Chapter 6: DNA Repair and Replication

DNA synthesis requires a multi enzyme complex -what are included (9)

1) DNA template 2) dNTP's (dATP, dCTP, dTTP, dGTP) Enzymes 3) DNA dependent DNA polymerase (DNA polymerase) 4) Primase (RNA polymerase) 5) Helices 6) Topoisomerase 7) Single Strand DNA Binding protein (SSB) 8) Clamp 9) Primer! (RNA polymerases do not need it)

Single-Strand Binding (SSB) Proteins

Bind to ssDNA and prevent the DNA double helix from re-annealing after DNA helicase unwinds it, maintaining the strand separation.

How is a DNA double helix is opened at replication origins?

DNA sequences at replication origins are recognized by initiator proteins (not shown), which locally pry apart the two strands of the double helix. The exposed single strands can then serve as templates for copying the DNA.

The Meselson-Stahl experiment

In each round of DNA replication, each of the two strands of DNA is used as a template for the formation of a new, complementary strand. DNA replication is "semiconservative" because each daughter DNA double helix is composed of one conserved strand and one newly synthesized strand. Labeled DNA with heavy nitrogen N15 and then allowed replication to take place with N-14 (regular N) and looked at the density of the DNA molecules that resulted (heavy strands and light strands)

DNA polymerase -are there variations? -processing speed (bp/sec) -synthesis direction

Many sub-types Highly processive (up to 1000 bp/sec) Synthesis 5'-3' direction

what identifies origins of replication? What proteins are required for initiation of replication? What does this does result in?

Specific sequences identify DNA origins of replication Initiation at specific A-T rich sites called Origins Proteins required: Initiator proteins Helicase Primase Bidirectional replication forks

Errors made during DNA replication must be corrected to...

avoid mutations. If uncorrected, a mismatch will lead to a permanent mutation in one of the two DNA molecules produced by the next round of DNA replication.

what happens to chemical modifications if left unrepaired?

they can produce mutations

nonhomologous end-joining

In nonhomologous end joining, the break is first "cleaned" by a nuclease that chews back the broken ends to produce flush ends. The flush ends are then stitched together by a DNA ligase. Some nucleotides are lost in the repair process, as indicated by the black lines in the repaired DNA.

Look carefully at the structure of the molecules. What would you expect to happen if dideoxycytidine triphosphate ddCTP was added to a DNA replication reaction in large excess over the concentration of deoxycytosine triphosphate (dCTP)? Would it be incorporated into the DNA? If it was, what would happen after that? Give your reasoning.

ddCTP has no open hydroxyl group so DNA replication would terminate after its addition

Clamp protein

- Clamp allows DNA polymerase to remain attached to DNA for long stretches -Clamp loader loads and unloads Clamp protein and DNA polymerase Clamp protein. DNA polymerase has the tendency to fall quickly from the DNA. This is advantageous when synthesizing Okazaki fragments, but is not so good when synthesizing the leading strand. The clamp protein is a ring like structure mounted behind the DNA polymerase that allows the DNA polymerase to remain attached to the DNA for a long time. An accessory protein called Clamp Loader, loads and unloads the clamp and the polymerase to the DNA. me: Attaches to DNA polymerase and holds it onto the DNA allowing it to process; clamp loader protein help facilitates this binding

On lagging strand, DNA is synthesized in fragments (Okazaki fragments) -synthesis steps

1. Primase makes short RNA primer 10 nucleotides long 2. DNA polymerase makes 200 nucleotide long Okazaki fragment 3. Rnase degrades RNA primers 4. DNA Ligase joins DNA strand

DNA Helicase

Also known as helix destabilizing enzyme. Unwinds the DNA double helix at the Replication Fork.

5'-3' direction of DNA synthesis results in asymmetrical replication forks: why?

At each replication fork, the lagging DNA strand is synthesized in pieces. Because both of the new strands at a replication fork are synthesized in the 5′-to-3′ direction, the lagging strand of DNA must be made initially as a series of short DNA strands, which are later joined together. The upper diagram shows two replication forks moving in opposite directions; the lower diagram shows the same forks a short time later. To replicate the lagging strand, DNA polymerase uses a backstitching mechanism: it synthesizes short pieces of DNA (called Okazaki fragments) in the 5′-to-3′ direction and then moves back along the template strand (toward the fork) before synthesizing the next fragment.

Single stranded DNA binding protein (SSB protein) -basic role

Binds to single stranded DNA Prevents hairpin loop formation Exposes DNA bases

Most common types of DNA damage (3)

Depurination can remove guanine (or adenine) from DNA. (Loss of a nitrogenous base (sugar-phosphate backbone stays in but not guanine or adenine)) Deamination converts cytosine to an altered DNA base, uracil Two adjacent thymine bases become covalently attached to each other to form a thymine dimer

error calculations -number of DNA pol errors per cell cycle (on average)

Human genome = 3.2 x 109 bp DNA Pol error rate 1x107 base *320 error per cell cycle*

Homologous recombination

If a double-strand break occurs in one of two daughter DNA double helices after DNA replication has occurred, but before the daughter chromosomes have been separated, the undamaged double helix can be readily used as a template to repair the damaged double helix by homologous recombination. This is a more involved process than non-homologous end joining, but it accurately restores the original DNA sequence at the site of the break. me:After replication you have sister chromatids so if there is a double stranded break in one of the sister chromatids, the other sister chromatid can be used for synthesis as a template strand which is called homologous recombination and is flawless

Why does the chain grows 5 ′→3 ′?

If chain growth proceeded in the opposite direction (3′→5′), proofreading mechanisms would terminate chain elongation

Chemical modifications of nucleotides, if left unrepaired, produce mutations. -deamination -depurination

(A) Deamination of cytosine, if uncorrected, results in the substitution of one base for another when the DNA is replicated. Uracil differs from cytosine in its base-pairing properties and preferentially base-pairs with adenine. The DNA replication machinery therefore inserts an adenine when it encounters a uracil on the template strand. (B) Depurination, if uncorrected, can lead to the loss of a nucleotide pair. When the replication machinery encounters a missing purine on the template strand, it can skip to the next complete nucleotide, as shown, thus producing a daughter DNA molecule that is missing one nucleotide pair. In other cases (not shown), the replication machinery places an incorrect nucleotide across from the missing base, again resulting in a mutation.

Base excision repair of a T•G mismatch. (short 3 steps and then longer word explanation) -glycosylase -APE1 -AP lyase -DNA pol B -DNA ligase

1) Specific proteins recognize damaged DNA (depurination, mispairing, dimers) and excise damaged strand 2) DNA pol synthesizes new DNA 3) DNA ligase seals nick word: Base excision repair of a T·G mismatch. A DNA glycosylase specific for G·T mismatches, usually formed by deamination of 5-mC residues, flips the thymine base out of the helix and then cuts it away from the sugar-phosphate DNA backbone (step 1), leaving just the deoxyribose (black dot). An endonuclease specific for the resultant baseless site (apurinic endonuclease I, APE1) then cuts the DNA backbone (step 2), and the deoxyribose phosphate is removed by an endonuclease, apurinic lyase (AP lyase), associated with DNA polymerase β, a specialized DNA polymerase used in repair (step 3). The gap is then filled in by DNA Pol β and sealed by DNA ligase (step 4), restoring the original G·C base pair. me: DNA glycosylase removes the base (the purine guanine) from the sugar in the molecule, APEI endonuclease nicks (cuts) the backbone, AP lyase removes the sugar phosphate backbone, and then DNA Pol 1 puts in the right nucleotide and DNA ligase seals the nick

Several members of the same family were diagnosed with the same kind of cancer when they were unusually young. Which one of the following is the most likely explanation for this phenomenon? It is possible that the individuals with the cancer have A) inherited a cancer-causing gene that suffered a mutation in an ancestor's somatic cells. B) inherited a mutation in a gene required for DNA synthesis. C) inherited a mutation in a gene required for mismatch repair. D) inherited a mutation in a gene required for the synthesis of purine nucleotides.

??

In which order are the following enzymatic activities required for DNA replication on the lagging strand in eukaryotes? Each activity may be used more than once or not at all. 1) DNA polymerase 2) RNA polymerase 3) DNase 4) RNase 5) DNA ligase A) 2 1 4 1 5 B) 1 2 3 4 5 C) 5 2 1 3 4 D) 3 2 1 4 3 E) 2 1 3 1 5

A

*Nucleotide excision repair in human cells. (4 steps)*

A DNA lesion that causes distortion of the double helix, such as a thymine dimer, is initially recognized by a complex of the XP-C (xeroderma pigmentosum C protein) and 23B proteins (step 1). This complex then recruits transcription factor TFIIH, whose helicase subunits, powered by ATP hydrolysis, partially unwind the double helix. XP-G and RPA proteins then bind to the complex and further unwind and stabilize the helix until a bubble of ≈25 bases is formed (step 2). Then XP-G (now acting as an endonuclease) and XP-F, a second endonuclease, cut the damaged strand at points 24-32 bases apart on each side of the lesion (step 3). This releases the DNA fragment with the damaged bases, which is degraded to mononucleotides. Finally the gap is filled by DNA polymerase exactly as in DNA replication, and the remaining nick is sealed by DNA ligase (step 4). me: Thymine dimers recongized by proteins that recruit protiens that cut it out, DNA opening, DNA resynthesis and DNA ligase Mutations in XP DNA repair proteins that repair mutations in skin skells leads to high rates of cancer

Incorporation of new nucleotides to growing chain

A new DNA strand is synthesized in the 5′-to-3′ direction. At each step, the appropriate incoming nucleotide is selected by forming base pairs with the next nucleotide in the template strand: A with T, T with A, C with G, and G with C. Each is added to the 3′ end of the growing new strand, as indicated. 5' end is attached to phosphate , 3' has a hydroxyl group at the bottom, next nucleoide brings 5' pyrophosphate to bottom of the previous nucelotide to bind with it's 3' hydroxyl in dehydration syntheis (Condensation reaction)

Topoisomerase's -in depth

As DNA helicase unwinds the DNA double helix, it generates a section of overwound DNA. Tension builds up because the chromosome is too large to rotate fast enough to relieve the buildup of torsional stress. The broken bars in the left-hand panel represent approximately 20 turns of DNA. DNA topoisomerases relieve this stress by generating temporary nicks in the DNA. Prevents supercoiling and tension in the twisted DNA; releases positive twists upstream (ahead) of replication fork and then DNA polymerase reseals it

Chapter 6: Essential Concepts

Before a cell divides, it must accurately replicate the vast quantity of genetic information carried in its DNA. Because the two strands of a DNA double helix are complementary, each strand can act as a template for the synthesis of the other. Thus DNA replication produces two identical, double-helical DNA molecules, enabling genetic information to be copied and passed on from a cell to its daughter cells and from a parent to its offspring. During replication, the two strands of a DNA double helix are pulled apart at a replication origin to form two Y-shaped replication forks. DNA polymerases at each fork produce a new complementary DNA strand on each parental strand. DNA polymerase replicates a DNA template with remarkable fidelity, making only about one error in every 107 nucleotides copied. This accuracy is made possible, in part, by a proofreading process in which the enzyme corrects its own mistakes as it moves along the DNA. Because DNA polymerase synthesizes new DNA in only one direction, only the leading strand at the replication fork can be synthesized in a continuous fashion. On the lagging strand, DNA is synthesized in a discontinuous backstitching process, producing short fragments of DNA that are later joined together by DNA ligase. DNA polymerase is incapable of starting a new DNA chain from scratch. Instead, DNA synthesis is primed by an RNA polymerase called primase, which makes short lengths of RNA primers that are then elongated by DNA polymerase. These primers are subsequently erased and replaced with DNA. DNA replication requires the cooperation of many proteins that form a multienzyme replication machine that copies both DNA strands as it moves along the double helix. In eukaryotes, a special enzyme called telomerase replicates the DNA at the ends of the chromosomes. The rare copying mistakes that escape proofreading are dealt with by mismatch repair proteins, which increase the accuracy of DNA replication to one mistake per 109 nucleotides copied. Damage to one of the two DNA strands, caused by unavoidable chemical reactions, is repaired by a variety of DNA repair enzymes that recognize damaged DNA and excise a short stretch of the damaged strand. The missing DNA is then resynthesized by a repair DNA polymerase, using the undamaged strand as a template. If both DNA strands are broken, the double-strand break can be rapidly repaired by nonhomologous end joining. Nucleotides are lost in the process, altering the DNA sequence at the repair site. Homologous recombination can flawlessly repair double-strand breaks using an undamaged homologous double helix as a template.

DNA Polymerase

Builds a new duplex DNA strand by adding nucleotides in the 5' to 3' direction. Also performs proof-reading and error correction. Many different types of DNA Polymerase exist, each of which perform different functions in different types of cells.

DNA acts as a template for its own duplication

DNA acts as a template for its own duplication. Because the nucleotide A will successfully pair only with T, and G with C, each strand of a DNA double helix—labeled here as the S strand and its complementary S′ strand—can serve as a template to specify the sequence of nucleotides in its complementary strand. In this way, both strands of a DNA double helix can be copied precisely. It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material. (Watson & Crick Nature 1953)

DNA ligase

DNA ligase joins together Okazaki fragments on the lagging strand during DNA synthesis. The ligase enzyme uses a molecule of ATP to activate the 5′ end of one fragment (step 1) before forming a new bond with the 3′ end of the other fragment (step 2).

DNA dependent DNA polymerase -Two catalytic sites: -synthesis direction

DNA polymerase contains separate sites for DNA synthesis and proofreading. DNA polymerase is shown with the replicating DNA molecule and the polymerase in the polymerizing mode (left) and in the proofreading mode (right). The catalytic sites for the polymerization activity (P) and error-correcting proofreading activity (E) are indicated. When the polymerase adds an incorrect nucleotide, the newly synthesized DNA strand (red ) transiently unpairs from the template strand (orange), and its growing 3′ end moves into the error-correcting catalytic site (E) to be removed. Delta delta G isnt big; its big enough so that theres a mistake only 1/10,000 base pairs > improved by proofreading *Polymerization site (pol) is in one domain of the molecule; exonucleosite (Exo) on other end and has exonuclease activity that cuts only the terminal nucleotide off a nucleic chain* So two catalytic sides: one to add (polymerizing site= Pol) and one to cut out mistakes (exo)

DNA synthesis is carried out by a group of proteins that act together as a replication machine. -clamp process for okazaki vs leading -clamp loader -helicase....etc -analysis of image A and B (see image)

DNA polymerases are held on the leading and lagging strands by circular protein clamps that allow the polymerases to slide. On the lagging-strand template, the clamp detaches each time the polymerase completes an Okazaki fragment. A clamp loader (not shown) is required to attach a sliding clamp each time a new Okazaki fragment is begun. At the head of the fork, a DNA helicase unwinds the strands of the parental DNA double helix. Single-strand DNA-binding proteins keep the DNA strands apart to provide access for the primase and polymerase. For simplicity, this diagram shows the proteins working independently; in the cell, they are held together in a large replication machine, as shown in (B). (B) This diagram shows a current view of how the replication proteins are arranged when a replication fork is moving. To generate this structure, the lagging strand shown in (A) has been folded to bring its DNA polymerase in contact with the leading-strand DNA polymerase. This folding process also brings the 3′ end of each completed Okazaki fragment close to the start site for the next Okazaki fragment. Because the lagging-strand DNA polymerase is bound to the rest of the replication proteins, it can be reused to synthesize successive Okazaki fragments; in this diagram, the lagging-strand DNA polymerase is about to let go of its completed Okazaki fragment and move to the RNA primer that is being synthesized by the nearby primase. Me: Replisome or replication machine is an assembly of proteins at the replication site that replicate the DNA quickly; multiple origins of replication on the 46 chromsomes in humans so DNA synthesis is proceeding at different places on the chromosome at the same time;

telomeres

Diresta: Telomeres allow completion of DNA synthesis at end of chromosomes. Without a special mechanism to replicate the ends of linear chromosomes, DNA would be lost during each round of cell division. DNA synthesis begins at origins of replication and continues until the replication machinery reaches the ends of the chromosome. The leading strand is reproduced in its entirety. But the ends of the lagging strand can't be completed, because once the final RNA primer has been removed there is no way to replace it with DNA. These gaps at the ends of the lagging strand must be filled in by a special mechanism to keep the chromosome ends from shrinking with each cell division. me: me: because there has to be a primer for DNA, the end primer has no open 3' end so it cannot be replaced and thus DNA should get shortened with each replication; but telmoeres prevent this. Telomeres are non coding sequences (G rich) that can be degraded at the end of each DNA replication

Proofreading by DNA polymerase. -how much does it decrease the error rate?

Editing site (exonucleolytic proofreading) - 3'-to-5' exonuclease Stalling mechanism in P brings misincorporated nucleotide to E site Decreases error rate 100 times to 1 error every 1x10^7 bases. (from 10^5)

editing site of DNA pol decreases error rate to

Editing site: Decreases error rate 100 times to 1 error every 1x10^7 bases.

steps of homogenous recombination (A-F) image included

Homologous Recombination Can Flawlessly Repair DNA Double-Strand Breaks A) Homologous recombination most often occurs shortly after a cell's DNA has been replicated before cell division. B) A nuclease chews back the 5′ ends of the two broken strands at the break. C) One of the broken 3′ ends "invades" the unbroken homologous DNA duplex and searches for a complementary sequence through base-pairing. D) Once an extensive, accurate match is found, the invading strand is elongated by a repair DNA polymerase, using the complementary strand as a template . E) After the repair polymerase has passed the point where the break occurred, the newly repaired strand rejoins its original partner, forming base pairs that hold the two strands of the broken double helix together . F) Repair is then completed by additional DNA synthesis at the 3′ ends of both strands of the broken double helix, followed by DNA ligation. The net result is two intact DNA helices, where the genetic information from one was used as a template to repair the other.

Holliday junction

Homologous strands Cut Exchange and Ligation Branch migration Resolution Holliday junction, cross-shaped structure that forms during the process of genetic recombination, when two double-stranded DNA molecules become separated into four strands in order to exchange segments of genetic information. This structure is named after British geneticist Robin Holliday, who proposed the original model for homologous (or general) recombination in 1964. Homologous recombination occurs during meiosis and is characterized by the exchange of genes between a maternal chromatid and a paternal chromatid of a homologous chromosome pair. The two parent DNA molecules, which have long stretches of similar base sequences, are separated into single strands, resulting in base pairing that leads to a four-stranded DNA structure. The Holliday junction travels along the DNA duplex by "unzipping" one strand and reforming the hydrogen bonds on the second strand.

How many Origins of Replication in Human DNA? Assuming that there are no time constraints on replication of the genome of a human cell, what would be the minimum number of origins that would be required? If replication had to be accomplished in an 8 hour S-phase and replication forks moved at 50 nucleotides/second, what would be the minimum number of origins required to replicate the human genome? (Recall that the human genome comprises a total of 6.4x109 nucleotides on 46 chromosomes)

If there were no time constraints on replication, one origin would be required for each chromosome; thus, a minimum of 46 origins, equal to the number of chromosomes in a human cell, would be needed. In 8 hours (28,800 seconds) the two replication forks from one origin would synthesize 2.88x106nucleotides [(2 forks) x(50 nucleotides/second)x(2.88 x104seconds)]. To replicate the entire genome would require about 2200 equally spaced origins [(6.4 x 109nucleotides)/(2.88 x 106nucleotides/origin) = 2222 origins]. It is estimated that the human genome has about 10,000 origins of replication, more than enough to finish replication within the time allotted in the cell cycle.

Telomerase

Lengthens telomeric DNA by adding repetitive nucleotide sequences to the ends of eukaryotic chromosomes. This allows germ cells and stem cells to avoid the Hayflick limit on cell division

DNA repair enzymes preferentially repair mismatched bases on the newly synthesized strand using the old DNA strand as a template. If mismatches were repaired instead without regard for which strand served as the template, would mismatch repair reduce replication errors? Would such an indiscriminate mismatch repair result in fewer mutations, more mutations or the same number of mutations as there would have been without any repair at all?

Mismatch repair normally corrects a mistake in the new strand, using information in the old, parental strand. If the old strand were 'repaired' using the new strand that contains a replication error as the template, then the error would become a permanent mutation in the genome. The old 'correct' information would be erased in the process. Therefore, if repair enzymes did not distinguish between the two strands, there would be only a 50% chance that any given replication error would be corrected. Overall, such indiscriminate repair would introduce the same number of mutations as would be introduced if mismatch repair did not exist. In the absence of repair a mismatch would persist until the next replication. When the replication fork passed the mismatch, and the strands were separated, properly paired nucleotides would be inserted opposite each of the nucleotides involved in the mismatch. A normal, nonmutant duplex would be made from the strand containing the original information; a mutant duplex would be made from the strand that carried the misincorporated nucleotide. Thus, the original misincorporation event would lead to 50%mutants and 50% nonmutants in the progeny. This outcome is equivalent to that of indiscriminate repair: averaged over all misincorporation events, indiscriminate repair would also yield 50% mutants and 50% nonmutants among the progeny.

Cells can repair double-strand breaks in one of two ways (name them)

Nonhomologous end-joining Homologous recombination

DNA polymerase P site -error rate -double check mechanism

Polymerizing site Error rate of 1 error every 1x105 bases. "Double check" mechanism: Base pairing leads to conformational change Conformational change leads to bond formation

primase

RNA primers are synthesized by an RNA polymerase called primase, which uses a DNA strand as a template. Like DNA polymerase, primase works in the 5′-to-3′ direction. Unlike DNA polymerase, however, primase can start a new polynucleotide chain by joining together two nucleoside triphosphates without the need for a base-paired 3′ end as a starting point. (In this case, ribonucleoside triphosphates, rather than deoxyribonucleoside triphosphates, provide the incoming nucleotides.)

DNA Ligase

Re-anneals the semi-conservative strands and joins Okazaki Fragments of the lagging strand.

Topoisomerase

Relaxes the DNA from its super-coiled nature. Relieves strain of unwinding by DNA helicase; this is a specific type of topoisomerase

telomerase -dependent on what -templates -rich in what nucleotide? -repeats

Telomerase: RNA dependent DNA polymerase Telomerase carries its own RNA templates Generates guanine-rich, 6-8 base repeats Primase and DNA polymerase complete synthesis

Mismatch excision repair in human cells. -what does it do? -what three main proteins are involved in recognition and binding of an endonuclease -proteins and steps involved

The mismatch excision-repair pathway corrects errors introduced during replication. A complex of the *MSH2* and *MSH6 proteins* binds to a mispaired segment of DNA in such a way as to distinguish between the template and newly synthesized daughter strands (step 1). This triggers binding of *MLH1* and the resulting DNA-protein complex then binds an endonuclease that cuts the newly synthesized daughter strand. Next a DNA helicase unwinds the helix, and an exonuclease removes several nucleotides from the cut end of the daughter strand, including the mismatched base (step 2). Finally, as with base excision repair, the gap is then filled in by a DNA polymerase (Pol δ, in this case) and sealed by DNA ligase (step 3). me: Proteins recognize and bind to site of mismatch, recruit excision proteins (endonuclease) that cuts out the a whole strand, then DNA helicase and DNA exonuclease somes in, gap is repaired by DNA polymerase and ligase

bidirectionally of replication forks + (see image)

The two replication forks move away in opposite directions at each replication origin. (A) These drawings represent the same portion of a DNA molecule as it might appear at different times during replication. The orange lines represent the two parental DNA strands; the red lines represent the newly synthesized DNA strands. (B) An electron micrograph showing DNA replicating in an early fly embryo. The particles visible along the DNA are nucleosomes, structures made of DNA and the protein complexes around which the DNA is wrapped

Strand directed mismatch proofreads newly synthesized DNA strand -decreases error to what -how is it done (VERY GENERAL) -error rate in human genome (number of errors per round of replication)

they increase the fidelity by another 100 fold; can recognize mismatches and fix them whehter it's a nick or a mismatch; will cut out that strand and synthesize with DNA polymerase and then seal with DNA ligase Further decreases error incorporation 100 times to 1 x 10^9 bases Error rate in human genome 3.4 bases per round of replication

During DNA replication in the cell, DNA primase makes short primers that are then extended by the replicative DNA polymerases. These primers ... A) more frequently in the leading strand than the lagging strand. B) are made up of DNA. C) generally have a higher number of mutations compared to their neighboring DNA. D) are joined to the neighboring DNA by DNA ligase. E) provide a 3′-phosphate group for the DNA polymerases to extend.

?

DNA clamp

A protein which prevents DNA polymerase III from dissociating from the DNA parent strand.

Which type of DNA damage is considered the most deleterious to the stability of the genome? A) abasic sites B) double-stranded breaks C) point mutations D) single-stranded breaks E) thymine dimers

B

how can cells further decrease error rate?

Cell can also fix mistakes via mismatch repair mechanisms;

endonuclease vs exonuclease

Exonucleases are enzymes that work by cleaving nucleotides one at a time from the end (exo) of a polynucleotide chain. ... Its close relative is the endonuclease, which cleaves phosphodiester bonds in the middle (endo) of a polynucleotide chain.

mismatch repair

Mismatch repair eliminates replication errors and restores the original DNA sequence. When mistakes occur during DNA replication, the repair machinery must replace the incorrect nucleotide on the newly synthesized strand, using the original parent strand as its template. This mechanism eliminates the mutation.

DNA Helices -requires what? -directionality specifics

Opens dsDNA Requires ATP 5'-to-3' and 3'-to-5' helicases

Before proofreading (just natural) # of polymerization errors

Polymerization site: 1 error every 1x10^5 bases.

okazagi fragments -RNA primer intervals (# of nucleotides) -length of primers -DNA pol -DNA ligase

Multiple enzymes are required to synthesize Okazaki fragments on the lagging DNA strand. In eukaryotes, RNA primers are made at intervals of about 200 nucleotides on the lagging strand, and each RNA primer is approximately 10 nucleotides long. Primers are removed by nucleases that recognize an RNA strand in an RNA/DNA helix and degrade it; this leaves gaps that are filled in by a repair DNA polymerase that can proofread as it fills in the gaps. The completed fragments are finally joined together by an enzyme called DNA ligase, which catalyzes the formation of a phosphodiester bond between the 3′-OH end of one fragment and the 5′-phosphate end of the next, thus linking up the sugar-phosphate backbones. This nick-sealing reaction requires an input of energy in the form of ATP me: Primer is laid down, polymerase adds onto a primer 5' to 3' then somewhere downstream of that another primer binds and DNA polymerase fills in the gap; when the primers need to be removed, Rnase degrades RNA primer, DNA polymerase replaces it with DNA, and DNA ligase seals DNA strand

mutations in any of the XP-A through XP-G genes

Mutations in any of at least seven different genes, called XP-A through XP-G, lead to inactivation of this repair system and cause xeroderma pigmentosum, a hereditary disease associated with a predisposition to cancer. Individuals with this disease frequently develop melanomas and squamous cell carcinomas if their skin is exposed to the UV rays in sunlight.

cancer and mismatch repair

Predisposition to a colon cancer (hereditary nonpolyposis colorectal cancer) results from an inherited loss-of-function mutation in one copy of either the MLH1 or the MSH2 gene. The MSH2 and MLH1 proteins are essential for DNA mismatch repair

Primase

Provides a starting point of RNA (or DNA) for DNA polymerase to begin synthesis of the new DNA strand.

The basic mechanism of DNA repair (3 steps)

Step 1 (excision): the damage is cut out by one of a series of nucleases, each specialized for a type of DNA damage. Step 2 (resynthesis), the original DNA sequence is restored by a repair DNA polymerase, which fills in the gap created by the excision events. Step 3 (ligation), DNA ligase seals the nick left in the sugar-phosphate backbone of the repaired strand. me: In step 1 (excision), the damage is cut out by one of a series of nucleases, each specialized for a type of DNA damage. In step 2 (resynthesis), the original DNA sequence is restored by a repair DNA polymerase, which fills in the gap created by the excision events. In step 3 (ligation), DNA ligase seals the nick left in the sugar-phosphate backbone of the repaired strand. Nick sealing, which requires energy from ATP hydrolysis, remakes the broken phosphodiester bond between the adjacent nucleotides

more on telomerase -why is it necessary and how is it done -uniqueness of telomerase

To complete the replication of the lagging strand at the ends of a chromosome, the template strand is first extended beyond the DNA that is to be copied. To achieve this, the enzyme telomerase adds more repeats to the telomere repeat sequences at the 3′ end of the template strand, which then allows the lagging strand to be completed by DNA polymerase, as shown. The telomerase enzyme carries a short piece of RNA (blue) with a sequence that is complementary to the DNA repeat sequence; this RNA acts as the template for telomere DNA synthesis. After the lagging-strand replication is complete, a short stretch of single-stranded DNA remains at the ends of the chromosome, as shown. Telomerase is a reverse trancriptase; it uses a guide of its own RNA (G rich sequence) to extend the template strand of the chromsome which allows RNA plymerase to add a primer and then DNA polymerase to come fill that in, allowing for extension of the telomeres

shape of replication forks

are assymetrical