Molecular Biology

Front 1

Strand invasion

Back 1

the nucleoprotein complex (composed of the broken single-strand DNA and the recombinase) searches and identifies a region of homology in intact duplex DNA

Front 2

the Holliday junction

Back 2

a mobile junction between four strands of DNA

Front 3

homologous chromosomes

Back 3

a. corresponding chromosomes, one from each parent, which contain the same genes;
b. Homologues look alike; they have same length and centromere position; have similar banding pattern human karyotype
c. A locus (location) on one homologue contains the same types of gene which occur at the same locus on the other homologue

Front 4

crossover

Back 4

exchange of genetic material during meiosis

Front 5

homologous recombination

Back 5

the exchange at(of) similar DNA sequences from different DNA molecules

Front 6

gene knockout

Back 6

a genetically engineered organism that carries one or more genes in its chromosomes that have been made inoperative (have been "knocked out" of the organism)

Front 7

genetic maps

Back 7

a way of measuring the relative distance between two genes based on the frequency of DNA exchange

Front 8

branch migration

Back 8

movement of the Holiday junction

Front 9

resolution

Back 9

cleavage of the Holliday junction

Front 10

strand invasion

Back 10

in where short stretches of DNA become base paired. Holliday junctions occur when both strands are base paired.

Front 11

what does branch migration do?

Back 11

to increase the length of DNA exchanged

Front 12

at which site resolution may occur?

Back 12

two different sites: vertical and horizontal cuts, yielding crossover and patch products

Front 13

major steps of homologous rec

Back 13

1. alignment of two homologous DNA molecules;
2.introduction of breaks in the DNA;
3. base pairing between short stretches of the two molecules to form a Holliday junction; 4. migration of the junction; 5. resolution of the junction by DNA cleavage.

Front 14

enzymes for pairing the homologous DNA molecules?

Back 14

RecA in E. coli;
Rad51 and/or Dcm1 in eukaryotic cells.

Front 15

enzymes for introduction of reccombinogenic breaks?

Back 15

unknown for E. Coli
eukaryotes: Spo11 and HO, Spo11 during meiotic recombination and HO during yeast mating strand switching.

Front 16

enzymes for the base pairing between the single-stranded sequences

Back 16

RecBCD (in bacteria);
MRX (in eukaryotes), which process double-strand breaks to generate 3' single-stranded tails, and strand invasion itself is mediated by the RecA/Rad51/Dcm1 proteins.

Front 17

enzymes for migration of the Holliday junction

Back 17

the RuvAB complex in E. coli; and presumably by an analogous (but as yet unknown) complex in eukaryotes.

Front 18

enzymes for Holliday junction resolution?

Back 18

RuvC in E. coli;
Mus81 in eukaryotes.

Front 19

What two consequences of errors in replication and DNA damage?

Back 19

1. permanent changes tot eh DNA (mutation), changing coding sequence or regulatory sequence;
2. chemical alternation to the DNA prevent its use as template for replication and transcription.

Front 20

what are the two sources of mutation?

Back 20

1. inaccuracy in DNA replication;
2. chemical damage to the genetic material.

Front 21

simple mutations

Back 21

1. transition
2. transversions
3. insertion
4. deletion
(a NT or small # of NT)

Front 22

transition

Back 22

pyrimidine-to-pyrimidine substitutions

Front 23

transversion

Back 23

pyrimidine-to-purine or purine-to-pyrimidine

Front 24

point mutations

Back 24

alter a single nucleotide

Front 25

complex mutations

Back 25

1. extensive insertion
2. extensive deletion
3. gross rearrangements
(of chromosome structure)

Front 26

overall mutation rate

Back 26

10^-6 to 10^-11 per round of DNA replication

Front 27

hotspots

Back 27

the sites on the chromosome where mutations arise at high frequency.

Front 28

DNA microsatellites

Back 28

mutation-prone sequences that are repeats of simple di-, tri- or tetranucleotide sequences
eg. CA, CGG, and CAG

Front 29

Triple Repeats Causes Dsiease

Back 29

1. adult muscular dystrophy
2. fragile X syndrome
3. Huntington's disease

Front 30

how much proofreading improve the fidelity of DNA?

Back 30

a factor of 100

Front 31

how many possible mismatch can occur?

Back 31

12
3x4

Front 32

mismatch repair system

Back 32

1. MutS scans the DNA
2. MutL activates MutH
3. MutH nicks the strand
4. UvrD (helicase) unwind DNA
5. exonuclease cuts a big piece
6. Pol III synth. new DNA
7. DNA ligase seals
(E. coli)

Front 33

dam methylase

Back 33

E. coli enzyme that methylates adenine on parent strands of the sequence 5'-GATC-3'
(Georgia Tech)

Front 34

how might the mutation rate be regulated?

Back 34

While the rate at which new mutations arise partly reflects uncontrolled environmental factors such as exposure to mutagenic compounds or radiation, the rate is also dependent on factors that can be largely controlled. For example, the DNA repair enzymes present in the cell could be expressed at a higher or lower level, and the repair enzymes and the proofreading activity of DNA polymerase could be more or less accurate. Accordingly, natural selection could act upon the mutation rate by influencing the conservation of genetic changes that affect proofreading or DNA repair activity in any of these ways.

Front 35

selective pressures acting on an organism that'd increase or decrease mutation rate

Back 35

The mutation rate in a species reflects a balance between the cost of having deleterious mutations appear too frequently and the cost of having too little genetic diversity.evolution exerts downward pressure on the mutation rate

Front 36

two major mechanisms to ensure the fidelity of DNA replication

Back 36

1. proofreading
2. mismatch repair system

Front 37

how E. coli identify which strand is the newly synthesized one

Back 37

newly synthesized strands are identifiable based on the fact that they contain the unmethylated half of the hemi-methylated GATC sites that are scattered throughout the genome.

Front 38

how Eukaryotes identify which strand is the newly synthesized one

Back 38

Eukaryotes rely on the fact that newly synthesized lagging strands have frequent nicks (reflecting the discontinuous synthesis of Okazaki fragments). The mismatch repair system brings another two or three orders of magnitude to the fidelity of replication. this process is not yet well understood.

Front 39

what features of DNA might influence the rate of mutation at any particular site?

Back 39

sequences including di-, tri-, and tetra-nucleotide repeats are especially unstable. During DNA replication, these repeats can cause slippage of the replication machinery, leading to an alteration in the number of repeats.

Front 40

rate of spontaneous mutation

Back 40

ranging from about 10-7 to about 10-11 per round of replication for any given site within the genome.

Front 41

why are methylated C's especially prone to mutation?

Back 41

because spontaneous deamination of methylated cytosine gives rise to a thymine. Because thymine is a natural base, the cell does not recognize it as a product of a deamination reaction and will therefore not repair it. Adjacent thymine residues are also relatively mutagenic, as they can form thymine dimers when exposed to ultraviolet light. Thymine dimers are incapable of base pairing and can block the replication machinery, potentially leading to mutation.

Front 42

why does male chromosome Y have higher mutation rate?

Back 42

because genomic regions that lack a homologous region—such as parts of the male Y chromosome—may be prone to increased mutation rates because they lack any available homologous regions that could be used for recombinational repair of damaged DNA.

Front 43

Examples of diseases caused by trinucleotide repeat expansion

Back 43

1. adult (myotonic) muscular dystrophy
2. Fragile-X syndrome
3. Huntington’s disease.

Front 44

what can happen during DNA replication that can lead to the expansion of tirbucleotide?

Back 44

The number of trinucleotide repeats can change during replication if the DNA polymerase “slips,” losing its place within the repeats and restarting at the wrong position. Such slippage can produce an increase or decrease in the number of repeats.

Front 45

how might the trinucleotides expanded lead to disease?

Back 45

by altering gene expression and by introducing deleterious changes into the coding sequences of genes. In the Fragile-X syndrome, for example, expansion of a repeat at the 5’ end of the gene affects the promoter and leads to silencing of the gene. In a number of other cases, including Huntington’s disease, the trinucleotide repeat expansion occurs within the coding sequence, introducing a string of glutamate residues into the protein and rendering it toxic.

Front 46

Bleomycin

Back 46

Double-strand breaks (repairable by the double-strand break repair pathway)

Front 47

nitrosamines

Back 47

Alkylation of the DNA (repairable by direct removal of the alkyl group from the base, by base excision repair, or by nucleotide excision repair)

Front 48

x-rays

Back 48

Double-strand breaks (repairable by the double-strand break repair pathway)

Front 49

water

Back 49

Deamination of the bases (repairable by base excision repair or by nucleotide excision repair)

Front 50

methyl-nitrosoguanidine

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Alkylation of the DNA (repairable by direct removal of the alkyl group from the base, by base excision repair, or by nucleotide excision repair)

Front 51

gamma radiation

Back 51

Double-strand breaks (repairable by the double-strand break repair pathway)

Front 52

reactive oxygen species

Back 52

Damage to bases, such as the conversion of guanine to oxoG (repairable by base excision repair or by nucleotide excision repair)

Front 53

UV light

Back 53

Thymine dimers (repairable by photoreactivation or by nucleotide excision repair)

Front 54

what effect does ethidium bromide have on DNA?

Back 54

Ethidium intercalates between the bases of DNA, causing the double helix to partially unwind.

Front 55

what are the potential biological consequences of ethidium bromide?

Back 55

The presence of ethidium can have serious biological consequences, because the distortion that it makes in the DNA can lead to the introduction of small insertions or deletions during DNA replication. The addition or deletion of even a single nucleotide can change the reading frame of a coding sequence and thereby destroy the function of the encoded protein.

Front 56

what mutations does ethylmethane sulfonate (EMS)produce?

Back 56

EMS tend to produce point mutations into the DNA, because alkylated bases are subject to mispairing events, which can lead to permanent nucleotide changes if the damaged bases are not repaired prior to DNA replication

Front 57

what mutations does X-rays produce?

Back 57

Unlike EMS, X-rays produce double-stranded breaks in the DNA. Accordingly, X-rays are generally used to induce chromosomal rearrangements such as deletions, inversions, and translocations.

Front 58

what's the role of base flipping in DNA repair?

Back 58

Because damaged bases are buried within the double helix, they cannot be easily accessed for repair by the DNA glycosylase enzymes of the base excision repair system. Glycosylases get around this problem by flipping the bases out so that they project away from the double helix and into the active site of the enzyme. Apparently, the DNA can withstand this flipping of the bases without suffering excessive distortion, which might otherwise make the mechanism prohibitively costly in terms of energy.

Front 59

base excision repair

Back 59

1. recognition of the damaged base by glycosylase enzymes 2.the glycosylase (which is specific for the particular type of damage) removes the base by cleaving the glycosidic bond connecting it to the sugar component of the nucleotide.
3.leaving an abasic sugar, which is subsequently removed by endonuclease enzymes.
4. the gap left in the DNA by the endonucleolytic cleavage is repaired by DNA polymerase (using the undamaged strand as a template) and DNA ligase.

Front 60

what if the excision repair system fails to remove a damaged base prior to DNA replication would it result in a mutation?

Back 60

If the excision repair system fails to detect a damaged base prior to replication, a backup mechanism exists that can intervene before a point mutation is permanently established in the genome. Specifically, the cell contains certain glycolases that can recognize the mismatches that result from the replication of damaged bases and selectively remove the nucleotide opposite the damaged base. For example, one such glycosylase can recognize base pairs between oxoG and A and specifically remove the incorrectly incorporated (but undamaged) A. Another glycosylase targets T:G base pairs, a potential consequence of the spontaneous deamination of methylated cytosines.

Front 61

nucleotide excision repair

Back 61

in E. coli: four proteins are involved:
UvrA, UvrB, and UvrC, work together as a complex to scan the DNA for damaged nucleotides and to make incisions in the DNA surrounding any identified lesions.
Within this complex, UvrA detects distortions in the DNA that are indicative of lesions, UvrB melts the DNA surrounding any identified lesions, and UvrC cleaves the lesion-containing strand on either side of the defect. The fourth protein, a helicase called UvrD, then unwinds the cleaved DNA to release the short lesion-containing fragment. This leaves a gap in the DNA that is then filled in by DNA polymerase and DNA ligase. Eukaryotes use a similar system, although it is more complex and involves a greater number of proteins.

Front 62

which of nucleotide excision repair step is affected in the xeroderma pigmentosum?

Back 62

Individuals with xeroderma pigmentosum are especially sensitive to sunlight, and are particularly susceptible to developing skin lesions such as cancer. This disease can be caused by mutations in any number of genes involved in the nucleotide excision repair pathway, including genes involved in the detection of lesions, in the melting of the DNA at the site of any detected lesions, and in the cleavage of DNA at positions flanking the lesion.

Front 63

which is the last resort to deal with lesion?

Back 63

Translesion synthesis is the option of last resort, because its reliance on polymerases that carry out base pair-independent synthesis means that it involves a high mutation rate. Still, it is better than nothing because a failure to complete replication (a likely consequence if the replication fork were stalled indefinitely) would introduce even greater problems when the cell attempts to pass through mitosis.

Front 64

what options exist for getting past the lesion?

Back 64

DNA polymerases that are blocked by unrepaired lesions in the DNA have a couple of options for bypassing the damage and continuing replication. One possibility is use recombinational repair, which allows the polymerase to replicate past the lesion using the other daughter molecule of the replication fork as a template. A second option involves translesion synthesis, a process that uses specialized polymerases to synthesize DNA in a template-dependent but base pair-independent manner. Translesion synthesis allows the replication fork to bypass the lesion and to continue replicating using the lesion-containing strand as a template. One disadvantage of this system, however, is that the nucleotides incorporated by the specialized polymerases are not selected using Watson-Crick base pairing rules, making the process highly error-prone.

Front 65

Nick

Back 65

A nick is a point in a double stranded DNA molecule where there is no phosphodiester bond between adjacent nucleotides of one strand typically through damage or enzyme action

Front 66

hydrolysis of bases

Back 66

1. deamination
2. depurination
3. depyrimidination.

Front 67

Double-strand breaks (DSBs)

Back 67

both strands in the double helix are severed, are particularly hazardous to the cell because they can lead to genome rearrangements.

Front 68

The SOS response

Back 68

a postreplication DNA repair system that allows DNA replication to bypass lesions or errors in the DNA. The SOS uses the RecA protein. The RecA protein, stimulated by single-stranded DNA, is involved in the inactivation of the LexA repressor thereby inducing the response. It is an error prone repair system.

Front 69

the role of RuvABC in Branch migration

Back 69

1. RuvA recognizes and binds the Holliday junction.
2. Once bound to the junction, RuvA recruits RuvB, a hexameric protein that uses energy obtained from ATP hydrolysis to promote the migration of the branch.
3.The endonuclease RuvC then resolves the Holliday junction by cutting two of the strands within the sequence 5'-A/T-T-T-T-G/C

Front 70

what are the genetic consequences of branch migration occuring during recombination?

Back 70

1. increases the amount of DNA that is exchanged between recombining molecules, thereby increasing the likelihood that gene conversion will occur. 2.strand exchange occurring during branch migration produces mismatches at every position where the two recombining molecules differ in sequence. The cell can repair these mismatches, although to do so it must remove one of the two mismatched nucleotides and replace it with one that is complementary to the base on the other strand. This has the result of eliminating sequence information from one of the recombining molecules and replacing it with sequence from the other, leading to gene conversion

Front 71

4 different biological processes that rely on homologous rec

Back 71

1. repair of double-stranded DNA breaks (prokaryotes and eukaryotes
2. rescuing stalled replic. forks (prokaryotes and eukaryotes)
3. recombination with foreign DNA (prokaryotes and eukaryotes)
4. meiotic recombination (eukaryotes), and mating type switching (eukaryotes).

Front 72

the Holliday model

Back 72

1. a single-stranded nick is made on each of two homologous double-stranded DNA molecules, 2. short stretches of single-stranded DNA from each of the two molecules are exchanged.
3. produces a Holliday junction, which can move in either direction along the DNA by branch migration.
4. the Holliday junction is resolved, giving rise to either crossover or noncrossover products.

Front 73

the double-strand break repair model

Back 73

1. a double-stranded break on one of the two involved DNA molecules
2. the break is then processed so that both sides contain a 3' single-stranded tail, and then one or both of these 3' tails invades the other DNA molecule where it serves as a primer for DNA synthesis using the invaded DNA as a template.
3. produces a structure that includes two Holliday junctions, which can migrate and which are ultimately resolved to produce either crossover or noncrossover products.

Front 74

comparison of the Holliday and the DSB model

Back 74

1. both invoke a break in the DNA to initiate recombination
2. both involve strand invasion, and they both include at least one Holliday junction.
3. They differ in the Holliday model posits that a single-stranded DNA initiates recombination, whereas in the double-strand break repair model it's a double-strand break that starts the process, leaving a ssDNA tail ending in 3'
4.the DSB repair model requires DNA recombination, strand invasion of 3'ssDNA tails and elongation, in contrast to the Holliday model.
5. the double-stranded break repair model involves two Holliday junctions, whereas the Holliday model only involves one.

Front 75

how do chi sites protect E. coli against foreign DNA?

Back 75

foreign DNA usually contains few chi sites and is therefore rarely able to trigger the transformation in RecBCD activity

Front 76

how do chi sites protect E. coli against foreign DNA?

Back 76

Chi sites (or, more precisely, their absence) help E. coli protect itself against foreign DNA because foreign DNA usually contains few chi sites and is therefore rarely able to trigger the transformation in RecBCD activity.

Front 77

what are the steps involved in the recognition of chi sites by the rec machinery?

Back 77

Accordingly, RecBCD acting on foreign DNA will tend to progressively destroy both strands of the DNA, moving away from a double-strand break until it destroys the entire DNA molecule.

Front 78

how does the process of homologous rec during mating-type switching differ from that which occurs during meiotic rec.?

Back 78

meiotic recombination is initiated by Spo11, which cuts DNA in a sequence-independent fashion on either of the two involved homologous chromosomes.
mating-type switching is initiated by HO, which cuts DNA in a sequence-specific way, and always on just one of the two recombining DNA partners.
meiotic recombination uses the Rad51 and Dmc1 proteins to carry out strand exchange, whereas mating type switching only uses Rad51.
the mechanism underlying strand invasion differs between the two processes, with meiotic recombination involving a Holliday junction that is resolved to give rise to crossover or noncrossover products, and with mating type switching involving an alternative mechanism that never produces crossover products.

Front 79

why did scientists suspect that the mechanism involved in these rec. events were different?

Back 79

the absence of crossover products following mating type switching that indicated to researchers that the mechanism underlying this process must be different from that acting during meiotic recombination

Front 80

how is the unidirectionality of mating-type switching ensured in S. cerevisiae?

Back 80

a consequence of the fact that HO always induces the double-strand break at the MAT locus, and never at HMR or HML. The gene conversion occurs when the double-strand break at MAT is processed by the MRX complex to produce a gap in the DNA, and the gap is then repaired using sequence information obtained from either HMR or HML.

Front 81

why does the linear and cicular forms of the DNA affect transformation efficiency?

Back 81

Linear molecules are often much more effective at promoting homologous recombination when introduced into cells because they already contain double-stranded breaks. As a result, they can be immediately processed by enzymes such as MRX before undergoing homologous recombination with the genomic DNA. In contrast, circular DNA molecules must first be cleaved before they can recombine with genomic DNA, an additional requirement that can limit the efficiency of the transformation

Front 82

when Spo11 induces DSBs in DNA, what happens to the 5' and 3' ends at the cleavage site?

Back 82

Spo11 induces double-strand breaks in the DNA by using a tyrosine side chain to attack the phosphodiester backbone, cleaving the DNA and leaving the protein covalently attached to the 5' end of the cleaved DNA. While the 3' DNA ends at the break are left intact, the exposed 5' ends are then digested by the MRX complex (named after its component proteins Mre11, Rad50, and Xrs2) to generate recombinogenic 3' single-stranded tails.

Front 83

3 activities of the RecA

Back 83

1. bringing homologous DNA molecules together by binding to single-stranded and double-stranded DNA
2. promoting various types of strand exchange
3. ATP hydrolysis.

Front 84

when is sequence homology is required?

Back 84

1. the alignment of homologous DNA molecules at the beginning of recombination occurs when RecA-like proteins bind to single-stranded DNA and scan for homologous sequences on other DNA molecules.
2. Strand invasion also depends on sequence homology, as it requires base pairing between the invading and the invaded strands.
3. once strand invasion has occurred, branches can only migrate through regions that contain continuous sequence homology between the strands.

Front 85

how is the homology identified?

Back 85

RecA identifies homology between DNA molecules by first forming a filament on a single-stranded region of a DNA molecule and then scanning other DNA duplexes for homology to the sequence. It is thought that the scanning occurs by flipping out A and T bases from the scanned DNA to see if they are complementary to the single-stranded region. If a preliminary match is found, the three strands form a joint molecule, in which the single-stranded region initially interacts with the minor groove of the double-stranded DNA. Subsequently, strand exchange produces an energetically favorable substrate for RecA binding, helping to drive the reaction forward.

Front 86

what happens if mismatches exist between recombining strands?

Back 86

mismatch can often be repaired by the mismatch repair system, which replaces one of the mismatched nucleotides to restore a proper base pair (and produces gene conversion in the process). If not, the sequence difference is made permanent when the DNA replicates, giving rise to two double-stranded daughter DNA molecules that differ at that position.

Front 87

what effect do the hot/cold regions have on the relationship btw physical and genetic maps in an organism?

Back 87

Recombinational "hot spots" and "cold spots" distort genetic maps by locally altering the relationship between physical and genetic distance. For example, a hot spot present between two neighboring genes will lead to frequent recombination between the genes, making them appear to be farther apart than they really are. Similarly, a cold spot can produce the illusion that two distant genes are actually close together on the genome, because even though they may be physically far apart recombination rarely occurs between them.

Front 88

what might influence if a spot is hot or cold?

Back 88

the accessibility of the DNA to the Spo11 endonuclease. For example, regions that are especially good targets for to Spo11 cutting, such as promoters, will tend to undergo recombination more frequently and therefore contain hot spots.
regions of the DNA that are tightly packaged into chromatin may be "cold" if they are less accessible to Spo11.
Chi sites are another potential source of hot spots, because they stimulate recombination by provoking a transformation in the RecBCD complex that causes the complex to produce recombinogenic single-stranded 3' DNA tails.

Front 89

proteins paring homologous DNAs and strand invasion

Back 89

RecA (E.coli)
Rad51 (eukaryotes)
Dcm1 (in meiosis)

Front 90

proteins for introduction of DSB

Back 90

unknown (E.coli)
Spo11 (eukaryote in meiosis)
HO (for mating-type switching)

Front 91

proteins for processing DNA breaks to generate single strands for invasion

Back 91

RecBCD(E.coli helicase/nuclease)
MRX (Rad50/58/60 nuclease)

Front 92

proteins for assembly of strand exchange

Back 92

RecBCD and RecFOR (E.coli)
Rad52 and Rad59

Front 93

proteins for Holliday junction recognition and brand migration

Back 93

RuvAB complex (E. coli)
Unknown for eukaryotes

Front 94

proteins for resolution of Holliday junctions

Back 94

RuvC (E.coli)
Mus 81 (eukaryotes)

Front 95

RecBCD pathway

Back 95

1. generates the ssDNA ends and helps load RecA.
2. Three subunits both helicase and exonuclease
3. use ATP hydrolysis to track along the DNA and unwinding and degrading it
4.once a chi site is encountered the nuclease activity stops for the 5'-3' strand leaving a 3' termini for RecA assembly

Front 96

four main types of damage to DNA due to endogenous cellular processes

Back 96

1. oxidation of bases [e.g. 8-oxo-7,8-dihydroguanine (8-oxoG)] and generation of DNA strand interruptions from reactive oxygen species;
2. alkylation of bases (usually methylation), such as formation of 7-methylguanine
3. hydrolysis of bases, such as deamination, depurination and depyrimidination.
4. mismatch of bases, due to DNA replication in which the wrong DNA base is stitched into place in a newly forming DNA strand.

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Main DNA damage caused by exogenous agents

Back 97

1. UV light causes cross linking between adjacent cytosine and thymine bases creating pyrimidine dimers
2. Ionizing radiation such as that created by radioactive decay or in cosmic rays causes breaks in DNA strands.
3. Thermal disruption at elevated temperature increases the rate of depurination (loss of purine bases from the DNA backbone) and single strand breaks. For example, hydrolytic depurination is seen in the thermophilic bacteria, which grow in hot springs at 85-250oC.[3] The rate of depurination (300 purine residues per genome per generation) is too high in these species to be repaired by normal repair machinery, hence a possibility of an adaptive response cannot be ruled out.
4. Industrial chemicals such as vinyl chloride and hydrogen peroxide, and environmental chemicals such as polycyclic hydrocarbons found in smoke, soot and tar create a huge diversity of DNA adducts- ethenobases, oxidized bases, alkylated phosphotriesters and Crosslinking of DNA just to name a few.

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DNA repair mechanisms

Back 98

1. Direct reversal
2. Single strand damage
3. Double-strand breaks
4. Translesion synthesis

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Direct reversal (DNA repair)

Back 99

1. not require a template, because the damage can only occur in one of the four bases.
2. are specific to the type of damage incurred.
3. photoreactivation process directly reverses this UV damage by the action of the enzyme photolyase,
4. methylation of guanine bases, is directly reversed by the protein methyl guanine methyl transferase (MGMT)
5. The third type of DNA damage reversed by cells is certain methylation of the bases cytosine and adenine.

Front 100

Single strand damage repair

Back 100

the other strand can be used as a template
1. Base excision repair (BER), which repairs damage due to a single nucleotide caused by oxidation, alkylation, hydrolysis, or deamination;
2. Nucleotide excision repair (NER), which repairs damage affecting longer strands of 2-30 bases. This process recognizes bulky, helix-distorting changes such as thymine dimers as well as single-strand breaks (repaired with enzymes such UvrABC endonuclease). A specialized form of NER known as Transcription-Coupled Repair (TCR) deploys high-priority NER repair enzymes to genes that are being actively transcribed;
3. Mismatch repair (MMR), which corrects errors of DNA replication and recombination that result in mispaired nucleotides following DNA replication.

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Double-strand breaks repair

Back 101

1. non-homologous end joining (NHEJ) and
2. homologous recombinational repair

Front 102

non-homologous end joining (NHEJ)

Back 102

DNA Ligase IV, a specialized DNA Ligase that forms a complex with the cofactor XRCC4, directly joins the two ends.
To guide accurate repair, NHEJ relies on short homologous sequences called microhomologies present on the single-stranded tails of the DNA ends to be joined. If these overhangs are compatible, repair is usually accurate.
NHEJ can also introduce mutations during repair. Loss of damaged nucleotides at the break site can lead to deletions, and joining of nonmatching termini forms translocations. NHEJ is especially important before the cell has replicated its DNA, since there is no template available for repair by homologous recombination. There are "backup" NHEJ pathways in higher eukaryotes. Besides its role as a genome caretaker, NHEJ is required for joining hairpin-capped double-strand breaks induced during V(D)J recombination, the process that generates diversity in B-cell and T-cell receptors in the vertebrate immune system.

Front 103

Recombinational repair

Back 103

requires the presence of an identical or nearly identical sequence to be used as a template for repair of the break.
The enzymatic machinery responsible for this repair process is nearly identical to the machinery responsible for chromosomal crossover during meiosis. This pathway allows a damaged chromosome to be repaired using a sister chromatid or a homologous chromosome as a template. DSBs caused by the replication machinery attempting to synthesize across a single-strand break or unrepaired lesion cause collapse of the replication fork and are typically repaired by recombination.

Front 104

Translesion synthesis

Back 104

allows the DNA replication machinery to replicate past damaged DNA.
This involves the use of specialized translesion DNA polymerases that can insert bases at the site of damage.
Some mechanisms of translesion synthesis introduce mutations, but others do not.
From the cell's perspective, the potential for introducing mutations during translesion synthesis is less dangerous than continuing the cell cycle with an incompletely replicated chromosome.