We start our discussion on nucleic acids and
their components before we understand what
are nucleic acids. So far we have studied
all the other molecules. Basically for life
molecule life meaning the carbohydrates, the
lipids that form the cell membranes and other
components such as amino acids and then proteins.
When we go on to nucleic acid, we will see
how important they are in their manifestation
in the formation of the proteins that we have
studies so long. When we look at the central
dogma of biology which goes as follows: It
is DNA to RNA to protein.
This is known as the central dogma of biology
going from DNA to RNA. The process known as
transcription, RNA to the protein is process
known as the translation and we know that
all the information is stored in the DNA.
That is the storage medium. It is then formed
or rather transferred to RNA forming the transmission
medium that then forms the proteins, expressing
a protein what we mean by the protein formation
that ultimately is required in all the activities
that go on in the body in terms of enzymes
so and so forth.
What we are going to do in nucleic acid? We
looked at some of the components but we will
see how the structures are related and how
actually some of these processes are going
through. If we look at just some idea of the
biological length scale, we looked at chemical
bond something that you looked at for a long
time now. They are in the order of Angstroms.
If we look at the amino acids, they are in
the order of 10 Å. When we look at proteins
they are in the order of 100 Å. As we go
higher and higher, you see how this lengths
scale actually goes on and ends at DNA which
is actually 10 cm which is pretty long, if
you look it from protein point of view considering
that you have a globular protein that is still
in the Å realm where you have it in the order
of 100 Å. We have chromosome DNA that is
around 10 cm.
Now the fact that you have DNA replication
and DNA processing going on extremely fast
in the body. It is extremely important to
understand how structurally it is place in
the body, how it is located and what holds
the two as will study later on the 2 trans
of the double helix together? This is something
we have looked at before when we are doing
vitamins and coenzymes. We consider what are
called nucleotides. DNA is deoxyribonucleic
acid and RNA is ribonucleic acid. Now in the
formation of these RNA, this nucleic acid
DNA and RNA, there are certain terminologies
that we have to go through once more to understand
their structure their bonding.
In nucleotides, we have what is called nitrogenous
base, a sugar and a phosphate. We now know
what is sugar and phosphate. We will just
go through what are nitrogenous base. When
we consider the nitrogenous bases, there are
2 types of bases that we consider purines
and pyrimidines. This is something that we
consider when we did vitamins and coenzyme
just to revise what we were studied there.
When we consider this nucleic acid, we have
the sugar and phosphate which is actually
called a sugar phosphate backbone. The primary
structures of both DNA and RNA are similar.
They have a sugar phosphate backbone.
The difference is in the type of sugar because
one is ribose sugar and another one is deoxyribose
sugar. The difference again also lies in the
type of base that is attached to the ribose
or the deoxyribose ring. So what we have is,
we have a phosphate that is shown in yellow
color here and sugar shown in blue. We have
different kinds of bases depending on nucleic
acids that is attached to the sugar. What
we have basically is the sugar phosphate backbone
and we have the bases attached to the sugars
in the backbones. The sugars that are used
in these two types of nucleic acids are ribose.
Ribose is used in RNA and the essential difference
between ribose and deoxyribose is the missing
OH at the 2 prime position of the sugar ring
in DNA.
When we have this linkage, this fine end will
attach to a phosphate, this end will attach
to a base and the bases will be different.
The sugar is essentially different when we
link these together as will see RNA. If we
link these together we will have DNA and there
is basic difference between structure, this
is also going to be reflected in the stability
of RNA and DNA. We have a nucleic acid that
is made up of polymers of four different nucleotide
residues. We will see what these are in a
moment. We have A C G U that makes of the
alphabets of RNA. A C G T that make up the
alphabets of DNA because the ribose sugar
in a deoxyform, we have d prefixed to the
AMP, CMP, GMP and TMP.
If you write AMP it is a ribose sugar, you
have to specify the deoxy type of the sugar
by writing d, which means that are the two
prime functional groups and you do not have
the OH attached to it. What are these base
families? We have the nitrogen base purines.
What are these purines? They are fused 6 and
5 membered rings a hetro carbon nitrogen ring
system and two commonly used ones in DNA and
RNA are adenine and guanine, these are the
purines.
The pyrimidines are 6 membered carbon nitrogen
rings that are usually unsaturated. There
are three common purines that are formed in
biological systems, C and T are used in DNA
that is cytosine and thymine. Cytosine and
uracil are used in RNA. We have the purines
and pyrimidine that are going to form the
nitrogenous bases of the nucleotides that
are going to be attached to the sugars in
the nucleotide structure to form nucleic acid.
These are our purines and pyrimidines and
they are in definite structures. This is the
numbering system that you have the purines
and pyrimidines. And as we mentioned before
that when we are looking at DNA we have A
G C and T.
When we have RNA, we have A G C and U instead
of T. These are our different base families.
This is something that we looked at before
when we are forming the nucleotide, we have
a sugar. The sugar in this case is a ribose
sugar because the OH at the position is present.
We have a -N glycosidic bond. We know why
it is , why it is glycosidic and why it is
N. Why is it because, it is cis to the CH2OH.
Why is it N because it is linking with the
N of the purine or the pyrimidine base? Why
is it glycosidic because you are linking a
sugar? Any time you link a sugar it becomes
the glycosidic linkage. This is the -N glycosidic
linkage i.e. linking the sugar ring to the
purine or the pyrimidine base at the position.
At the position, you have either OH or H being
the ribose sugar or a deoxyribose sugar. Then
you have the phosphate attached to the position
where you have either this 1 phosphate or
you can have 3 phosphates as we looked at
the structure of ATP. So when we have the
OH we have the ribose, when we have just the
H we have deoxyribose. We know how to designate
these by writing either a d or without the
d when we want to specify a ribose sugar.
When we have the base attached to the sugar
we have the nucleoside, as soon as the phosphate
is attached to the position we have a nucleotide.
We have a nucleobase and adenine. We have
nucleoside which is the base attached to the
sugar. The sugar in this case is deoxy which
means there is the H and there is no O.
Then we have our nucleotide which is deoxyadenosine
mono phosphate but I could have just written
is this as DAMP. Just writing it as DAMP you
know that this is the structure. The D is
signifying no OH, the A signifying this and
MP signifying the mono phosphate. Each of
these, so the DAMP you know exactly how you
have to write it even for the DCMP or the
AMP, ATP. Considering that we have the sugar
of the phosphate attached to one another and
we have this bases basically sticking out
that is what I showed in the first picture.
It means that we have to look at conformational
configurations. Conformational considerations
in terms of the backbone just like we had
in the protein.
What did you have in the protein? You had
certain angles that mention how the backbone
would be oriented. What did we have sticking
out the backbone; the side chains the different
R groups sticking out from the amino acids.
See groups of the backbone and we had different
orientation possible. We can have the same
here. What is that? This backbone of RNA and
DNA consists of the alternating phosphate
ribose or deoxyribose deoxyribose chain. We
have alternating phosphate and ribose. You
see how that is formed, once we understand
the torsional angles. When we have conformational
variation this arises from restricted bond
rotation. Where are we going to get restricted
bond rotation? We have a sugar ring. We have
single ring, what do we have? We have something
like this. This is our sugar ring.
You have to remember all of these are single
bonds. What is possible puckering? You do
not call rotation because it is restricted
in its rotation. If we have just single bond
we know that we can rotate all the way through.
We have something like this. You cannot have
all the way gone through, because it is going
to twist the molecule. This twisting is what
is known as puckering. What kind of observation
can we have? We can have the oxygen go up,
go down with respect to this bond here. So
we can have restricted bond rotations within
the sugar ring because of ring system. We
are not free to rotate all through. This give
rise to different ribose ring pucker and torsional
angle, that bonds connect the phosphate to
the ribose. Let us see what we mean by that.
We have here, just look at one of these figure.
First this red sphere is the oxygen of the
sugar ring. We have 5 membered sugar ring
so if you just look at the different, so this
is 1, 2, 3, 4 and 5 atoms that we have to
connected to form the 5 membered ring. This
is carbon atom that is usually attached to
the phosphate. See this is what we are looking
at, you have something like this. This is
attached to the phosphate and where is this
attached? This is attached to the base so
this is . This is the prime position. This
is the , this is our , this is our . When
we go back to this we recognize what is happening
here. We have 1, what is attached to the base.
So the blue circle represents the base nitrogen
that is attached to it.
We remember that this is the N glycosidic
bond, so the sugar is attached to the base
by the nitrogen. It is because these are cis
to one another. We have the phosphate or let
us look at the carbon, so we have basically
the carbon here which is essentially attached
to the phosphate and we have the base. Now
we are looking at what can happen to the sugar
ring? The sugar ring can bend in such a way
that we can have some thing that is called
endo, endo and O- endo. If I look at each
of these structures, you will see that the
bond or rather the atom that is forming the
endo conformation is getting close or rather
in the same direction as the phosphate and
the base.
When you have the endo, pucker up to be cis
to the phosphate and the nitrogen. You have
an endo, if the takes up towards in the same
direction as the base and the phosphate, you
have the endo. If the oxygen picks up in the
same direction, you have an O endo. The endo
configuration is, when you have some carbon
atom or the oxygen atom pushed to a position
or pucker to a position that is in the same
direction as the base attachment and the phosphate
attachment. Only if these are in the same
positions would you have, what is called endo
conformations so you would have the opposite
if you have the exo, you can a have exo which
would mean that the carbon is away from the
phosphate and the nitrogen.
The exo means that you would have carbons
away from the phosphate and the base. If you
look at just away and trace the numbers here
or we just trace the carbon here, if you go
down this way then up this way and down this
way. You are basically tracing S in this way.
So you are going down this way because you
are endo going up coming down again. What
you are looking? You are looking at what is
called then S twist. The twist is such you
can just imagine the twisting of the wire.
If you just connect wires, where you have
one as oxygen then you connect these wires
by twisting it in such a way that you have
the S twist. You can twist in the opposite
direction that is going to give you what is
called N twist. If you just trace the atoms
here we will have in this case endo. You follow
the direction from this carbon up we have
this and this. What it is tracing? Something
like the N.
If you go in this way, the exo we would have
this up down again, up again tracing some
thing look like a N. This is why just two
name given to this called S twist and N twist.
This is how we represent the sugars. This
is important because now if you look at the
orientation of the phosphate and the nitrogen
there is a slide change in how the bases are
connected. How we are going to see? How this
is going to help later on in the overall structure?
Basically we are going to have a polymerization.
A polymer formation based on the orientations
of the sugar rings and their conformational
considerations you can position the base.
The reason why we need to position the base
is to form favorable interactions. So this
is what we have here a dinucleotide. Why do
I call this as a dinucleotide because I have
2 nucleotides? You recognize here that this
red, we know the oxygen of the sugar. This
is the nitrogen of the base. This is the other
sugar and what do we have here? We have this
again this that is linked to the phosphate.
This is the phosphate. Try and recognize this,
so does have the oxygen. Does it have the
oxygen? It does not have the oxygen, so what
is this? It is a deoxyribose. We are just
going to look at this for torsional configurations
because now we know that when we have a single
unit.
Let us consider a single unit, when we have.
This is one unit here, the top this is one
unit and this is another unit. We have linked
these units together we will see how they
linked together later on. First we have to
understand how actually the conformational
considerations are going to play an important
role? What we have learnt from the previous
slide is we can have puckering in this sugar
ring and also in this sugar ring. Due to the
puckering, what is going to happen? The positions
of this phosphate and position of the base
are going to change. This is the overall backbone
and we can have positional changes due to
the ring puckering.
We can also have positional changes due to
rotations about this bond and this bond. Remember
the angles of proteins, if we just consider
that we have the sugar the phosphate, we have
here single bonds. What do single bonds allow?
They allow free rotation. Because of this
rotation we can have again changes in the
position of the bases and the position of
the phosphate. What happens if I rotate about
this angle, the base is going to come up all
the way on the other side.
What is going to happen if I rotate about
this? This base can shift, so we have a rotation
and we have a rotation or rotation. We have
to give another reason why we are looking
at this is? If we look at the previous slide,
what happened here is? You had changes in
the orientation of the bases. Why, because
of the puckering due to the sugar. What is
that doing? That is changing the position
or the orientation of the base and also the
phosphate. But you can also have rotations
about these two angles. What is that going
to lead? That is also going to lead to changes
in the positioning of the bases. The changing
in the positioning of the bases is going to
help in the bond formation that we are going
to see later on.
We have to look at another angle. What we
are looking at here? What is this? Is this
a nucleoside or nucleotide? It is a nucleoside
why because it does not have the phosphate
attached to it. It has the base attached to
the sugar. You have to remember again we have
the single bond here so what is possible?
Rotation about the bond is possible. This
is a syn orientation, because in the purine
case we have the 6 membered and 5 membered
ring fused to one another. When you have the
syn orientation, you have the base and the
pentose that is this part of the sugar on
the same side. It is anti when it goes on
the other side. This rotation is also possible.
What are you getting? Since we have to study
nucleic acid structure and its components
we are looking at all the different structural
aspects are possible because you have the
single bond. These single bonds allow a rotation
and sugar ring for puckering.
All these put together is going to get this
into very-very flexible structure but the
structure in the sense is not flexible. We
will see later on. What we have? We can have
syn-adenosine and anti-adonsine where we have
the sugar ring and base away from the pentose
sugar. When we have the purine nucleosides
we have an anti-configuration. When this oxygen
is away from this part, we can have the syn
orientation. When there is rotation about
this and this oxygen comes about this part.
Usually the pyrimidine adopts this anti-configuration
because we obviously going to result in some
steric clash if the oxygen comes here. Again
we have phosphate attached to it so on and
so forth. We rather have an anti orientation
but since this allows rotation, it may be
possible that in some cases you might have
a syn orientation also.
Before we get into how these are formed, we
need to look at the structure of ATP once
more. What we look at, what is this now? As
soon as attached to the phosphate it becomes
nucleotide. Now that we have the nucleotide,
a position. This is the phosphate that forms
AMP. When we have the phosphorylation here,
it is ADP but when we have at the position
as well, it is ATP. So we have AMP, ADP and
ATP. We are going to see how we can actually
form these? We are synthesizing a nucleic
acid polymer. Essentially what we have is
something like this. This is our sugar. What
do we have attached to this position? A base.
What do we have? We have something else attached
here. If we have the phosphate then we have
the nucleotide, OH and this here. What is
this? Deoxy, now we have ATP.
ATP means, I have O here. What do I have here?
I have A basically here and OH. This is then
what? dATP, then I have O P O P O P. What
do I have? dATP, O O– so I have one part
here and ATP here. What is happening now?
Is this lone pair attacks this phosphate.
What we are going to form then? What is going
to happen? What I am going to have here? If
we have this A here, this oxygen is now linked
to this phosphate. This oxygen is now linked
here and we have this part released. So what
are all essentially done? I have linked this
with this. Where I have linked it at this
position?
What is this position? The position. The 3
is going to 5. What can this do? This has
now its position open. What can it do? It
can now attack another one with another base.
You can have GTP. What I am going to have
here? Then my base is going to be different.
I am actually linking or polymerizing into
forming a linear chain of the sugar phosphate
backbone. You recognize why this is now a
sugar phosphate backbone? Let us back to the
slides here.
What we are looking at? We have the hydroxyl
group here. What it is doing? Attacks this
phosphate of the triphosphate, releasing the
pyrophosphate with the cleavage of anhydride
bond. What do you have? What is this? It is
dinucleotide. This OH can do what? It can
go and attack another triphosphate that has
another base attached to it. It is going to
then form a linear chain of the nucleic acid
polymer. What do we have? We have a sugar,
phosphate, a sugar if we have another linkage
and another phosphate and so on and so forth.
We have the polymer made by linking nucleotides
by phosphodiester bonds. How are these formed?
Synthesized by the attack of the alcohol residue
from the ribose on the phosphate to release
diphosphate residue. What is the diphosphate?
We are talking about the pyrophosphate that
comes up after, this attack takes place. Then
what do we have? We have this linkage.
What are we looking? We are looking at a phosphate.
It was this OH that attack to CTP and you
linked T and C together. What you are essentially
doing? You have a phosphate. You have the
base attached by this glycosidic linkage to
the sugar. What is happening? This OH is free
to attack the triphosphate. What happens then
the formation of dinucleotide. This OH is
again free. What can that do? It goes and
attacks other triphosphate. In this case now
has attacked ATP. This is adenine. It has
attacked ATP. We have now a trinucleotide
which has the sequence T C A. Just like, we
did in proteins. What do we need to know?
We just need to the bases that are attached
because we need to know the type of sugar
that’s all the information. If we looked
at I am sure you have seen books or just the
DNA sequence.
When you look at the DNA sequence, what do
you see A C T G and so on and so forth. What
does it mean? It means that, the structure
is like this because you have the specific
sugar. You have it linked by phosphodiester
bond in the sugar phosphate back bone. The
difference lies in the types of bases that
are attacked or attached to the sugar. Just
like in the protein. Do we like the peptide
bond? We don’t, we just write V A G T what
ever. What does it mean? We have valine, alanine,
glycine and threonine but we know that they
are linked by peptide bonds. Is the same thing
here, we have the sugar phosphates. All the
information I need to know is just what are
the bases, because I know that. I need to
know the deoxyribose or ribose. The strand
has a direction that is referred to as to
. This then can attach another one and so
on so forth.
You can have the increasing length of the
nucleic acid based on this. So if you look
at this base families, all the information
you need to know is, what is attached to what.
If I say A G C T, you know what it means.
I have a linear polymer that is A T G C. What
do we know? First of all you know its DNA,
why do you know it is DNA because I have included
T instead of U. You know what sugar you have
to draw, you know what bases you have to draw
and you know that they are linked by the phosphate
backbone. Let me just go back to one structure,
last picture now. We know how does it looks
like? You understand that you have the phosphodiester
linkage so with the phosphodiester linkage
what can happen apart from the sugar puckering
which is going to change the orientation of
the base. We can also have rotation about
this, which is also going to change the directionality
of the bases. How does that help? What can
that do?
Once we have these bases we can have base
pairing. What is base pairing? Base pairing
is when we have say now that we have a sugar
phosphate backbone. What is that sugar phosphate
backbone mean? It means you have this and
you have your P. You have your base attached
here. Again what you are going to have? One
strand here and another strand here. You are
going to look at how does this happen later
on.
In base pairing, what you have is you have
specific hydrogen bonds interactions between
bases, between the two transfers. What is
this strand made up? This is just the sugar
phosphate backbone. This is the sugar phosphate
back bone and we have basically the bases
sticking out. When you see DNA is actually
drawn, it is just drawn with A G and so on
and so forth. Where, this is coming from?
This linked to the sugar that forms part of
the sugar phosphate backbone. What do we have
here in between? We have phosphodiester linkages,
sugar, the base attached to the sugar. This
is the strands of DNA. This is another strand
of DNA. What do we have? We have linkages
between the bases, these linkages and these
pairings is extremely important in the structure
of DNA. We have here what is called Watson
Crick base pairing.
Now if you notice, you have here guanine and
cytosine. What is guanine? Guanine is a purine,
cytosine is a pyrimidine. You have a link
between a purine and the pyrimidine. If we
look at the other base pairing, we have adenine.
What is adenine? It is a purine. What is thymine?
It is a pyrimidine. We have purine pyrimidine
pairing. In the paring you will notice which
is extremely important for the structure of
DNA. We have hydrogen bond formation. These
red thick lines are actually representation
of the hydrogen bonds. What we are talking
about? We are talking about an oxygen, hydrogen
and nitrogen, here is one hydrogen bond here
we have another hydrogen bond. We have three
hydrogen bonds in the pairing of G and C.
We have two hydrogen bonds in the pairing
of T and A.
If we look at the structure what is going
to happen when we have T at this position.
We are going to have linking of the T with
an A of the other strand. If we are looking
at C, we are going to have this link with
G of the other strand. If we have a on this
strand it is going to link with the T on the
other strand. If you look very carefully the
pairing as I mentioned is between the purine
and pyrimidine, there is one member of the
pair that has a fused 6 and 5 membered rings,
being part of the purine family. And we have
just 6 CN ring that is part of the pyrimidine
family. This is extremely important if we
look at the distance between these two. What
kind of pairing we are going to have?? We
can have A T. That is represented like that
or we can have G C. What is that means? It
means you have two hydrogen bonds here and
three hydrogen bonds here.
We have purine and pyrimidine base pairing.
We have another type of base pairing, the
one that I mentioned before is Watson Crick
base pairing. There is another type of base
pairing that is called Hoogsteen Base pairing.
What do you notice here? What is this? What
is that? This is it purine or a pyrimidine
purine. What is this? It is also purine. In
this case you not only have purine pyrimidine
base pairing but you can have purine purine
base pairing also.
We will see how this is not seen in the double
stranded DNA, because double stranded has
to form a uniform distance between the helices.
You have to have purine pyrimidine fit in
every case. When we have the two strands of
the DNA come together so we have one strand
this way and one strand this way. Then the
length between has to be the same. We have
one purine that is the 6 and 5 membered fused
together. We have the one pyrimidine that
is the 6 membered. Both of them coming together
gives the exact distance that is the distance
between the strands but in the Hoogsten base
pairing what happens you can have the base
between 2 purine. You also have pruine pyrimidine
base pairing.
Since this is also possible you do not see
in the double stranded DNA. There are some
cases this is observed. So far what we need
to know is the basic pairing between the purines
and the pyrimidines, that form the bases of
double stranded DNA, where we have three hydrogens
between G and C and two hydrogen between A
and T. If we look at basically adenine here,
we have this face. How many hydrogen bonds
are in the adenine 2 A T in the Watson Crick
face, here also in the Hoogsteen face that
is part of the 5 membered rings and the NH
of the NH2. If we go back and look at where
this is formed, you see this is a Hoogsteen
pairing.
This is between the 6 membered ring here and
the 6 membered ring here. In this case when
we look at the A, this is between what? 2
is 2 purines, when we look at normal Hoogsteen
base pairing the difference between the Watson
crick pairing is, let us look at the Watson
crick pairing, we have where is the hydrogen
bonding? It is all from the 6 membered ring.
In the Watson Crick base pairing, the base
paring that adenine forms with thymine is
from the 6 membered ring. But you have a fused
5 membered ring in Hoogsteen base pairing
so what did we have? We had this face that
was forming the hydrogen bonds in the Watson
Crick base pairing.
In the Hoogsteen base pairing we have 1 hydrogen
from the 6 membered rings and the nitrogen
taking part in the hydrogen bonding from the
5 membered ring. What we essentially have
is we have what is called Hoogsteen face where
it is the 5 membered ring nitrogen and the
NH of the 6 membered ring taking part in,
what it is taking part in? The hydrogen bonding.
The Watson Crick face it is only the 6 membered
rings that is taking part, this is the essential
difference in the pairing that occurs so when
we have a Hoogsteen. This is the normal purine
pyrimidine linkage A and U. U is found where
in RNA? So when we have an A and U linkage,
and you see that the linkage is between obviously,
we are going to have it between 2 electronegative
atoms but you see it between, What or what
is taking part that is what you have to look
at. What do you see taking part here? We see
the 6 membered and 5 membered ring taking
part.
What kind of base pairing is this? Hoogsteen
base pairing that is essentially what we have
to recognize. When we see Watson base pairing
in this case what would happen? The linkage
would be on this side because it would be
the 6 membered ring that would be involved
in the base pairing. So we have a Watson Crick
face, we have Hoogsteen face. By far it is
the Watson crick base pairing that is the
most important. Essentially what we have done
is we have looked at how we have the linkages
of the 2 base pairs. We have A T G C, two
hydrogen bonds and three hydrogen bonds. These
are coming from where? They are coming from
our sugar phosphate backbone that is essentially
what is happening. We have essential rotations
about the backbone. Where are these rotations
possible? We have the phosphodiester. If we
have the phosphate atom here, we have rotations
about this. We have the sugar ring somewhere
here. We have puckering about the sugar ring
now. What is going to happen? This puckering
is going to orient this G in a specific position;
this rotation is going to orient the overall
backbone in the specific position. What is
that is going to assistance that is going
to assist in orienting the bases in such a
manner that you can have the specific hydrogen
bonding possible.
Without this slide flexibility it would not
be possible to have the hydrogen bonding.
You understand that because you have to have
the nitrogen and oxygen in the specific orientation,
specific distance requirements for this to
occur. If you look at this result, we have
A and T. What is this base pairing? What base
pairing are we looking at here? Only the 6
membered ring involved, so it is Watson Crick,
if we look at the A and T pairing and the
G and C pair, you see the distance from the
C here. What is the C ? It is where it is
attached to the sugar the N glycosidic linkage
that is what C is. What is the distance 11
Å? When you have A, T, G and C it is also
the 11 Å, so you see how nature has sort
of designed it in such a manner that you would
have a purine and pyrimidine link together.
You would have a constant distance here that
would give you actually 11 Å that holds the
bases together. You have base pairing in such
a way that not only the distances but also
the hydrogen bonding is complementary and
also you have the flexibility possible that
makes it feasible for the hydrogen bonding.
We continue our discussion on nucleic acids.
What did we learn last time was how we have
these specific bases the purines and the pyrimidine’s
interact to form with double bonded, a hydrogen
bonded structures, how they form complementary
bases basically. What we have here is if we
look at nucleic acids, we know that they are
now comprised of this pentose sugar of phosphate
and a nitrogen containing base.
We know that this pentose sugar can be of
2 kinds either a deoxy kind or a ribose. Deoxy
ribose or a ribose depending on the type of
nucleic acids that you are considering. Obviously
we have these 2 types the deoxyribonucleic
acid where what is missing at the position.
The OH is missing at the prime position and
we have the sugar and the phosphate. The base
families are the purines and the pyrimidines
and what do they do they interact with hydrogen
bonding, a purine and a pyrimidine to form
a basic. Two basis coming together in hydrogen
bonded network. There is an additional factor
that has to be considered here that is a tautomerization
possibility of the bases.
If we look at the adenine consideration here,
what do you have here? You have a NH2 group.
What can happen to that NH2 group? You all
know about keto enol tautomerization. What
happens in keto enol tautomerization? What
happens there, you have a C=O and that is
converted to an OH from an adjacent HCH2.
You have a keto enol tautomerization. We are
having an amino type and an imino type.
The basic idea is the same where you are shifting
this hydrogen in the case of adenine to the
adjacent nitrogen. In the keto enol tautomerization,
what do you do? You have the H shifted from
the carbon to the oxygen, where you have a
keto enol tautomerization but what we are
talking about here is an amino and imino case
which is possible in adenine and cytosine.
It has 10.5 bases per turn. just like we had
a certain pitch of the helix. It is similar
to some thing like that. When we have the
A DNA, this is formed when B DNA is chemically
treated. Basically it doesn’t have those
water molecules in the water spine. That is
what A DNA is and it has 11 bases per turn.
The Z DNA as it is called is a left handed
helix with 12 bases per turn and it usually
plays a role in gene expression.
These are the 3 forms of the DNA and the most
common by far is the B DNA. These are some
of the features of the A DNA, B DNA and Z
DNA. We have a pitch. What is this pitch?
The distance covered by one rotation so the
A DNA pitch is 2.8 nano meters, the B DNA
is 3.4 nanometers and the 4.5 nanometers for
Z DNA. The base pair repeats are 11 bases
per turn, 10 bases per turn and 12 bases per
turn. The twist per base pair, you realize
that there is a slight twist. As you have
a base pair like you would also have angle
disposition for the helix. These are the twists
per base pair. We have slight base pair tilt
which is not very much in the B DNA just 40.
That is a slight base pair tilt. The 3DNAs,
the B DNA, the A DNA and the Z DNA and the
most common structure that we will be considering
is just the B DNA.
That has about 10 base pairs per turn 3.4
nanometers and base pair tilt of 40. The double
helix of DNA is actually well, it wouldn’t
be a secondary structure that is the structure
of DNA but there are other forms of DNA also.
Studies have shown that the native intact
form of DNA can be linear and circular. If
you look at the double helix here, if it goes
straight up and straight down we would have
a linear structure.What happens is if the
2 ends are covalently joined together so if
we chuck this up we are going to get a linear
form,
