Information

Does DNA polymerase I require a $3^prime$ end?


DNA polymerase III adds nucleotides in the $5^prime ightarrow 3^prime$ direction because it can only add nucleotides to the $3^prime$ end of the previous nucleotide. This is why it requires a primer. However, does DNA polymerase I operate by the same criterion? Does it require a $3^prime$ end of a previous nucleotide in order to bind successive DNA nucleotides?

If it does, then how can it do so for Okazaki fragments when each Okazaki fragment is unbonded to each other? It is the DNA ligase that finally catalyzes the phosphodiester linkage between the $3^prime$ end $5^prime$ beginning of two Okazaki fragments, isn't it?

If it does not, then what's the deal with telomeres? After each DNA replication event, the DNA gets shorter and shorter at the very ends because that final primer can be removed but not replaced by DNA via DNA polymerase I, correct? This suggests to me that DNA polymerase I requires a previous nucleotide's $3^prime$ end to work with, and it has confused me regarding its action on Okazaki fragments in conjunction with DNA ligase.


DNA Pol I does require the 3' end of a previous nucleotide to initiate elongation.

Regarding Okazaki fragments, this is accomplished by the annealing of small RNA primers to the lagging strand part of a replication fork. DNA Pol I extends the lagging strand off of the 3' end of those primers, generating the Okazaki fragments. The RNA primers are later removed, leaving gaps between Okazaki fragments, which are later filled in through the combined actions of DNA PolI and DNA ligase.

In the case of telomeres, the final RNA primer can't be filled by DNA Pol I because polymerase requires a free 3' hydroxyl for the attachment of the first DNA nucleotide. After removing out the RNA primer, the first DNA nucleotide would need to be attached to the DNA nucleotide preceding it, which is not found in the case of the final primer on a telomere, since it's all the way at the end of the linear strand.

In this case then, DNA Pol I removes the RNA primer is removed by DNA pol I but cannot replace it with DNA, leaving a gap (step 6 in the following image).


5 Prime And 3 Prime Dna

Decoding the genetic code from dna to mrna to trna to amino acid duration. In this video you will learn what 5 and 3 in a dna strand stands for in details.

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That is a forms two hydrogen bonds with t.

5 prime and 3 prime dna. Normal dna polymerases are 5 to 3 polymerases. For both dna shown above and rna the 5 end bears a phosphate and the 3 end a hydroxyl group. The 5 5 prime and 3 3 prime ends.

Polymerases would never work because the energy required would be way too high. The replication from a 3 prime end to the 5 prime end is on the lagging strand. In any nucleic acid rna or dna 3 refers to the 3rd carbon of sugar ribose or deoxyribose which is linked to oh group and 5 linked to a triple phosphate group.

It is also essential to know that the two strands of dna run antiparallel such that one strand runs 5 3 while the other one runs 3 5. Hope you like this videos. The 5 prime and 3 prime ends r the two ends of a dna.

Tail of the dna molecule but it synthesizes 5039 to 3039. The replication in this direction is on the leading strand. And replication in a dna occurs from the 5 prime end to the 3 prime end.

So these 5 and 3 group provide a directional polarity to the dna or rna molecule. A key feature of all nucleic acids is that they have two distinctive ends. After this you will know why its called 5 and 3 and what its used for.

A phosphate group attached to the 5 end permits ligation of two nucleotides ie the covalent binding of a 5 phosphate to the 3 hydroxyl group of another nucleotide to form a phosphodiester bond. In most cases the two stranded antiparallel complementary dna molecule folds to form a helical structure which resembles a spiral staircase. Advances 3 prime to 5 prime down template strand to create an mrna that is 5 prime to 3 prime.

This terminology refers to the 5 and 3 carbons on the sugar. At transcription stop site polymerase releases completed rna and dissociates from dna. Now a good question would be y 3 and 5 not 3 and 5.

Elizabeth godwin 407240 views. Polymerase binds promoter sequence closed complex unwinds part of dna open complex catalyzes phosphodiester bond of 2 initial rntps. C forms three hydrogen bonds with g.

The 5 end pronounced five prime end designates the end of the dna or rna strand that has the fifth carbon in the sugar ring of the deoxyribose or ribose at its terminus. The enzyme dna polymerase synthesises strands in the 5 prime to 3 prime direction and as dna is antiparallel the replication of the leading strand occurs from the 3 prime end of the template to. Dna polymerases extend the 3039.

At each nucleotide residue along the double stranded dna molecule the nucleotides are complementary.

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DNA Replication in Eukaryotes | Genetics

In this article we will discuss about the DNA replication in eukaryotes.

In eukaryotes there are only two different types of DNA polymerases in contrast with DNA polymerase I, II and III of prokaryotes. Furthermore the DNA of eukaryotes is a long linear molecule with several replication units. A diploid mammalian cell contains on an average about 6 pg of DNA in the G phase. This much DNA is equivalent to a length of 2 metres of a linear DNA molecule.

If a single replication unit were to move along this length of DNA, it could complete replication within the 8 hour S phase only if its rate of movement is about 4 mm/min. This is obviously a very fast rate.

The replicating fork actually moves at a slower speed (0.5 to 2.0 micron/min.) in eukaryotes adding about 2,600 bases per minute. In E. coli it moves faster adding about 6,000 bases per minute. It is, therefore, necessary that in eukaryotes replication be initiated at several points of origin.

Auto-radiographic studies on labelling patterns of individual metaphase chromosomes have shown that multiple adjacent units initiate replication simultaneously. The most convincing demonstration however, came from similar observations in giant polytene chromosomes.

Here tritiated thymidine is incorporated simultaneously into a large number of different bands. By the same technique the egg in Drosophila is shown to have 6,000 replication forks and all the DNA synthesis is completed within 3 minutes.

The unit of replication is the replicon. The size of the replicon is estimated from the distance between adjacent initiation points (centre-to-centre distance). By autoradiography it has been found that units within the same cell are not uniform in size but fall within the range of 15-60 micron.

Replicons in rapidly growing cells with short S phases are smaller than those in cells growing more slowly with longer S phases. Blumenthal (1973) has estimated that in Drosophila melanogaster replicons in embryonic cells are as short as 3-4 micron, whereas in a cell line of the same species they were about 13 micron long.

Experimental studies on cultured mammalian (Chinese hamster) cells have shown that the rate of DNA synthesis is not constant throughout the S phase, Kleveroz (1975) found that synthesis is slow at the beginning of S phase, thereafter it increases. About 50% of replication occurs during the last hour of the 5.5 hour long S phase.

The occurrence of multiple adjacent units has led to the concept that replication units exist in clusters. All units in a cluster do not replicate simultaneously, some being late replicating. In mammalian cells there are about 100 replicating units in a cluster.

The essential features of DNA replication are similar in eukaryotes and prokaryotes. After replication begins at a central point of origin in each unit, it proceeds in both directions away from the initiation site. Chain growth occurs by means of fork-like growing points. Electron micrographs therefore show a number of ‘eyes’ or ‘bubbles’, each formed between two replicating forks along the linear molecule.

It appears that there are no specific term in DNA for stopping replication. The forks travel towards each other and the newly synthesised chains meet and fuse with chains synthesised on adjacent units (Fig. 14.11). In this way long DNA duplexes characteristic of eukaryotic chromosomes are produced.

As in prokaryotes, the first step in DNA synthesis in eukaryotes is the formation of a primer strand of RNA about 10 nucleotides in length—catalysed by the enzyme RNA polymerase. After that DNA polymerase takes over and adds deoxyribonucleotides to the 3′ end of the primer RNA.

The Okazaki fragments thus formed are shorter in eukaryotes (about 100-150 nucleotides long) than in prokaryotes (1,000 to 2,000 nucleotides). The gaps between the fragments are filled up against the parent DNA template and their ends are joined by DNA ligase enzyme. The RNA primer is digested, starting from its 5′ end by the exonuclease activity of DNA polymerase.

Significance of the RNA Primer in DNA Synthesis:

Why should DNA replication be initiated by the enzyme RNA polymerase and formation of RNA strand take place? Detailed analysis of DNA polymerase enzymes have revealed the fact that each polymerase enzyme can add nucleotides only to an already existing polynucleotide chain.

These enzymes are not able to initiate new DNA chains. The point of origin in a DNA duplex is perhaps recognised by RNA polymerase, the enzyme which catalyses the synthesis of RNA on a DNA template. In other words, RNA polymerase is required for both RNA and DNA synthesis.

Synthesis of RNA primer on the DNA template continues until a stop signal is reached. The enzyme is then released and the RNA chain serves as a primer for addition of DNA nucleotides by DNA polymerase enzyme. However, the molecular mechanism which initiates DNA replication is not fully known.


Quantum Leaps in Biochemistry

Anil Day , Joanna Poulton , in Foundations of Modern Biochemistry , 1996

Mitochondria Import RNA

The discovery that the RNA primer used to initiate DNA synthesis in mammalian mitochondria was imported from the nucleus dispelled the belief that only proteins but not RNA could cross organelle membranes ( Chang and Clayton, 1987 ). Evidence that tRNAs are imported from the cytosol into mitochondria is derived from elucidating the coding capacity of mitochondrial DNA and also characterizing tRNAs present in isolated mitochondria. Mammalian mitochondria use 22 unusual tRNAs to decode the 61 sense codons. The linear 15.8-kb genome of C. reinhardtii only encodes three tRNAs ( Michaelis et al., 1990 ). Although liverwort mitochondrial DNA encodes 27 tRNA species, two species necessary to read leucine and threonine codons are absent. C. reinhardtii, plant, and trypanosome mitochondria appear to import nuclear-encoded tRNAs to make up a complete set for protein synthesis. Eleven of the 31 tRNA species present in potato mitochondria are encoded by nuclear DNA and are imported from the cytosol ( Dietrich et al., 1992 ). The plastid genome of Epifagus virginiana, a nonphotosynthetic parasite of beech trees, lacks 13 tRNA genes found in green plastids. This suggests tRNA import can also occur in plastids ( Wolfe et al., 1992 ).


What is DNA Polymerase 2?

DNA polymerase 2 (Pol 2) is a prokaryotic enzyme which catalyzes the DNA replication. It belongs to the polymerase B family and is encoded by the gen polB. It was first discovered from E Coli by Thomas Kornberg in 1970. Pol 2 is a globular protein composed of 783 amino acids. It has both 3’ to 5’ exonuclease activity and 5’ to 3’ polymerase activity. It interacts with DNA polymerase 3 enzymes to maintain the fidelity and processivity of DNA replication. Pol 2 also has the ability to proofread the newly synthesized DNA for accuracy.

Figure 03: DNA Polymerase 2


DNA Polymerase I

DNA polymerase I participates in the DNA replication of prokaryotes. DNA chain growth is in the 5&rsquo to 3&rsquo direction with addition at the 3&rsquo hydroxyl end. The new chain is base-paired with the template, and the new chain and template are antiparallel. DNA polymerase I is the most abundant polymerase and functions to fill gaps in DNA that arise during DNA replication, repair, and recombination.

DNA polymerase I was discovered by Arthur Kornberg et al. in 1956. His initial results were first presented at the 1956 annual meeting of the Federation of American Societies for Experimental Biology (FASEB) in Atlantic City, New Jersey. Reviewers of his initial paper suggested that the authors refer to the product as &lsquopolydeoxyribonucleotide&rsquo rather than &lsquoDNA&rsquo &lsquoDNA&rsquo was only approved after an appeal to the editor-in-chief, John Edsall (Friedberg 2006). Two more papers were published in 1958 by Lehman et al. and Bessman et al., which definitively established DNA polymerase was performing DNA replication. Kornberg was awarded the Nobel Prize in 1959 for his discovery of DNA polymerase I.

In 1969, Jovin et al. elucidated the amino acid composition (Jovin et al. 1969a, b). That same year, DeLucia and Cairns isolated an E. coli strain with a mutation that affected the DNA polymerase and surprisingly found that the mutant synthesized DNA normally. This discovery cast doubts on the role of DNA polymerase in replication and led groups to search for other replication enzymes. At the same time, Klenow and colleagues showed that the treatment of DNA polymerase with the proteolytic enzyme subtilisin (type Carlsberg) resulted in an increase of polymerase activity and decrease of exonuclease activity. The resulting DNA polymerase was isolated and was named the &ldquoKlenow fragment&rdquo (Klenow and Henningsen 1970a, and Klenow and Overgaard-Hansen 1970).

In 1970, DNA polymerase II of E. coli was isolated and characterized by Arthur Kornberg&rsquos son, Thomas Kornberg (Kornberg and Gefter 1970). DNA polymerase II was also independently reported on by Knippers and by Moses and Richardson in 1970 (Moses and Richardson 1970b). A year later, Thomas Kornberg and Gefter identified DNA polymerase III (Kornberg and Gefter 1971).

Recent work with DNA polymerase I has included investigating the molecular basis of substrate specificity through thermodynamic studies (Wowor et al. 2010) and single-molecule FRET experiments (Santoso et al. 2010). Hastings et al. have investigated the interactions of the five E. coli DNA polymerases during cellular stress (Hastings et al. 2010), and Kukreti et al.&rsquos studies have aimed to determine which residues are important for 3&rsquo-5&rsquo exonuclease activity (Kukreti et al. 2008).

Specificity:

DNA synthesis requires a primer strand with a free 3&rsquo-hydroxyl terminus annealed to a DNA template strand and the deoxynucleotide triphosphates form base pairs with the template. Addition is in the 5&rsquo to 3&rsquo direction with release of pyrophosphate. The enzyme is active with DNAs containing single stranded gaps and also with DNAs with single-strand breaks or nicks. Under some conditions, RNA-DNA hybrids and an RNA duplex may serve as template-primer (Setlow 1972).

The 5&rsquo to 3&rsquo exonuclease activity associated with DNA polymerase I degrades both single and double stranded DNA in the 5&rsquo to 3&rsquo direction, yielding 5&rsquo-mononucleotides. The 5&rsquo to 3&rsquo exonuclease activity is specific for double stranded DNA, yielding 5&rsquo-mononucleotides and oligonucleotides. DNA polymerase I can also excise mismatched regions in DNA (Setlow 1972).

The similar structure of DNA polymerases has indicated that most DNA polymerase enzymes use an identical two metal ion-catalyzed polymerase mechanism. One metal ion activates the primer&rsquos 3&rsquo-OH for attack on the a-phosphate of the dNTP. The other metal ion stabilizes the negative charge of the leaving oxygen and chelates the b- and g-phosphates (Steitz 1999).

The Klenow fragment is a proteolytic product of E. coli DNA polymerase I that retains polymerization and 3&rsquo to 5&rsquo exonuclease activity, but has lost 5&rsquo to 3&rsquo exonuclease activity.

Composition:

DNA polymerase I is the predominant polymerizing enzyme found in E. coli. It contains a single disulfide bond and one sulfhydryl group (Jovin et al. 1969b). Five distinct DNA polymerases have been isolated from E. coli and have been designated I, II, III, IV, and V. DNA polymerase I functions to fill DNA gaps that arise during DNA replication, repair, and recombination. DNA polymerase II also functions in editing and proofreading mainly in the lagging strand (Kim et al. 1997, Wagner and Nohmi 2000). DNA polymerase III is the main replicative enzyme. DNA polymerase IV and V have large active sites that allow for more base misincorporation, and are therefore more error-prone. They also lack proofreading-exonuclease subunits to correct misincorporations (Nohmi 2006, and Hastings et al. 2010). DNA polymerase V is present at significant levels only in SOS-induced cells and over-expression restricts DNA synthesis (Marsh and Walker 1985).

The domain shape of all polymerases whose structures are known has been described as a &ldquoright hand&rdquo with &ldquothumb&rdquo, &ldquopalm&rdquo, and &ldquofinger&rdquo domains (Kohlstaedt et al. 1992). The palm region is thought to catalyze the phosphoryl transfer, and the finger region is thought to interact with the incoming nucleoside triphosphate and the template base it is paired to. The thumb is believed to help in positioning the DNA and in translocation (Brautigam and Steitz 1998).

Molecular Characteristics:

The gene encoding DNA polymerase I (polA) contains approximately 3,000 base pairs and encodes approximately 1,000 amino acid residues in a simple polypeptide chain. Even organisms separated by a billion years of evolution (such as Deinococcus-Thermus genera and E. coli) have approximately 35% amino acid identity and approximately 50% homology (Patel et al. 2001).

Protein Accession Number: P00582

Molecular Weight:

  • 109 kDa (Jovin et al. 1969a, b)
  • Klenow fragment: 70 kDa (gel filtration, Klenow and Overgaard-Hansen 1970)
  • Maximum activity is obtained at pH 7.4 with potassium phosphate buffer for native DNA or poly dAT template-primer systems (Richardson et al. 1964)
  • Klenow fragment: Maximal activities are obtained at 7.4 with phosphate buffer and at 8.4 with Tris-HCl buffer

Isoelectric Point:

Extinction Coefficient:

  • 81,030 cm -1 M -1 (Theoretical)
  • E1%, 280 = 7.86 (Theoretical)
  • In 10 mM sodium bicarbonate, the A280/A260 ratio is 1.81 and the absorbance at 280 nm of a 1 mg/ml solution is 0.85 (Jovin et al. 1969a)
  • A divalent cation is required for activity
  • Mg 2+ at a concentration of 7 mM yields optimum activity under the conditions of the standard assay (Richardson et al. 1964)
  • Mn 2+ can partially fulfill the metal ion requirement
  • Enzyme activity is also influenced by concentrations of monovalent cations such as K + , Rb + , Cs + , and NH4 +
  • Kanchanomycin, mitomycin, bleomycin, phleomycin, ananthramycin, plurmycin A (Tanaka et al. 1965), and neomycin (Lazarus and Kitron 1973)
  • Actinomycin inhibits only when guanosine and cytosine nucleotides are present (Cohen and Yielding 1965)
  • Dideoxynucleoside
  • Arabinosyl nucleotide triphosphate
  • Deoxyuridine-5&rsquo-triphosphate and analogues of uridine and deoxyuridine with 5&rsquo-hydroxy or amino substituents (Kornberg 1974)
  • Chloroquine and some of its analogs (Cohen and Yilding 1965)

Applications:

  • High percentage incorporation of radioactivity for nick translation assays
  • Standard reference material for the study of DNA polymerases
  • Manufacturing of alternating copolymers such as poly d(A-T) and homopolymers such as poly dG-poly dC
  • Klenow fragment: DNA sequencing (Sanger et al. 1977), fill-in of 5&rsquo overhangs and removal of 3&rsquo overhangs to form blunt ends (Sambrook 1989), and second strand synthesis in mutagenesis (Gubler 1987)

DNA Polymerase I

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How does DNA polymerase work?

The DNA polymerases are enzymes that create DNA molecules by assembling nucleotides, the building blocks of DNA.

These enzymes are essential to DNA replication and usually work in pairs to create two identical DNA strands from a single original DNA molecule. DNA polymerase “reads” the existing DNA strands to create two new strands that match the existing ones.

DNA polymerase’s rapid catalysis is due to its processive nature . In the case of DNA polymerase, the degree of processivity refers to the average number of nucleotides added each time the enzyme binds a template.

As I said, the main function of DNA polymerase is to make DNA from nucleotides.

Look at the first picture:

When creating DNA, DNA polymerase can add free nucleotides only to the 3' end of the newly forming strand. This results in elongation of the newly forming strand in a 5'-3' direction. No known DNA polymerase is able to begin a new chain it can only add a nucleotide onto a pre-existing 3'-OH group, and needs a primer at which it can add the first nucleotide. In DNA replication, the first two bases are always RNA, and are synthesized by another enzyme called primase.

Since DNA polymerase requires a free 3' OH group for initiation of synthesis, it can synthesize in only one direction by extending the 3' end of the preexisting nucleotide chain. Hence, DNA polymerase moves along the template strand in a 3'-5' direction, and the daughter strand is formed in a 5'-3' direction. This difference enables the resultant double-strand DNA formed to be composed of two DNA strands that are antiparallel to each other.


Module 1 : Synthesis of Oligonucleotide and the Central Dogma

The topic of last lecture was oligonucleotide synthesis.

Now will see how this central dogma of biology involves different processes, different

chemistry for the flow of information from DNA to the protein. What is central dogma in

biology? It has to do certain processes and the first process is the replication, i.e making

the copy of the existing DNA in the cell. So, one cell is converted to two cells, both

having the pieces of the DNA.

After DNA replication, DNA has to be transcribed into RNA. There are several classes

of RNA- m RNA i.e messenger RNA, t RNA transfer RNA, r-RNA i.e. Ribosomal

RNA. Conversion of DNA to RNA is called transcription- it could be ribosomal RNA/ or

it could be t-RNA the transfer RNAs and that is what is called transcription.

Then messenger RNA with the help r RNA and the t-RNA forms the protein and this

process is called translation. Possibly you have read this in the biology books at the

lower level, but we will talk about the chemistry involved in these three processes -one is

replication, another is transcription and translation. At first, I will discuss replication.

DNAs are replicated to form new two double stranded DNA.

One possibility is that the new strand is forming double strand with the old strand. It

could be that the old strands remain coupled with each other hand the new ones are

forming double strand with each other. Third possibility is that a portion of the strands

consists of part of the old and the new strand.

So, basically three possibilities exist. At first, we have to sort out the pathway by which

DNA is being copied. What is called conservative replication? If it goes through

conservative pathway the old strands remain with each other and the new strands are the

daughter strands remain with each other. In case of semi conservative pathway, the old

and the new combine with each other to form a double helix. For dispersing pathway,

each strand is composed of old and the new one.

If your initial DNA is red colored one and the and the new DNA is when it is made that

new DNAs it is colored like that ok.

So, what will happen in conservative replication?

Old DNA are together to form the double helix and new DNA strands are also forming

double helix with each other. DNA contains a lot of nitrogens in the bases and each base

contains more than one nitrogen. If bacteria are allowed to grow in ammonium chloride

medium then nitrogen comes from ammonium chloride. So, if you use ammonium

chloride containing only 15N level (heavy isotope of nitrogen) the DNA that will be

formed here only consists of 15N initially. Now this bacteria are transferred to normal

ammonium chloride medium containing 14N i.e lighter isotope of nitrogen and allowed to

grow one cycle. What will happen?

In conservative replication, you will have a heavy DNA consists of only 15N and a light

DNA consists of only 14N. There will be significant molecular weight difference between

this two DNA double helix. If you do centrifugation the heavier DNA will collect at the

lower side of the centrifugation tube and the lighter one should be at the higher level than

the heavier one. If it follows conservative pathway this should be observed. But actually

Now consider all the three possibilities. If it is semi conservative then what will happen?

After the first cycle (i.e when the 2 strands become 4 strands), it is expected that we will

have two DNA strands each having one 14N strand and one 15N strand. It will be

intermediate between this fully 14N and fully 15N and we expect a band in between these

two. So, there should be only one band after first cycle if it is semi conservative.

If it is dispersive 50 percent of this will be coming from the old strand and 50 percent

from the new strand. So, the molecular weight of these DNA will be similar because this

is 50 50. So, there will be no difference between you have 4 strands and out of which 50

percent is the light blue. Here also you will have 4 four strands and out of that 50 percent

is the light blue. So, their molecular weight will be same. After the first cycle, you

cannot distinguish between the dispersive and the semi conservative modes.

But it can distinguish the conservative ones. The famous experiment was done by

Meselson Stahl. After the first cycle, it was found that there is only one band and it is

higher than the first band which was made from only 15N leveled DNAs. You have to go

for the second cycle in order to distinguish between the dispersive and the semi

There is another schematic diagram where the old strands are mentioned by D and the

new strand is mentioned by L. For conservative pathway, there should be two types of

DNA-DD and LL. So, you should see two bands and that is written here. After first

cycle, the initial band consisting only DD should have two bands.

For the semi conservative, if you have this DD you should see a DL at little higher level

than DD. Because it will have lower molecular weight than DD. In case of dispersive

pathway, 50 percent is covered by the new DNA strands and the rest 50 percent is the old

When you go to the second generation of now semi conservative pathway, you will have

If it is like dispersive then what will happen? Here after the second generation, there are

8 strands. Out of this eight, 6 strands actually comes from the newly generated strands

and two are the old strands. So, the molecular weight of this DNA is basically 25

That will be 75 percent of L and 25 percent of D. It will be little bit on the lower side of

the LL, but it will be in between LL and DL. For semiconservative pathway, after second

generation this is pure LL, this is LD and this is 75 percent L and 25 percent D. So,

Meselson Stahl came to the conclusion that it is actually semi conservative.

So, DNA replication is semi conservative.

As DNA replication is semi conservative, these two strands need some separation to start

replication. The old strand will act as the template for making new strand. Inevitably the

question will arise that there is a stability of this double helix. So, you have to unwind

the double helix. Remember one strand goes from 5 prime to 3 prime, another strand

goes from 3 prime to 5 prime.

So, the first step that is needed is the DNA needs to be unwind. That means, double helix

have to be converted into single helix single strand and then new strands can be

synthesized on this. Why is it semi conservative? You have to separate the new strands

from the old strands. As the synthesis is progressing you make these new strands. That

new strand will be already tied up to the old strand. Now who actually does that?

Who disrupts this helix? There must be something which is this double helix breaker.

You have to break the double helix at a particular point and then that needs to be

progressed further. Once this part is over the synthesis has to progress. Helix breaker

which is an enzyme breaks the hydrogen bonds between two strands and slowly proceeds

ahead. As it proceeds new strands of DNA are synthesized and this point is called the

An enzyme called helicase is acting as helix breaker. After breaking the helix, these two

strands again want to really go back to the double strand by forming the same Watson

Crick Base pairs. So, something also needs to be there that will hold these two separated

strands. Because this has to be separated for some time till the synthesis of the other new

strands are complete otherwise they will come back and will rejoin or reanneal again. So,

the next question is who holds the single strands generated out of the double strand?

SSB proteins i.e Single Strand Binding proteins stabilize these single strands. So, these

strands do not fall back and anneal with itself or hybridized with the other one which is

If you try to unwind the double helix the gaps between the major and minor grooves will

decrease. This will create super coiling ahead of the helices. At some point of time, it

will be very difficult to open up the portion of the DNA which you want to replicate.

Why there is supercoiling? Whenever a helix is taken out, you will always see that there

will be supercoiling in the front. That means, a force which does not want to allow

further unwinding of the DNA. Now this super coiling can be prevented by an enzyme

called DNA topoisomerase. When there is super coiling the best way to reduce that strain

due to super coiling is to nick or cut the chain at certain point. The chain which is going

below the in order to avoid or reduce the super coiling that will turn around and go to the

As the helicase progresses you need to cut or nick the DNA at difference points to reduce

This topoisomerase will cut DNA. When copying of the DNA is done, that cuts can be

sealed again. So, topoisomerase takes care of the super coiling. Number 4 is that in case

of DNA when you have a DNA strand like this 5 prime to say 3 prime direction. The

other strand also has to be copied but that is a little bit complicated. We will go to that

aspect later on. The enzyme called DNA polymerase adds the oligonucleotide one after

another. Remember DNA polymerase works by from 5 prime to 3 prime direction.

3 prime OH is attacking the 5 prime triphosphate.

There is a problem of copying this or making the complementary DNA for the original

DNA which runs from 5 prime to 3 prime direction. There is no such problem for the

other strand. For the other strand, the DNA synthesis can start from here because now

this is the 5 prime and going to the 3 prime. What happens to the complementary strand?

It has got a 3 prime OH and there is another oligonucleotide which will have a 5 prime

triphosphate. So, this OH will attack the phosphate, that goes out and that forms the

phosphodiester linkage. So, one synthesis is not a problem because there is no the

direction of the movement of the helicase or the replication fork. In this case, as the

helicase moves, the replication also moves to the right side. The synthesis of the

complimentary DNA strands which runs from 3 prime to 5 prime direction is not a

problem. Because the synthesis of the complementary DNA strand moves towards 5

Also in complementary strand, DNA polymerase does the synthesis of oligonucleotide.

There is a problem of directionality. Our system the biological system has sorted out this

problem. DNA synthesis cannot take place on a strand where there is no portion of any

double strand. So, you cannot start the DNA synthesis. If you can remember the Sangers

method there is something is called a primer. DNA polymerase can work towards 5

prime-3 prime direction in presence of a slight piece of double stranded DNA i.e primer.

Tiny portion of the double strand could be a DNA RNA double strand or it could be a

In this case the primers are RNA primers and these RNA primers are synthesized. There

must be some enzyme which will synthesize this primer. So, these primers are

synthesized by an enzyme called primase. Both the strands are now called template

strands because the strands are separated and they act as a template to synthesize the new

complementary strands. In one case, there is no problem because the direction of

synthesis matches with the progress of the replication fork or the helicase.

Let us consider this strand where there is no problem because here continuous DNA

synthesis takes place. By the way this is called the leading strand and this is what is

called the lagging strand. This is called lagging strand because the DNA synthesis cannot

be continuous. So, they are made in pieces. However, in the leading strand DNA

synthesis is continuous because in that case the 5 prime to 3 prime DNA synthesis is

matching with the progression of the replication fork and the helicase.

Now this is the reaction where the 3 prime is attacking the 5 prime phosphate and then

This is your leading strand that runs from 5 prime to 3 prime direction and here is the

lagging strand which runs from 5 prime to 3 prime in this direction. There is this RNA

primer that has to be synthesized by primase and then the DNA polymerase comes and

adds the nucleotide according to the sequence in the leading strand. In case of the

lagging strand, the synthesis has to go towards this direction i.e 5 prime to 3 prime. At

first, RNA primer is made and then The DNA polymerase strats the synthesis.

Remember this is going in the opposite direction of the replication fork.

The origin of the replication fork is at the point where it started, but then this is

continuously moving. This is done along with this RNA primer. So DNA polymerase

now jumps back here to make this strand from the 5 prime to 3 prime direction because it

is lagging behind. So, it has to go back and then again start synthesizing the piece of

DNA. But every time when it starts it needs a piece of double strand here.

Only one RNA primer serves for the leading strand, but you need several RNA primers

for the lagging strand. The RNA primers are by the way joined here join to the new piece

of DNA. Every time it comes to the starting point of the RNA. So, it again goes back and

makes the RNA. The RNA primer is again made and then the DNA synthesis takes place

up to this point. At the end of the game, you have the DNA which are made in pieces or

fragments. So, this is called the lagging strand because here the DNA synthesis is not

completed. In leading strand, DNA synthesis is almost complete except the first primer

that has to be removed later on.

But here the DNA is synthesis takes place in fragments. This fragments are called

Okazaki fragments. So, once the synthesis is all done then what will happen? For the

leading strand, you have to take this RNA primer out and this should be replaced by a

DNA. This portion should only contain DNA. So, this RNA has to be chopped out

making it free and then you add the piece of DNA to ligate this position.

So, you get the complete daughter strand which is complementary to the leading strand.

For the lagging strand, there are more work to do. This has to be chopped off because

here are the DNA. So, that will be chopped off and that will be replaced by DNA. But

the problem is that there is no connectivity between this part of the DNA and this new

DNA will replace the RNA. So, you have to do ligation.

DNA ligase comes and adds these DNA pieces. You need joining of these two ends and

that is done by DNA polymerase. DNA polymerase can only do addition from one end

and then add all the time. But if there are two DNA fragments facing each other then

there is a nick in between. The nick can only be repaired by DNA ligase.

So, basically what we have learned? We have learned that first of all DNA replication is

semi conservative. As DNA replication is semi conservative in nature, the two old

strands have to be separated. For separation, you need an enzyme called helicase.

You need SSB protein i.e single stranded binding proteins which will stabilize the two

isolated strands. Then what happens? Primers are synthesized. Primers are made of

RNA. The RNA primers bind to the old DNA strand. The DNA synthesis starts because

DNA polymerase cannot work on it. RNA primers have to be synthesized and then the

synthesis starts. Who does the synthesis? That is the DNA polymerase.

There are different categories of DNA polymerase. DNAs polymerase 3 does the

synthesis by putting the oligonucleotides one after another depending on the sequence of

the leading strand. Now there is this leading strand. There is no problem because it runs

the synthesis runs from the 5 prime to 3 prime. It runs towards the direction of the

movement of the helicase or movement of your replication fork. The lagging strand runs

from 5 prime to three prime. So, the complementary strand should run from 5 prime to 3

prime in a direction opposite to the movement of the replication fork or the helicase.

In that case, the synthesis has to be taken in fragments, the RNA primer is synthesized

and then the DNA synthesis starts by the DNA polymerase. But as it goes to the origin of

the replication, it has to go back and then again synthesize the RNA primer to start the

synthesis of the DNA. When it comes to the almost end it stops. Here again goes back

and then synthesize RNA primers by primase.

Now RNA primer and this DNA are joined. This DNA at this point is not joined to the

RNA because the synthesis started from this and ended here. So, there is a gap i.e nick

between these portions. Now what will happen? DNA polymerase 1 takes care of this

replacing the RNA primers by DNA.

But there is still nicking. DNA piece is not connected to the new the DNA piece which is

replacing the primer. As here it runs in the opposite direction there will be a

disconnection between these fragments. These are called Okazaki fragments. So, after

this the DNA ligase comes and joins these nicking points. So, that is the whole story

about the replication of DNA. So, you see it is a very complicated process. Many

enzymes are involved here- helicase i.e a helix breaker, topoisomerase that takes care of

This primase makes the primers. There are two types of DNA polymerase. One is DNA

polymerase 3 which does the synthesis. DNA polymerase 1 takes care of the replacement

of the RNA primers and replaced by the DNA. Finally, DNA ligase joins the Okazaki

fragments to make the complete piece of the complementary strand. So, that is the story


Importance

From the time of their discovery DNA polymerases have paved the way to new understandings of how DNA is replicated and how it is transcribed. They have also been crucial to the development of DNA sequencing and PCR, upon which much of modern biotechnology is built. Today polymerases are the core tools for DNA labelling, sequencing and amplification. DNA polymerases are also intrinsic components for the development of molecular diagnostics for personalised medicine. They are at the forefront, for example, of techniques to detect genomic alterations that can cause diseases like cancer or cause patients to experience adverse reactions to drugs.


Why are nucleotides added to 3' end?

The DNA is only copied in the 5' to 3' direction because eukaryotic chromosomes have many origins for each chromosome in keeping with their much larger size. If some were copied in the other direction, mistakes will happen. It keeps every cell division on the same page, so to speak.

Because DNA synthesis can only occur in the 5' to 3' direction, a second DNA polymerase molecule is used to bind to the other template strand as the double helix opens. This molecule synthesizes discontinuous segments of polynucleotides, called Okazaki fragments. Another enzyme, called DNA ligase, is responsible for stitching these fragments together into what is called the lagging strand.

The mechanism of DNA ligase is to form two covalent phosphodiester bonds between 3' hydroxyl ends of one nucleotide, ("acceptor") with the 5' phosphate end of another ("donor"). The two "sticky ends" have to be in opposite directions for replication of the entire DNA molecule to be complete.

The average human chromosome contains an enormous number of nucleotide pairs that are copied at about 50 base pairs per second. Yet, the entire replication process takes only about an hour. This is because there are many replication origin sites on a eukaryotic chromosome. Therefore, replication can begin at some origins earlier than at others. As replication nears completion, "bubbles" of newly replicated DNA meet and fuse, forming two new molecules.


Watch the video: La duplicazione del DNA HD (November 2021).