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which what does it mean? What does it implies
that that the fractional molecule 2 point
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3 4 molecules of oxygen are reacting with
hemoglobin? Of course not. What it means is
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that, there is co-operativity between the
molecules. Co-operativity means that the affinity
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towards.
So, first say for example, so, you have the
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hemoglobin which consists of four 4 4 of molecules
and you have together cluster of it and you
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have one oxygen combining with the first hemoglobin
molecule. What follows is after that, there
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is a certain change of affinity of that process
of the of the system which means that the
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second oxygen molecule may not be attracted
towards the second hemoglobin molecule as
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much as the first one was or may be more attractive.
Whatever is the case. In most cases this is
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less attracted or more more attractive whatever
is the case, but the point I am trying to
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make over here is that there is a change in
the affinity.
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So, for the first molecule, it reacts as if
it is simple bimolecular. The second molecule
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in in the presence of the first reaction having
occurred does not react any further as if
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it is simple biomolecular. There is an alteration
in the thermodynamics of the process. There
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is the alteration in the basic which leads
to the alteration in the basic kinetics of
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the process.
So, this is known as co-operativity that having
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one of them having reacted in the vicinity
if you have gone to have the second reaction.
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The second reaction may be more favoured or
less favoured. And they form, that you come
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up with the concept of the positive cooperativity
or negative cooperativity. Positive cooperativity
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means there is a positive that is a more favoured
effect. So, positive co-operation between
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the groups of molecule.
So, the presence of the first group for the
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first complex actually facilitates the formation
of the second complex. Negative cooperativity
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means the presence of the first complex actually
retards the formation of the second complex.
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So, this is the concept of co-operativity
is that clear? I think for many of you it
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might be a new concept, but the point fact
of the matter I am trying to you know trying
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to convey over here is simple. It is that
you have for most cases we assume a simple
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bimolecular kinetics. But, when clusters of
molecule react with each other then simple
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bimolecular kinetics is no longer preserved,
not necessarily preserved. You can have the
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first reaction the first set of reactions
favouring or hindering the second reactions.
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Similarly, once two clusters, two 2 reactions
have occurred for example, in the case of
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haemoglobin once two reactions have occurred,
the second set, third set of reactions may
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be favoured or hindered. So, depending on
that you can have positive cooperativity or
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negative cooperativity. And if I go back to
this screen now you will see that this is
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the case. So, central line in the middle this
is the scatchard plot again N C over C L naught
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versus N C. This is the case where there is
just normal bimolecular b m represents bimolecular
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reversible R e v b m binding reversible bimolecular
binding.
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So, this is the case, but simple bimolecular
binding is there. This is the case where positive
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cooperativity is there. K D K K D is the dissociation
rate constant. So, if I go with I think K
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minus 1 let us see I think yeah K minus 1
over K 1 is a dissociation rate constant the
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dissociate not the rate constant the equilibrium
rate constant, dissociation equilibrium rate
14:19.420 --> 14:23.730
constant.
So, positive cooperativity as you will, as
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you as I just as I showed you for the case
of hemoglobin is what did I say that positive
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cooperativity means that the second reaction
is more favoured than the first reaction.
14:32.220 --> 14:36.870
The third reaction is more favoured by then
the second and so on. So, for earlier formations
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actually favour the later ones fine which
means that the K D apparent the dissociation
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rate constant apparent dissociation rate constant
is less than the bimolecular rate constant.
14:48.790 --> 14:55.529
Is that clear? Because if the dissociation
dissociation dissociation is what backward
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or forward?
So, if the backward rate constant is actually
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lower than the than the previous case then,
it means that dissociation is less favoured
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and association is favoured. That is formation
is favoured you know binding is favoured.
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So, this reaction oxygen plus hemoglobin over
here is is pushed in the forward direction
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more for the second case. So, the third case
it would be even more pushed in the even more
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forward direction. So, I gave you this example
of hemoglobin to explain this. So, how these
15:23.079 --> 15:27.540
four molecules bind one after another is it
clear. So, that is the concept of positive
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cooperativity.
So, positive cooperativity means that, the
15:30.279 --> 15:35.870
dissociation rate constant, apparent dissociation
rate constant is less than the bimolecular
15:35.870 --> 15:39.139
dissociation rate constant. Similarly, you
can have negative co-operativity and I will,
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we will have examples here itself where the
apparent dissociation rate constant is greater
15:44.999 --> 15:51.939
than the bimolecular rate constant which or
in other words it means that the dissociation
15:51.939 --> 15:58.329
is being favoured. Now, dissociation is being
favoured, the association is being less favoured.
15:58.329 --> 16:03.519
So, this is this is what we have the negative
cooperativity. So, what we want to study is
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that as I said that what we did in the last
class was a simplistic case, a simplistic
16:08.850 --> 16:13.079
case of simple bimolecular kinetics. And if
you have a simple bimolecular kinetics as
16:13.079 --> 16:19.059
I showed that the slope of this is simply
going to be minus 1 over K D and the intercept
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is going to be N R T over K D. And you can
evaluate K D O and N R t, but if it is if
16:23.619 --> 16:27.740
it is positive cooperativity or negative cooperativity
then what happens? These slopes are these
16:27.740 --> 16:31.819
are convex or concave in shape and then you
cannot make such simplistic or straight forward
16:31.819 --> 16:35.629
calculations from the slope.
So, you have difficulty in figuring out what
16:35.629 --> 16:42.629
the apparent, even the apparent the K D apparent
that you have is. So, what we will try to
16:42.790 --> 16:47.689
understand is how these what is the real chemistry
behind this question of co-operativity, how
16:47.689 --> 16:52.249
does it happen? and and try and understand
and quantify these phases.
16:52.249 --> 16:56.740
So, we will look at each of these cases slowly.
So, the first case is Ligand depletion and
16:56.740 --> 17:00.939
I already talked about this in the previous
class. What did we talk about in the Ligand
17:00.939 --> 17:06.350
depletion case? What is that? You assumed
that the Ligand is present in present enough
17:06.350 --> 17:12.459
yeah that is present in excess, but that is
not necessarily true. The Ligand is obviously,
17:12.459 --> 17:16.809
going to be depleted unless you have lot of
Ligands. So, in real cases the Ligand depletion
17:16.809 --> 17:22.010
can occur. So, that is case A.
So, that can lead to this kind of deviations
17:22.010 --> 17:29.010
from simple bimolecular kinetics right. The
second possibility is that the multiple receptor
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populations of different affinity. So, which
means that, the same Ligand can bind to multiple
17:36.640 --> 17:41.120
receptor. You can have 2 kinds of receptors
2 populations of receptors; R 1 and R 2 and
17:41.120 --> 17:45.630
the same Ligand can bind to these multiple
receptors. So, in which case there is; obviously,
17:45.630 --> 17:49.399
a deviation from the simple bimolecule still
might be bimolecular, but each of these might
17:49.399 --> 17:54.510
be same bimolecular, but the simple one Ligand
one receptor bimolecular kinetics is no longer
17:54.510 --> 18:00.929
there. The third case is multivalent Ligand
binding which means that earlier we had one
18:00.929 --> 18:05.149
Ligand binding to one receptor one Ligand
molecule you can have more than one Ligand
18:05.149 --> 18:09.220
molecule binding to one receptor molecule.
So, that is the third possibility and the
18:09.220 --> 18:13.970
fourth possibility is receptor aggregation.
So, in receptor aggregation what you have
18:13.970 --> 18:18.590
is receptors themselves can aggregate in form
of complex before they bind with the Ligand.
18:18.590 --> 18:22.620
So, receptors may form complex with each other
before binding to the Ligand. So, as you see
18:22.620 --> 18:29.340
over here an example is given. So, two receptors
R and R they bind to form RR and then they
18:29.340 --> 18:36.340
react with the Ligand L to form LL LRRL and
then they again react to find another Ligand
18:36.760 --> 18:43.279
L to form LRRL. So, what. So, this is exactly
similar to the hemoglobin thing you know.
18:43.279 --> 18:47.370
So, hemoglobin hemoglobin what happens is
essentially if I can write this or if I have
18:47.370 --> 18:49.580
to do this four 4 of them.
18:49.580 --> 18:56.580
So, this is how it happens. H b 4 O 2 then,
H b 4 O 2 reacts with O 2 again this is exactly
19:03.750 --> 19:10.750
that that thing. So, it is like a Ligand aggregation
thing or receptor aggregation thing is O 4
19:12.220 --> 19:19.220
then H b 4 O 4 plus O 2. So, this is how it
happens. So, you have the, say let us say
19:28.510 --> 19:32.929
this is the receptor for your case. So, you
have four of those receptors aggregated together
19:32.929 --> 19:35.309
and the Ligand come one by one and bind to
them.
19:35.309 --> 19:42.309
So, this is receptor aggregation. So, so go
back to the screen. So, this is this is what
19:47.529 --> 19:50.830
you have. So, these receptors have aggregated
and it does not have to be two. I gave you
19:50.830 --> 19:54.340
an example of two. It could be three receptors,
four receptors here. For haemoglobin you can
19:54.340 --> 19:58.020
imagine the four receptors being aggregated.
So, four receptors may be aggregated then
19:58.020 --> 20:04.580
the Ligand will come one by one and bind and
finally, we will still have the one receptor
20:04.580 --> 20:08.440
for one Ligand.
So you will have R N L N that kind of molecule.
20:08.440 --> 20:13.570
So, the molecule formed over here is R N L
N where N equals two, but you can have N equals
20:13.570 --> 20:17.510
3 4 and so on. But the point that I am trying
to make is that when you have these kinds
20:17.510 --> 20:22.980
of in terms of molar ratio you can still have
the one is to L ratio of R N L. In the final
20:22.980 --> 20:27.559
thing, but the kinetics changes whenever you
do something like this. This kind of aggregation
20:27.559 --> 20:31.870
occurs the kinetics is no longer simple bimolecular
kinetics. So, that is the thing that I am
20:31.870 --> 20:36.620
trying to convey you over here. So, let us
look. So, what we will do is we will you had
20:36.620 --> 20:37.659
these four different cases.
20:37.659 --> 20:43.070
So, I will try to look 1 by 1 at all the cases
and I might leave out some and which might
20:43.070 --> 20:48.649
give give an assignment. So, the first case
I gave was, talked about was the Ligand depletion.
20:48.649 --> 20:54.470
So, let us assume that Ligand depletion does
not occur we and does occur. So, we had an
20:54.470 --> 20:58.750
initially assumed that Ligand depletion does
not occur or what how did it help us not assuming
20:58.750 --> 21:03.230
that Ligand depletion occurs.
Reduce the order of.
21:03.230 --> 21:08.510
Reduce the order of kinetics from second order
to first order easier for us to solve now
21:08.510 --> 21:12.250
what will happen? It is very straight forward.
We will simply have a second order kinetics
21:12.250 --> 21:19.250
if I go back to the screen. So, my earlier
assumption was that C L equals C L naught
21:19.429 --> 21:24.240
because this term was very valid, very small
and this is no longer valid. So, what I do
21:24.240 --> 21:31.240
over here is let us go back to the equation
otherwise you will not remember here sorry
21:31.840 --> 21:36.860
here this is this is my equation after I had
I had converted all the concentration into
21:36.860 --> 21:40.640
numbers.
So, this is the equation I have. So, what
21:40.640 --> 21:45.549
I want to do to this equation; I want you
to you know do this on your copy, may be you
21:45.549 --> 21:50.519
write down this equation and unless you do
this quick you would not be able to follow
21:50.519 --> 21:56.500
what I am trying to say. So, write this equation
for and D d T of N C equals K 1 N R T minus
21:56.500 --> 22:03.500
N C times C L naught minus N over N a N C
minus K minus 1 N C done.
22:07.980 --> 22:14.980
So, So, this becomes the second second order
if C L naught if I do not make any assumption.
22:25.730 --> 22:30.490
But before that I want to make want you to
make this dimensionless and this is something
22:30.490 --> 22:36.620
that we need later. So, now, I want to make
this dimensionless using these dimensionless
22:36.620 --> 22:43.620
groups which is, u equals N C over N R T.
Tau equals K inverse T K K minus 1 T K minus
22:49.279 --> 22:56.279
1 is first order rate constant. So, the tau
is dimensionless. Eta is N R T over N A C
22:59.390 --> 23:06.149
N n N R T over N a C L naught or n N R T over
N a because that has the units of concentration
23:06.149 --> 23:13.149
right. N N R T, N is number of receptors per
unit sorry N R T is number of receptors per
23:13.860 --> 23:20.100
unit cell N is number of number of cells per
unit volume. So, N times N R T would be number
23:20.100 --> 23:25.490
of receptors per unit volume that divided
by Avogadro number would be molar, molar concentration.
23:25.490 --> 23:32.490
So, that you divide by C L naught it is dimensionless
and alpha is C L naught over K D. Is that
23:34.140 --> 23:41.140
dimensionless, C L naught over K D? Yes that
is because K D has K is K minus 1 over K 1
23:42.029 --> 23:49.029
K k minus 1 has units of inverse time and
K 1 has units of inverse concentration inverse
23:49.200 --> 23:54.820
time. So, K D has units of concentration fine.
So, what you what do you get if you put that?
23:54.820 --> 23:59.019
If you, so, what I want you to do now is,
put these dimensionless constants in to your
23:59.019 --> 24:06.019
equation the equation four that you wrote.
24:13.679 --> 24:20.679
Equation 4 was so, then u equals naught over
K d. So, this is N R T del u del tau N R T
25:14.309 --> 25:21.309
over K minus 1 equals K 1 N R T 1 minus u
and C L naught minus K minus 1 times N R T
25:43.159 --> 25:50.159
times u. Then it should be alpha K naught
K D. Then, if I cancel if I cancel N R T K
26:21.350 --> 26:27.690
minus 1 all through equation.
26:27.690 --> 26:34.690
Then you will get del u del tau equals 1 minus
u u alpha K. Alpha will come in the outside
26:38.649 --> 26:45.649
here if we checked it. So, you have K 1 N
R T is this what I is this right or is there
27:25.409 --> 27:30.389
some term missing here?
will be K minus.
27:30.389 --> 27:32.740
Which one? This one? This side.
Yes sir.
27:32.740 --> 27:38.139
Into K minus 1. No yeah that is why it will
come in the denominator then and then you'll
27:38.139 --> 27:45.139
have K D over here from that yeah into K minus
1 and this is correct right alpha alpha D
27:46.210 --> 27:53.210
over this. So, basically what I have to do
is I have to actually it might be it might
28:01.090 --> 28:05.820
be better to just keep this as C L naught
instead of alpha D and then divide the whole
28:05.820 --> 28:12.820
thing by C L naught yeah. So, so then I think
you will get alpha yeah because. So, you divide
28:15.529 --> 28:22.460
the whole equation everything by both sides
by K minus 1 N R T C L C L K minus 1 N R t.
28:22.460 --> 28:26.720
So, this side it will cancel out you’ll
get K minus 1 N R T over here. So, let me
28:26.720 --> 28:33.720
do that. So, you’ll get del del u del tau
equals K 1 K 1 over K minus 1 which is 1 over
28:34.039 --> 28:41.039
K D into 1 minus u and this is C L naught
minus N n R T C L naught I can take out. So,
28:45.309 --> 28:52.309
this will have 1 minus eta times u right minus
u fine.
28:56.870 --> 29:03.870
So, and this will give me alpha. So, this
is 1 minus u times 1 minus eta u times alpha
29:05.929 --> 29:12.929
minus u all right. So, then you will have.
So, this is the basic equation and I now need
29:17.220 --> 29:21.340
to solve. So, what is going to be my initial
boundary conditions and so, on initial or
29:21.340 --> 29:28.190
boundary condition whatever it is. So, initial
my initial condition earlier was that N C
29:28.190 --> 29:35.190
equals N C naught at T equals 0. So, now,
if I put back into my dimensionless variables,
29:37.350 --> 29:44.350
I will have u equals u naught equals N C naught
over N R T at T equals zero fine. So, this
29:52.100 --> 29:59.100
is my equation that I now need to integrate.
So, these are you need the steps. These are
30:04.620 --> 30:07.500
the steps. So, so these are the steps and.
So, these are what I got and this is what
30:07.500 --> 30:12.490
I need to integrate now.
So, how do I do that? We discussed this already
30:12.490 --> 30:19.370
in the last class. You can use that use partial
fractions and integrate it and because this
30:19.370 --> 30:23.309
is the second order and it is already in the
form you can break this up and get one full
30:23.309 --> 30:28.850
partial fractions in u and then it would be
two exponential fine and we will do that in
30:28.850 --> 30:34.649
a minute. Before we do that we are going to
do something little above which is the steady
30:34.649 --> 30:38.700
state solution of this. So, this thing that
you have over here we will look at the steady
30:38.700 --> 30:45.700
state. So, the steady state is simply 1 minus
eta u alpha equals u steady state fine. So,
30:48.909 --> 30:55.909
why are we looking at the steady state? So,
that we can do they do the scatchard plot.
31:00.659 --> 31:07.659
So, let us now go to the screen yeah.
31:09.950 --> 31:16.950
So, so what we have over here is the scatchard
plot. So, this is u u over alpha equals 1
31:18.759 --> 31:24.000
minus u times 1 minus eta u is the scatchard
plot and this is how it varies. So, as you
31:24.000 --> 31:31.000
can see over here, what we have here is that
two variables are here. So, one is alpha and
31:32.450 --> 31:36.820
the other one is eta. Two parameters sorry
u is a variable and we have two parameters
31:36.820 --> 31:42.649
in the system. So, earlier we had how many
parameters we had we had plenty of parameters
31:42.649 --> 31:48.830
actually we had the forward rate constant
backward rate constant N n R T N a C L naught
31:48.830 --> 31:51.389
plenty of parameters. So, six parameters we
had.
31:51.389 --> 31:55.799
What we had been able to do through these
dimensionless these dimensionless numbers
31:55.799 --> 31:59.590
and groups over here is that we have been
able to reduce the number of unknown parameters
31:59.590 --> 32:04.159
from six to two because it is always hard
to be able to evaluate these unknown parameters
32:04.159 --> 32:07.409
and now we have been able to reduce them from
six to two.
32:07.409 --> 32:14.409
Now, let us look at this expression carefully
u over alpha equals 1 minus u times 1 minus
32:16.500 --> 32:22.470
eta. You got this expression one of these
say steady state plus. Now what is alpha?
32:22.470 --> 32:29.470
Alpha is is. So, this is as I told you just
a little while back that this is scatchard
32:29.919 --> 32:34.389
plot and there is no need that to assume that
scatchard plot is simply going to be N C over
32:34.389 --> 32:40.789
C L naught versus C L naught. All steady state
plots that involves the complex concentration
32:40.789 --> 32:45.139
and the C L naught that is a Ligand concentration
and the complex concentration are scatchard
32:45.139 --> 32:48.350
plots. So, this is also a scatchard plot and
there is no need to presume the scatchard
32:48.350 --> 32:52.379
plots are only going to be N C over C L naught
versus C L naught. So, if you look at this
32:52.379 --> 32:57.120
plot what is u over alpha look at the dimensionless
variables given on the screen. So, u over
32:57.120 --> 33:01.570
alpha is N C over C L naught times K D over
N R T.
33:01.570 --> 33:06.580
So, it is essentially again N C over C L naught.
So, the original scatchard plot had N C over
33:06.580 --> 33:11.139
C L naught versus N C. So, this is also N
C over C L naught times some parameter and
33:11.139 --> 33:15.610
what is u? U is N C over N R t. So, simply
it is the same kind of thing. Now, what is
33:15.610 --> 33:21.570
eta? That is something you have to pay a little
attention to it. Eta is N n R T over N a over
33:21.570 --> 33:28.570
C L naught. So, what does this eta signify
to you just look at the screen and tell me
33:35.490 --> 33:41.929
what does this eta signify to you? And the
answer is right there in front of you on the
33:41.929 --> 33:48.929
screen. So, you just have to look and tell
me. So, just look at these plots over here
33:53.769 --> 34:00.769
for different values of eta what does it signify?
It signifies about the.
34:05.470 --> 34:05.960
About what?
Concentration of Ligand.
34:05.960 --> 34:10.399
Yes. So, essentially look at eta eta is N
R T N a over C L naught. So, N R T N a N n
34:10.399 --> 34:15.050
R T N a over N a is fine, denominator a of
C L naught which means that if C L naught
34:15.050 --> 34:19.409
is very large if the Ligand concentration
is very large eta goes decreases and it goes
34:19.409 --> 34:24.310
to 0. So, as you see over here that is why
I purposely drew this for eta going to 0 point
34:24.310 --> 34:30.870
01 it goes back it collapses to the linear
first order kinetics and for eta larger larger
34:30.870 --> 34:35.450
and larger more is a deviation from the first
order kinetics. That that is all I wanted
34:35.450 --> 34:39.050
to say over here. That is the reason I drew
this scatchard plot. As I just wanted to convey
34:39.050 --> 34:44.390
the fact that if you go away from eta from
eta equals 0 your deviation from the simple
34:44.390 --> 34:48.010
linear kinetics is more and you have this
kind of things.
34:48.010 --> 34:54.070
So, that is the scatchard plot now can we
this model that we over here yes here. sorry
34:54.070 --> 34:57.730
This model that we have over here we can we
integrate it in time of course, we can as
34:57.730 --> 35:01.870
I said through partial fractions and we can
get the temporal variation. Also, we are always
35:01.870 --> 35:06.630
more interested in steady state concentrations
of scatchard plots. The reason I just mentioned
35:06.630 --> 35:10.970
this because they are easier to quantify.
Experimentally scatchard plots are easy to
35:10.970 --> 35:15.780
draw because you can just leave the system
as far as long as you want in your, let let
35:15.780 --> 35:20.290
it I you know let let the experiments happen
for four hours, six hours, eight hours, ten
35:20.290 --> 35:23.850
hours. And at the end of which you can go
and collect where as if you want to do a dynamic
35:23.850 --> 35:28.190
study you have to collect at every time point
you know, every thirty minutes or twenty minutes
35:28.190 --> 35:30.630
you have to collect and it makes it difficult.
35:30.630 --> 35:37.630
So, this, but again you know you can always
do that. So, again this is this is this is
35:41.080 --> 35:46.550
the solution of the unsteady state model and
what I have done over here is, this is the
35:46.550 --> 35:53.550
case for eta equals zero which means that
simple linear kinetics and these are the deviations
35:54.070 --> 35:58.570
from that and as you see that as you keep
increasing eta, the deviation keeps increasing
35:58.570 --> 36:05.570
more and more. And also, the deviations in
dependent alpha two, but here I think this
36:07.340 --> 36:14.340
is small typo over here. excuse me errors
are are the errors errors means basically
36:16.940 --> 36:19.590
deviation. These are not errors, these are
deviations as you look on the screen at the
36:19.590 --> 36:22.570
bottom of the screen here I think this is
typo.
36:22.570 --> 36:29.060
So, for eta equals 0 point 1 the error is
1 percent and eta equals point 1 no I think
36:29.060 --> 36:36.060
it is sorry the typo is eta equals zero point
zero 1 the the deviation is 1 percent and
36:36.140 --> 36:42.540
for eta equals point 1 the deviation is around
3 percent. So, these numbers and then as you
36:42.540 --> 36:46.940
keep increasing your eta is going to see,
that the huge difference this is almost what
36:46.940 --> 36:51.080
forty percent difference and this is error
is around 40 percent not error is essentially
36:51.080 --> 36:55.320
these are differences. So, this forty percent
and this could be around seventy percent or
36:55.320 --> 37:00.960
so. So for so, what this is trying to show
is that the that near eta equals 0. So, this
37:00.960 --> 37:07.120
I am sorry this this 1 eta should be eta equals
0 point 01. So, for that the deviation is
37:07.120 --> 37:11.680
1 percent for eta equals 1 and the deviation
is 3 percent.
37:11.680 --> 37:15.730
So, the next case we look at. So, if you go
back and looked at the four, look at four
37:15.730 --> 37:21.980
different cases was. First one was Ligand
depletion, second was the multiple receptor
37:21.980 --> 37:27.540
and third one is multivalent Ligand. So, the
next case that we look at is the multiple
37:27.540 --> 37:31.060
receptor two or more receptor populations.
Now what do you what does your intuition tell
37:31.060 --> 37:35.830
you you know. So, the first case we did was
very straight forward where we had the equation
37:35.830 --> 37:40.370
and the model and all we needed to do is fiddle
around with the C L a term. But this we do
37:40.370 --> 37:45.130
not have a model for for multiple receptors.
So, if I am to if I have to ask you to write
37:45.130 --> 37:51.500
a model. So, what does your intuition tell
you? How would you write a model for multiple
37:51.500 --> 37:58.500
receptors? You have one set of Ligands, but
let us say two set of sets of receptors. How
38:05.010 --> 38:12.010
would how would you do that? One possibility
and the simplest possibility is that the receptors
38:13.740 --> 38:19.240
would not interfere with each other where
they bind which means that receptor one binds
38:19.240 --> 38:24.380
independently with Ligand and receptor two
binds independently to Ligand. That is the
38:24.380 --> 38:26.150
that is a possibility.
38:26.150 --> 38:31.360
So, this is this is the possibility that we
explore over here where receptor one and receptor
38:31.360 --> 38:35.420
two bind independently to each other. And
if they bind independently, how do you, how
38:35.420 --> 38:41.260
do you model it? The model is still not there
on the screen. So, how do you model it? Let
38:41.260 --> 38:48.260
us go back to the old model and then you can
tell me how you are going to model it. This
38:52.490 --> 38:57.870
is this is the old model I have right. Now
I tell tell you that very very clearly that
38:57.870 --> 39:01.170
the receptors do not interfere with each other.
So, what will happen?
39:01.170 --> 39:08.170
So, you will have these two sets of reactions
L plus R 1 giving C 1 and L plus R 2 giving
39:11.430 --> 39:18.430
C 2 right. So, D C 1 D T would equal. Let
us say this is K 1 1 and K 1 minus 1 and K
39:23.800 --> 39:30.800
2 1 and K 2 minus 1 fine. So, then this will
be my K 1 one C L C R minus K minus 1 1 C
39:39.290 --> 39:46.290
C 1 and D C 2 D T would be K 2 1 C L C have
R 2 minus K minus 1 2 C 2 over.
40:05.960 --> 40:12.960
Now, if the receptors so, from these set of
systems are these two inter dependent on each
40:13.800 --> 40:19.390
other? This is a question you have to answer
a and under what circumstances they are not
40:19.390 --> 40:26.390
really inter dependent? b Two questions. Are
these two equations that I wrote in a curly
40:26.690 --> 40:33.690
brackets here are they dependent on each other?
Yes they are because through C L what conditions
40:33.810 --> 40:39.860
does they not inter dependent?.
Large if there is no Ligand depletion if Ligand
40:39.860 --> 40:45.450
is in excess then C L each of these C L you
can write it as C L naught over here. So,
40:45.450 --> 40:52.450
for C L not very large, replace C L by C L
naught fine. We can replace C L by C L naught
40:59.380 --> 41:03.820
and then what happens is these two systems
become decoupled. So, typically the Ligand
41:03.820 --> 41:07.410
concentration is more or less large if it
is not large then what what you are suppose
41:07.410 --> 41:11.040
to do? You are suppose to solve these two
coupled equations together. The two coupled
41:11.040 --> 41:15.020
equations that you have on you have here you
have to solve these two together that is the
41:15.020 --> 41:19.170
possibility. I might give you a question like
that, but if the Ligand is large enough then,
41:19.170 --> 41:22.680
you can solve, decouple them you can solve
them separately and what will happen you will
41:22.680 --> 41:27.890
get the same solution for each of them right.
So, the maximum N C max that you that you
41:27.890 --> 41:34.890
got previously depended only on one Ligand
concentration and one receptor concentration
41:35.210 --> 41:37.840
concentration.
Now, they are going to be two of them and
41:37.840 --> 41:41.760
you can add them up if these two are completely
independent then what you see over here is
41:41.760 --> 41:45.660
not complete independent. But if the Ligand
concentration is very large then, they become
41:45.660 --> 41:49.860
completely independent you allow each of them
to react in their own way get the steady state
41:49.860 --> 41:55.180
value and you add them up. That is that is
a possibility, but that is not the always
41:55.180 --> 42:02.180
the possibility that is one of the possibility
that can happen. So, this is the case where
42:02.340 --> 42:06.360
each receptor acts independently.
So, as you see on the screen the first part
42:06.360 --> 42:11.390
C L naught. So, this is this is the case where
everything is C L naught. So, there is no
42:11.390 --> 42:18.390
Ligand depletion at all and C L naught N R
T 1 over C L naught plus K D 1. That is the
42:18.680 --> 42:25.000
first one and then C L naught N R T 2 over
C L naught plus K D 2. Now, I I tell you one
42:25.000 --> 42:29.400
thing I give you an assignment you can do
it in and submit it you are suppose to submit
42:29.400 --> 42:35.320
one assignment today.
No whatever. You submit the assignment you
42:35.320 --> 42:40.680
know, whatever you did you submit the assignment
and I gave you this assignment for the, for
42:40.680 --> 42:46.040
next week. I I probably add in another problem
tomorrow. So, this assignment is this is the
42:46.040 --> 42:51.750
case where we consider two or more receptor
populations. Previous case what did we assume
42:51.750 --> 42:55.720
Ligand depletion. This case there what we
have done here there is no Ligand depletion.
42:55.720 --> 43:00.500
What I want you to do is, couple these two
cases a and b. That is Ligand this depletion
43:00.500 --> 43:05.920
plus two or more receptor population’s fine.
And then get the model for that and the solution
43:05.920 --> 43:10.300
and come up with the steady state. Is that
clear? You can write, all of you down yes.
43:10.300 --> 43:14.700
So, first case we did was Ligand depletion.
Second case we we are doing now is two or
43:14.700 --> 43:19.680
more receptor population I want to, want you
to couple these two and come up with the model.
43:19.680 --> 43:24.350
I model, I already wrote, but you solution
of the model essentially and the steady state
43:24.350 --> 43:31.350
value of two or more receptors in the presence
of Ligand depletion. Clear?
43:33.160 --> 43:40.160
So, then the receptors cannot act completely
independently like it does over here fine.
43:42.480 --> 43:48.670
So, even here when you have a steady state
solution, steady steady state solution or
43:48.670 --> 43:52.350
steady state value for this as you see, given
on the screen what happens you know, you were
43:52.350 --> 43:56.670
doing this whole experiment in a beaker with
two sets of receptor, one set of receptor
43:56.670 --> 44:02.000
and let us say two sets of one, two sets of
receptors and one set of Ligand at the end
44:02.000 --> 44:09.000
what you can you probably have? How do you
how do you how do you measure? how do you
44:09.770 --> 44:16.520
measure how do you measure I mentioned this.
By labeling the Ligands right. You label the
44:16.520 --> 44:20.040
Ligands with some sort of fluorescence or
something and once the complex is formed you
44:20.040 --> 44:26.060
have the labeled complex right. So, you have
one one set of Ligands and two sets of reactor
44:26.060 --> 44:28.940
receptors and you have labeled your Ligands
for example.
44:28.940 --> 44:33.610
Now, the complex is formed you can get the
total amount of complex that is formed at
44:33.610 --> 44:39.440
steady state right. But it is going to be
very hard to be able to separate out the kinetics
44:39.440 --> 44:43.630
of these two. Why is that? Because what you
will see at the end of the day is a labeled
44:43.630 --> 44:47.790
labeled Ligand. Now, there is a way when you
draw the scatchard plot you may be able to
44:47.790 --> 44:52.250
separate out these two the kinetics of these
two reactions. That is reaction of Ligand
44:52.250 --> 44:56.440
with receptor one as against reaction of Ligand
with receptor two. You may be able to separate
44:56.440 --> 45:03.360
it out through scatchard plot, but only in
the case when these two are very dissimilar
45:03.360 --> 45:08.620
or the kinetics of these two are very dissimilar.
What I mean by dissimilar is that for example,
45:08.620 --> 45:11.940
one of them for example, is very slow, the
kinetics of one of them is very slowest compare
45:11.940 --> 45:17.430
to the other then only can you can you separate
them out and we will look at this now.
45:17.430 --> 45:22.400
So, is as I said and it is the on screen that
it may be may not be possible to distinguish
45:22.400 --> 45:27.050
between the two kinetic constants, the dissociation
constants here unless they are very different
45:27.050 --> 45:31.530
from each other and by very different at least
by one order of magnitudes. So, one has to
45:31.530 --> 45:35.810
be ten times larger than the other at least.
45:35.810 --> 45:42.810
So, this is this is the this is a this is
the thing you know, with the assumption that
45:44.920 --> 45:48.110
this is not what you are suppose to write
the model. But this is the model with the
45:48.110 --> 45:52.660
assumption that Ligand depletion does not
occur. So, then these two equations become
45:52.660 --> 45:57.950
independent with each, to each other and you
can solve them and this is the solution you
45:57.950 --> 46:02.110
get. So, this is the I want you to get same
kind of solution, but with these two coupled
46:02.110 --> 46:07.610
systems and yeah.
So, yeah I think these are the second order
46:07.610 --> 46:14.560
coupled equations. You probably have to use
what do you think would you be able to solve
46:14.560 --> 46:21.560
this analytically the equation that I if I
can go to this page over here? This this thing
46:21.970 --> 46:26.370
that I gave you this these two equations do
you think you can solve it analytically? Here
46:26.370 --> 46:29.780
we could solve it analytically the one that
we did. But do you think these these two we
46:29.780 --> 46:36.780
would be able to solve analytically? Just
as a hint you know to help you with the process.
46:39.740 --> 46:46.740
Do you think you can solve this analytically?
Or numerically of course, you can solve second
46:48.170 --> 46:54.770
order second order system, but what about
analytically can you solve this? Once you
46:54.770 --> 46:58.340
put you have to go back and put your N a and
all that stuff you know, convert this into
46:58.340 --> 47:05.340
numbers like we did did all the process, but
after that what if it is a single second ordered
47:06.070 --> 47:10.650
equation you can solve it analytically.
Now, I have coupled the system then what?
47:10.650 --> 47:17.080
I showed you last the single Ligand depletion
case. Can you or can you not? Anyway, let
47:17.080 --> 47:21.330
us not debate over that you do that as an
assignment, but for this case where Ligand
47:21.330 --> 47:26.020
depletion does not occur you can get an analytical
solution. So, each of them becomes like what
47:26.020 --> 47:31.700
we got before and this is an analytical solution.
Remember the initial condition is given as
47:31.700 --> 47:38.700
N C 1 equals 0 and N C 2 equals 0. So, that
any other term that was say drop out and you
47:38.860 --> 47:43.270
have these two.
Now, so, but what I am, I was trying to tell
47:43.270 --> 47:48.550
you is experimentally you cannot separate
out N C 1 from N C 2. Is that clear? When
47:48.550 --> 47:52.370
you are measuring you can only measure N C
because you would be able to get the labeled
47:52.370 --> 47:57.070
Ligand. Now, how do you separate these two
rate constants? Then, if you can only measure
47:57.070 --> 48:04.070
and see and how then what is the way to separate
out these two rate constants? The way would
48:10.720 --> 48:15.340
be to look at the slopes and we will do that
in in the next few minutes. So, look at the
48:15.340 --> 48:19.920
slopes and look at if there are differences
in the in the slopes and what you will find
48:19.920 --> 48:25.180
is that, if if one of those dissociation constants
is ten times other then there are differences
48:25.180 --> 48:28.870
in the slopes.
So, this is how it looks like the diagram
48:28.870 --> 48:34.650
that you see on the screen. So, this is the
dynamics of the process. So, this is this
48:34.650 --> 48:41.500
equation the final model that I wrote over
here that one solved. So, this is the case
48:41.500 --> 48:48.500
of a single receptor. This is the old solution
solved sorry. This this one solved and and
48:58.740 --> 49:00.870
this this this one is the last one, the two
points.
49:00.870 --> 49:05.060
So, as you can see over here that with two
receptor model what we have done over here
49:05.060 --> 49:09.120
is K D we have taken a system where there
is a large difference in the dissociation
49:09.120 --> 49:14.260
constant. We have purposely taken this where
there is a two orders of magnitude difference
49:14.260 --> 49:20.460
one is hundred times the other. As the result
what happens you see a very distinct slope,
49:20.460 --> 49:27.080
difference in slope. Can you explain why this
is happening? The and This curve is the x
49:27.080 --> 49:33.020
axis is time, the y axis is the total amount
of complex that has been formed and x axis
49:33.020 --> 49:35.460
is time.
Now, you have to tell me. So, given the value
49:35.460 --> 49:41.760
the K D 1 and K D 2 values are given. So,
you have to tell me that what is happening?
49:41.760 --> 49:48.740
May be we stick to the go back to the screen
and look at the screen here. So, what is happening
49:48.740 --> 49:53.590
over here you know, these two slope being
very different, why are the two slopes very
49:53.590 --> 50:00.590
different? And what can you infer from the
difference of slopes? K D 1 is K D 2 is hundred
50:09.730 --> 50:16.730
times K D 1 what does that mean? you know
Not that complicated.
50:21.570 --> 50:28.040
Which 1 you have to tell me specifically not
one of the term. K D 2 is dissociation rate
50:28.040 --> 50:35.040
constant. So, if K D 2 is higher which means
what? K D 2 is higher; much much higher than
50:36.080 --> 50:40.430
K D 1. Which means out of receptor one and
two what is the process what is really happening
50:40.430 --> 50:44.740
here? Let us forget the, look at the curve
because it is important to look at the curve.
50:44.740 --> 50:48.750
But from the curve, let us try and understand
what is happening, what is the process that
50:48.750 --> 50:55.750
is happening here? Say K D 2 is much, much
higher than K D 1. K D K 2 K minus 1 let us
51:08.580 --> 51:15.580
say 2 over K 1 and K D 1 is K minus 1 1 over
K 1. So, if K D 2 is much, much higher than
51:21.520 --> 51:28.520
K D 1. It implies that K 1 2 let us call this
and K 1 1.
51:32.260 --> 51:39.260
So, one possibility is that K 1 2 is much,
much lesser than K 1 1 or K minus 1 2 is much,
51:45.990 --> 51:52.990
much greater than K minus 1 1 whichever way.
What does it imply? That what does it implies
51:55.420 --> 52:02.420
here that the second if second association
rate constant for the second is much, much
52:04.700 --> 52:08.820
lower than the first. So, as soon as you put
the whole stuff in there you know, into your
52:08.820 --> 52:13.700
beaker the receptors are there and you put
your Ligands in there. What, what starts to
52:13.700 --> 52:20.700
happen is that receptor one starts to react.
Receptor one starts to react with the Ligand
52:21.420 --> 52:25.860
and then if there is a large difference in
magnitude. So, receptor two hardly reacts
52:25.860 --> 52:31.290
with the Ligand. So, receptor one starts to
react dynamically. This is what happens receptor
52:31.290 --> 52:34.850
one starts to react with the Ligand and then
it saturates out.
52:34.850 --> 52:40.960
So, all the receptor ones receptor one that
are there are being combined or reacting with
52:40.960 --> 52:45.720
the Ligand. They are being taken up and complexes
are been formed still a point where no more
52:45.720 --> 52:50.650
receptor one is available is that clear? So,
till now till a point where no more receptor
52:50.650 --> 52:57.650
one is available and then the receptor two
starts to react with receptor one with Ligand.
52:58.070 --> 53:02.580
So, if there is a hundred to 1 ratio then
if I go to the screen and we will see if there
53:02.580 --> 53:08.060
is a 100 to 1 ratio then what happens? When
this this is reacting for the first receptor
53:08.060 --> 53:13.660
one is reacting. So, ninety nine percent of
the receptors that are reacting and are receptor
53:13.660 --> 53:18.750
one and only 1 percent are receptor two and
that continues more or less till steady state
53:18.750 --> 53:21.950
is b.
So, this time over here, whatever the time
53:21.950 --> 53:26.340
is let us say 0 point 1 five minutes of a
something one 0 point 1 minute whatever the
53:26.340 --> 53:31.730
time is, is a time when the first process
that is receptor one combining with the Ligand
53:31.730 --> 53:38.640
has more or less reached the steady state
and then the second process starts. So, ideally
53:38.640 --> 53:42.940
what would happen is, if there is a one receptor
only receptor one would be there. This will
53:42.940 --> 53:47.090
saturate out over here, flatten out over here,
but that does not happen because that is second
53:47.090 --> 53:51.580
receptor staying there, standing there and
that starts to react and form this.
53:51.580 --> 53:57.890
So, this difference in the slope can help
you figure out what the constants are and
53:57.890 --> 54:04.130
as you can see over here look at this numbers
N R T 1 over N C max is point 3 and N R T
54:04.130 --> 54:10.980
2 over N C max is point seven. This is very
intuitive. The reason being that N R T 1 plus
54:10.980 --> 54:14.760
N R N C max is what the maximum number of
complex being can that can be formed. The
54:14.760 --> 54:19.340
maximum number of complex equals the total
number of receptors that are there if there
54:19.340 --> 54:26.240
are, no other resistance is in the system.
The total number of receptor one at the limit
54:26.240 --> 54:30.010
at best can what can happen is total number
of receptor, all the receptor one can react
54:30.010 --> 54:36.330
which it does and also all the receptor two
can also react which may or may not happen,
54:36.330 --> 54:40.970
but all receptor one will definitely react.
So, if that happens then N C max will be N
54:40.970 --> 54:45.830
R T 1 plus N R T 2. As a result N R T 1 over
N C max is zero point 3 and this is point
54:45.830 --> 54:49.500
seven and the summation of these two would
be one or in other words N C max would be
54:49.500 --> 54:54.030
N R T 1 plus N R T two. So, the total number
maximum number of receptor complexes that
54:54.030 --> 54:57.000
can be formed is a sum total of the two kinds
of receptor.
54:57.000 --> 55:03.510
Now, the last thing we will do today is look
at this Scatchard plot over here sorry not
55:03.510 --> 55:08.060
this scatchard plot. This is the still the
unsteady state plot, but this is written done
55:08.060 --> 55:14.480
in terms of 1 minus N C over N C max why 1
minus N C over N C max because, if you if
55:14.480 --> 55:21.480
you look over here this there is this terms
over here this constant terms over here and
55:21.570 --> 55:26.410
if you do the 1 minus N C minus N C max and
what will happen is that you will get the
55:26.410 --> 55:28.870
exponential parts and you can take the log
of that.
55:28.870 --> 55:33.840
So, is the same kind of variation as you see
exactly the same thing it is only that the
55:33.840 --> 55:37.770
plot has reversed itself because it is in
the log scale and you know minus and all that.
55:37.770 --> 55:42.500
So, you have 2 diff, varying slopes and from
these two varying slopes you could be able
55:42.500 --> 55:48.270
to you would be able to separate this thing
out. Now what I want you to do is quickly
55:48.270 --> 55:52.020
of again we have a class 1 more class today.
55:52.020 --> 55:56.840
But write write down problems for now and
this should be a second problem for assignment.
55:56.840 --> 56:03.840
Show by rearranging the equation N C equals
C L naught N R T 1 over C L naught plus K
56:07.810 --> 56:14.810
D 1 plus C L naught N R T 2 over C L naught
plus K D 2.
56:15.270 --> 56:22.270
So, this is essentially N C max 1 plus N C
max 2. Show that by rearranging this equation
56:22.900 --> 56:29.450
you can determine the four parameters N R
T 1 N R T 2 K D 1 K D 2 for two receptor populations
56:29.450 --> 56:34.810
from the scatchard plot. Assume that one of
them is matching. So, you can assume one to
56:34.810 --> 56:39.360
be hundred times I had in the other. So, what
I did I showed you the scatchard plot and
56:39.360 --> 56:45.590
I all I am trying to tell you is that how
can you exploit this this steady states steady
56:45.590 --> 56:52.590
state equation that is there to get how can
you exploit this to get the four parameters
56:52.760 --> 56:59.760
N R T 1, N R T 2, K D 1 and K D two. So, we
will stop here and we will continue later
57:00.950 --> 57:07.950
today.