Practice English Speaking&Listening with: Lecture - 4 Protein structure II

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Welcome, 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 loose 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 loose their protons at different times. So, for example

if we have a pH = 2 then we have all the side chains are protonated as well as the amino

terminal and the carboxylic acid terminal are protonated. We cannot say which is which.

We know that one of this has to be the terminal and we know that one of this has to be the

terminal but this may belong to Lysine and this may either belongs to an Aspartic acid

or a Glutamic acid. Now as we increase the pH we will loose the carboxylic acids protons

first. In this case pH = 7 so the net charge on the protein is zero. So this would correspond

to Isoelectric point of such a protein.

Now here you may have another protein that has an extra NH3+ to reach a net charge of

zero or have to go even further high in the pH to achieve a net charge of zero. So we

will reach the pI of the protein by once achieving the net charge of zero. So as I keep on increasing

or rather deprotonating all these protons that are available it will come to a very

high pH where it has lost even all the protons that were associated with the amine groups.

So, if you were to have a basic protein means a protein that has a large number of Lysines

or Arginines associated to it then the pI of the protein will be high. If I have an

acidic protein rather then pI will have a low value because it will reach a negative

charge or a negative charge after the deprotonation of the carboxylic acid groups will achieve

at a lower pH. So the protein will have a net charge is zero at a lower pH then would

a basic protein. It means when we consider the pI of the protein it depends on the number

of charged amino acids residues. So it will depend upon the number of Lysine, Arginine,

Aspartic acid, Glutamic acid that you have in the protein. Because you are going to associate

the charge that are going to be essential in understanding how you can actually determine

the pI of the protein. So, if you have a polypeptide you can determine the pI from just knowledge

of the content of the polypeptide.

You know the number of charges you know you have to come to a net charge of zero to attain

the pI. Now we will consider the importance of the charge or the importance of the protein

structures, a very slight change in the structure mutagen.

The red blood cells which are originally circular in shape they become sickle shape. This is

the ß chain of hemoglobin. Let us consider it as a polypeptide sequence of one of the

chain that is presented in hemoglobin. Here you can see everything is the same except

this Glutamic acid. It has become Valine, the rest of the protein chain is exactly the

same. It has a hundred and seventy four amino acid residues and this is the sole change

that makes this round red blood cell as this. Now here essentially you have a Glutamic acid,

a Glutamic acid is a charged amino acid residue.

It would prefer to be on the surface of the protein. Because it has a negative charge associated to it.

And it would like to associate with the solvent around it. Now when we change it to a Valine

we can make it a hydrophobic amino acid residue that does not like to be on the surface at

all. So here you can see this is hemoglobin, this is the beta chain. The blue ones are

the beta, the yellow ones are the alpha. So here I have the quaternary structure of the

hemoglobin comprised of four sub units in which two of them are a type, two of them

are ß type. The one that is marked in red here is the Valine.

Now this Valine is on the surface because the Glutamic acid was on the surface. So another

hemoglobin molecule comes here and this bottom one of the next hemoglobin molecule sticks

with this one means it forms a hydrophobic interaction because it ones to be away from

the solvent.

So we will get a fiber because we can have other hemoglobin that will stick to the bottom

of this one. So you are eventually get fibers and these fibers are in the form of the sickle

cell. This was the first molecular so called disease that was determined to be solely on

the amino acid sequence of the protein. You see the importance in understanding the whole

protein structure. It is just a single property of that amino acid that results in such a

detrimental effect to red blood cell forming the fibers to form your sickle cell anemia.

And the reason being is that the Glutamic acid was on the surface which is no longer

there. You have Valine which would prefer to be away from solvent. So we have the amino

acids that are linked by these peptide bonds. And we have a partial double bond character

to the peptide bond. In this case we have the Ca, this is the Ca of the next amino acid

and both of them are attached side with chains. This is trans in nature because the Cas

are on two opposite directions.

Here we can have a cis peptide also. But in the case of cis peptide the Cas are on

the same side. And if we have Cß’s on top of this so there might be a steric clash.

This is the reason why you have more trans peptide bonds then you have cis peptide bonds.

Also we have prevention of rotation about this single peptide bond due to the formation

of this partial double bond. You have the lone pair on nitrogen so it forms this partial

double bond and because of its partial double bond character it imposes the rigidity, it

is not as flexible as the other single bonds in the peptide chain.

These are the other single bonds will be amount for the rotation around these single bonds.

Here we have the peptides that form two rigid planes. We are calling these as rigid planes

because of the partial double bond character restrict rotation about the CN peptide bond.

Here we have free rotation about the ø and the ? angles. We define it by four atoms.

For example we can define ø angle by the C, the N, the Ca and the C. It means that

if I construct a plane that has the points C, N and Ca on it and I construct another

plane that has N, Ca and C on it then the angle between those two planes will give me

the ø angle. Similarly I can get the ? angle that will be the angle between the planes

that have N Ca C and Ca C N. So we have ø, ? rotations and this gives rise to the Ramachandran


We will understand how these locations are restricted in their ø, ? values based on

the geometry that they have. This is a beta sheet, when you look at a beta sheet you can

straight away tell in which direction the polypeptide chain is moving because you know

that you always begin with an N so you have N, a Ca followed by a C. So this amino acid

is a Glycine. Here we have N Ca C and N Ca C so the protein chain is in this direction.Then

we move to the next one, if I go from here I have N C Ca that is a wrong direction. The

right direction is N Ca C and it the back bone, N Ca C. so the direction of the polypeptide

chain is opposite to the previous one, it is anti parallel.

Then we have two strands at the bottom. We have N Ca C so it is going from left to right.

We also have N Ca C in the bottom strand so it is also going from left to right. So we

have two parallel strands here and we have two anti parallel strands here. So it is just

the direction of the beta strand that is going to determine whether it is in anti parallel

beta sheet or it is a parallel beta sheet.

Now very carefully we can look at the hydrogen bonding between the sheets. There are two

types of secondary structure are important. They are the alpha helix and the beta sheet.

Both of these have hydrogen bonds that associate with them. Here the anti parallel beta sheet

and the parallel beta sheet have different types of hydrogen bonds. If you look carefully

you will be able to distinguish the difference. The difference this N Ca C is a single amino

acid. If we look at this hydrogen bonding it is in the opposite direction from right

to left. So I have N Ca C, it is the same amino acid.

So we have the N of this amino acid link with the O of another amino acid on another strand

and the same amino acid will have its amino NH link with the carboxylic O of the other

amino acid. So we have an amino acid N Ca C on top. On the second strand we also have

N Ca C and the hydrogen bonds are between the same set of amino acids.

Looking at the difference in the parallel set, we have chains in the same direction

and the strands are also in the same direction in the parallel set. So we have N Ca C on

the top strand and we also have N Ca C in the bottom strand. This has a hydrogen bond

with this amino acid and the CO of this amino acid here has a hydrogen bond with another

amino acid.

In the first case the hydrogen bonding is the NH and CO bond with the same amino acid.

In the second case the NH and CO bond with two different amino acids. So when you have

the amino acid in a beta strand will form a hydrogen bond with two different amino acids

then it is a parallel beta strand. When you have the amino acid forming hydrogen bonds

with the same amino acid then it is an anti parallel beta strand. So this is the hydrogen

bonding difference between the two irrespective of alpha helices.

When we speak about alpha helices that is another type of secondary structure that also

have specific ø, ? angles that we have to consider and they are ø = -57 and ? = -47

which on the Ramachandran plot.

So this would be a typical Ramachandran plot for a protein where you see some ahelices,

some in the ß-sheet region and some here which are usually Glycine. The ones on the

right hand side with positive ø are usually Glycine.

Here when we look at the chain you can say from the location of the nitrogen, Ca and

the C. We know this is C N Ca C so I am going from the bottom to the top. And we know the

type of angles we are considering. We have an ø angle, we have a ? angle and the ? angle

is the angle of the peptide bond. So this ? angle is usually trans so it is usually

a hundred and eighty degrees. You can at times get a cis peptide bond at that point ? will

be close to zero. Now let us look at the hydrogen bonding pattern of helices. When we have a

helix we turn it round to form a spring. The n is 3.6 for a normal a-helix means that this

is the number of residues per turn. You have understood that the helix looks like a spring

so it will have a number of turns associated to it. These are features of secondary structure

conformations. And you already know about the parallel beta sheet and anti parallel

beta sheet. And these are the ø, ? locations where you might see them.

But usually we know that a right hand helix is here, a ß sheet is here and this is a

left hand a-helix or mostly as we mentioned it is Glycine because it is most flexible

amino acid. And these are the residue conformational preferences. Now we will come to the secondary

structure called the helices. Here we have a specific individual amino acid. If we were

to follow a direction of the polypeptide chain we can say in which direction it is going.

The nitrogen atoms are in blue, the oxygen atoms are in red and these are the R groups

that are circled in gray. So we have the nitrogen here, this is the Ca because we have the R

group connected to it. This is the CO so we are going from the top down because this is

the amino acid N Ca C after that N Ca C and then N Ca C then next N Ca C then after that

N Ca C.

So you are traversing a helix. Now you will go from one point to the next point as you

traverse a helix. So as you go form one point to the next point 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. You have to follow the

polypeptide chain and have to follow its direction of propagation by looking at the ends of Ca

C. the N Ca C and N Ca C and N C alpha C. Look at which direction it is going by following

the whole chain. Once you follow the spring you can determine in which direction it is

going and you can find out that this happens to be a right handed alpha helix and it has

specific connectivity. We can see connectivity in the form of hydrogen bonds much likely

saw in beta sheets but these hydrogen bonds sort of link the spring together. So we do

not have an extended polypeptide chain we have a helix. These are the different types

of helices you can have.

You can have a-helix, a 310 helix and a pi helix. These just tell you how many turns

there are because they look kind of different. You can see here there are about four atoms

are in red, here there are three and here there is just about five one picking out there.

Now again we are looking at the same thing we have N Ca C. So this is the traversing

of the chain. So the polypeptide chain is now shaped like a helix. And we have the nitrogen of the amino acid, we have the

Ca of the amino acid and we have C of the amino acid along the helix. So we have the

two carbons and we have the nitrogen. Now this is the pitch. The pitch is for one complete

rotation the height that the helix or the polypeptide chain has traversed. The helix

should have a pitch because it is also a spring just like a screw. So it is basically a rotation

of the polypeptide chain and you have the pitch. Now if we look at the pitch of these

helices, the difference between a 310 helix and a-helix and a p-helix is the hydrogen


They all form hydrogen bonds and it is just with whom they are forming the hydrogen bonds.

So we have a hydrogen bond formation from the ith residue to the (i + 3)rd residue in

the 310 helix and the 310 tells you that you have three residues for turn and there are

ten atoms in those in that turn around. So in the 310 the 3 stands for three residues

in a turn and then 10 stand for ten atoms in the turn. And the hydrogen bonding of the

helix is from the ith residue to the (i +3)rd residue. Now we will look at the normal a-helix.

In this a-helix the hydrogen bonding pattern is slightly different. It has the hydrogen

bonding from the ith residue to the (i +4)th residue. So you have basically a polypeptide

chain that is rotating like a spring. You have a hydrogen bonding between the CO and

NH of the back bone. In the 310 case it is from i to (i +3), in the normal a-helix it

is from i to (i + 4) and in the p-helix which is extremely rare it has from i to (i +5).

Now the nomenclature for the 310 helix can also be written for the a-helix. In this case

it is 3.6 residues per turn and thirteen atoms. So if you were to write it in the same manner

instead of writing 310 we would write 3.613. You need to understand the difference in the

different types of helices is solely due to the difference in the hydrogen bonding pattern.

So basically this makes the sort of a tighter helix, this is slightly loser than that and

this is even loser because basically you are increasing the pore. You will have a tighter

helix if the hydrogen bonding is from i to (i + 3). So the inside cylindrical part is

smaller. If you have it from i to (i + 4) then it is slightly larger and if it is from

i to (i + 5) is even larger. So you can see how this diameter gets larger. This shows

the hydrogen bonding pattern even better.

The normal alpha helix is from i to i + 4 which is the hydrogen bonding pattern. You

can see the C double bond O here. So we have N C double bond O, N C double bond O. The Ca is in between these

here.Here the amino acid is N Ca C because there is the amino terminus on one side, the carboxylic acid terminus on

the other side and in between is Ca. So the representation is in this fashion to show

the hydrogen bonding pattern. So here if I have the first amino acid then fifth one is

supposed to link to it. So if this amino acid is one then the remaining are 2, 3, 4, 5.

Here we have a turn which is helix.

So the number 2 should have a link with number 6 then 2, 3, 4, 5 and then 6. We have the

hydrogen bonding between the NH of this and the CO of 2. It is the second amino acid number

2 that is what is contributing to the hydrogen bonding here is the CO of residue number 2

and NH of residue number 6. If I go to one it is the CO of 1 and the NH of 5. If I go

to number three it is the CO of 3 and the NH of 7. Here you notice something that all

COs are pointing in the same direction. It means this alpha a-helix can have a dipole

associated with it which imparts a polarity to this. Due to the fact that all the COs

in this helix have to point in the same direction if they are to form this hydrogen bond which

is specific for an amino a-helix.

So we have basically the N terminus of the helixes to the C terminus of the helix. We

have an over all dipole that is from the N term to the C term. Why do we have it? We

have it because we have the hydrogen bonding pattern of the helix. This hydrogen bonding

pattern of the helix is going to impart a polarity to the helix. And because of this

polarity in the helix we have all the oxygen atoms pointing in the same direction, we have

a dipole associated to that and we have the helix dipole in that direction.

Usually when you have two helices they are anti parallel. The reason being is the dipoles

cancel one another out because you would not want the protein to be Polar in that sense.

So usually when you have helices they are anti parallel in nature so that it can counteract

the dipole that is associated with the a-helix. So whatever I have shown here in the pink

are the alpha helices which are connected by just some part of the polypeptide chain.

Now we will learn more about a helical wheel. You are going to construct a helical wheel.

Now, looking at this diagram, this is the picture of a protein here I have a strand

this which is a strand of a particular ß-sheet and these are just connectors. This is the

a-helix. If I look at the a-helix I know that I have solvent surrounding here, I have particular

residues that are actually sticking out from the helix that all the R groups are sticking

out from the helix. If I look at the helix I can say that this part is associated to

the solvent and this part is associated to the core of the protein. These amino acids

should be hydrophobic nature and ones that are sticking out on this side should be hydrophilic

in nature.

So if I consider every turn say here I have one residue and the fifth residue here i and

i + 4, so if I have a hydrophilic residue sticking out here then the fifth residue should

also be hydrophilic in nature and so on and so forth because I my helix is a surface helix

and here I expect hydrophobic types of residues because this is facing the central core of

the protein. I can expect hydrophilic amino acid residues on the surface because it is

facing the solvent.

Now I am looking down the helix and these are all back bone atoms. The ones that are

in white are all the Ns, Cas and Cs and so on, the green ones are the R groups.

Now what happens to the R groups is they are surfaced around here. So what I should expect

is this set should be one type, this set should be another type because I am looking down

the helix. If I am looking down the helix, all the ones that are one side here are facing

inside the protein and all of them should be hydrophobic in nature and all these ones

should be hydrophilic in nature.

This is an example for a typical helical wheel that you could construct. The ones that are

in blue in the helices are hydrophilic in nature. How can you say that? Here we have

Asparagine, here we have Glutamic acid, here we have Aspartic acid here we have Lysine

and these are all hydrophilic in nature. And here these are Isoleucine, Isoleucine, Methionine

and Tyrosine to some extent and Alanine.

These are all mostly hydrophobic in nature. So having a picture of the helical wheel we

can straight away say which part is going to face the solvent because we know the characteristics

of the amino acids. So all these blue ones will be outside because they are hydrophilic

in nature and they will interact with the solvent. All the red ones will be inside because they are hydrophobic

in nature. So how do we construct the helical wheel? This is how you construct the helical

wheel. Now you have 3.6 residues per turn or usually we can say there are four residues

per turn.

If you want to construct the helical wheel we would put residue number one say on the

top, for every residue there is a rotation about 100º so the next residue is will come

here that is residue number two. So if we want to construct a helical wheel we are going

to have residue one here which is on the top, we have the center. So the residue number

one is here, residue number two is here. Where is residue number three going to be? This

is the 180º but residue number three is going to be around 200º that will be some where

here. Where residue number four will be? I have this is at 100º, this is at 200º, so

the fourth is going to be at 300º. This is approximately 300º so this is residue number

four. Then you can go on.

Now we want to construct a helical wheel for this so let us construct it. The first one

I have is Lysine and the sequence we have here is Lys His Gly Val Thr Val Leu Thr Ala

Leu Gly Ala Ilc Leu Lys Lys.Now, where is residue number 1 going to be? It is going

to the top. The residue number two is Histidine. so this will be Histidine, residue number

three is Glycine so this will be Glycine and residue number four is Valine. And then residue

number five will be at 400º which is basically forty degrees after this so it could be some

where here so this is going to be residue number five which Threonine.

So the residue number 6 is going to be somewhere here which is Valine, residue number seven

is going to be some where here which will be Leucine. So what you can do is you can

construct the whole helical wheel. What do you need to know? All you need to know is

this sequence and you need to know for every residue the rotation is 100º that is sufficient

information for you to construct a helical wheel. Then once you construct a helical wheel,

from the wheel itself you can predict which part is going to be outside, which part is

going to be inside if it were to fold into a protein. So you can actually say which part

will be hydrophilic in nature and which part will be hydrophobic in nature.

Thus if we do whole construction for a helical wheel then it will be look like this. Now

if we look this picture it should have some of these at least Lysine, Histidine, Glycine,

Valine, and Threonine and so on and so forth. So now you know how to construct a helical

wheel. When you construct this helical wheel we have a dotted line here where it tells

that these are mostly hydrophilic in nature so this is going to be the outside, these

are mostly hydrophobic in nature so they are going to be on the inside. Even though it

is a very simple construction it gives a lot of information.

All you need to know is just the sequence of the protein. Once you know the sequence

of the protein you can construct a helical wheel. For example, as I said this sequence

forms a helix then you can say which part of the helix is inside and which part of the

helix is outside. All you need to know is for every residue that you consider you have

to go 100º. And then obviously you have to know types of amino acid residues to know

which one can forms hydrogen bonds and which one can not forms hydrogen bond. Now what

happens is this line does not have to be at the center as you can see it is not at the

center. It may be just up here, just saying that most of it is bedded or most of it inside

but a little part is outside.

So when we are looking at this helical wheel, we can straight away say that we have an outside

region and an inside region based on the types of amino acid that we have. Now we are considering

a helical wheel for say a membrane protein. What do we have in a membrane? We have some

thing like this. If I have a helix here what should I get when I construct a helical wheel? If I construct a helical wheel I will have different residues for say one,

two, three, four and five and so on and so forth. What do I expect on the surface here?

Here these are hydrophobic and these are Polar.

If I look at this side so I have interactions here and here in the helix, the interactions

have to be with the hydrophobic tales of the lipids. So what do I expect the helical wheel

to comprise of all hydrophobic residues? Does that make sense? It does because you should

have hydrophobic interactions. If you do not have hydrophobic interactions then this helix

will not stay here at all because the helix need to have a favorable interaction with

a lipid tales which are hydrophobic in nature. The surface of this helix has to be hydrophobic

in nature. So if I construct a helical wheel for this helix I would not have a distinction

where I would get an outside and inside I would only get all hydrophobic. So if you

do get all hydrophobic it means that your helix is part of membrane. Now if we have

it such that I have four helices here.

If I am looking from the top to down these are my polar heads and I have the helix here. That is also possible

because you have to remember that your surface actually looks like this. If I look at the

membrane from the top it looks like this. This is a cross section. If I look from top

then I can look at the surface which is actually all polar heads of lipids. I have four helices

in the protein that is embedded in that. if in construct a helical wheel for A, B, C and

D then helical wheel will be such that this part is going to be hydrophobic because it

is interacting with the hydrophobic tales of the lipids that are down there. Then the

inside region is hydrophilic which is exactly what you see in such types of proteins. And

it is convenient to have this part as hydrophilic so that it can transport ions. And this part

is hydrophobic. So what we learned today was we have rotations about the bonds in the polypeptide

chain that give rise to ø, ? angles and we know that the ? angle is present in proteins.

But we rather have a Trans peptide than a cis peptide due to the restricted rotation

because of the partial double bond character of the peptide bond. Then we have talked about

the secondary structural elements where we looked at the features of the ß-sheets and

a helices. What are the features of the ß-sheets that we looked at? They are parallel and anti

parallel in their differences in the hydrogen bonding patterns. Then we looked at the hydrogen

bonding pattern for a helices. And we learned how to construct a helical wheel that is going

to exactly tell us where a helix is located. So today we will stop here.

Thank you.

The Description of Lecture - 4 Protein structure II