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cy12-noc19 lec37 Understanding Protein ligand interaction by NMR Chemical exchange


[noise]
[noise]. So, we have started looking at ah, [noise] the concept of chemical [noise] shift
perturbation. Ah this is a method where we want to study how protein and ligand interact
with each other. So, we saw briefly this in the last class.
That if we have a protein and a ligand. [noise] Ah you can have a complex where it binds to
a particular region of the protein. The ligand binds to a particular region of the protein.
[noise] And we saw that this interaction can be studied by [noise] H S Q C NMR spectrum.
So, typically we take N 15 proton H S Q C not carbon. Carbon also can be used. ah In
case of carbon HSQC normally we look at the methyl [noise] ah signals. But in ah 915 HSQC
we look at this signals coming from the amide protons and nitrogen [noise].
So, for each ah peak in the NMR spectrum. We know in HSQC corresponds to one amino acid.
So, what happens is, those amino acids which are at the interface this is called this here.
This is called a binding interface. Where the ligand binds with the protein.
So, the amino acids on the binding interface, are expected to undergo chemical shift changes
[noise] because, the chemical environment around this residues change when the ligand
binds to the protein [noise]. So, this is interface is amino acids are expected to therefore,
shift ah in their peaks. [noise] And that is something which we capture by HSQC [noise]
and that is what we looked at last time. how do we do that? [vocalized-noise].
So, we saw that you can have basically ah interaction with a specific set of residues
on the interface. And that is now shown here, ah this is called a Chemical shift perturbation
plot [noise]. So, what is plotted on the y axis here, is the amount of shift. That means,
how much shift has taken place for a given amino acid from the original free protein.
Mean so, what we do is we take a free protein. And we add the ligand, [vocalized-noise] normally
we do what is called titration. Is slowly add ligand in a small amount and as the reaction
proceeds the binding takes place, and we can monitor [noise] the chemical shift changes
the difference here between the red and the blue is plotted on the y axis [noise].
But as we saw last time that this red and blue are not just changes in the amide proton.
This is [vocalized-noise] x axis. It is also shifting in the y axis that is nitrogen. So,
you have to combine the shift [noise] of nitrogen and proton, and give a single number.[noise].
So, how do we combine these two shifts or changes? Usually we use this [noise] a formula
like this, where for a given amino acid we saw that we can actually look at the hydrogen
amide and nitrogen shift and combined together [noise] and that is a number which is plotted.
Now, here one more thing which we saw [vocalized-noise] in the last class. Is that region which binds
can be identified based on how much significant shift they are showing upon binding. So, in
this ah [noise] particular protein, which is a real case example. [noise] We see that
residues number 150 to 160 [noise] or 165, ah somewhere in this region they are binding
ah sufficiently strongly to the protein. Because those amino acids are undergoing a significant
change [noise] in their backbone amide and nitrogen chemical shift. So, this is how we
can find out which region of the protein ah is binding to the ligand.
So, remember we are not, we are trying to characterize a ligand binding in two ways.
Number one, we are trying to find out where in the protein on what particular part of
the protein does the ligand bind? And the second thing which we are trying to characterize,
[noise] is what is the strength of the binding? Means how strong is the binding? ‘Is it the
binding very weak? [noise] or sufficiently in strong [noise]?
So, that is measured by a number known as dissociation constant. And the dissociation
constant K d will give us the information whether it is a strong binding or a weak binding
[noise]. So, this is something again we saw in the last class, that you can have what
is called as Electrostatic interactions. [noise] Where in the positive charge surface or a
negative charge surface of this ligand. Can interact with the [noise] positive or negative
charge surface or oppositely charged surface [noise] on the protein.
So, we saw one real case example of a protein ubiquitin, [noise] binding to a nano particle.
And we [vocalized-noise] and I mentioned that the nanoparticle which is silver nanoparticle,
[noise] is [vocalized-noise] negative in overall is charged. And therefore, its interacts to
the with or with the by positive charge surface [noise] on the protein.
So, this blue color patch, what you see here [noise] is a positive charge surface on the
protein. And the red colour means negative charge [noise]. So, these are the ways with
NMR can help us to identify where is the binding region. But what is not coming out from these
studies is, how strong is the interaction. We want to measure also [vocalized-noise]
the strength of the interaction. And strength of the interaction is normally [noise] measured.
So, when we write the strength, [noise] of interaction. Means how strong is this interaction
[noise] that is measured in terms of a number called [noise] K d. So, K d is called dissociation
constant. [noise] Dissociation constant [noise] ok. So, if it is a weak interaction, [noise]
weak binding [noise] then K d will be in the range of somewhere between millimolar [noise]
to micromolar values. And strong binding [noise] will be, a K d
will be in the range of micromolar [noise] to [vocalized-noise] picomolar [noise] [vocalized-noise].
So, these are the typical ranges [noise] of the binding affinities. So, this is also called
binding affinity [noise]. So, let me write down binding affinity [noise] [vocalized-noise].
So, NMR can also be used [noise] for characterizing the binding affinity, of a protein and ligand
interaction. And that as I said is used this this number.
So, let us see what happens in case of whether a ligand is strongly or weekly bind bound.
How can we understand that idea? And this [vocalized-noise] lead us to a very important
concept [noise] called Chemical exchange [noise]. So, based on Chemical exchange a phenomenon
called Chemical exchange we can find out whether there is a strong [noise] interaction or a
weak. And this is a very important concept because, in most of the drug discovery process
[noise] where we [vocalized-noise] look at drug molecules.
It initially the the initial screening of the drug molecules are done ,based on this
interactions [noise] and it is normally found that there are not very strongly bound. So,
the initial hits what we get in drug discovery are normally weak binders. Then typically
it is further improved with the synthesis method and different approaches. And slowly
it becomes a strong binder. So, the drug molecules are typically supposed to have nanomolar to
picomolar affinities. So, that they can work [noise] very efficiently in our body.
So, let us see how weak [noise] interaction shows up in NMR and how does a strong interaction
work [noise]. So, now, for that we have to look at this concept of chemical exchange.
[vocalized-noise]. So, we will go here ah looking at this model. So, imagine that there
is a protein molecule ok and there is hydrogen or a some [vocalized-noise] proton in some
this molecule. Let us take any hydrogen it doesnt matter
this is just a schematic. The point here is imagine that the protein, is an state A. Means
one particular state. [noise] Now let us say that because of something, that is interaction
with a ligand or change in temperature or anything it goes into another state B [noise]
ok. So, this conversion from A to B [noise] happens because of some interactions or some
reason. So, suppose this conversion is characterized
by this constants of rate constants. So, rate constant means how fast it goes from here
to here. And rate constant for how fast it goes back from here to here [noise]. So, typically
this kind of a change of a protein state from A to B is reversible. [noise] Meaning suppose
I have a ligand which binds to protein and [noise] in the ligand is is not very strong
binder then, the ligand will not always be bound to protein. So, the bound form can be
thought of as a [noise] state B and the free state free form of the protein can be thought
of as state A [noise]. Similarly, for the ligand we can think of
this molecule as a ligand. The ligand in the free form, [noise] and ligand in the bound
form [noise]. So, whenever there is an interaction normally between ligand and protein. It will
be a dynamic or a equilibrium interaction. So, there will be a conversion always happening
from free [vocalized-noise] form to bound form and bound form to free form [noise] ok.
So, this can be written [noise] more [vocalized-noise] clearly here. Suppose I have a protein [noise]
plus ligand a protein ligand always as a reversible [noise] [vocalized-noise] interaction. [noise]
They are all. So, if it is very strong binding then of course,
this reversibility that is going back is very slow [noise]. But if it is a weak interaction
going forward and backward is equally fast [noise]. So, this is captured by this numbers
create constants k A and k B ok. So, let us say that this [noise]. So, this is what is
shown the different possibilities of state A and state B [noise]. So, state A for example,
a state A could be a free ligand, [noise] and state B could be a ligand bound to protein
ok. So, therefore, the [prot/protein]- ligand has now two possible states. [noise]
And it interconverts, means it goes from one to another and back in a reversible manner.
[noise] Or it could be that the protein is slowly undergoing a conformational change
[vocalized-noise]. So, many proteins for example, when the [vocalized-noise] in helix, helical
or beta sheet. It is possible that the helix is slightly opening up [noise] or closing.
So, this kind of a motion is also represented [noise] in this manner. That means, a protein
is going from one confirmation structure to another structure. Similarly, in organic chemistry
we know, this concept of Keto-enol tautomerism. [noise]
So; that means, in the the [vocalized-noise] number of molecule that is a proton is in
the kito form in one state. And then it undergoes conversion to an enol form in the state B.
So, we can there the enol form can be represented or considered as state B, and the keto form
can be considered as state A [noise]. Another example of states ah A and B can be hydrogen
exchange. For example, this hydrogen [noise] can exchange with solvent and become deuterated.
Suppose a solvent is deuterated solvent [noise] ah solvent proton can change, and go to become
deuterated [noise]. So, H D exchange [noise]. So, in thats case the protonated form, [noise]
can be considered as one state [noise] and the deuterated form considered as second state
[noise]. So, essentially in chemical exchange what we are trying to do [noise] we are trying
to represent the molecule in two states. [noise] And then we assume there is a [vocalized-noise]
reversibility [noise] and then we try to find out what happens in such cases in by NMR spectroscopy
[noise]. So, let us see from NMR what do we what information
we get [noise]. So, let us say that [noise] this state A, has a chemical shift delta A.
[noise] And the state B has a chemical shift delta B [noise]. So, we are assuming that
there is a difference, in the chemical shift between these two states. This is not necessarily
true always, [noise] ah if it is not same then it is difficult to study.
But sometime fortunately they are changes in [noise] chemical shift because the structure
changes slightly. So, whenever there is a change in structure NMR chemical shifts will
change ah. So, let us assume that in our example whatever we are studying now, that there is
a two the two different chemical shifts between state A and state B [vocalized-noise].
Now, how do we characterize the exchange [noise]? So, in NMR the exchange is always with respect
to the difference in this chemical shifts [noise]. So, what matters to us is these difference
[noise]. We use our delta delta [noise]. So, these [noise] difference is very very important
for us, all the concept of exchange in NMR depends on this. [noise] Now ah [vocalized-noise]
remember one thing very [vocalized-noise] crucial here is this difference is not we
do not measure in ppm for exchange we measure in Hertz. [noise] Not in ppm [noise] because
in ppm scale [noise] it does not change. Whether you go [noise] from 600 to 800 or 900 mega
Hertz [noise] your ppm difference will not change, but the Hertz will change [noise].
So, what really matters in NMR is how much is the difference in the chemical shift [noise]
ah between the two states A and B in the scale of hertz [noise] ok. So, now, these are the
different possibilities. [noise] For example, this exchange is very slow meaning the rate
of these exchange what is shown on this here, [noise] is much much less [noise] than the
difference absolute difference in the chemical shift [noise].
That means, the K value suppose this is about 100 Hertz. [noise] Example [noise] let us
say, this this is delta-delta [noise] this is called delta-delta [noise] ok. So, if delta-delta
is 100 Hertz. [noise] and your K is 1 Hertz [noise]. Then, it becomes a slow exchange
[noise] ok. The example like this [noise]. So, this [vocalized-noise] rate constant is
much smaller [noise] than the difference in the chemical shift, [noise] which is measured
in Hertz scale [noise] ok. So, let us see how that ah second possibility.
So, that ah earlier case was the slow exchange case. [noise] Now [vocalized-noise] there
can be another scenario, [noise] another possibility [noise] where the rate constants of the conversion
from A to B and B to A is almost [noise] [vocalized-noise] nearly same as the difference in chemical
shift between the two states. Again measured in Hertz [vocalized-noise]. So, if such [vocalized-noise]
if that happens then, the peak of the two peaks which are separated earlier start merging
[noise] ok. And they become a broad single peak. So, this is called Intermediate exchange.
So, there are [noise] different regimes or different types or different categories of
exchange [,noise] in NMR. So, there are three as mentioned here the second is the first
one was slow exchange and second is called Intermediate. [noise] The third scenario [noise]
or possibility is that, the rate of inter conversion between A to B and B to B is much
higher than the difference absolute. Again remember, we are looking at the absolute difference
[noise] in Hertz scale between the two states [noise].
So, what happens in such a case [vocalized-noise] we again see a peak. So, here also we are
getting one peak, but it was very broad [noise] here it is a very sharp peak [noise]. But
where does it appear it appear somewhere between A and B. So, it is neither A nor B. It has
come in between and what is this value? This value is obtained or is based on this formula.
So, here it says, that delta C where the final that merged peak comes, this peak comes, [noise]
is equal to population. Means how much percentage of a is present in the sample? [noise] Multiply
by that chemical shift, [noise] plus percentage of B population that is how much of B is present
multiply with the chemical shift of B [noise]. So, you see here what is happening is? We
are dividing the population into two parts. Because remember we are looking at either
A or B. That means, the total population we will assume to be constant. [noise] If A going
to B, A will [vocalized-noise] reduced and B will increase. So, therefore, total population
should remain constant [noise]. So, we say partial populations, fractional population.
So, its amount is 1, fraction [vocalized-noise] adds up to 1. Say this is 50-50 percent [noise]
and this will be 0.5 this will be 0.5 [noise]. [vocalized-noise]. So, if its 50-50 [noise]
then, the delta c will come at the half midway between this and this. [noise] Because it
will be half of that plus half of that [noise] ok. But if the population of a is very high,
then it will be close to the A. So, we can see this line here seems to be not at the
centre, it seems to be towards a because in this example it is possible that A is P A
is more than P B. So, it get shifted or dominated by the chemical shift of A. So, one thing
you will notice from this equation, is that if I keep changing the populations. [noise]
I will keep moving this peak this peak will shift here here based on how much of A or
B I add [noise]. So, this is something which is the interesting
thing which we will be using later on. I will [vocalized-noise] show you for characterizing
fast or slow exchange [noise]. Now, again which I mentioned earlier this [noise] whole
concept of chemical exchange whether it is fast or slow, depends on the field strength
[noise]. So, if I go from 400 megahertz to 800 megahertz my field will double. Means
my delta will double this delta- delta will be doubled. Because, this let us say this
is 3 ppm difference, 3 ppm on a 400 megahertz in proton, is 1200 Hertz.
But the same 3 ppm if I go to 800 megahertz will be 2400 Hertz [noise] ok. So, this changes.
But this number this rate constant does not change with the field strength. Means it is
not different whether it is 400 or 800. That remains the same because, this is not in NMR
parameter. [noise] This is a parameter which is inherent to the reaction or change of A
to B. It does not depend on what magnetic field under what strength of magnetic field
I am using or I am doing the experiment [noise]. So, this is not an NMR [vocalized-noise] related
parameter [noise]. So, therefore, this rate constant is constant. So that means, whether
I call it as a fast or slow all depends on what is the field strength I am using.
So, it may happen that at 200 megahertz or 300 megahertz. What I call it as slow [noise]
ok or what I called as fast, may become slow in terms of in 800 megahertz [noise]. So,
this is very important concept chemicals exchange regimes are based on [noise] the field strength.
So, this is what is [vocalized-noise] mentioned here a fast exchange on a low spectro field
spectrometer, means suppose it is 200 or 300 megahertz. [noise] There this k value may
be bigger than the difference. But when I go to a high field spectrometer, high field
sorry this is a typo here. So, high field spectrometer it may become reverse. This k
may become less than this, and then [noise] it becomes a slow exchange. So, we can change
from slow to fast based on [noise] the NMR spectrometer.
So, now let us see how slow exchange [noise] is manifested or identified? How do we find
out whether ah particular ligand interaction or a change of protein from A to B is it slow
or is it undergoing a fast exchange [noise]? So, typically when a [noise] ligand binds
strongly with protein. So, protein ligand A is our ligand B is our protein [noise] ok.
So, if a ligand binds to a protein strongly than, slow exchange is the scenario possible.
So, this is typically happens when there is a strong interaction. Again remember what
is strong and weak? It is based on the K d value [noise].
So, typically if your K d of interaction is somewhere from micromolar to nanomolar. [noise]
We can say that it is a strong interaction. And in such a case [noise] your delta free.
That is the ligand, free ligand [noise] and bound ligand [vocalized-noise] we will have
different chemical shifts and as we increase the population of A. Means [noise] when you
[inc/increase] add slowly A to B [noise] let me let us say initially that there is no protein
added. Ligand is only present in that case, your peak off only ligand will come there
is no bound form because there is no protein been added. [vocalized-noise] Such a case
[noise] the [prot/protein]- ligand will have a particular free volume.
Now, when we slowly add the protein, we add ah where is 10 percent 0.1 is to 1 ration
[noise]. Then the bound population will start coming now. Because this complex is getting
formed and this is decreasing now. So, this A will decrease in intensity, but the A to
B the A B complex is increasing in intensity. Now because there is a slow exchange, these
two peaks are well separated. So, there is no question of a broad peak in
centre or a population weighted average which I mentioned in the previous slide for fast
exchange. So, this is the slow exchange scenario, [noise] which means strong binding [noise].
So, now suppose I increase further. The [populatio/population] I mean, I add more of B [noise]. So, I increase
the population of A B complex. So, you can see here A to B ratio is 1 is
to 0.5. Meaning 0.5 of A has reacted with 0.5 of B and formed 0.5 of A B [noise]. So,
the A B population now is 0.5 [noise] in some units. Whatever be the units But a also is
0.5. Because half of A has reacted with [vocalized-noise] the B. And half of A is free [noise]. So,
now, the intensities of these two peaks become equal. [noise] Because they have same population
and NMR [vocalized-noise] basically reflects the population and therefore, that is equal
in intensity. Now, I can continue this further [noise] and I add equal amount of A and B.
So, the entire will get reacted with B [noise] and you will end up with A B as 1.
If [vocalized-noise] let us say I add 1 of B and 1 of A. Then complete A B complex is
formed and a zero [noise]. So, a will almost be zero. Again remember, it wont be fully
zero. Because this is a reversible reaction [noise]. So, there will be always a small
population of A present in the solution [noise]. So, that is what is depicted here. You may
have zero or very small intensity. But what has happened is the bound form is come to
full intensity and strong peak [noise] ok. So, this is the typical scenario, which can
be noticed when there is a slow exchange [noise] in NMR spectrum. [noise]
And that is what is basically that this type of exchange depends on the chemical shift
[noise]. This is something I have been mentioning that the exchange is fast or slow, all depends
on the chemical shift value of A and [noise] B.
So, now let us see how does a weak or a fast exchange manifest? So, in a fast exchange
it will be a slightly different scenario, let us look at that [noise]. So, again imagine
this free ligand, which is binding [noise] to protein. So, B is our protein and forming
a complex. [noise] Now here this cause we are assuming a fast exchange condition, it
means the rate constant between conversion from A B to A this complex. And this back
[noise] is [vocalized-noise] very much larger than the difference in the chemical shift
between the A in the free form and A in the bound [noise].
So, delta B is not B [noise]. So, please remember this [noise] this actually [noise] is this
chemical shift, and this is this chemical shift [noise] ok. So, I should mention this
in the previous slide as well. It should be [noise] free is this chemical shift delta
A and delta B is the bound. Yeah as mentioned here ok. So, in this case, in this first exchange
scenario. [noise] Let us [ass/assume]- start from assuming that no protein is there. So,
the ligand is only present. So, it is the free form. [noise] Now [vocalized-noise] the
whole thing can also be assumed from the protein point of view [noise].
So, I can [sta/start]- assume that A is a protein, [noise] and B is a ligand. So, it
does not matter because I will then be monitoring the peaks of protein and not ligand. So, remember
it does not matter I consider [noise] A or B. I just have to consider one [vocalized-noise]
species and I have to considered the chemical shift of that in the bound form. So, I can
considered A as a [vocalized-noise] ligand molecule or I can consider A as a protein
molecule and the B becomes a ligand [noise]. So, for time being let us say that A is the
ligand. Which we thought saw in the flow exchange also. So, then now is no protein added. So,
it is zero [noise] the bound form is not present [noise]. But now as I increase the population
of B [vocalized-noise]; that means, I am increasing P B. [noise] Because remember P A plus P B
is [noise] 1. So, if P A was earlier 1, P B was zero. Now I am adding PA as 0.9, I am
[vocalized-noise] I mean the reaction population of P A is decreasing [noise] because it is
decreasing reacted with B. But P B is now increasing [noise].
So, but what happens in a fast exchange, in a fast exchange you dont expect this peak
to come. Because remember I showed in the earlier case, we get a average chemical shift
which is somewhere in between A and B. [noise] And it depends on the population of A and
B. So, therefore, we can see this peak is come in somewhere close to A, because right
now A population is more [noise] than B. Now if I increase the population or [vocalized-noise]
concentration of B. Then peaks starts moving even more it has come to the centre. Now if
I go to very high 1 is to 1. Where a large fraction of A B is expected
to be present [noise]. So, the P B. So, P B again remember is a bound form. P B is not
the [noise] ah the population of B it is the population of the bound form [noise]. So,
[noise] I should make it clear again, This is bound form [noise] ok. So, that has increased.
So, therefore, the peak has shifted towards the bound form [noise]. So, this is how we
can find out ah if there is a fast exchange. Because that will slowly gradually shift the
peaks [noise] ok. So, the peak shifts because the population of the bound form [noise] increases.
This is exactly what I was saying. So, one thing is I again remember this depends on
the field strength. What I call as fast, can become slow if I go to a higher field. So,
in a higher field suppose this is done at 300 megahertz, and if I go to 900 megahertz,
three times bigger in field strength [noise]. Than the this may not appear like this it
will become like the previous case of [noise] slow exchange [noise] this is 0.1. And second
is again repeat if that A and B it does not matter what you consider I can considered
a as A protein and B as a [noise] ligand, in such cases I will monitor the protein spectrum.
If I look at a as a ligand then, I will be monitoring the ligand spectrum. So, both ligand
and protein will show fast exchange because both are fast with respect to the chemical
shift [noise]. So, here it does not matter whether you consider A or B. Either 1, but
you have to look at the bound form. So, if I am in the bound form of A and I should
look at the free form of A. Similarly if I am looking at B in the free form then I should
look at the B in the bound form [noise]. So, in the next class we will take up a protein
scenario. Means what happens when I look from the protein point of view and we will see
how [noise] the chemical shifts of protein change when a ligand is added.

Randall Smitham

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