Welcome, we continue our discussion on the
Acid Base chemistry of Amino acids which we
were discussing in our last class. We understood
that if we have structure such as Glycine
we will have basically protonated NH3+ and
we have a COO-. Now it has zero charge because
it has reached the point where it has lost
carboxylic H+ the proton belonging to COOH
group and it is yet to loose a proton which
attached to the amino group of the amino acid.
When calculating the pI you have to go from
+1 charge to a -1 charge through the zero.
So for each structure that you are considering
if I want to go from the +1 to the zero to
the -1 these are the two steps that I am going
to follow. So If I want to calculate the pI
of Glycine I have to sum up the pK1 and the
pK2 or pKa of the both acids then divide them
by 2 to get the pI value corresponding to
Glycine.
Now, if we go to the Glutamic acid initially
if all of these are protonated so the charge
is +1. When I loose one proton from the COOH
group I am going to loose the acid proton
belonging to the amino acid first but not
the side chain. When I lose this proton then
it has a plus charge here and a minus charge
here and the side chain is still protonated.
So now this is at the zwitterionic form. So
the pI is going to be sum of 2.1 and 4.07
divided by 2.
In the next case, suppose we have a lysine.
Then initially the charge is +2. When I loose
the first proton here then this becomes COO-
but it still has a charge of +1. Now I have
to loose the next proton. So the next proton
that is going to be lost is the amino group
proton of the amino acid. Then to calculate
the pI in this case going to a +1 through
a zero to a -1, this going to be (9.18 +10.79)
/ 2. So we just considering is in calculating
the pI’s you have to go from a charge of
+1 to -1 obviously through the zero. And the
zero is the zwitterionic form of the amino
acid. So basically if you want to calculate
the isoelectric point of the amino acids,
if we just consider the simplest one the Glycine
then I have an NH3+ here, I have a COOH here.
This carboxylic acid H+ is going to be lost
first because it can is easily removed. So
when going from HAH+ I am going to an HA0
+ H+.
Here it is a +1 form and to go to zero form
by losing this hydrogen first. After loosing
this hydrogen based on the Henderson Hesselbach
equation you can calculate the pI form for
both Ka1 and Ka2. When you are calculating
it from pKa2 you go to the A- form because
you are loosing this proton as well you have
already lost this proton. Here the pI is actually
going from the +1 to the -1 through the zero,
the pI is the summation of the pKa values
divided by 2. We have to consider the +1 form
to the -1 form going through the zero form
for which we calculate the pI value of you
specific amino acid.
Now going back to the properties of amino
acids where we have consider the Size and
shape, the Charge, the Polarity, some of the
Hydrophobicity, part of the Aromaticity and
the Confirmation that is determined by the
side chain due to the different R groups that
are present and obviously the properties of
these R groups. Next we are will consider
protein structure in general how we have learnt,
how these amino acids are linked together
by peptide bonds. So, if we consider just
the protein structure we have is a peptide
bond.
The difference between a peptide, a polypeptide
and a Protein is a peptide is usually a short
chain of amino acids, a polypeptide is a longer
chain a set of amino acids. A protein is a
polypeptide that occurs in nature and folds
into a defined three dimensional structure.
This means all Proteins are polypeptides but
all polypeptides are not Proteins. The reason
being is all polypeptides may not fold into
a definite three dimensional structure which
is true for a Protein. So a protein is a polypeptide
chain of set of amino acids that are linked
together by peptide bonds and it has a definite
three dimensional structure. So we have a
peptide, we have a polypeptide, and we have
a Protein.
Now if we look at these linkages you can say
that at when we are at a pH = 7 we have lost
the carboxylic acid proton. So a protein is
always represented in this form. This is a
try peptide, the try peptide at physiological
pH which is pH = 7 we have not lost the proton
that belongs to the amino group but we have
lost the proton that belongs to the carboxylic
acid group because this is represented in
this form. We always begin the protein with
the amino terminus. So we have an amino terminus
and we have a carboxylic acid terminus. We
have the different R groups which are going
to behave differently based on the different
properties that it has.
So, if we consider one of these proteins called
Lysozyme. If I have a protein in its three
dimensional structure and I just open it up
into its polypeptide chain I will find that
it opens up into a random polypeptide chain.
Here these bonds are covalent peptide bonds
linking the beads that are amino acids together.
Here we have these amino acid residues which
are linked by the peptide bonds to form the
polypeptide chain. This polypeptide chain
will form a Protein because it is going to
fold into a definite three dimensional native
structure.
So the sequence of amino acid residues of
this protein is it start with a protonated
amino terminus and end with a carboxylic acid
terminus which does not have the proton attached
to it and this is the sequence of my protein.
In this sequence we know that Lysine is followed
by Valine followed by Phenylalanine followed
by Glycine and so on and so fourth. Now you
can recognize the 3-letter codes for the whole
polypeptide chain that is comprises the sequence
of Lysozyme. This is going to fold into a
definite structure. Therefore when we consider
the protein structure we have to consider
the physical properties of the protein that
influence the stability of the protein and
determine its fold.
Here there is a specific Rigidity of the backbone
that is essential for the structure, there
are interactions of the amino acids with water
depending on the Hydrophobicity of the amino
acid. If the side chain has oxygen or a nitrogen
atom so it is possible to form a hydrogen
bond but if a side chains contains only carbon
and hydrogen it is not possible. so you have
a specific index called a hydropathy index
which actually tells how the amino acids interact
with water and how we can actually determine
different regions of the protein that are
going to be embedded in the protein because
it has a stretch of amino acid. If you have
a stretch of hydrophobic amino acid they will
not be on the surface of the protein.
Then we have to consider the interactions
among the amino acids. There are electro static
interactions because we have charges on the
amino acids. For example, Lysine which is
a positively charged amino acid can interact
with the Aspartic acid or Glutamic acid which
is negatively charged amino acids. So we can
have these acidic amino acids interact with
the basic amino acids. We also have the hydrogen
bonds between the polar amino acid residues
and the solvent or between the polar amino
acid residues themselves.
The next one that we already mentioned was
the S-S bonds. The S-S bonds are the disulfide
linkages. They arise from Cysteine residues
coming together and forming Cysteines. So
we have these S-S bonds and these are the
only other covalent bonds apart from the peptide
bonds that are present in proteins. And we
also have volume constraints. Volume constraints
means we have the sizes of the different amino
acid side chains and obviously there is going
to be a steric hindrance, a steric hindrance
into the accommodation as to how they occurred
in the folded protein. If we look at a Hydrophobicity
scale you can see the Kyte Doolittle Hydropathy
index.
It is nothing but the scale for Hydrophobicity
values. Here a positive value indicates a
hydrophobic residue which we have for Alanine
and the Alanine is just a methyl group so
it hydrophobic in nature. We know that Valine
is also a hydrophobic residue. The hydropathy
index for valine is 4.2. It is a high positive
value indicating that it is hydrophobic in
nature. Then looking at the Lysine, the Lysine
is a positively charged amino acid residue
but has a negative hydropathy index indicating
that it is a hydrophilic residue. It would
rather remain on the surface of the protein
than embedded in the protein. Then Aspartic
acid, Arginine, Glutamic acid, Glutamine all
of these are hydrophilic residues. And the
once that have positive values here are hydrophobic
residues.
Later on we will see how this index actually
helps us in determining which stretch of amino
acids are on the surface of the protein and
which stretch of amino acids are embedded
in the center of the protein. So let us look
at the structures of the two proteins. The
right hand side one is a membrane protein.
You can see this is the cell membrane which
is a lipid bilayer. You have polar head groups
and hydrophobic tails in a lipid bilayer.
We have a cell membrane that happens to be
a lipid bilayer, a lipid is basically a polar
head group with hydrophobic tails.
If we look at the structure of a solvent protein
or protein in a polar solvent then there will
be specific characteristics of the residues
that are on the surface of the protein. For
example, if we look at this chain is a hydrophobic
chain. So we can expect the hydrophobic amino
acid residues will be on the surface of the
membrane protein. A polar residue on the surface
will not interact favorably with a hydrophobic
tail of the lipid. So you will have preferably
hydrophobic interactions here.
But if we are looking at this protein we can
expect hydrophobic amino acids residues to
interact amongst themselves in the center
of the protein. And I will expect polar residues
on the surface which is just reverse of what
I would expect for a membrane protein. So
any protein that is embedded in the Membrane
would rather have a hydrophobic surface to
it because it can interact with the hydrophobic
tails of the lipid bilayer. And a Globular
protein that would be in a solvent would rather
have its hydrophobic amino acid residues embedded
in the center of the protein. So here we have
a list. So we have Globular protein that is
this, we have a Membrane protein that is this.
I look at non polar residues. Here the single
letter codes for the amino acid are given.
So we have Valine, Leucine, Isoleucine, Methionine,
Phenylalanine, Tyrosine, Tryptophan. In a
Globular protein they prefer to be in the
interior to form the hydrophobic core. So
they are some way here. The polar charged
residues in this case are on the surface and
they form the catalytic sites. They are on
the surface of the protein. And the polar
neutral ones would preferably form the hydrogen
bonding network. So, if we look at a Globular
protein and we look at the specific types
of amino acids residues we know some extent
where they might be located in a protein.
On the other hand if we look at a Membrane
protein, the non polar residues are not in
the interior. They are on the surface because
they interact with the lipid because they
have to anchor themselves to the lipid and
this hydrophobic tail would not want to interact
with the polar residue. So the non polar residues
would preferably be on the surface in the
case of Membrane protein.
The polar charged groups would rather be in
the core of the protein because they want
to be as away from the surface is possible.
So instead of hydrophobic core which you find
in Globular proteins that are in solvents
you have a hydrophilic core for Membrane proteins.
And the polar neutral parts would form the
inside surface are the part of the channel.
Here a part of a channel means suppose on
the bottom we have the inside of the cell
and on the surface we have the outside of
the cell. This is a cell membrane. So if I
have a protein that actually integrates through
the membrane there has to be a channel that
goes in and out whether the salts inside the
cell or the salts from outside the cell. And
if you want to transport say sodium ions or
potassium ions would 
prefer a hydrophilic environment. So it makes
sense that for the lipid the inside surface
of the lipid has the pattern of the hydrophilic
amino acid residues that are eventually going
to allow the passage or the transport of ions
from the inside to the outside or the outside
to the inside.
If you have a hydrophobic core in here it
would be extremely difficult for you to allow
this transfer to occur. So we have hydrophobic
amino acid residues on the surface in the
Membrane proteins that are going to link up
or form interactions with the hydrophobic
tails of the lipid bilayer and inside we have
a hydrophilic core, we have a hydrophilic
surface that allows the transport of ions
in and out of the cell. So this is very clear
to understand once you know the characteristics
of the amino acid residues. And you also know
the characteristics of the protein whether
you are talking about the Globular protein
or you are talking about a Membrane protein.
So considering the protein structure there
is conformational flexibility. The protein
polypeptide chain is formed by the linking
of the amino acids residues. Now we have a
lot of geometrical considerations that we
have to make once we link the amino acid residues
together.
Here we have the Ca carbon atom which is linked
by the amino group and is linked by the carboxylic
group. Following the chain the one in black
is called the back bone that is where we have
our linkages. Suppose this is just part of
the chain, here we are looking at this particular
amino acid I have this NH which was part of
NH2 originally, this is the Ca which is the
one to which the amino group attached and
the carboxylic group is attached. So this
forms the part of one amino acid in the chain.
The R group has linked to the Ca. One side
you have the hydrogen atoms. You have hydrogen,
you have the R group, you have the NH2 group
and you have the COOH group so this makes
the asymmetric. So this is part of the amino
acid which is now present in the protein structure.
it is part of the protein structure. So now
we have torsion angles where we are looking
at rotation about bonds. These torsion angles
actually define the rotation about the single
bonds that are present. This is the amino
part of the chain, this is the Ca which is
attached to the R group. This is the C double
bond O. here it was attached originally an
OH group which has lost with this amino group
of H to form the peptide bond. Now this peptide
bond actually has a partial double bond character
to it.
This is because there is a lone pair on nitrogen
that actually can be donated to this bond
where this is looses a COO-. So here it has
a partial double bond character. Now in the
partial double bond character will impart
the rigidity to the polypeptide chain. The
rigidity will not allow a rotation about the
polypeptide chain about the peptide bond.
It will restrict the rotation. Now it is partially
double bonded in character. This bond is free
to rotate, it is a single bond, it can rotate
freely. Similarly this rotate freely, this
can rotate freely but this has some restrictions
due to the partial double bond character.
Here the lone pair on nitrogen will results
in a partial double bond character.
Now we can look at the disposition of the
Ca residues. This is residue number one R1,
this is residue number two R2. So this is
the Ca of R2, this is the Ca of R1. So this
is the Ca, this is C double bond O, this is
NH, this is Ca of the next. Now this is R1
because its tail is carboxylic acid. NH is
the beginning of the next one, it has to be.
So this is R2. So here the orientation is
Trans if this R1 and this is R2. If the rotate
is about the peptide bond so this becomes
Cis. So normally you can have a trans, in
very rare cases you have cis. It is because
you just have the Ca attached to the rest
of the side chain. So here might steric hindrance
or clashed of the side chains unless they
are very small. So we would expect the peptide
bond to be preferably in nature.
So now there are different definitions for
the torsion angles. If we have the geometry
of any compound for say to tell you about
the geometry of hydrogen the only information
you just need is the distance between the
two hydrogen atoms. Because if you know where
one is located you can find out where the
other one is just by knowing the distance
between the two atoms. So all you need in
this case is this distance.
If you look at the geometry of water where
you would look at H-O–H. it is not sufficient
to explain by knowing only the distances.
If I tell you this H is this distance away
from the oxygen and this H is this distance
away from the oxygen that does not suffice
because I have to tell you the angle between
HOH if you have to get the correct geometry
of the water molecule. So the distance is
fine in the case of a diatomic molecule. If
you have a tri atomic molecule you need two
distances and an angle. Now this is the dihedral
angle. Suppose you have four atoms A B C D.
You know that a plane is defined by a minimum
of three points. So I can define a plane using
the atoms ABC, I can define another plane
using the atoms BCD.
So now I have two planes. The dihedral angle
is nothing but the angle between these two
planes. For example let us take hydrogen peroxide
which is an example of ABCD. Here we will
have one plane with HOO and the other is with
BCD which is OOH. So we have two planes one
is this plane that we have here and one is
the plane behind that. We do have an angle
between the two planes. This angle is the
dihedral angle.
So when you are defining the geometry for
these two atoms you just need the distance,
when you defining it for three atoms you need
two distances and an angle, here we do not
need a dihedral angle because it does not
occur. And we do not need an angle in the
first case because it is not necessary. But
when we come to the dihedral case we need
everything we need the distance, we need the
angle, we also need the dihedral angle.
We need to know HOO angle on that plane, we
need to know the OOH angle on this plane.
We also need to know the dihedral angle to
exactly know the disposition of this H and
this H with respect 
to 
the oxygen and oxygen. So we have a torsion
angle that is defined by rotations. For say
we have the ø angle that is defined by rotation
about the N-Ca bond. It mean this is N, this
is Ca and it is similar to the hydrogen peroxide
where these are actually the two oxygen atoms.
Now when you mention a polypeptide chain you
can speak about the back bone. So the four
atoms are the C, the N, the Ca and the C that
belongs to the next or other this amino acid.
So this C belongs to this amino acid because
here this has an amino group, a Ca group and
a carboxylic group. So this has to be one
amino acid. This is the previous amino acid,
it has to be because it begins with an amino
terminal, has a Ca and the C of the carboxylic
group.
So when I am talking about in similar cases
that I mentioned about the hydrogen peroxide
I have to have four atoms that are going to
define the torsion angle or the angle between
the planes. Here the points are A B C D. I
have a plane that will define A B C and this
plane defines C N Ca. the next plane is N
Ca C. the ø angle is the angle between these
two planes.
If I want to define rotation about the N-Ca
I 
have to go one atom before it and one atom
after it, it is just like doing HOOH. we have
two planes, the plane that contain atoms number
one, two, three and another plane that contains
atoms number two, three, four. So here I have
C N Ca C that is defined an ø angle. The
C N Ca C means we have two planes in which
one plane is C N Ca, the other is N Ca C plane.
The angle between these two planes is the
ø angle. Similarly you can define a ? angle.
The ? angle is the angle between the Ca, C
atoms. So the four atoms we have here is N
Ca C N, the two planes we have here is NCaC,
CaCN.
Now here this is the back bone, and here the
marked planes are 
the peptide bonds. This also forms the plane.
I can also have a dihedral that defined this
but I am not considered that importantly because
it has a partial double bond character. There
are restrictions to its rotation. So it is
less flexible and it is more rigid. But does
not mean it can not rotate you do get trans
and the cis.
The angle between Ca C is ? angle. So here
N Ca C is one plane, Ca C N is the other plane.
The ø is the rotation about the N Ca. So
the four atoms in this case are C N Ca C and
two planes are C N Ca and N Ca C. So defining
specific ø, ? angles that can defines the
geometry of the polypeptide chain the amino
acid disposition. If we rotate this, the R
has also rotated along with it. So the whole
polypeptide chain is going to have a large
dependence of the ø, ? angles. So in the
whole polypeptide chain each amino acid will
have an ø, ? angles associated with it.
Now if you plot the ø, ? angles you will
get the Ramchandran plot which extremely important
plot in Protein structure. And we have to
proud that it was GN Ramachandran who actually
pointed this out. He has a land mark paper
where he showed the definite geometry of the
ø, ? angles that are defined here belonged
to the specific regions. Today this is called
as Ramachandran plot. And this is must for
any protein structure that is solved you will
have a correct Ramachandran plot for that
protein.
Here we have the primary structure, we have
the secondary structure, we have the tertiary
structure and we have the quaternary structure.
The 
primary structure is the amino acid sequence.
It is nothing but just the sequence of amino
acids which give you no information about
the structure. You only get the structure
once you fold it.
Next we have the secondary structure. Here
this is the primary structure and these linkages
are peptide linkages. This is a disulfide
linkage and these are the only two covalent
linkages that we have in protein structure.
And this is amino terminal and this is carboxylic
acid terminal. So this is the amino acid sequence
of the polypeptide chain.
Now this sequence you found out with the ø,
? angles can bend out the flexible in any
direction. So you can have some thing that
looks like a helix because you have flexibility
to rotate the polypeptide chain. So the polypeptide
chain can rotate and it can form a helix.
It can also form a ß strand. So when we want
to form a total polypeptide or rather a protein
structure it is just a linking of all these
types of different secondary structures together.
So we basically build up this chain from bits
and pieces of helices and some turns or random
coils at linked these two together. So if
these were in the same protein they would
be just linked by another part of the polypeptide
chain. So we get the tertiary structure which
is the side chain packing in the three dimensional
structure.
Now this could be a specific unit called as
a sub unit. It means that if you have a dimeric
protein which means the two monomer units
linked up together non covalently, so they
just associate with one another. So you have
a dimeric structure. For example Hemoglobin,
the Hemoglobin is a tertiary structure. Hemoglobin
has a tertiary structure, it has a quaternary
structure and it has four sub units. It means
that it has four such monomeric units that
form the protein. So all protein will have
a primary structure, a secondary structure
and tertiary structure but all proteins will
not have a quaternary structure. You only
have a quaternary structure if it is not monomeric
in nature. If the protein is monomeric there
is just primary, secondary and the tertiary.
But if there are a number of sub units that
have to link together then you have a quaternary
structure. So this is the primary structure,
this is a secondary structure. The two major
types of secondary structure are a helices
and ß sheets.
The ß sheets will form by linking the ß
strands. Apart from the primary structure
all the other structures do not have any covalent
linkages. So we will see a-helix or some thing
that looks like a-helix but there is no covalent
linkage between the polypeptide chains informing
the a-helix which is extremely important.
When forming the secondary structure that
is a-helixes or ß sheets the only non covalent
interaction that is important in this case
are high hydrogen bonds. So we will consider
this when we linking them up. Here we have
a beta strand. This is the polypeptide chain.
So here the 
protein will begin with an amino terminus.
The arrow is pointing in upward direction
so this has to be amino terminus of the protein,
it is going up.
This is a ß strand and it will form an a-helix
then it forms another ß strand. So together
this is going to form a polypeptide chain
structure or folded structure.In the second
case we have an amino part, it goes up, it
comes down and it goes up again. So here we
have three ß strands that form a ß sheet.
Then in the third case we have again the protein
begin here then we have just one huge loop
that has no characteristic structure 
to it. But here we can have different sorts
of linkages. We already knew that the polypeptide
chain is just a linkage of the amino acid
residues one of the other. For example, if
I take a necklace or beads and I drop it on
the table what is it going to form? It is
going to have some structure. If I pick it
up and drop it again it is going to have some
other structure. But the protein will always
fold into the same structure every time.
We have the primary structure which is the
sequence of amino acid residues. The secondary
structure is the local folding of the a-helix.
It is maintained by short distance interactions
and in this case hydrogen bonds only. When
we consider the tertiary structure it is additional
folding that is maintained by more distant
interactions. It means suppose I have one
polar amino acid here and I have another polar
amino acid on this side then they are likely
to form a hydrogen bond so they will come
together. What about the disulfide linkages?
We can have a disulfide linkage between residue
number one and residue number seventy eight.
It means that distant parts of the protein
structure come together to form the overall
structure.Here we have different secondary
structure conformations. The secondary structure
conformation means you can have a-helix and
this is the preferred ø and ? of the a–helix.
We have a right handed alpha helix and left
handed alpha helix. What do we mean by a right
handed alpha helix? We mean that the helix
can go actually in two directions it could
go up or down. Now if this is the direction
of the propagation and this is the way your
polypeptide chain goes so this is a right
handed helix. If you consider your left hand
and this is the polypeptide chain and this
is the direction of the propagation the left
hand. Then this is a left handed a–helix.
Left handed a-helix is very rare in proteins.
You do not generally see them. You just have
right handed a helices. You can see the conformation
of amino acids is always L conformation.
So, if you look at a- helix usually when we
mention a- helix we do not say it is a right
handed a-helix because by default it is. So
this is the preferred region for the a-helix
in proteins, this is the preferred region
for the ß- sheets in proteins. These are
ß strands that form a ß sheet. Now in between
these ß strands in this blank space we could
have any thing we could even have a helix.
Because this is one part of the polypeptide
chain linking with the some other part of
the polypeptide chain. But the difference
between the a-helix and the ß helix is the
a-helix has to be contiguous. We cannot have
the part of the a-helix here and the rest
from the other part of the protein that is
not possible.
If an a-helix begins at residue number sixteen
and continuous to residue number thirty means
residues from sixteen to thirty will all be
part of the an a–helix. But if I have a
ß sheet whether it can have residue three
to seventeen or residue 40 to 46 it does not
matter they just form ß strands that are
part of the same ß sheet. Now when I consider
beta sheet here and right handed an a-helix
here then I have specific bonding characteristics
that are linking the secondary structures.
We have different types of secondary structure
conformations. We have considered a ß sheet
parallel or anti parallel. The strands are
might be parallel to one another. here we
have an anti parallel ß sheet because one
of the strand is in this direction, the other
is in this direction then the next one is
in that direction and then I have one in this
direction so this would be an anti parallel
ß sheet. Now 
some residue conformational preferences are
listed here. It means these residues would
like to be in a-helix, they preferred to be
in an a helices. These residues preferred
to be in strands and there are certain other
residues are preferred to be in turns. Here
turns are going to link the two secondary
structural elements together. So we did learned
today was how we can look at the geometry
of the polypeptide chain, we have learned
definite definitions of the polypeptide chains
in terms of different torsion angles.
These torsion angles are one was we learned
at the peptide plane would be planar in nature
because of the double bond characteristic.
Due to the partial double character of the
peptide bond makes it rigid. But we can have
rotation about the other single bonds that
are present. These can form the ø and ? angles.
Rotation about these gives the definite conformational
considerations that have to make and they
give rise to specific secondary structural
considerations. So I have my primary structure
that is linked by peptide bonds, I have my
secondary structure which we only learn the
a-helix and the ß sheet. And the a-helix
and the ß sheet will have definite ø, ? angles
that they correspond to and they would form
a hydrogen bonding network. Thank you We continue
our discussion on Protein structure and Protein
architecture. Here we have the primary structure
which is just the amino acid sequence of the
protein, the secondary structure that comprises
helices, sheets and turns.
Then we have the tertiary structure of the
protein that is the side chain packing in
the three dimensional structure. Finally we
have the quaternary structure of the protein
that is the association of sub units.
Here we have the primary structure followed
by the different elements of secondary structure.
Then we have the tertiary structure of the
protein. And this monomeric sub unit has associated
with another sub unit where you have a connection
that is not a covalent bond. It is just an
agglomeration or aggregation of these two
units together is forming the quaternary structure
of the protein. There is usually no covalent
formation between the quaternary sub units
in the quaternary structure.
All proteins will have a primary structure,
a secondary structure and a tertiary structure.
But all proteins will not have a quaternary
structure because all of them do not have
a polymeric or an oligomeric structure. Now
when we consider the specific elements of
secondary structure associated to form the
tertiary structure.
So from the primary amino acid sequence we
go on to the helix and then we go on to the
tertiary structure. We already know that each
amino acid will have its own Isoelectric point.
Each amino acid will have its own definite
point where it will lose its protons or it
would be zero in net charge.
The Isoelectric point also exists for the
protein. So, the pH at which net charge on
the protein is zero. A pH at which a protein
has a net charge of zero means the protein
will have a large number of amino acid residues,
some of them may be acidic and some of them
may be basic so each of them will lose their
protons at different times.
The Isoelectric point also exists for the
protein. So, the pH at which net charge on
the protein is zero. A pH at which a protein
has a net charge of zero means the protein
will have a large number of amino acid residues,
some of them may be acidic and some of them
may be basic so each of them will lose their
protons at different times. So, you have a
certain turn associated to it. Now for this
particular one the direction of propagation
is down. The helix is going in this direction
and down so it is a right handed helix. If
it were going up it would be a left handed
helix. You will have to follow the polypeptide
chain. In this case the nitrogen is here and
there is Ca then you have C so you are going
in that way. If you are going that way then
the direction of propagation is down so the
helix is a right handed helix. If the polypeptide
chain is turning in this way and the direction
of propagation is this so it is a left handed
helix. So this is a right handed helix.
