mod08lec18

mod08lec18


our third lecture on nucleic acids is going
to speak about the stability of dna okay and what are the certain factors that are going
to lead to the denaturation of dna or the renaturation of dna and how that is extremely
important for designing of drugs for the cleaving of dna as i mentioned in the last class where
we want to consider that the protein synthesis is inhibited or stopped okay if we go back to look at the double helical
structure of dna we have our sugar phosphate backbone and we have the specific bases linked
together now apart from the hydrogen bonding present here we are going to have certain
other interactions that are going to be responsible for the double helix to maintain its structure
in solution now the forces that actually result in the
dna structure are electrostatic forces that are mainly repulsion forces now can you guess
why we would call them repulsive forces why were the electrostatic forces in the case
of dna or the structure of dna that i showed on the previous slide we call repulsive what
do we have in the backbone we have sugar phosphate what is the charge on these phosphates they
are all negative so the strands were tend to be as far away as possible right which
was going to result in an electrostatic force that is going to be basically a repulsion
we have hydrophobic forces we have hydrogen bonds obviously between the bases and also
possible between the other hetero atoms present in the bases with the water that is around
and we have stacking interactions between the bases between the bases at different levels now considering the properties of the watson
crick base pairing the base pairings are planer okay so in the resulting planarity that we
have between the g c and the a t pairs we are going to see a stacking interaction that
again can be disrupted with any agent that penetrates this region between the bases so
we have these four types of forces that actually determine the dna structure now the base pair are usually found in the
interior of the helix okay which is going to result in the stacking interactions between
the bases the charged and hydrophilic sugar phosphate backbone is on the exterior which
results in the electrostatic repulsion of the strands the phosphates from the two strands
would obviously be as far away as possible why because they are negatively charged and
they would reduce the electrostatic repulsion if they would be as far apart as possible some of this is neutralised by polyamines
why would they neutralised by polyamines or magnesium because there would be positively
charged they would counteract the negative charge of the phosphate and result in stabilisation
usually we have magnesium and polyamines that would result in the stability of dna so there
are certain forces that keep the dna structure together keep it stable and we are going to
learn now of what agents can actually disrupt the structure of dna the stacking interactions stabilise the double
helix as much as probably close to the base pair hydrogen bonds okay because what do we
have in those cases we have a pi pi interaction okay and we have a series of such pi pi interactions
right so the stacking interactions actually stabilise the double helix as much as the
base pair hydrogen bonds the stacking interactions are ot sufficient
to overcome the electrostatic repulsion of the phosphates okay so the phosphates will
remain as far apart as possible that would be stabilised by some positively charged divalent
ions or polyamines and the hydrogen bond between the purines and the pyrimidines guarantees
that there is going to be complementarity of the strands this is extremely important in dna because
you understand when we go from a double helical dna to forming the daughter dna then what
happen you have to have perfect complementarity it is not like rna where you can have a single
strand and you can have complementarity for some region of the rna but not for the whole
region so we have these stacking interactions we
have the electrostatic interactions and we have the hydrogen bonding interactions and
this is what wholes are dna together we also have this is one picture we saw previously
where we also have an additional hydrogen bonding between water that goes through the
minor groove of dna right so we have the major groove that is this large gap here and we
have the minor groove and the minor groove also twists around with the double helical
structure so we have these hydrogen bonds also form in addition to the other types of
bonds or the other types of interactions that are observed now dna usually is quite stable it actually
resists attack in acid and alkali solutions now can you tell me why it is more stable
compared to rna why is dna more stable compared to rna what are we differing in this right we have
our base and we have our phosphates we have continuation here so what do we have in rna
we have this in dna you are missing that what is linked here you have the other strand here
you have the phosphate here then you have the phosphate again another sugar another
base and so on and so forth so what happens here is this is susceptible
to hydrolysis but dna can resist hydrolysis which is why ribonuclease does not work on
dna it cleaves only rna in the mechanism that we learnt in our enzyme mechanism classes so dna actually is quite stable and in mild
acid solutions at ph 4 in this case we have a different kind of hydrolysis whereby the
purine bases themselves a hydrolysed okay since the oh is absent that the 2 prime position
unlike rna what happens in this case the beta glycosidic bond to the purine bases are hydrolysed
and we also have protonation of the purine bases now what is going to happen if you are going
to have protonation of the purine bases at the acidic solution you are going to disrupt
the hydrogen bonding right as a result of which you are going to have this hydrolysis
which leaves your purines protonated that is going to act as a good living room and
you are going to have an isomerisation of the depurinated sugar this is the hydrolysed sugar where you are
going to have what is this 5 prime and what is this signify what is this 5 prime and 3
prime end signify it signifies that this is a part of a dna strand right you do not have
the oh here so this is part of dna strand and this occurs under mild acidic conditions
where there is a protonation of the purine the purine is lost and what happens is this
sugar now that is on the left hand side which is depurinated isomerises to an open chain
form so what have you done in mild acidic conditions
you have not only depurinated your dna you have also opened up the sugar right so that
where basically be a problem what happens in rna this is rna now right
in rna what we have here is we have the case where we have the oh at the 2 prime position
the oh at the two prime position is going to give you what this is your oh at the 2
prime position what is going to happen in that case you are going to have easy hydrolysis
possible with an oh minus you can have hydrolysis at the oh position now happen when we consider the ribonuclease
mechanism histidine 12 and histidine 119 where important in doing what donating a proton
so one acted as an acid and one as a base and the roles were reversed in what was called
the hydrolysis step so what do we have in a sense we have transphosphorylation so this
is what is happening we have what is this we have a cyclic phosphate intermediate which
results in the phosphate being transferred you see other phosphate has been transferred
where was it originally here okay so you have either it go back to its original form or
but it is cleaved remember it is cleaved and so we have this part which is what which is
the rest of the chain and this part is the previous part of the chain right and we either
have this phosphate remain here or go over to the 2 prime position is that clear so in this case now this is not possible with
dna you understand that why is not possible simply because you do not have that oh there
so it is not possible but this is easier for the case of ribonuclease so when we look at
our rna we have is very unstable in alkali solution basically because of that 2 prime
oh that is present it results in hydrolysis the possibility of hydrolysis is always of
the phosphodiester backbone and this renders your rna susceptible to strand cleavage so what is going to happen is your rna strand
that was a single strand is going to easily cleave right because of the oh being present
there even under alkali conditions it is going to break up and we already considered the
enzymatic hydrolysis of rna where there are ribonucleases that are going to cleave rna
in a similar fashion but in this case it is going to be the histidines that are going
to be important in the cleavage mechanism whereas in just considering an alkali solution
it is going to your oh minus that is going to attack the 2 prime oh which it cannot do
in the case of dna because it is absent we looked at this representation and as i
mentioned before when we were looking at enzymatic cleavage we were talking about nucleases okay
we were talking about ribonuclease which is going to cleave ribonucleic acid we were talking
about deoxy ribonuclease which is going to cleave deoxy ribonucleic acid there are two types of nucleases there are
two types the two types are exonucleases and endonucleases exonucleases chop of nucleotides
from the ends okay so when you have a nucleotide and you are chopping it off from the end then
you have an exonucleases you have an endonucleases when you remove the internal phosphodiester
bonds that is something that is within the chain again you have two types there because
you have two bonds you can remove at the 3 prime position or
you can cleave at the 5 prime position if you act on the 3 prime hydroxyl group of a
nucleotide it is type a if it acts on the 5 prime hydroxyl group it is type b okay so the two types of nucleases are exonucleases
and endonucleases exonucleases are like how would you what is an analogous case for proteins
an exonuclease and analogous enzyme for proteins remember when we did the c terminal you had
a carboxypeptidase what did that do chop off from the carboxy terminal of the polypeptide
chain but if you have trypsin or chymotrypsin what did that do chop off in the middle it
cleaved in the middle okay so an analogy for exonuclease in the case
of a polypeptide chain would be that for carboxypeptidase the endonuclease is for two types okay we have the 5 prime end and the 3 prime end
the 5 prime we have it acting a type a acts on the 3 prime hydroxyl group so we have our
ag what is this part this is the 3 prime of this right this is the 5 prime of c are you
following we have a g c t in this sequence we have a 5 prime end for each we have a 3
prime end for each right this is the 3 prime end of the g but it is
the 5 prime let’s go back one what do we have here the 5 prime end and the 3 prime
end this is the 3 so this phosphate what is this this is the 3 of what of c this is the
3 of g so if you cleave here first of all you are in endonuclease and type a is that
clear right so when you are cleaving this is the 3 prime
end and you have to remember this is the 3 prime end this is the 3 prime end of g what
is this of c right so if you now look at the cleavages what do we have here if the cleavage
is here it is cleaving at the 5 prime end so it is type b if it is cleaving here where
the red dotted lines are it is cleaving at the 3 prime end it is type a for example the
phosphodiester is present in snake venom is a type a endonuclease it just chops off the
rna okay so these enzymes are actually used for cutting
dna and rna into manageable sizes and there used a lot in microbiology molecular biology
where you have these specific genes that are tailored to what you want to make which protein
you want to make you want to make a mutation in a polypeptide chain you know the genetic
code you know what amino acid you want and you know what basis you want for that amino
acid to be made so what you have to do is in your dna sequence
you have to change that set of bases okay this is routinely done and it is called site
directed mutagenesis you call it recombinant dna technology where you have the set of bases
that you can change to change what to change the protein that you are going to synthesise
so if you want of change a specific amino acid the rest of the chain is all the same
you cleave at a specific position of your original dna you change it to whatever you wanted to be
and then you have the protein expressed in bacteria and ones is expressed in bacterial
then you have the mutated protein okay because the dna what you do is you use the machinery
of the bacteria to make the protein for you that is what you are doing okay you have the dna you have changed a particular
sequence of the dna once you change that what is going to happen a different protein is
going to be formed why because it is going to go from dna to rna to protein so the message
that the messenger rna is going to get is going to be different than the original case
because you have already changed the base or a set of bases now when you use the bacteria
to make the protein for you the protein is going to be the changed protein okay a lot of routine studies are done in protein
chemistry to understand the effect of certain amino acids for example in ribonuclease you
know that histidine 12 is important for your activity you change histidine 12 to alanine
and then you check for the protein activity you will not get any activity why because
histidine 12 is crucial for the ribonucleolitic activity of the protein in this way you figure
out which amino acid residues are important in determining what the mechanism of the reaction
is basically okay there are these different types of basically
nucleases we have rattle snake venom here or snake venom these are mostly nucleases
okay this cleaves dna and rna that is why its snake venom in the first place it cleaves
exo(a) what is that mean the 3 prime end chopping off one nucleotide at a time and there is
no base specificity so it will just chop off you dna or you rna okay rendering any protein
synthesis impossible you have pancreatic ribonuclease a that is
on the 3 prime end it has a preference for pyrimidine it’s the type is endo it is an
endonuclease because it cleaves in the middle and it is a b type okay this is as much as
we are going to do about the endonucleases or the nucleases in general because we studied
the mechanism of ribonuclease in detail okay now we are going to study what interactions
of dna can be disrupted by disrupting the chains or by separating of the chains so what
we have is there are certain terminologies that are used here we have dsdna what is this
mean ssdna dsdna double stranded dna going to single stranded dna okay because this is what is going to suppose we
add such an agent that is going to disrupt all hydrogen bonds what is going to happen
the chains are going to separate so we are going to go from dsdna to ssdna okay so we
have the strands hydrogen bonded separated the process is that we rather the terminologies
that we are going to see are melting denaturation strand separation then the terms that describe
the changing from ssdna to dsdna what are we doing then we are reforming so it is called
annealing renaturation and sometimes hybridisation okay we even have a process that is called zippering
you just have one you start linking one set of bases and the rest zip sub by itself okay now how can dna be denatured dna can be denatured
under extreme conditions of temperature of ph what do we do we want to disrupt any of
the interactions that are responsible for stabilising dna as simple as that so we want to disrupt either the hydrogen
bonding or the hydrophobic interactions or the stacking interactions whatever intercalation
just to separate the strands denatured dna is less viscous than native double helical
dna and the bases exhibit greater uv absorption we will see that in a minute what we have here is denatured dna is less
viscous why would that be when we have normal dna we have a double stranded dna that is
solution would render the solution more viscous why because you have two strands that have
to be kept together always because of the stacking interactions the hydrogen bonding
or whatever forces are holding it together so the solution is going to more viscous once
you separate the strands what can happen is within the strands you can get some coiling
so what is going to happen is you solution of the dna is going to be easier for it to
flow so it makes it less viscous the bases exhibit greater uv absorption let’s
go the analogy of a protein you have a protein a tryptophan say that has where do you monitor
the uv of proteins at 280 nanometres you have tryptophan that is embedded in the centre
of the protein you unfold the protein the tryptophan can be seen right so your absorption
is going to increase the same thing here the absorption the uv absorption that you see
for the nucleotides is due to the bases so if the bases are always involved in an
interaction within themselves you cannot see as much but as soon as you open up the strands
what is going to happen you will have greater uv absorption so we have the basis exhibit
greater uv absorption and the dna is less viscous and this transition from double stranded
dna to single stranded dna is very commonly called a helix coil transition okay because
you are going from the helix to the coil so this is exactly what is happening you have a double helix you are disrupting
it dna denaturisation and you have a coil so this is helix to coil transition okay a
dsdna going to an ssdna this is an actual picture of you see how this
strand this is actually the double helix and how the double helix has opened up here where
the arrow is can you see that the double helix (28:05) there is a strand going down here
then it is single helix again and then slightly double helix not well all of it is double
helix it looks like a single thread here but it is opened up here okay so what has happened here it has denatured
okay it finds usually dna repairs itself okay if there is some problem it will form the
double helix back again itself okay now we can follow the denaturation of dsdna
by spectroscopy the bases have a maximal absorption at 260 nanometres not 280 like proteins in
double stranded dna the absorption is decreased due to the base stacking interactions because
you cannot see as much as you would see when they are single stranded when dna is denatured
these interactions are disrupted and you see an increase this is called as hyperchromic
effect why because it is more and the extent of the effect can also be monitored by a function
of temperature so let’s see what we get so this is native
dna the blue line at 25 degree centigrade where is the maximum 260 if i did the same
for proteins what would it look like i would have something that it comes down here and
it goes up here and i would get a maximum at 280 right for a protein when the dna is denatured you increase the
temperature so you have disrupted the dna interactions they are now single stranded
so you have rendered a helix coil transition what has happened to the absorption it has
increased and you now know why it increases okay so you have a relative absorbance of the dna
that increases on dna denaturation okay because there are stacking interactions there are
hydrogen bonding interactions in the double stranded dna that are not going to allow an
absorbance as high as it could be in the case of a single stranded state so we have denaturation now what happens in
some cases is you cannot reverse the situation its irreversible denaturation now what happens
is if say the temperature is rapidly decreased then the change in the viscosity of the absorption
that is absorbed cannot be fully reversed and the change occurs over a broader range
of temperatures we will see what that means sometimes what happens is because now if you
just look at say the strand like this now what are you intending if you separate the
strands out altogether then what is going to happen and if you result in a cleavage
of the strands to it is unlikely that they will coming together right so what happens
is with increase in temperature with certain agents also it is not possible for the renaturation
of dna the overall renaturation actually depends
upon the average length of the dna segments okay if the segments are small there is a
possibility that they are going to find the complementary base strands and join up to
form the double stranded helix then it also depends upon the concentration of dna okay
if the strands are too far apart then it is unlikely that they are going to find the partner
dna and also the complexity of the dna what do we mean by the complexity of the dna
we have to remember look for complementarity in the bases right so for looking for complementarity
in the bases it might not always be possible to find the same stretch of dna that is going
to act or form the double helix back together again okay so we have the overall rate of renaturation
determined by these specific factors the concentration of the dna the average length of the dna segments
and also the complexity of the dna now under certain conditions dna can be renatured
when we say renatured in this case what do you mean how is it different from the renaturation
of proteins when we denature a protein we are unfolding the polypeptide chain right
but when we are denaturing dna we are separating the strands right but for the protein the amino acids are still
linked to one another so when we remove the denaturing agent for example urea whatever
has been used usually temperated denaturation is not always not renaturable but suppose
we have urea and the solution when we have denatured the protein we have just prepared
the polypeptide chain back again we remove the denaturing substance what is going to
happen the protein will fold back in this case the dna has to find its complementary
strand right so under certain conditions dna can be renatured
where the complementary strand can be brought back together why because only then are you
going to get the proper double stranded linear double helix right or rather the double helix
structure because you have to have that ladder formation first where you are going to have
complementary base pairs and you have to have the correct hydrogen bonding pattern reproduced
again only then can you renature the dna back to where you started from now this occurs at a temperature called the
annealing temperature very efficiently that is tm minus 25 degree centigrade now what
the tm is it is called the melting temperature of dna the melting temperature of dna is referred
to as tm what are you monitoring here we are monitoring the melting of dna the melting
of dna is basically the separation of the strands helix coil transition that is going
to render the dna structure disrupted so we have to have a specific measure of how we
can do the what can we measure here tell me what can we measure here what did we use when we measure the denaturation
at for the temperature cases we measured the relative absorbance so i can measure the absorbance
and now what nanometre what wave length am i going to use 260 right this is going to
be my temperature scale i am going to monitor the absorbance with temperature as i increase
the temperature what is going to happen to my double stranded dna it is going to come
apart right as it comes apart what is going to happen to the absorbance it will increase
that is exactly what happens it will increase but will it keep on increasing
it will come to a point where all the bases are exposed so what will happen to the absorbance
then it will not increase anymore basically so we will get something like this now so
what sort of a dna do i have here a helix a dsdna and what do i have up here a single
stranded dna a coil so i have a transition the midpoint of this
curve is what is your dna okay so basically somewhere here would be the tm of this so
what you are doing this when you increase now what did we do in the previous one that
i showed you we had the dna curve you are monitoring it at different wave lengths right
and we found out that the maximum absorbance was at 260 nanometres right now this maximum absorbance at 260 nanometres
increases when we increase the temperature so what is now done is we are increasing the
temperature and i am monitoring the absorbance at 260 we know that the strands are being
separated so because why because the hydrogen bonds are being disrupted between the bases
you are disrupting the structure you have this increase the midpoint of this is the
tm what we are going to look at is factors that
are going to affect the tm okay now before we get into that we have the renaturation
that we were talking about so we have a tm minus 25 degree centigrade that is going to
be an efficient annealing temperature where the strands are brought back together okay
so whatever the tm is minus 25 degree centigrade it is kind of a rule of thumb that is the
annealing temperature now in the there are two steps in the renaturation
first there is a nucleation this nucleation is where the two strands find a region of
complementarity and they form a short double helix then there is zippering in either direction
from the paired region of complementarity the double helix is elongated and you understand
that only if it finds a correct set otherwise what will happen it will form a bulge right suppose what are we looking at here we have
two strands we have two strands a t c g we have other strands floating around but say
it is found this part that is complementarity so we have t what do we have here a g c so
this part it forms the double helix within this part and then it zips up in both directions
right so then it is going to form something like that once it zips up but if it so happens that this part is complementary
but his part is not then what are you going to have here you are going have some part
that looks like a proper double helix but some parts that sticking out like that right
so then you would have a bulge you would not have a proper renaturation okay it has to find the nucleation step is important
that is rate limiting in finding the two strands where you have the complementary region and
then it zips up in both directions and if it finds a right kind of base pairs obviously
it is going to be a very fast and it is a first order reaction and it is pretty rapid
okay now we have different melting temperatures
this is what we would call then a denaturation curve like the one i showed you here what do we have here a denaturation curve
why is it called a denaturation curve what do we have here we have helix here we have
coil here we have double strand here we have single strand so we have denatured it so this
is a denaturation curve and the midpoint of that will give me the tm now look at what i have three denaturation
curves here so i have three corresponding melting temperatures of three different dna
samples that have differences in their g plus c content what is a g plus c content a g c
base pair content the g c percentage has increased from left to right and this has resulted in
an increase in the tm why what happens in the g c pairing in the g c
pairing i have triple bonds i have double bonds in my a t pairing so the higher amount
of g c that i have it is going to be more difficult for me to (43:18) the strands so
the melting temperature is going to be higher okay so the g c content is important in determining
what the melting temperature is okay so the higher that we are going to see other factors
also of dna but the higher the g c content the higher the tm why because it has to disrupt
in that case three hydrogen bonds to separate the strands so we have to look at the influences on the
melting temperature the g c content now we know why okay the higher the g c percentage
the higher the tm because the g c base pairs have three hydrogen bonds and are thus stronger
than a t base pairs that have two hydrogen bonds what are the factors do you think might be
important you know what interaction stabilise the dna right we know what interactions stabilise
the dna so the basic idea in this case is going to be disrupting those interactions
right so what else is stabilising in the dna we have electrostatic repulsion right so salt
what is salt going to do the salt is going to shield the charges lessen the repulsion
between the phosphates and the higher the salt concentration lesser the repulsion than
higher the tm what is going to happen the higher the salt
concentration suppose you have a large amount of magnesium what is going to happen the ions
are going to shield the phosphate charges on the backbone this backbone is going to
remain as far apart as possible right because you have the phosphate now if you have salt ions that are going to
stabilise this what is going to happen it will not be very easy for you to separate
the chains because this chains if say you have for the negative charge here what is
going to happen there is going to be repulsion between the chains right repulsion between the strands is what is going
to take it a part because it is essentially what you are doing is you are going you are
resulting in a helix coil transition that is what you are looking at if you are looking
at a helix coil transition what you want is you want destabilisation of your strands right now you want this repulsion to be minimum
what is going to happen to the repulsion here if you add salt if you add mg 2+ the higher
the salt concentration the higher the tm right because you are stabilising the strands when
you have ph what is going to happen when you have different ph at low ph what are you doing
you protonate the bases as you protonate the bases what is going to happen you are disrupting
the hydrogen bonding right again the same thing when you increase the
ph you are disrupting the hydrogen bonding how is that going to effect the tm it will
decrease the tm because you are resulting in a larger disruption all you have to think
about is what effect each of these is going to have on the forces that are stabilising
the dna right then we have organic compounds organic compounds
can act in different ways one thing is they can intercalate between the bases and what
are they going to do they are going to disrupt then in that case the stacking interactions
they can also form more favourable hydrogen bonds with the bases and in that case what
are they going to do again disrupt the stability of the dna disrupt the hydrogen bonding either
the electrostatic repulsion or the hydrogen bonding or the stacking interaction whichever type of interaction is disrupted
will result in the separation of the chains okay so what we learnt was we learnt how the
overall structure of dna and rna is formed by the different nucleotide basically coming
together we have the sugar phosphate backbone in both cases just the sugars being different
in the deoxy type and deoxy ribose type and the ribose type the bases are the purines
and the pyrimidines for both cases we just have a change from thymine to uracil when
we go from dna to rna then we looked at the hydrogen bonding patterns
in each cases where we found that we had two hydrogen bonds between a and t and three hydrogen
bonds between g and c and we saw how the stability of dna can be disrupted by certain factors
and what the melting temperature was and what effect the g c content and the a t content
had on the overall dna stability and other factors like we have temperature also that
we have to look at we have the g c a t content that we have look
at we have the ph we have the addition of ions and the addition of organic solvents
this completes our discussion in nucleic acid we will begin bioenergetics in our next class
thank you

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