Molecular

Question 1

The Polynucleotide

Answer 1

The DNA is a polymer(polynucleotide), formed by monomerd called nucleotides, made of 3 parts.

1. a base, that has an heterocyclic structure(carbon + nitrogen ring)
2. a pentose sugar( carbon from here connected to the base ring) 4 carbon + O on the ring and outised 1 carbon bindinf the phosphate.
   - Ribose has OH (RNA)
   - Dexoxyribose has H (DNA)
3. a phosphate

2 types of bases:
-Purines: Adenine and Guanine, double heterocyclic ring. The difference between the two are the side atoms placement.
Adenine: NH2 to the top
Guanine: O to top

-Pyrimidines: 1 heterocyclic ring, Uracil, Thymine and Cytosine
Uracil:
Cytosine: NH2 to the top of the chain
Thymine: CH3 group to the top

Question 2

Polynucleotide Chain

Answer 2

sequence of nucleotides, bonded together by phosphodiester bonds(Ester bond: bond betweed and alcohol and an acid group) : Ester bond+ phosphate.

Bonds happens at the 5’ end that has free phosphate and the 3’ end where is the the free OH.

Longer chain do not equal complex orgnanism.
our DNA has 3,2 billion nucleotides

Question 3

DNA bonding

Answer 3

DNA is the double strand polynucleotide chain.
The bonding happens with 2 polynucleotide chains connecting in reverse(antiparallel), one in 3’ to 5’ the other is 5’ to 3’ direction.
The polynucleotide chains bond together because of the H bonds between the bases. (between the Nitrogen or Oxygen of the other base)
Those weak bonds are made between a Purine and a Pyrimidine.
A + T = 2 H bonds
G + C = 3 H bonds

DNA has a double helix structure , the sugars and phosphate are more hydrophilic that the bases, so when it forms, sugars and phosphates are more exposed on the outside.
clock wise: right handed
anti-clock wise: left handed

Question 4

Type of DNA

Answer 4

There are 3 types of DNA

1. A-DNA: wider spiral with a shallow and minor groove and a deeper and narrower groove
2. B-DNA: most organized and common, helical axis in a right handed spiral
3. Z-DNA: most unorganized one, strands where basese where chemically modified , thinner, and left handed spiral

An inportant factor for distinguishinf the types of the DNA is the BP, base per turn, number of nucleotides pair per turn

• A type is founded in double strand RNA
• Z DNA: is present in regions odìf our bodies, Where we have only polypurines (strand only of purines) and on the other we have pyrimidines

if we apply heat or put DNA in an alkaline solution (high PH) to have a single strand, we have denaturation.

Renaturation is the inverse, we go back to 2 strands

Question 5

Denaturation

Answer 5

divide DNA into 2 single strands, we obtain a single stranded denatured state, we can go back to the previus state, so reform the double helix, the renatured state.
To perfom denaturation we need to provide to the system an amount of energy that is higher than the energy of the H bonds, for break them via:
physical methods: heating /cooling(renaturation)
chemical: raising the PH so the nucleotide will be modified and cannot form H bonds, we cannot have renturation

Question 6

LIGHT ASSORBENCE OF DNA

Answer 6

DNA is able to absorb light, through mesuring it’s light absorbance we could know:

1. Concentration
2. Single or double strand status
3. Complexity

The smaller the wavelight the higher the energy associated with it.

Question 7

LIGHT ABSORBANCE IN SOLUTIONS AND DNA

Answer 7

For mesure light absorbance we need the spectrophometer, that is ocmposed by:
1. a light source
2. monochromator(refracting prism)
3. Specific device with a slit that allow only a specific componet of light to pass through
4- test tube: contain the solution with the element that we have to analys, to define if the components of light will pass(if liht is absorbed or not)

After the tube we have:
-Photoelectric tube that collects energy when the light is not absorbed by the solutions, and activate a circuit of electric current, that will be detected by galvanometer.

at the end the spechtromer will tell us if the wavelenght has been absorbed.

if we put DNA, we will see that a wavelenght of 200nm it will be absorbed. The peak of absoption is at 260, because DNA its able to absorb only UV light.

There is a direct relatioship between the absorbance and the concentration(higher the absorbance, higher the concentration)

it can tell us if the DNA is a single or double strand: because UV light is absorbed by the electrons that run around the bases, n electrons present on the double bonds of the basis expecially. energy moves electrons to a basal state to an exiceted state. , vibration of molecules to dissipate and eliminate energy.

On double strands of DNA , vibrations are inhibits because the H bonds between the basese stabilize the structre.

single strand of DNA vibrate freely

Single strand absorb more energy that the double DNA strand

So we put DNA in the test tube, and see the amount of absorbition
we heat it then
if the absorbtion increase that the one before was a double strand, if not, it’s a single strand.

same thing but cooling the DNA to see if it was a single strand.

cooperative functioning: the strenght of DNA is given by the cooperation of all the base pairs which have a strenght together. so to break the basepairs, at the beginninh we need more energy because bases stay all together closed, but once that we’ve broken the first 3/4 it become easier

Question 8

THE MELTING TEMPERATURE (TM)

Answer 8

it refere to the temperature at whivh 50% of the DNA has been denaturate

1. The greater the DNA lenght, the higer the melting point.
2. More G and C bases it will have an higher melting point.
3. the power of H bonds is greater than the negative chatrges repulsion

The higher the saline concentration of DNA solution is, the higher the ,melting point.

Question 9

Complexity of DNA

Answer 9

consider a solution at high temperature with a DNA comletely dissociated that has to be reassociated by cooling it in order to obtain a renatured DNA

reassociation od DNA and Cot analysy: denatured DNA and its reassociation to oduble strand status is used to study DNA complexity.
Cot is based on DNA reassociation Kinetics, that mesure how much repetitive DNA in present in the DNA.

CoT is a value that depends on concentration (co) and time(T) and mesure the time that a single stranded DNA needs to return in a double strand status, and depends on the lenght of the DNA molecule; the longer the DNA the longer the amount of time to reassociate it.

The curve of Eukaryotic DNA reassociation is complex because it has repetitive regions.
it will have 3 types of components
1. highly repeated DNA; hundres lenghts repeted thousand of times,fast components with high velocity of association
2. moderly repeated DNA, size of thousand, repeted houndred of times, intermediate componetes, intermeduate velocity of association
3, no repeated DNA, slow component, lot of time to be reassociated

Question 10

➢ In a nutshell, what is meant by the complexity of eukaryotic DNA?

Answer 10

It means that eukaryotic DNA has 3 different regions: non repetitive regions, moderately repetitive regions and highly repetitive regions.

Question 11

➢ How do you know if a DNA is complex or not?

Answer 11

You must do the C0t analysis with its graph of reassociation. If, in the graph, there is only one time of reassociation and a single curve, we’ll have
a not complex DNA molecule, if we have more than one time of reassociation, the DNA will be complex.

Question 12

Why highly repetitive DNA reassociate faster than the other non-repetitive sequences?

Answer 12

Because its easier. If we have a DNA molecule with non-repetitive sequences (many different nucleotides on one strand), during the process of reassociation, the other strand must find the corresponding complementary nucleotides. On the other hand, in highly repetitive DNA, during the reassociation, it will be easier the combination of complementary nucleotides because the sequence is always the same (because of the availability of numerous complementary sequences).

Question 13

RNA

Answer 13

Also RNA has double strand helices
However, there is a substantial difference between

DNA and RNA structure:
the two RNA strands derive from the same molecule, indeed it is an intramolecular (intra=within) double strand, while
DNA has two strands between two molecules and so it is an intermolecular (inter=between) double strand. (There are exceptions).

In RNA there are regions that are in double strand and regions that are in single strand.
So it is important to remember that the molecules are not built in a simple way.

Question 14

DNA REPLICATION/DUPLICATION

Answer 14

DNA replication is one of the most important events of life.
An overview of DNA replication:
• During cell division in eukaryotic cells, the replicated genetic material is divided equally between two daughter cells.
• It is important that each cell gets an exact copy of the parent cell’s DNA.
• In replicating, the DNA double helix unwinds (H-bonds between the strands are broken) and each single strand acts as a template for a new strand.

But how does the DNA replicate?

Possible mechanism of DNA replication
At the beginning of the last century, there were three different hypotheses about the mechanism of replication:

1) The semiconservative hypothesis proposed that the two strands of a DNA molecule separate during replication. Each strand then acts as a template for synthesis of a new strand. Both daughter cells will have a DNA that contains an original strand and a new synthesized one.
2) The conservative hypothesis proposed that the entire DNA molecule acted as a template for the synthesis of an entirely new one. So, one daughter cell receives both original DNA strands, while the other one will have a completely new DNA molecule.
3) the random dispersive hypothesis proposed that DNA replication results in two DNA molecules that are mixtures, or “hybrids,” of parental and daughter DNA. In this model, each individual strand is a patchwork of original and new DNA.

The experiment done by Meselson and Stahl demonstrated that DNA replicated semi-conservatively, meaning that each strand in a DNA molecule serves as a template for synthesis of a new, complementary strand.

Question 15

Meselson-Stahl experiment

Answer 15

Meselson-Stahl experiments:
In lab Meselson and Stahl:
- cultivate bacteria such as Escherichia coli in a flask. In order to grow a bacteria cultivation, it’s fundamental to give it the necessary nutrients to live (glucose, vitamins, amino acids, nucleotides etc). In nucleotides there are bases that normally contain 14N (the most common isotope of nitrogen).

• Extract the DNA from these cells and put it in another test tube containing a caesium chloride solution (a salt).
• Centrifuge the solution: during the centrifugation, the DNA will move in the test tube and reach the region of the solution that has a density equivalent to the DNA density

Meselson and Stahl demonstrated in a definitive way that the semi-conservative hypothesis
is the right one.
We previously talked about how to grow bacteria in a lab, like Escherichia. coli (E. coli), which
is a very used bacteria in labs.

When this growth occurs in a flask, we have to add nutrients
to the bacteria, such as sugars, amino acids, lipids, etc. as well as nucleotides in order to help
them prepare the DNA. Within the nucleotide we have bases, which are made of heterocyclic
rings within which we find nitrogen.

The most common isotope form of nitrogen present in the
molecules we have in our body is the isotope 14. (¹⁴N)

Remember: N is not only present in DNA, but also in amino acids, that have COH and NH3⁺
residue
We add the nutrients, we let the cell grow and then with
some molecular biology procedure, we extract the DNA
from the cell.
Then we put the DNA in a test tube on top of
a solution of caesium chloride.
You then start to centrifuge,
where there is rotation of the test tube and there will be the
so called centrifuge force that will push the molecules
towards the bottom.
During the centrifugation, the caesium chloride solution
will form a gradient: this means that this solution will be
more dense and more concentrated at the bottom, and less
dense at the top, so less concentrated. The salt molecule
of caesium chloride will move from the top to the bottom
so you have a different gradient of concentration going
from less to most dense.
While the caesium chloride is
performing the gradient, the DNA molecule that was isolated
will move to the region of the gradient density which is
equal to its own density.

First of all, they did a control experiment, allowing E. coli to grow for many
generations in ¹⁴N medium then they extract the DNA from the cell, centrifuge
the cell and observed where the DNA was.
Then they did another experiment where instead of using ¹⁴N, they used ¹⁵N which
is a heavier isotope. They grew the bacteria in this ¹⁵N medium, during many
generations, allowing it to duplicate many times.
They extract the DNA and then
centrifuge the DNA, and since the DNA included ¹⁵N and not ¹⁴N, the DNA was
more dense and so was found more towards the bottom of the test tube.

Then they did the two most important experiments. First of all, you have to know that the time
needed to wait to get a new generation by duplication is around 30 minutes.
This means if I
have 1000 bacteria in a flask, and wait 30 minutes in optimal growing conditions, I will have a
doubling of the bacteria giving 2000 bacteria, as every cell will divide to give 2 daughter cells,
each containing their own DNA.

They grew the bacteria for many generations, in ¹⁵N. Then, they transferred this bacteria to
¹⁴N.
The cells replicate once, to produce the first generation of daughter cells, having to use
the ¹⁴N in the DNA, having used ¹⁵N in the previous medium.
If the semi-conservative
hypothesis is true, I will have 2 daughter cells that each contain a DNA molecule: each
molecule will have one strand containg ¹⁵N and the other strand ¹⁴N. We would expect that
after DNA extraction and centriguation, we would have a band in an intermediate level due to
the presence of both isotopes.
After this first generation, if the semi-conservative hypothesis
was not true, instead of having these intermediate bands, and the conservative is true, what
would appear? If this were true, we would have one DNA molecule made of 2 ¹⁴N strands, and
another made of 2 ¹⁵N strands.
This was not the case.
In order to be sure, they continued the experiment and maintained the bacteria
for another generation in ¹⁴N.
Remember now, the starting material is the one
seen on the right of the picture seen above.
Each of these strands will be a
template for the new strands.
After centrifuging the second generation that will
have grown in ¹⁴N, you obtain what can be seen in the image on the right.

Question 16

DNA replication machinery

Answer 16

replication= formation of new strand
we open the double strands and each strand will be a template for a new one.
things needed:
1.DNA polymerase in an ezyme needed for polymerase the DNA, it need a DNA template and deozynucleotides.
2. Primer: RNA fragment that is the starting point fot the formation of new DNA. it gives to the DNA polymerase the indication on where to start.

The formation of the new strand form the Comlementary rule( A +T, C+G).
DNA polymerase catalyze the reaction:
1. reads the nucleotides on the template
2. adds complementary nucleotides. The added one is bound to the previous one by a phosphodiester bond between 5’P of the the new one and the 3’OH of the other.
The new one has 3 P, DNA polymerase use only 1 and delete the other 2.

Polymerisation runs from 5 to 3, if the growing DNA stands goes from 5 to 3the template must be 3-5 because the DNA strands are antiparallel and complementary.

The energy necessary for the formation of the phosphodiester bond it’s given by the DNA polymerase,:
- the new nucleotide, arrives with a nucleoside triphosphate with PPP, having energy for bonding. DNA polymerase breaks the bond between the first and the second P, producing energy that is used for the formation of the phosphodiester bond.

DNA polymerase starts by adding the new nuclotide to the primer, and will have an 3OH, at which the DNA polymerase will start the polymerisation.

The primer is produced by the enzyme PRIMASE, that is an RNA polymerase, that adds nucleotides that are complementary to a template strand. RNA polymerase need a DNA template and will polymerase to 5’ to 3’. RNA polymerase DO NOT use deoxyribonucelotides but ribonucelotides.
DNA polymerase need a primer, RNA polymerase do not.

Steps for DNA duplication:
1.unwinded of the double helix of DNA
2, each helix is the template for a new one
3. there is one or more replication origins
4. replication goes in 5’-3’
5. It goes in both ways, it’s bidirectional
6. Besides DNA polymerase, there are alsp, DNA helicase, topoisomerase, single-strand binding proteins, primase, ligase, telomerase.
- DNA polymerase need a primer to polymerize.
-DNA polymerase need deocynuclotides for DNA replication

1. DNA unwinding occur at specific DNA region called replication origins (in bacteria is only OriC), we need to break the H bonds, the OriC region is full of AT, that are easier to break, For each reaction on DNA or RNA, we have 3 protein that cooperate:
2. recognize which is the region on the DNA where the reaction need to occur. This proteins binds this region and recruit another protein
3. This new protein bind to duplicate the DNA so DNA polymerase is needed.There is a protein that binds to the OriC region and recruits DNA polymerase
4. there is a third proteins that help the enzyme do the work
5. Initiator protein that binds the OriC
6. helicase: that open the DNA
7. SSB: single strands binding protein: that recognize and binds the single strand avoinding the re-binding of the two

In eukaryiotic cells, there are hundred of replications origins, that have some regions that are recognised by a protein called DnaA.

1. DnaA recognized the repeated sequences(replication origin)
2. recruits DnaB(helicase) and DnaC.
3. DnaB + ATP unwind the DNA.
4. DNA polymerase duplicate the DNA in bi directional growth of both strands.

Thw origin or replication region hase 2 regions called GROWING FORKS, polymerisation runs in one way and in the other. Both DNA will be used as a template, one from left to right and the other to right to left so we have an opnening, REPLICATION EYE OR REPLICATION BUBBLE

topoisoerase prevent the formation of nodes in DNA, so eliminates supercoling,

LAGGING STRANDS: old strand that appears to be getting polymerised in 3-5, during the polymerrization there are difficulties and in this strand we have the formation of the long frangment.

Question 17

Fork Growth e strands formation

Answer 17

for every fork growth we have a
1. Leading strand: that is synthesized quickly and in long fragments, in 5-3 direction as the fork.

2. lagging strand: it’s slower and has discontinous frangments, called OKAZAKI FRANGMENTS, it’s belived that it 3-5

The reason why the DNA polymerase is capable of synthesizing the leading strand “properly” and the lagging strand in a more “defective” way, is because the polymerase will ONLY work in 5’ to 3’.
this happens:
1. lagging strand turn 180°, the polymerase 5-3 will polymeraze both strands simultaneously.
2.lagging strand growing is fragmented, because as the polymerase ytavel across the flipped one, a loop will increase ans it will not allow the polymerase passage anymore.
3. Polymerase stops, eliminate the loop and resume.
4. by stopping it creates a fragment of DNA previousy synthesized = OKAZAKI FRAGMENT.
5. at this point, the Primers at the end of the Okazaki are removed by Exonuclease, and the fragments are bound together by the ligase.

6. Nuclease: remove nucelotides:
7. one that goes in the same direction as dna polymerase 5-3
8. Exonuclease: remove from 3-5
9. DNA polymerase III elongates the RNA primers by adding more RNA nucleotides
10. DNA polymerase I removes 5’ RNA at the end of the frangments and fills the gap with DNA nucleotides. It hase both exonuclease and polymerase activity.

So, DNA polymerase I firstly performs its nuclease function to remove the ribonucleotides found on the fragments and then “fills the gaps” by adding deoxyribonucleotides to replace them.
Because we still

9. we must join the 3’ of each fragment to 5 ends of the next one. To do it it need ATP and ligase.

In bacteria:DNA polymerase III add new nucleotides following the synthesis of primers, then DNA polymerase I bind the 3’ end of each Okazaki fragment, removing the primer and replacing RNA with DNA, Then ligase links these fragments. Primers can also be removed by other enzymes, called nuclease.

In prokariotes: DNA polymerase I, II, III
Eukaryotic: 12 DNA polymerase.

In Eukaryotic cellls: removal of primers is done by RNAase H. RNAase H are nuclease, they degrade nucleotides, in this case, ribonucleotide.
The H in its name stands for Hybrid, because it’s an hybrif of DNA or RNA, this enzyme may recognize the hybrid and degrade it.

DNA polymerase alfa: synthesis of nuclear DNA. conplex with primase, begins synthesis at 3’ of an RNA primer for both strands

DNA polymerase gamma: mithocondrial DNA synthesis

DNA Polymerase delta: leading strands synthesis e proofreading
DNA Polymerase epsilo; lagging synthesis e proofreading.

Question 18

proofreading and editing

Answer 18

DNA and RNA polymerase can make mistakes during the replication stage.
DNA polymerase can wrongly pair two non complementary nucleotides, creating a mutation.
To prevent this we have proofreading and editing

-Proofreading exonuclease domain is intrinsic to most DNA polymerases, it allow it to check each nucelotide during DNA synthesis and excise mimathched nucelotide in 3-5 (opposite of DNA polymerase direction)

Question 19

Tautomerism

Answer 19

some compounds are capable of interconversion, and it’s the main reason why DNA polymerase makes mistake.

nucelotides are usually found in keto form or amino form.

the Keto can switch to enol, and amino to imino, called rare forms, that is not recognised by DNA polymerase.

Question 20

Replicating the end of DNA molecules

Answer 20

• limitations of DNA polymerase create problems for the linear DNA of eukaryotic chromosome
• The usal replication machinery provides no way to complete the 5’ ends, so repeated rounds of replication produce shorter DNA molecules with uneven ends
• Not a problem for prokariotes,that have circular chromosome

1. Replication fork moving to the end of the parental DNA
2. RNA primer at the beginning of each Okazaki fragment.
3. primers are removed, and replaced with DNA where 3’ is avaible

At the end of the first round of replication, the primers are removed and the fragments are shortened. With other rounds of replications occurring, the fragments

DNA will countinously shorten because of this mechanism, can have negative ripercursion.
To prevent any negative thing, we have Telomerase that is an enzyme that add DNA, they postpone the erosion of genes near the end of DNA molecules.
this lets the duplication of DNA to occur without its shortening.

Question 21

Telomeres and Telomerase

Answer 21

Telomerase: enzyme which name derives from the ending parts of chromosomes that are Telomeres, that protect the chromosomes from degradation.

They have repetitive non coding nucelotide sequences, that delay the erosion of genes near the ends of DNA molecules, the shorteningof telomerase is connected to aging.

For certain replication the telomeres are not removed because they are preserved by Telomerase, but only if the enzyme is present in the cell during duplication. after each reolication it will be shorteed by 1.

Telomerase function: To prevent the shorteneing, they uses RNA template called TERC to lengthen the repeat units.
1. Telomerase binds to DNA and beging polymerization of the 3’, so it shyntetize another REPEAT, that will be added to the DNA end that need elongation.

After the work of telomerase, we will have 4 repeat units, so that after the replication, the original number of 3 repeat units will be maintained

Question 22

what is a gene

Answer 22

sequence of DNA that is trascribed into RNA.
The RNA that contains the instructions to produce proteins is called “Coding RNA” and the RNA that does not have informations is called “non-coding RNA”

Non coding RNA: 95%
coding RNA: 2/5 %

Question 23

RNA

Answer 23

coding: 2/5%
non-coding: 95%

different types of RNA:
CODING:
-mRNA: messenger, participate in protein synthesis.

NON CODING:

• rRNA: ribosomal
• tRNA: transfer, interface between mRNA and amino acids.
• snRNA: small nuclear RNA, RNA that form the spliceosome(eukaryotic cells)
• snoRNA: small nucleolar RNA, found in nucleolus, involved in modification of rRNA(eukaryotic cells)
• miRNA: microRNA, smallRNA involved in regulation and expression mechanism
• siRNA: small interfering RNA, active molecules in RNA interference.

Question 24

mRNA

Answer 24

Eukatyotic one has 3 different regions:

• coding region : translated into protein
• untranslated region: can be
  1. before coding region: 5’ (UTR) untranslated region it starts with a cap of 7-methylguanosine, that do not have the typical phospate bon, but has 3 P group for link the nucelotides. The cap protect the mRNA from degeneration. it permits the mRNA to go from the nucleus to the cytoplasm and ribosomes. In order to have ribosomes to start their work of proteins synthesis thy need the cap.

2. After: 3’UTR: we have a poly A tail. it protect the mRNA amd help ribosomes with protein synthesis.

Question 25

rRNA

Answer 25

protein: ribosome must link the specific order of amino acids given by mRNA.

rRNA:

• small subunit: read the order the order
• large subunits: link subunits together

prokaryotes :

1. smaller subuinit in prokaryotes is made by 1500 nucelotides with 16S(svedberg coefficient= density indicator). The smaller + ribosomal proteins = 30S
2. larger subunits: have two RNA molecules, one 3000, 23S and the other 120, 5S. larger + proteins= 50S

Eukaryotic: 
ribosome made of 40S and 60S  subunits. 
two short RNA molecules, 5S and 5,8S
two longer RNA, 18S and 28S
ribosome has a sedimentation of 80S

have DNA also in mithocondria and chloroplast.
ribosomes associated with endoplasmatic reticulum.

In the bacteria:

• two subunits
• 5,.8 not present
• 28 S and 18 S are there but smaller as 16S and 23S

Question 26

tRNA

Answer 26

unique structure: 3 stem(double strand) and loop (single strand) regions.

• Dihydrouridine UH2 in the D loop.
• pseudouridines are present in the T£C loop.

They can form hydrogen bonds with mRNA.
They can also form ester linkages with amino acids and can bring mRNA and amino acids together during the process of translation.
they pair with tRNA in a complementary and antiparallel way.
tRNA can base pair with a stretch of 3 nucleotides of mRNA.
mRNA= codons
tRNA= anticodons loop

Question 27

Prokaryotic Transcription

Answer 27

Transcriotion: from DNA to RNA, one of the two strands of the DNA is the templates

1. Every gene has a promoter, when it’s supposed to start, RNA polymerase binds to start the trascription.
2. core enzyme binds with another subunit, a sigma factor, to form holoenzyme, DNA will properly open and the sigma factor will be lost as the RNA chain is elongated.
3. the sigma factor must bind to the promoter region before the starting point. The region in between -35 and .10 in each gene, is the called SIGMA FACTOR STICK REGION.
   At 10. the promoter nucleotides are similar: TATA BOX
   at 35. the region is TT.

In bacteria the RNA polymerase has also an helicase activity so it does not need an helicase to open the double helix of DNA.
In this case we will have the trascription bubble and not the eye.

For each temlate strands there is a complementary RNA strand.

DNA polymerase can correct the mistakes. RNA polymerase cannot and we will have more mistakes, because istakes there are less dangerous. RNA polymerase transcribe the DNA strand

5. How does it stop?
   there is a region of RNA that is transcribesd, this region will self-assemble, thanks to complementarity, formin stem and loop region, that will stop RNA polymerase(steric hindrance) In some cases only this regions is needed for the detachement of DNA from the new RNA strand.

In other cases stem and loop regions are not enough, so ther is the RHO protein that is attached to the stem and loop detaches the DNA from RNA polymerase.
The termination of tge trascription is based primarly on the stem and loop and rho protein, so there are two terminations.
1. Rho dependent
2. Rho independent

Question 28

Eukaryotic Transcription

Answer 28

We have different types of RNA and also RNA polymerase I, II.III
phases:
1. Initiation
2.Elongation
3.Termination
4. Maturation (capping, splicing, poly adenylation)

RNA Polymerase II: transcribe the mRNA and microRNA.
we have a gene a promoter, the polymerase goes to the promoter, recognizes the specific region on the promoter, opens the gene and the RNA is transcribed in direction 5 to 3.
in eukaryotic soon after the trascription the RNAs transcribbed are not ready to work , there is a process called maturation that is fundamental to produce a functional mature RNA. Without maturation in a eukarytic cell the RNA cannot work.
Also ther is this case there is a promoter, in bacteria it was the TATA box, in -10 nucleotides, before the first nucleotide transcribed. it’s also possible to have TATA BOX at 26.

initiation region: binding with, The RNA polymerase II,, close to the starting point or the BRE.
not all genes have the same promoter, itis possible to have different promoters like TATA or DPE. In ,many cases there is the TATA BOX, but also called TATA less.

1. INITIATION: RNA polymerase in eukaryotic cells does not work alone, there are some other proteins helping to do their job

Trascription Factors
RNA polymerase need trascription factors (in bacterdia RNA polymerase and holoenzyme directly bind DNA)

1.TFIID is the first factor attached to the TATA,
2. TFIIB and TFIIA lock strengthining the binding of TFIID on promoter.
Only after Tfs RNA polymerase II arrives.
3. the last factor TFIIH is important: it has helicase activity that opnes the DNA, moreover it activates the RNA polymerase with the kinase activity adding some phosphate group on some amino acids of the RNA P II, the phosphorylation indicates and activates the RNAPII and the trascription starts.

SUmmary:
1- TFIID: first one, attach to TATA, start sign
2. TFIIB and TFIIA: lock stregnthsdg the binding of TFIID on the promoter to be sure that TFIID remains stick attached to the promoter
3. TFIIH: helicase activity opend the DNA
4. Activation of RNA polymerase with kinase activity adding some phosphate groups
-Promoter melting: open of the promoter
- Formation of the first phosphodiester bond

The polymerase with all the trascriprion factors moves away from the promoter going along DNA and we have the so called promoter-clearance

ELONGATION:
in the ideal trascripion there is no stop nor pause, but in the real one, the RNA polymerase can run into obstacles, such as nucelosome that wrap the DNA

• pause: non definitive, RNA can continue the transcription
• stop: definitive, dangerous, the RNA polymerase detaches from DNA and resulting in an incomplete transcription.

There are several proteins that have the role to avoid stoping and pausing:

1. ELL: elonghin inhibitor pausing of RNA P II
   2- SII: inhibits stop of RNA P II
2. FACT: modifies nucelosome structrure and/or removes histones and facilitates RNA P II activity

Termination- Polyadenylation:

during elongatio we have a sequence necessary for cleavage AAUAAA. SO soo after the trascription, a nuclease enzyme will cleave a phosphodiester bond.
A enzyme, nuclease, cut the new RNA, and after that we have a specific polymerase that is only capable of adding A. It is called a Poly-A-polymerase. It does not need a template, in only adds A at the end of RNA.

when there are reactions ouccuring on the nucleic acids, we have several proteins that cooperate.

1. Identifying the region: CPSF ( clevage polyadenylation factor) recognize the sequence of AAUAAA,
2. does the job needed: CPSF recruit CFI and CFII, that are clevage factors + a stimulating factor. Those are endonuclease that will cut the RNA, The PAP, Polya polymerase add the PolyA tail.
3. cooperates

CPSF recognize the region, so CFI CFII arrives and cut the DNA and PAP cause polyadenylation.

For protecting RNA from degradation caused by exonuclease, there will be some protein that will bind to this PolyA region, and when exonuclease arrives it has some difficulties in breaking because there is another protein called PAB, PolyA binding protein, bindend to the RNA so exonucelase cannot bind it.

Splicing: the spicing occure after the capping and the polyadenylation

Is the process in which the introns are removed by nucleases that will cut out the beginning and the end of each intron and the remaining exons are ligated. it is called splicing because you cut and join.

How it occur:
1- see which factors that allow the removal of introns and the ligation of exons. Each intron has specific feuters.
Every intron begins with G and U and end with A and G.
Every intron has a Branch point, where there is an A.
Introns hase a secondary and tertiary structure, exon 1 and 2 are close, So when the introns are remobed is easy to ligate the exons together because they are close.

Factors that participate in splicing:
snRNA: non coding RNA important for formation of mRNA, important for the splicing of pre-mRNA.
snRNA is rich of Uridine:
1. U1 attaches to 5’ splice site
2. U2 attaches to the branch point
3. U4, U5, U6 join thr RNA
4. U4, U5 and U6 allow the first cleavage at 5’
5. U4, U5 and U6 allow branching
6. U4. U5 and U6 will allow the second cleavage

U1 and U2 have complementary regions, to the areas that are supposed to bind to.
U1 complementary to 5’ splice site
U2 complementary tp the branching point
snRNA are associated with proteins, so they are snRNA complexes.

Altermative splicing:

we have a gene with exons 1,2,3,4. The introns will be removed. So there is the trascription, we have the primary trascript, the premRNA that will have introns+ exones.

In alternative splicing we have alternative typeso of mRNA, that are produced from one single premRNA, In this case, protein A’s mRNA will have exon 1, 2 and 3, but no 4.
protein B mRNA will have 1,2,4 but no 3.

different mRNA produces different proteins, because the coding sequence will change so, the translation will change and the protein will change

in alternative splicing we have: 
1. one gene
2. one premRNA
3. different mRNA
4, different proteins.

Question 29

Classical view of transcription events

Answer 29

1-chromatin that opens, we find genes.

2. genes have promoters
3. RNA polymerase + trascription factors binds the TATA BOX.
4. RNA polymerase perfom elongation and termination.
5. we have a pre-mRNA, where we have a capping in 5’ and polyadenylation on 3’ and splicing.
6. mature mRNA, goes into cytoplasm, translated into protein

All the steps of trascription will not occur one after another but almost at the same time. During trascription we have the maturation.

Question 30

New theory of trascription events.

Answer 30

1. RNA polymerase stay on the promoter and ther eis a region of the RNA polymerase that must be phosphorylated for the activation.
2. TFIIH phosphorylates the RNA polymerase, the RNA polymeraseis activated and start trascription
3. First nucelotides are trascribed, polymerase perform a temporary stop for the capping. Capping happens as soon after the beginnig of trascription, for protecting the new RNA.
4. transcription continues where introns and exones are transcribed, splicing happens.
5. Poly A tail will also occur during transcription, termination of it.

RNA polymerase II transcribes the mRNA. In eukaryotic cells we have a pre-mRNA and then, after the maturation we have mRNA.

in bacteria: transcribed mRNA is already consideed to be mature and doesn’t undergo the conformational changes.

RNA POLYMERASE I

transcribe 1 out of 4 rRNAs.
We have a promoter that has 2 regions, CORE and UCE (upstream core element).
We also have transcription factors, cause polymerase isn’t able to bind to DNA in eukaryotic cells and it need UBF1 and SL1 and polymerase can bind.

UBF1 binds the 2 promoter regions
UFB1+ SL1 allow the binding of the polymerase.

The CORE promoter is called like that because is the starting point.

Polymerase need this factors because DNA has a tridimentional shape.

Maturations of rRNAs:
involves cleavage of multiple rRNAs from a common precursor.
The eukaryotic transcription unit, includes the genes for the 3 largesr rRNA, found in multiple copies and these units are arranged in tandem arrays with non-transcribed spacer between them.

each unit is transcribbed by RNA polymerase I into a single long transcript, with a sedimentation coefficent of 45S.

tandem array: succession of coding and non coding parts.

in rRNA we have multiples copies of the same genes due to the fact that our cells need an abbundance of ribosomes.

EVERY transcription units is trasncribed by RNA polymerase I, pre-rRNA 45S, maturation, rRNA after cleavage and procession.

After cleavage snoRNA (nucleolar) perform modifications.

1. snoRNA have partially complementary zones to rRBA, induce methylation of 2-ribose in the rRNA.
2. snoRNA can also modify uridine and pseuridine

RNA polymerase III

need promoters and transcription factors.
use 2 different promoters.
1. first one has two regions, box A and box C
2. the second promoter also has 2 regions but they are called box A and box B

Transcription fators will bind to the promoters and then the polymerase will arrive, attach and start.

The promoters of RNA polymerase III are differend because aren’t at the starting point, but are internal promoters, founds along the strand, but still allow the trascription of first nucleotides, because RNAP III will bind the RNA where polymerase should start.

RNA P III will polymerize the transfer of RNA, tRNA.
1. RNA polymerase II synthesize the pre-tRNA, which like rRNA, undergo maturation

in mature tRNA we find:

1. at 3’OH a ACC sequence that must be added to the pre-tRNA, because amino acids will binds to the 3’ end
2. in pre t-RNA we see a long sequence, the intron sequence that must be removed through splicing.
3. creation of D loop: change the U nucleotides into D
4. zone T U CG: where U became pseudoridine)

Question 31

Protein Synthesis/Translation

Answer 31

DNA is
transcribed and the synthesized
mRNA, which does not undergo
further processing and is already
mature, directly binds to a ribosome
and translation occurs.
But, in
eukaryotic cells, transcription occurs in the nucleus and the synthesized RNA, called pre-mRNA, has to undergo
further processing.
Following maturation, mRNA will leave the nucleus and will bind with a
ribosome to start protein synthesis, also known as translation.

Protein Synthesis

We have a piece of information (genes) that is in our DNA , Which is passed to RNA which is used to synthesize protein.

Proteins are made up of Amino acids which are linked together by peptide bonds , There are 20 different types of amino acids they form chain called polypeptide

The Pathway :

DNA→ mRNA (Messenger)→tRNA (Transfer)→Protein

Making The Protein

1. to translate from nucleic acid language to protein language. To do that we need a code that relates nucleotides with amino acids. The language of nucleic acids has 4 different nucleotides , these 4 nucleotides correlate to 4 different letters of nucleic acid language.
   In Protein language we have 20 different amino acids which corresponds to 20 different words , which consists of a common group known as amino acids.

The nucleic acid language and the protein language are incompatible, because the nucleic acid language has 4 nucleotides while protein language has 20 amino acids. Since the protein language has more words (amino acids) than nucleotides , our cells must have some combinations (codon) of nucleotides to produce more words which will be more than or equal to the words of the protein language.

If we try the combination of 2x2 (for eg. AT , GC) on 4 nucleotides the resulting combination would only have 16 words , which is still less than the words required for protein language. Furthermore if we try 3x3 combination (eg. AUG,ATG) we get 64 combinations , which is more than enough to be compatible with protein language , this gives birth to a code which is a triplet containing 3 nucleotides each (the codon).

Representation of Steps of Synthesis

1. Transcription- The one strand of DNA is transcribed to produce one mRNA
2. Translation – The codon is read on the mRNA , forming the polypeptide chain. Only the coding region of mRNA is translated into amino acids , the unread or the region which is not translated is called UTR’S (Untranslated regions)

Question 32

Genetic code

Answer 32

the genetic code is a triplet containing 3 nucleotides each.
This code in the form of triplet is present in DNA.
Which means that every 3 nucleotides on DNA stands for 1 amino acid.
The triplet copied on mRNA after transcription from DNA is called codon.
This codon represents the compatibility with the protein language , hence every codon on mRNA is responsible for 1 amino acid, which also means that codon represents the genetic code.
Since there are 64 codons for 20 amino acids , most amino acids have more than one codon.
This code is universal , which means every organism has the same corresponding codon for the same amino acid. Methionine is the only amino acid which has only one codon.

One particular amino acid can be coded by many codons , for example, the amino acid leucine (abbreviated as “Leu” in table) has 6 codons , these 6 codons only code for leucine.

From 64 codons , 3 codons which do not code for anything are called STOP codons , the rest 61 code for the corresponding amino acids which are important for the termination of protein synthesis.

Important- one codon codes only for one amino acid , but one amino acid can be coded by many codons

Question 33

What are the RNA’s that make proteins synthesis?

Answer 33

1.mRNAs: messenger RNA, contains the start codon and the stop codon at the terminating end, the codon is read thoroughly unless it is intercepted by stop codon, which terminates the translation process by forming peptide bonds.

2. Transfer RNA: 75-100 nucleotides, the job is to bring the right amino acids from the cytoplasm and transport amino acids to mRNA.
   tRNA has 2 sites,
3. amino acid site where amino acid attache the other
4. anticodon site that pais with mRNA codons.

They have anticodons that are complementary to codons in the mRNA, They recognize the appropriate codons on mRNA and bond them with hydrogen bonds, thus brungung them the corresponding amino acid.

Accurate translation requires 2 steps:
1. A correct match between tRNA and amino acid, done by enzyme, aminoaclyl-tRNA-synthetase.

2. a correct match betwee tRNA anticodo and a mRNA codon, pr the tRNA will bring wrong amino acid, hence affecting the resulting protein.

• Association between amino acids and tRNA(mechanismo)
  1. amino acids intercats with aminoacyl-tRNA-synthetase, and gets activated, with an expense of ATP, hence amino acid combines with the enzyme and AMP is released from the enzyme.

2. AMP is relesade, tRNA binds with the Aminoacyl-tRNA-synthetase and forms bonds with amino acid present on the enzyme.
3. the tRNA is now charged t-RNA. AS soon as t-RNA gets linked with amino acid it detached from the enzyme carrynig the correspondin amino acid
4. rRNAs 100-3000 nucleotides. They associate with proteins to form ribosomes. The role of ribosomes is to facilitate specific cuopling of t-RNA anti codons to m-RNA codons in protein synthesisi. Ribosome are protein factories because they are the sites of protein synthesis.

ribosomes have 2 subunits made of proteins + rRNA:

1. large: 3 cavities A P and E
2. small: has mRNA to whuch the m-RNA binds

m-RNA has 3 binding site for t-RNA

1. P site( peptidyl t-RNA BS); holds t-RNA that carries the groeing polypeptide chain.
2. A site (aminoacyl t-RNA BS): holds the t-RNA that carries the next amino acid to be added to the chain.
3. E site( Exit site): exit side, where discharged t-RNA leave the ribosome

Question 34

Building a polypetide

Answer 34

3 stage of translation:
1. Initiation
2. Elongatio
3, Termination.

Need of translations factors:

1. Initiation Factors IF
2. Elongation FActors EF
3. Release Factors (Termination factors)

(e added in front of eukaryotick factors)

1. Ribosome association and initiation of translation:
   - initiations stage brings together mRNA, a tRNA with the first amino acid and 2 ribosomal subunits.
   1- a small ribosolam subunits binds with mRNA and a special initiator tRNA.
   2- small subunits move along the mRNA unitl it reacges the start codon (AUG)
   3- initiation factors bring large subunit that complete the translation initiation complex.

• Prokatyotic association and initiation of translation:
  1. small subunits of ribosome bind with mRNA :
  
  • the initiatio codon require the Shine-Dalgarno sequence, that is complementary with 16S rRNA in the small subunit, that use this sequence to binf the small subunits to the mRNA

-Eukatyotic association and initiation of translation:

• Shine-Dalgarno sequence is NOT present.
• So the small subunits will bind to the mRNA with the helo of initiation factors. All these initiation factors will help the subunit to bind to mRNA.
• the initiation factor bind to the cap and poly A tail. (role to protect and promote synthesis)
• there is tRNA in P site, that is associate with methionine bound to the start codon
• with GTP, the large subunit will join the complex.

ELONGATION OF POLYPEPTIDE CHAIN.
1- During the elengation stage, amino acids are added one by one to the preceding amino acid at the C-terminus of the growing chain.

2. Each addition involves proteins called elongation factors and occurs in 3 steps:
3. codon recognition
4. peptide bond formation
5. translocation
6. Translation proceed along the mRNA in 5’ to 3’

Elongation: first amino is present in P site, after the arrival of another tRNA to A site.
The amino acid on the second tRNA bind the first amin o acids, and on the first tRNA there will be no amino acid.
the mRNA moves, the first tRNA will be discharged amd the second goes into P site, this continues until the STOP codon.

3. TERMINATION OF TRANSLATION:
4. occurs when a stop codon in the mRNA reaches the A site of the ribosome.
5. the A site accepts a protein called release factor
6. the release factor cause addition of water insted of amino acid
7. this reaction releases the polypeptide chain and the trabslation assembly then comes apart.

start codon; AUG
stop codon: UAG, UAA, UGA

A correct translation requires two steps:

1. correct match between a tRNA and an amino acid, done by enzyme Aminoacyl-tRNA-synthetase.
2. coorect match between tRNA anticodon and mRNA codon.

POLYRIBOSOMES:
A number of ribosomes can translate a single mRNA simultaneuosly forming a polyribosome or polysome. that enables a cell to make copies of a polypeptide very quickly.

COMPLETING AND TARGETING THE FUNCTIONAL PROTEINS:

1. often translation is not sufficient to make a functional protein
2. polypetide chain are modified after translation or trageted to specific sites in the cell.

PROTEIN FOLDING AND POST-TRANSLATION MODIFICATION:

1. During and after synthesis, a polypetdie chain spontaneously coils and folds into its three dimensional shape.
2. Proteins may also require post-translational modifications before doing their job.
3. some polypeptided are activated by enzymes that cleave them.

for example: insulin in synthetixed as along polypeptide chain, called pre pro insulin,there is cut and we have pro insulin, then more cuts and we have mature insulin.
Instad of a long polypeptide we have small pieces, 2 polyeptides + disulfide bond.

4. Other polypeptides come together to form the subunits of a protein: glycosylation of proteins, in endoplasmatic retuculum or in golgi, or the adds of lipids into the endoplasmatic.
   There are a lot of modificarions that a protein must receive before it becomes funcional.

Question 35

DNA mutations

Answer 35

What is a Mutation?
A mutation is a modification in a sequence of DNA or modification of a nucleotide in DNA.
There are two types of mutation: point mutations and chromosomal mutations. When a large part of the DNA is modified, it is known as a chromosomal mutation.

Question 36

Chromosomal mutations

Answer 36

When a large part of the DNA is modified, it is known as a chromosomal mutation.

Types of Chromosomal Mutations

Deletion: The segment of a chromosome is deleted and the resulting chromosome will be of shorter length.

Duplication: The segment of chromosome is duplicated and the resulting chromosome will be of longer length.

Inversion: The segment of chromosome is inverted (rotated by 180 degrees) but the resulting chromosome will have the same length

Insertion: The chromosome has a segment of another chromosome inserted in it.

Translocation: two chromosomes exchange their segment, resulting in translocation mutations.

Chromosomal mutations can occur spontaneously, or we have induced mutations.
The spontaneous mutation can occur if there is alteration during meiosis or mitosis resulting in chromosomal mutations.

Question 37

Point Mutations

Answer 37

Point mutations are modifications of a single nucleotide or a few nucleotides in a DNA segment.
Point mutations can be classified as:

1. Transitions and Transversions – A single nucleotide will be substituted, but there will be no extra or deleted nucleotide, hence the nucleotide number remains the same.
2. Insertions and deletions- A nucleotide will be added or deleted, hence affecting the nucleotide number present in DNA.
3. Transition Mutation – In a transition mutation, a purine (A or G) is replaced by a purine, or a pyrimidine is replaced by a pyrimidine (T or C), the same type of nucleotide is substituted.
4. Transversion Mutation- In transversion mutations, a purine is replaced by a pyrimidine or vice versa. Different types of nucleotides are substituted.
5. Insertion Mutation- in the insertion mutation, a nucleotide is added and the resulting nucleotide number will have an extra nucleotide.
6. Deletion Mutation – In deletion mutation, the nucleotide number is deleted and the resulting nucleotide number will have one nucleotide less.Point Mutations

Generally chromosomal mutations are more dangerous because they affect larger amounts of DNA.
But mutation of one or a few nucleotides may also be dangerous because they can affect protein structure and function, the change of a single nucleotide in a DNA template strand can lead to the production of an abnormal protein. The point mutation should be in a coding region of m-RNA to produce an abnormal protein.

sickle cell hemoglobin is an example of point mutation

Question 38

consequences of Transversion or Transition Mutations

Answer 38

1. Silent Mutations
2. Missense Mutations
3. Nonsense Mutations

1.Silent Mutations –They have no effect on the amino acid produced by the codon because of redundancy in the genetic code.
consider wild type (Normal) m-RNA which contains codons for methionine, lysine , phenylalanine and glycine.
The glycine region is coded by GGC.
Now if there is a mutation in the coding region of DNA, like instead of G there is A (refer the diagram), The m-RNA will have an altered codon in which the cytosine (C) in GGC is replaced by U in the glycine region making it GGU.
Luckily the codon GGU also codes for glycine, hence there is no change in the sequence and a normal protein will be produced. This type of mutation is a silent mutation. Silent mutations are common in our DNA and they happen almost every day!

2.Missense Mutations – In missense mutations, the point mutation of one nucleotide results in change in codon thereby changing the corresponding Amino Acid producing an abnormal protein.
For example, we consider the wild type m-RNA with the same sequence in the previous example, the coding region of DNA has T instead of C (glycine), the m-RNA for the corresponding amino acid gets altered to AGC instead of GGC (glycine region). The m-RNA which coded for glycine now codes for serine hence, affecting the sequence of amino acids in the protein.
Missense mutations can be dangerous or non-harmful depending on the nucleotide substituted and amino acid produced.

3.Nonsense Mutations- It changes the codon to a stop codon which leads to a shorter protein.
For example, we consider the wild type m-RNA with the same sequence in the previous example again, in the coding region of DNA we have an A instead of T and a T instead of C. So, instead of having AAG (Lysine) on the mRNA we have UAG which is the stop codon.

Question 39

Deletion and insertion

Answer 39

insertion and deletions are additions or losses of nucleotide pairs in a gene.
These kinds of mutations have a disastrous effect on the resulting protein more often than substitutions do, because insertion or deletion of nucleotides may alter the reading frame, producing a frameshift mutation.

Insertion:
For example, we consider the wild type m-RNA with the same sequence in the previous example. I have AUG (Met) and then I should have AAG (Lys) but I have an extra nucleotide so the codon will no longer be AAG, it will be UAA (stop codon). After the insertion, all the codon sequence will be changed and this is called frameshift mutation.

Deletion:
If we consider the same example as before, in the third codon I have UUU. If I lost a U, the codon will be changed to UUG. All the amino acids after the deletion will be changed. This is very dangerous.

Question 40

Causes of Mutations:

Answer 40

• Spontaneous Errors
• Chemicals
• Physical Agents

There are a lot of mutations happening all the time in our bodies. Luckily, we have a DNA Repair System that repairs the DNA. Otherwise, we would accumulate a lot of mutations which would make us sick and eventually lead to death.

But on the other hand, we have to avoid overloading our DNA Repair System.
If we expose ourselves to chemicals or physical agents, these mutations will add to the spontaneous ones and in many cases our system cannot cope with all of these mutations. For example, we all know that we have to stay away from radiation because it induces mutations and alterations.

Question 41

Spontaneous mutations:

Answer 41

Spontaneous Mutations are due to two mistakes that the DNA Polymerase makes: the slippage of DNA Polymerase and wrong base pairing due to tautomerism.

Slippage: there are repetitive DNA sequence so imagine that for example this is a triplet of a repeated section like CCG.
Each arrow represents a repetitive DNA sequence.
The number one black arrow will pair with number one pink arrow, number two black arrow with number two pink arrow and so on.
It may occur that for example number two black arrow will not pair at all and the number three now will pair with number two so the number four will pair with number three and so on.
I have 8 repetitive regions here so in the end I should have 8 new repetitive sequences.
If I have a situation like this in the end, the new strand (black one) will have more than 8 repeats, 9 to be exact.
At the end of replication, you will have a longer DNA which will result in genetically inherent neurological diseases such as myotonic dystrophy and spinocerebellar ataxia.

Tautomerism:

keto - enol
amno - imino

the enol-imino are rare and DNA polymerase do not recognize it

Question 42

Induced mutations

Answer 42

We can have mutations induced by chemicals or physical agents.
Chemicals: 
1Insertion of Base Analogs, 
2.Deamination, 
3.Alkylation, 
4.Intercalation, 
5.Free Radicals

1.insertion of Base Analogs:

Base Analogs have a structure similar to normal nucleotides, but they don’t pair with the proper nucleotides which makes them dangerous.
For example, we have the structure of thymine (as seen in photo below). When there is a bromo group instead of a methyl group, we obtain 5-Bromouracil. It is quite similar to Thymine but instead of pairing with an Adenine, it may pair with another nucleotide causing a mispairing. And notice that these are kinds of pollutants produced by factories etc.

2.Base Deamination:

Deamination means the loss of an amine group. It’s caused by human pollution. For example, a Cytosine becomes a Uracil when deaminated and an Adenine becomes Hypoxanthine when deaminated.

Nitrous Acid is a deamination agent. If cytosine is deaminated and becomes uracil, it will no longer bind with guanine. With base deamination, you may not have correct base pairing just like base analogs.

3.Alkylation (alkylation):
Alkylation is the adding of a methyl group or an ethyl group to the nucleotide.

4.Intercalating Agents:
Intercalating Agents such as ethidium bromide have a structure with aromatic rings. Since they have a structure similar to the bases they can insert themselves among the bases (intercalate).
The intercalating agents produce mutations by sandwiching themselves between adjacent bases in DNA. They distort the three-dimensional structure of the helix causing single-nucleotide insertions and deletions in replication.

5.Reactive Oxygen Species (Free Radicals):

They are molecules with an unpaired electron. For example Oxygen loses an electron to produce Peroxide which is a free radical. They are very eager to take back the electron they lost and so a reactive oxygen species will extract an electron from another molecule in order to recover the loss.

What causes these Free Radicals to form?

Free Radicals and DNA Damage:

a free radical needs an electron, so it steals an electron from a DNA. This will result in a damaged DNA and a free radical that becomes a stable molecule.

For example, a Guanine nucleotide damaged by a free radical will become Oxoguanine.

2. Physical Agents:
3. UV Radiations,
4. Ionizing Radiation
5. Heat.

1.UV Radiations and DNA Damage:
When you stay under the sun, the UV light will reach the DNA and if in our DNA there are two Thymine, one after another, the UV light will join them together and a Thymine dimer will be formed and the cells will be damaged. Luckily, we have a repair system or else we could not stand under the sun at all.

Ionizing Radiations: Light from sun, radiations from rock and all radiations in our environment travel as waves. Some radiations have long wavelengths(lower energy) and some have short wavelengths(higher energy). Radiations having higher energy are more dangerous than radiations having lower energy. UV, X-Ray and gamma rays are capable of ionization as they have high energy.
How does a radiation ionize DNA? Radiations having higher energy react with atoms in DNA and extract electrons from them and form cations.

Types of ionizing radiations:
1- Alpha particles (2 protons and 2 neutrons)
2- Beta particles (1 electron)
3- Gamma rays (1 photon)
4- X-Ray (1 photon)
5- Neutron
These particles are either produced by Nuclear fission or Nuclear fusion reactions.

In fission reaction a neutron is absorbed by the nucleus of
U-235 turning it briefly into an excited U-236 nucleus with the
excitation energy provided by kinetic energy of neutron plus
force that binds the neutron. U-236, then turns into two fast
moving lighter elements plus three free neutrons. At the
same time gamma rays are produced.

Fusion is what powers the sun. Atoms of Tritium and Deuterium
( isotopes of hydrogen; H3 and H2 respectively) unite under
extreme pressure and temperature to produce a neutron and a helium isotope. Along with this an enormous amount of energy is released which is several times the amount produced from fission.

Fusion is environment friendly as it does not produce nuclear waste material.

Penetrating power: The extent of ionization also depends on the penetrating power of radiations. Alpha particles are absorbed by only a sheet of paper while beta is absorbed by plates of wood or aluminum. X rays and gamma rays can penetrate through wood or aluminum plates and are absorbed by lead, iron or other thick metals. Neutrons have the highest penetrating power and can penetrate lead. Neutron is absorbed by concrete or water.
Interaction with DNA: Ionizing radiations interact with DNA in two ways: Directly or Indirectly.

When ionizing radiations hit DNA directly, they destroy both
strands of DNA and this is more dangerous than indirect interaction.

On the other hand, indirect radiations react with water molecules to produce highly reactive free radicals(species having unpaired electrons) which further react with DNA and cause damage to its structure.

Heat: The last physical agent which causes change in DNA structure is heat. When we raise the temperature around 80 degree celsius, DNA denatures. The bond between purine and sugar is less stable than the bond between pyrimidine and sugar.So energy associated with heat will break the glycosidic bond between purine and sugar (depurination) and bring denaturation.
Heat rarely causes mutation as for depurination we have to stay at around 80 degree celsius, much higher than normal surrounding temperature.

Question 43

DNA Repair:

Answer 43

There are such a huge number of DNA mutations and in order to survive we must have a DNA repairing system.
DNA is repaired by directly or indirectly; indirect method further includes excision, mismatch and recombination repair.

1.Direct method:
Direct method means we repair damaged DNA without replacing it with new one.
For example if we have methyl group (alkylation) on guanine nucleotide and we repair it directly. We use
Methyl transferase which remove methyl from guanine,
converting O6-Methylguanine nucleotide to Guanine nucleotide.

In bacteria and lower eukaryotes, direct repair is done
by Photoreactivation (direct reversal). Photoreactivating enzymes (photolyase) :
1.recognize damage,thymine dimers,
on DNA caused by UV light.
Photolyase is activated by light and breaks thymine dimers. Photolyase is not yet
found in mammals and is only found in bacteria and in lower eukaryotes.

2.Indirect method: In this method repair is done by replacing damaged with new one.

1.Base Excision repair: DNA bases can be modified bydeamination or alkylation. Only one nucleotide is repaired at one time. When there is any damage in DNA, glycosylase
detects and removes it.
But glycosylase removes only base not sugar and leaves sugar with phosphate group.
Following glycosylase endonuclease(break phosphodiester bond)
removes the rest of the nucleotide and creates a gap.
This gap is filled by polymerase which adds new nucleotide and
ligase that joins two fragments.
Then endonuclease removes nucleotide and polymerase and ligase do rest of work

2.Nucleotide Excision Repair: NER can remove more than one damaged nucleotides at a time.
NER is more versatile than BER since it does not recognize a specific DNA damage rather it recognizes double helix distortions induced by damage.In this way almost all kinds of DNA damage may be repaired.

In prokaryotes base dimers created by UV can also be
removed by NER.
This is done by many proteins namely;
UVrA, UVrB & UVrC.

1. UVrAB complex scans DNA and binds at damaged site (kinked or distorted).
2. UVrA leaves DNA when UVrB joins the lesion;
3. UVrB recruits the UVrC which is basically an endonuclease.

4.Endonuclease cutsthe DNA and this cut is not only of damaged DNA rather it is done at a little bit before and a little bit after the
damaged part.

5.UVrB and UVrC work together; UVrC cuts
DNA strand and UVrB removes the cut part; acts as helicase.
UVrB generally kinks or denature DNA strands.

6.After the removal of DNA fragment, there comes DNA polymerase1 which adds corresponding nucleotides and then DNA ligase joins the fragments.

BER cannot remove UV damaged nucleotides as it remove
only one base or nucleotide at a time while removal of a base
dimer requires elimination of two nucleotides at a time.

Mutations in transcribed region produce:

1: Formation of mutated RNA
2: Stall of RNA polymerase & production of truncated DNA
3: Degradation of RNA polymerase

For this reason, transcribed regions are double checked by
DNA repair system.
We have GGR (global gene repair)& TCR (transcription coupled repair).

In eukaryotes more proteins take part in NER.
These are called XP proteins (in GGR) due to rare genetic disorder xeroderma pigmentosum in which the ability to repair damaged DNA caused by UV is deficient.

Whole process in eukaryotes is also the same as
prokaryotes but with different proteins.
Eukaryotic repair mechanism follows the same process; structure distortion, detection of damage, incision of enzymes, excision of
fragment,addition of nucleotides and joining of fragments.

1. Mismatch Repair: Mismatch generally means ¨not associated¨. We know that sometimes DNA polymerase adds wrong nucleotides and some wrong insertions may escape proof reading (polymerase goes back, exonuclease removes wrong nucleotides and then polymerase again move forward to add new nucleotide) which results in mismatch between correct nucleotides on template strand and wrong nucleotides in new strand.
   Wrong insertions happen when base is in its rare form rather than frequent form and this happens due to methylation, deamination or tautomerism.
   Wrong insertion may be the main cause of evolving new variants each day like delta and omicron.
   When there is mismatched nucleotide and proof readings miss it,it will form mutated RNA and then mutated protein.

RNA polymerase can’t remove mismatched nucleotides as it lacks proofreading activity.So if there is no proof reading, the number of mutations will be higher and we will come across new variants each day.

How is it possible to eliminate mismatches?
Methylation and alkylation are not always harmful like O6-Methyl Guanine.
Sometimes methylation is a physiological process and we have methylated DNA strands.
After replication, for a short period,the new strand is
unmethylated and is different from the template strand.
We have hemi-methylated DNA.This allows us to identify
wrong insertion by DNA polymerase as both strands are
easily distinguishable.
If we know that there is a mutation or wrong insertion during replication,it will surely on new strandbecause it has DNA polymerase and is unmethylated.
So it is easy to distinguish both strands.After a few minutes the new strand gets recognized by the repair system and is methylated.
Now both strands are no longer distinguished.
Mismatch repair must work quickly after replication because an
unmethylated DNA strand can be recognized at this time.
If unmethylated strand form double helix,it will be impossible to perform methylation or repair DNA.

In prokaryotes, mismatch is recognized by the enzyme MutS.
Consider we have prokaryotic DNA with two strands one is template(methylated) and other is new growing strand (non methylated).
By mismatch we don’t mean any modification or alteration rather it is simply not a complementary base pair.
MutS recruits MutH & MutL;MutH cleaves DNA strand and removes the cut fragment.
This arises a gap in the growing strand which is filled by polymerase and ligase.
Then a methyl group is added to the new strand.

In eukaryotes the process is a little bit different because in addition to MutS,MutH & MutL there are several other enzymes that take part in the process.

-Homologous recombination:

All previous repair systems involve repair or replacement of single base or nucleotide on one of two strands.
High energy radiations such as UV and X rays can break both strands of DNA and it is more complicated than single nucleotide repair.
In single base repair, template/complementary strand will decide which base or nucleotide has to be added.
In case of high energy radiation, we don’t have complementary strand which gives information to add corresponding base.
This repair is done by homologous recombination and may occur only in dividing cells.

During division of cells, chromosomes associate into fibers and are held as homologous.
The only possibility to repair both strands of DNA is to pair up with a homologue.
Homologous chromosome acts as a template and the damaged region is repaired via recombination, using sequences copied from the homologue.

• Following slide is the complete demonstration of homologous recombination.Professor said it will not be included in the exam.*
• Non homologous recombination

As homologous recombination occurs mainly in dividing cells then how can we repair DNA at non dividing state?

1.In non homologous chromosomes Ku proteins bind to the free ends of DSB(double stranded broken) DNA .
2.recruit DNA-PKcs (protein kinase).
There is also recruitment of XRCC4 & Ligase 4 to re-ligate the DNA ends but we are not sure whether the recombination is perfect or not.

Consider we have regions A,B &C on a DNA strand broken by ionizing radiations. In order to repair DNA two ends must pair in a short region and it is possible that A joins with C instead of B and creates a problem.

            A                                    B                                            C — - - –  - - - - - —--          — - - - - - - - —-----              —-------------------

There is another problem associated with non homologous recombination for which microhomology is necessary.
Imagine we have a broken DNA and Ku proteins bind to its free ends.
The interaction between two Ku proteins will form synapse.
Helicase activity of Ku proteins unwinds the both ends in order to create micro homologous region.
It is a single stranded region in which base on one fragment pairs with homologous region on other.
When strands are closed, extra nucleotide has to be removed and we lose some informations.

So non homologous recombination creates problems; we don’t know whether the fragments are joined in the correct sequence or not and we also lose some fragments.