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Chapter 7: DNA Repair and Recombination I. Introduction A. Introduction: Reasons For DNA Damage 1. DNA damage can occur due to internal or external cues a. Internal cues generally refer to mistakes that directly arise due to the processes DNA replication or DNA repair b. External cues include many types of mutagens that can cause DNA damage 2. A common cause of DNA damage is mistakes due to DNA replication 3. DNA replication is generally a process that occurs with high fidelity a. Over many cell divisions (i.e. many rounds of replication) errors do occur because the replicative DNA polymerases are not perfect b. Oftentimes the proofreading mechanisms of the replicative polymerases pick up the errors c. If the replicative DNA polymerases miss the error, then it becomes fixed in the DNA sequence 4. A variety of different external agents can also cause DNA damage a. Chemical agents
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Chapter 7: DNA Repair and Recombination

I. Introduction

A. Introduction: Reasons For DNA Damage

1. DNA damage can occur due to internal or external cues

a. Internal cues generally refer to mistakes that directly arise due to the processes DNA replication or DNA repair

b. External cues include many types of mutagens that can cause DNA damage

2. A common cause of DNA damage is mistakes due to DNA replication

3. DNA replication is generally a process that occurs with high fidelity

a. Over many cell divisions (i.e. many rounds of replication) errors do occur because the replicative DNA polymerases are not perfect

b. Oftentimes the proofreading mechanisms of the replicative polymerases pick up the errors

c. If the replicative DNA polymerases miss the error, then it becomes fixed in the DNA sequence

4. A variety of different external agents can also cause DNA damage

a. Chemical agents

b. Ionizing radiation

c. Oxygen free radicals

B. Introduction: Definition of Mutation

1. A mutation results from the process of DNA damage

a. Internal Cues (considered spontaneous)

b. External Cues (considered induced)

2. Mutation can have two definitions depending on the type of Biologist that is studying mutations

3. Geneticists define a mutation as a heritable change in DNA sequence that leads to a specific phenotype

a. This is a very specific definition in that the change in DNA sequence must result in an overt physical change for the organism

b. This definition does not encompass changes in DNA sequence that do not lead to phenotypic changes

4. Molecular Biologists define mutation as a change in DNA sequence (there is no concern with regards to phenotype)

C. Introduction: The Effects of Mutation

1. Mutations have three important effects in nature, and one major use in the laboratory

2. Effect 1: Mutations are an important source of genetic variation that drives evolutionary change

3. Effect 2: Mutations may have deleterious or (rarely) advantageous consequence to an organism or its descendants

4. Effect 3: Mutations in germ cells lead to heritable genetic disorders, whereas those in somatic cells may lead to acquired disease

5. In the laboratory, organisms carrying mutations in specific genes allow for characterization of those genes

D. Introduction: Locations of Mutations Within Genes

1. Mutations can occur within the regulatory regions of a gene or within the protein coding region

2. Mutations within the regulatory regions generally affect the amount of protein produced

a. Promoter regions

b. Sequences that will encode the 5’ UTR or 3’ UTR of an mRNA

3. Mutations in the protein coding region will affect the corresponding amino acid sequence of the peptide that will be produced

II. Types of Mutations

A. Types of Mutation: Introduction

1. There are several different types of mutations that can occur to the DNA

2. Some types of mutations have the opportunity to be more or less deleterious to the organism than others

3. Types of mutations that can occur are as follows

a. Base Substitutions

b. Deletions

c. Insertions

d. Chromosomal rearrangements

B. Types of Mutations: Base Substitutions

1. Base substitutions result from a change of one nucleotide base in the DNA sequence to another

2. Base substitutions are a type of point mutation as there is a change in the DNA sequence at one specific point

3. With respect to the base changes anywhere in the gene, there are two types of point mutations

a. Transitions: Base change from purine to another purine

b. Transversions: Base change from a purine to a pyrimidine and vice versa

4. Transitions are more common than transversions

5. Base substitution mutations can arise spontaneously or be induced

6. An example of a spontaneous base substitution (transition) occurs during a round of DNA replication in which a G is substituted for an A

a. Purine/Purine substitution

b. Creates a mismatched G-T base pair

7. In the subsequent round of replication two products will be created

a. One product will contain an A-T base pair at the position where the mismatch occurred

b. One product will contain a G-C base pair where the mismatch occurred

c. This is the point where the transition is fixed

d. Each new daughter cell will receive DNA with different sequence

8. Base substitutions can occur either in regulatory regions of genes or in the actual coding regions of genes

9. With respect to a mutation specifically in the protein coding region, there are three different definitions for base substitution mutation

a. Missense mutation: The single base change results in a change in the corresponding amino acid sequence of the protein

b. Nonsense mutation: The single base change results in the formation of a premature stop codon-the corresponding protein will be truncated (also, note the mRNA becomes unstable)

c. Silent mutation: The single base change results in no change in the amino acid sequence of the corresponding protein (Note: silent mutations usually occur in the third base of a codon)

10. Phenylketonuria is a disorder that can be caused by a base substitution in the phenylalanine hydrolase gene

a. The mutation that leads to PKU is a transversion from a G to a C at codon 413

b. This mutation is also a missense mutation in that it will result in an amino acid substitution in the corresponding protein from Pro413 to Arg 413

11. This mutation will result in a PAH enzyme that is no longer functional, and will not be able to metabolize phenylalanine

12. This leads to a variety of developmental defects including severe mental retardation

13. One can manage the disorder by feeding the patient a diet low in protein, and with amino acid supplements lacking phenylalanine

C. Types of Mutations: Insertions and Deletions

1. Insertions and deletions are generally rare as compared to base substitutions

2. The size of insertion or deletion can be from just nucleotide to thousands of nucleotides

a. Insertion or deletion of just one nucleotide is also called a point mutation

b. No matter the size of insertion/deletion, it can possibly lead to a deleterious phenotype

3. Insertions and deletions, if they occur in the protein coding region of a gene, can affect the reading frame, which is read in multiples of three bases (codons)

4. If the length of the insertion or deletion is not an exact multiple of three nucleotides, then the mutation causes a frameshift

a. The frameshift mutation will shift the phase in which the ribosome reads the triplet codons that are 3’ to the insertion/deletion

b. A peptide with a completely different amino acid sequence is produced

5. Insertions and deletions of multiples of three will have no effect on the frame

a. The codons 3’ the deletion/insertion are unaffected, and thus, other than the insertion/deletion, the other amino acids in the corresponding protein are unaffected

b. Other than the amino acids that are encoded by the deleted codon(s), the rest of the amino acid sequence for the peptide

6. Insertions/Deletions that alter frame generally have stronger phenotypes than those that do not

D. Types of Mutations: Expansion of Trinucleotide Repeats Leads to Genetic Instability

1. A select number of genes within the human genome have trinucleotide repeats

a. The FMR1 (Fragile X Mental Retardation) gene normally has 6-50 repeats of the sequence CGG in the sequence encoding the 5’UTR

b. The Huntington’s gene normally has 10-26 CAG repeats located in the coding region

2. These trinucleotide repeats can be expanded due to errors in DNA replication

3. The trinucleotide expansion is really just a type of insertion

4. If the CAG repeats in the Huntington’s gene is expanded beyond approximately 40 repeats can develop Huntington’s disease

a. Neurodegenerative disorder

b. Generally has an onset after 40 years of age

c. Progressive motor disorders

d. Progressive cognitive disorders

e. Behavioral disturbances as the disease progresses

f. Eventually paralysis and death

5. The trinucleotide expansions can adopt triple helix conformations and assume unusual DNA secondary structures that interfere with transcription and DNA replication

III. DNA Damage

A. General Classes of DNA Damage: Introduction

1. As we previously talked about, a mutagen is any agent that causes an increase in the rate of mutation above the spontaneous background

a. Chemicals

b. Radiation

c. Free Radicals

2. Damage to DNA consists of any change introducing a deviation from the usual double-helical structure

3. There are three classes of DNA damage

a. Single base changes

b. Structural distortion

c. DNA backbone damage

B. Classes of DNA Damage: Single Base Changes

1. Single base changes affect the DNA sequence, but generally does not have a large affect on overall DNA structure

2. One of the most common causes for a single base change is a deamination reaction

a. Can occur spontaneously by the interaction with water or by the action of a chemical mutagen

b. The deamination reaction replaces the amino group on cytosine to an oxygen

c. This converts the cytosine to a uracil, leaving a G-U base pair in the DNA

d. This G-U base pair can cause a minor distortion in the DNA structure, but will not affect transcription or replication

3. In vertebrates, the DNA frequently contains methylated cytosine residues (5-methyl cytosine)

a. If these undergo a deamination reaction, then the cytosine will be converted to a thymine

b. This will ultimately result in a change from a G-C base pair to an A-T base pair

4. Other causes of single base changes are:

a. Alkylation

b. Oxidation

c. Radiation

5. Alkylating agents such as nitrosamines lead to the formation of O6-methylguanine

a. This base mispairs with thymine

b. This leads to a change from a G-C base pair to an A-T base pair

6. Ionizing radiation and chemical mutagens can act as potent oxidizing agents

a. Reactive oxygen species can generate 8-oxoguanine (oxoG), which is a damaged guanine base containing an extra oxygen atom.

b. OxoG can form a Hoogsteen base pair with adenine

c. Will give rise to a G-C T-A conversion

d. This conversion is one of the most common mutations found in cancers

C. Classes of DNA Damage: Structural Distortion

1. Structural distortions are caused by agents that that can grossly effect the structure of DNA, which may in turn inhibit various important cellular process

a. DNA replication

b. Gene expression

2. Both chemical agents as well as radiation can cause structural damage to occur

3. There are three classes of agents that can lead to structural distortions

a. Ultraviolet Radiation

b. Intercalating agents

c. Base analogs

4. The nitrogenous bases in DNA optimally at a wavelength 260 which falls within the UV spectrum

5. Absorption of light at a wavelength of 260 nm can cause significant structural changes in the DNA

a. Can introduce pyrimidine dimers in the DNA between neighboring thymines

b. Dimers are also termed cyclobutane-pyrimidine dimers (CPD) because a cyclobutane ring is formed between carbon atoms 5 and 6 in the thymine rings

c. More rarely pyrimidine neighboring cytosines and thymines

6. The covalent bonds formed between the two thymines disrupts the normal base pairing with adenines and distort the overall physical structure of the DNA

a. Disrupts transcription

b. Disrupts replication

7. Due to the structural distortions, pyrimidine dimers cannot remain within the DNA and are removed by the repair machinery

D. Classes of DNA Damage: Structural Distortion-Intercalating agents

1. Intercalating agents (chemical) are also a source that commonly causes structural distortions in DNA

2. By definition, an intercalating agent will cause the distortion by chemically inserting itself within the base-paired structure

3. A common intercalating agent that is used in molecular labs is ethidium bromide

a. Contains polycyclic rings which are common amongst intercalating agents

b. The polycyclic rings are flat and are able to insert within the bases

c. The ethidium bromide will insert in a manner which allows it to stack with the other base pairs causing significant structural distortions

4. This distortion will result in either deletion or insertion mutations in the following round of DNA replication

E. Classes of DNA Damage: Structural Distortion-Base analogs

1. The last class of agents that cause structural distortions are the base analogs

2. Base analogs are by definition similar in structural to the commonly found nitrogenous bases in DNA

a. Can be taken up by normal cells

b. Can be incorporated into DNA

c. Can form base pairs, albeit inappropriately

3. 5-Bromouracil is a base analog of thymine

a. 5-bromouracil does not pair with adenine as thymine does

b. 5-bromouracil base pairs with guanine

c. These mispairs cause structural changes to the DNA, and result in mutation

F. Classes of DNA Damage: DNA Backbone Damage

1. DNA backbone damage results from either double strand breaks or the formation of abasic sites from DNA (loss of a base from the DNA)

2. DNA backbone damage can be induced by a number of different agents

a. Ionizing radiation (X-rays and radioactive materials)

b. Action of water

3. Abasic sites are generated spontaneously by the formation of unstable base adducts due to the action of water

4. The DNA then becomes depurinated by hydrolysis of the N-glycosyl linkage

a. Normally the sugar-purine bonds are relatively labile

b. Hydrolysis of the N-glycosyl linkage removes the purine base and leaves a hydroxyl group in its place

5. The double strand breaks are caused by ionizing radiation

6. The ionizing radiation attacks the deoxyribose sugar by directly or indirectly generating reactive oxygen species

7. Since the double strand breaks affect both strands of DNA, they can be among the most severe form of DNA damage

IV. Cellular Responses to DNA Damage

A. Cellular Responses To DNA Damage: Introduction

1. DNA damage is a problem for both prokaryotes and eukaryotes

2. Mechanisms have evolved in both types of organisms to deal with DNA damage

3. There are three main categories of responses to DNA damage

a. Those that bypass the damage

b. Those that directly reverse the damage

c. Those that remove the damaged section of DNA and then replace the sequence

B. Cellular Responses To DNA: Bypassing the DNA Damage (Lesion Bypass)

1. Within the process of replication, high fidelity DNA polymerases function to remove errors

a. Recognize mismatched base pairs

b. Use exonuclease functions to remove them

2. The replicative DNA polymerases are unable to bypass damage that causes larger structural distortions, leaving two possible consequences

a. The replicative DNA polymerase can fall off

b. The replicative DNA polymerase traslocates downstream of the lesion to continue replication

3. In either case the appropriate replicative DNA polymerase is incapable repair the error

4. In order to repair the damage, specialized low-fidelity “error prone” DNA polymerases transiently replace the appropriate replicative DNA polymerase

5. The error prone DNA polymerases copy past the damaged DNA by a process called translesion synthesis (TLS)

6. The error prone DNA polymerases are able to copy past damaged DNA templates efficiently

7. Due to the fact the error prone DNA polymerases can copy past the damage efficiently, they do not do this very accurately

a. They generate many mutations

b. The error prone polymerases insert incorrect nucleotides opposite the lesions giving rise to base substitution mutations

c. Error rates ranges from 1 in 10 bp to 1 in 1000 bp

8. Even though the bypass creates mutations, it is still a better alternative than leaving the damage which can block replication and lead to death

C. Cellular Responses To DNA Damage: Translesion Repair

1. The goal of translesion synthesis is to somehow get the replicative DNA polymerase beyond the damage

a. The replicative DNA polymerases will not be able to replicate through the damage

b. Alternative DNA polymerases will be switched in to repair the damage

c. This gets the replicative polymerase beyond the block caused by the damage

2. Translesion Synthesis (TLS) is used to repair thymine dimmers which cause significant structural distortion

3. Translesion synthesis is carried out by a series of Y-family DNA polymerases

a. DNA polymerase η

b. DNA polymerase ι

c. DNA polymerase κ

d. Rev1

4. Supplemental Figure: Cellular Responses To DNA: Bypassing the DNA Damage (Lesion Bypass)

5. Translesion Synthesis (TLS) is used to repair thymine dimers which cause significant structural distortion

6. The Y-family DNA polymerases all share homology in domains 1-4

7. Structurally, Y-family DNA polymerases have a more wide open structure that gives them more flexibility to accommodate structural distortions in their active site

8. The structural changes due to the thymine dimer or abasic site are generally too large for the replicative DNA polymerase

9. In order to deal with the lesion, a DNA polymerase switch occurs

a. Switch from the replicative polymerase to one of the translesion repair DNA polymerases

b. The translesion DNA repair polymerases have significantly large active sites which can wholly incorporate the damaged DNA (in in the case of DNA polη, will actually incorporate the template strand as well)

c. The polymerase switch is mediated by the polymerase binding to PCNA

d. The Y-class DNA polymerases have very low processivity, and thus can induce errors as they repair the DNA

10. Translesion DNA repair starts when the replicative DNA polymerase reaches the appropriate DNA damage

11. At that point, the replicative DNA polymerase is switched for a Translesion Repair DNA polymerase which has a larger active site

12. If DNA polη is used, then it can incorporate the entire damage in its active site, and will first place two new adenine bases opposite the thymine dimers before removing the thymine and placing new thymines in its place dimers (This is where the bypass takes place)

13. Then the appropriate replicative DNA polymerase will re-associate with PCNA to continue with replication

14. However, DNA polη works with low processivity and sometimes can incorporate guanine instead of an adenine across from the dimer-after repairing the thymine dimer a mutation is left

15. Note: Evidence suggests that DNA polι cannot bypass thymine dimers

D. Cellular Responses To DNA Damage: Direct Reversal Of DNA Damage

1. In DNA damage reversal, the damage is repaired in a manner in which a mutation is not left afterwards

2. There are two instances by which DNA damage reversal occurs

a. Photoreactivation of pyrmidine dimers

b. Removal of methyl groups

3. In photoreactivation, pyrimidine dimers are removed, leaving the two neighboring thymine residues in place afterwards

a. Photoreactivation is mediated by an enzyme photolyase

b. Photolyases use the energy from blue light (visible spectrum) to break the covalent bonds between carbons 5 and 6 of each of the thymine residues

c. Most organisms contain a gene that encodes the photolyase enzyme, however placental mammals do not

4. DNA damage due to the process of methylation can also be reversed by removing the methyl groups

5. Enzymes known as methyltransferases catalyze the removal of the methyl groups

a. Methyltransferases are strongly conserved as they are found in organisms from prokaryotes all the way up to mammals

b. O6-methylguanine, which is produced through methylation of guanine can be reversed back to guanine by removal of a methyl group by methyltransferase

c. A sulfhydryl group of a cysteine residue of the methyltransferase accepts the methyl group from the O6-methylguanine, thus reversing it back to guanine

6. In the reaction, the methyltransferase enzyme is consumed by accepting the methyl group

E. Cellular Responses To DNA Damage: DNA Damage Removal

1. In DNA damage removal, the DNA damage will be recognized, then removed with new sequence possibly being filled in where the damage occurred

2. DNA damage removal is used to repair the following types of DNA damage

a. Damaged bases

b. Mismatched bases (left from replication)

c. Double stranded breaks

d. Pyrimidine dimers

e. Bulky adducts on bases

3. Below are the types of DNA damage repair that fall under the class of damage removal

a. Base excision repiair

b. Mismatch repair

c. Nucleotide excision repair

d. Double-strand break repair

4. An important feature of DNA damage removal is that in each of the removal pathways there is an intricate and dynamic array of proteins and protein-protein interactions that mediate the process

a. There will be an ordered hand-off of DNA from one protein complex to another

b. The DNA repair proteins are modular, composed of multiple structural domains with distinct biochemical functions

5. To remove base substitutions, base excision repair or mismatch repair is used

6. To remove structural distortions, nucleotide excision repair is used

7. To remove double strand breaks, Double-strand break repair is used

8. All the DNA damage removal pathways follow the same general procedural paradigms

a. The DNA repair machinery must sense the damage

b. The DNA repair machinery must gain access to the damaged DNA

c. The DNA damage must be removed

d. The resulting gap from the removal must be filled in by the appropriate DNA polymerase

9. All the pathways share the second step to allow the repair machinery access to the DNA

a. Upon sensing the DNA damage, the chromatin is loosened by a combination of chromatin remodeling and by histone acetylation

b. After the chromatin is loosened the appropriate DNA repair machinery (depending upon the damage and the pathway being used) can gain access

F. Cellular Responses To DNA Damage: Base Excision Repair

1. In base excision repair, the damaged base is going to be cleaved out of the DNA and then replaced

2. Base excision repair is mediated by a group of enzymes called DNA glycosylases

a. DNA glycosylases cleave the N-glycosidic bond between the 1’ carbon and the damaged nitrogenous base

b. The damaged base is excised to form an abasic site within the DNA

3. Base excision repair can be used for a single nucleotide (short patch) or anywhere from 2-10 nucleotides (long patch)

4. Below are two examples of base excision repair

a. Removal of uracil from UA and UG base pairs

b. Removal of oxoG from the DNA (8-oxoguanine)

5. oxoG removal is catalyzed by the enzyme hOGG1 (8-oxoguanine DNA glycosylase 1)

6. hOGG1 binds the DNA non-specifically and searches for oxoGbasess within the DNA

7. When hOGG1 finds an oxoG base paired with cytosine, the enzyme uses its G-specific pocket to remove the oxoG base

a. The enzyme can remove both guanine and oxoG bases

b. The enzyme can only transiently interact with guanines, and thus these get placed back within the DNA

c. The enzyme’s active site is specific can bind oxoG much more strongly allowing for its removal from the DNA

8. As stated earlier, base excision repair can span from 1 nucleotide to 10 nucleotides

a. Single nucleotide repair is also known as short patch repair

b. Repair 2-10 nucleotides is known as long patch repair

9. In both short patch and long patch repair the following paradigm is followed

a. The damaged base(s) are excised by endonucleases

b. The abasic sites are recognized by specific enzymes (e.g. β-lyase and APE in short patch repair)

c. DNA polymerases and DNA ligases repair the DNA

d. DNA ligase 4-XRCC1 complex seals the gap between in the sugar phosphate backbone

G. Cellular Responses To DNA Damage: Base Excision Repair: Short patch repair

1. The steps in short patch repair are as follows

2. The β-lyase enzyme (in association with PCNA) can recognize the abasic site and will nick the DNA and produce a 3’ nick or the 3. APE1 (in association with PCNA) will recognize the abasic site and cause a 5’ nick

4. The APE1 will then excise the damage nucleotide

5. Repair synthesis will be mediated by DNA polymerase β

6. DNA ligase3-XRCC complex will then fill in the gap in the DNA backbone (re-constitute the phosphodiester bond where the nick was produce)

7. Figure 7.8: Cellular Responses To DNA Damage: Base Excision Repair

H. Cellular Responses To DNA Damage: Base Excision Repair: Long Patch Repair

1. The steps in long-patch repair are as follows:

2. The damaged, abasic sites are recognized by APE1 which will then cause a 5’ nick (i.e. the nick is 5’ to the abasic site

3. Then DNA polymerase δ or ε binds PCNA, and displaces the strand 3’ to the nick to produce a 2-10 nt flap

4. The 2-10 nt flap is then excised by the FEN-1 endonuclease

5. A repair patch is then synthesized by DNA polymerase δ/ε, leaving a nick which is then repaired by DNA ligase I

6. Cellular Responses To DNA Damage: Base Excision Repair: Long Patch Repair

J. Cellular Responses To DNA Damage: Mismatch Repair

1. In mismatch repair, we going to repair damage caused by mispaired bases

a. Mismatched bases can cause structural changes in the DNA

b. The mismatch repair mechanism is conserved from bacteria to humans

2. Mismatch repair depends on a number of different protein/protein complexes

a. MutSα (MSH2-MSH6 heterodimer)

b. MutSβ (MSH2-MSH3 heterodimer)

c. MutLα (MLH1-PMS2 heterodimer)

d. EXO1 (exonuclease)

e. DNA polymerase δ

f. PCNA

g. RPA

h. Non-histone chromosomal protein HMGB1

3. It is possible to reconstitute mismatch repair in vitro

a. In vitro reconstitution has shown which proteins are absolutely necessary and which ones are not

b. All the proteins in the list above are necessary except for MutSα, DNA polymerase δ and Exo1

4. In order start mismatch repair, the error must first be recognized

a. Mispaired bases are recognized by MutSα/Mutlα

b. One base on one strand will be correct and the other strand will actually contain the error

c. The newly synthesized of the two strands is the one that will have the mismatch (generally)

d. The newer of the two strands is not methylated in prokaryotes

e. However, in eukaryotes the mismatch repair does not depend on methylation, and so how the system actually recognizes which strand is the newly synthesized is unknown

5. Next, EXO1 associates with PCNA and RFC

a. EXO1 can generate a single strand break either 5’ or 3’ to the mismatched base

b. EXO1 can initiate the break either 5’ or 3’ to the mismatch

6. Once the mismatch is recognized, the repair is not involve the simple removal of a single nucleotide, instead a large region of DNA is going to be removed, which will include the mismatched base

a. Removal of DNA is carried out by EXO1 which has both 5’ 3’ and 3’ to 5’ exonuclease activities

b. The specific exonuclease activity used depends on the location of the single-strand break

7. Next, EXO1 dissociates from the PCNA sliding clamp and is replaced by DNA polymerase δ

8. Repair synthesis is mediated by DNA polymerase δ in the 5’ 3’ direction leaving a gap in the DNA backbone at the point where DNA polymerase δ adds the last nucleotide

9. Ligation of the gap is carried out by DNA ligase I

K. Cellular Responses To DNA Damage: Mismatch Repair and HNPCC

1. The mismatch repair system is incredibly important to human health, as mutations in genes encoding enzymes within the mismatch repair system result in development of certain types of cancers

a. Only need to inherit one copy of the defective gene

b. Inheritance of the disorder is in an autosomal dominant fashion

2. Individuals who inherit mutations in genes encoding mismatch repair enzymes have a marked predisposition to hereditary nonpolyposis colon cancer (HNPCC)

3. HNPCC is a familial form of colon cancer characterized by the following

a. Early age of onset

b. Autosomal dominant mechanism of inheritance with high penetrance with about 80% actually developing HNPCC

4. Although patients with HNPCC develop polyps similar in number to those in the general population, they develop them at younger ages

5. The types of tumors that generally develop are adenomas and carcinomas with about 70% of them occuring between the splenic flexure and the ileocecal junction

6. In addition to colorectal cancer, patients may develop other HNPCC-related cancers

a. Stomach

b. Small Bowel

c. Pancreas

d. Kidney

e. Endometrium

f. Ovaries

7. Management of the disease requires regular colorectal screenings starting at age 25 or prophylactic surgical removal of the colon

a. Regular screenings increase life expectancy by 13.5 years

b. Surgical removal can increase life expectancy by 15.6 years

8. 60% of all mutations causing HNPCCC are in the MLH1 gene (which encodes one of the peptides present in the MutLα heterodimer)

9. 35% of of all mutations causing HNPCC are in the MSH2 gene (which encodes one of the peptides present in the MutSα heterodimer)

10. Other mismatch repair genes containing mutations may also lead to HNPCC

L. Cellular Responses To DNA Damage: Nucleotide Excision Repair

1. The purpose of nucleotide excision repair is to remove structural distortions from the DNA

a. This pathway can be used to remove bulges in the DNA caused by thymine dimers

b. This pathway is of particular interest in humans because this is how thymine dimers are removed in humans

2. Six repair proteins mediate nucleotide excision repair

a. RPA (also has a role in DNA replication)

b. XPA

c. XPC

d. TFIIH (also has essential role in transcription)

e. XPG

f. XPF/ERCC1 (excision repair cross complementing rodent repair deficiency) – also has essential role in homologous recombination)

3. The first step in nucleotide excision repair is to recognize the DNA damage

a. The damage is recognized by the cooperative binding of TFIIH, XPA, RPA and XPC

b. The proteins that make up the complex bind in random order

c. Please note that TFIIH is composed of both the XPB and XPD peptides

3. TFIIH has 5’ 3’ helicase activity and can unwind the DNA double helix

a. The XPB peptide has helicase activity

b. The XPD peptide has helicase activity

4. XPG, which is an endonuclease, will then make an incision approximately 6 nucleotides 3’ to the damage

5. XPF/ERCC, which is also an endonuclease, will make an incision 20 nucleotides 5’ to the damage

6. Once both incisions are complete, then the damaged DNA is released

a. A total of 24-32 nucleotides of only ONE strand are removed

b. The stretch of single stranded DNA that is removed contains the damage

7. Once the damaged DNA is released, there is a subsequent hand-off of the DNA to DNA polymerase ε or DNA polymerase δ

a. The choice of DNA polymerase that will be used is dependent on cell type

b. The DNA polymerase that is used will associate with RPA/PCNA

c. The DNA polymerase that is used will mediate the repair synthesis

8. Repair synthesis will leave a gap in the DNA backbone which will be repaired by DNA ligase I

9. Nucleotide excision repair and transcription can be coupled, and the repair process proceeds most rapidly on an actively transcribed gene

10. Figure 7.11: Cellular Responses To DNA Damage: Nucleotide Excision Repair

L. Cellular Responses To DNA Damage: Nucleotide Excision Repair and Xeroderma Pigmentosum

1. Xeroderma pigmentosum (XP) is a genetic disease, which is inherited in an autosomal recessive manner

2. Xeroderma pigmentosum is a somewhat rare disorder

a. Has an incidence in North America of 1/1,000,000

b. Has an incidence in Japan of 1/100,000

3. Patients with Xeroderma pigmentosum have the following symptoms

a. Photosensitivity (sensitivity to UV light)

b. Sunburns result with very minimal exposure to sunlight

c. Extreme freckling of the skin (especially the face, throat, neck, arms and hands)

d. The surface of the skin is also extremely dry

e. Patients with xeroderma pigmentosum have a significant increased risk of skin cancer as compared to the general population (greater than 1000 fold)

f. In a low percentage of patients, neurodegeneration occurs

4. The mean age for developing skin cancer is age 8, with two-thirds of patients dying of the disease before adulthood

5. Xeroderma Pigmentosum is caused by recessive mutations in genes that encode proteins involved in DNA damage repair

a. Many are involved in the nucleotide excision repair system

b. A mutation in one of the genes will disrupt the process of nucleotide excision repair and can lead to the development of the XP phenotype

6. Mutations in the genes that encode the following enzymes in nucleotide excision repair can lead to XP

a. XPB (Rare)

b. XPD (intermediate)

c. XPC (Third most common)

d. XPA (most common)

7. There is no cure for XP

8. Management is limited to reducing skin exposure to the sun, wearing protective clothing, protective sunglasses, chemical sunscreens, careful screening and excision of cancerous lesions

M. Cellular Responses to DNA Damage: Double Strand Break Repair

1. Double-strand breaks are a particularly lethal form of DNA damage

a. For a unicellular organism, this may mean death

b. For humans double strand breaks are linked to an increased susceptibility to breast cancer

2. Double strand breaks can be induced by the following agents:

a. Ionizing radiation

b. Reactive oxygen species

c. Chemicals that create reactive oxygen species (free radicals)

3. Double strand breaks are repaired by one of two mechanisms

a. Homologous recombination

b. Non-homologous end joining

4. In mammalian cells, non-homologous end joining is primarily used to repair double strand breaks, whereas in prokaryotes and single celled eukaryotes, homologous recombination is used

N. Cellular Responses to DNA Damage: Double Strand Break Repair (Homologous Recombination)

1. Besides double strand break repair, homologous recombination has several other purposes in vivo

a. Meiotic recombination, which leads to genetic diversity

b. Mating type switching in yeast

c. Important in transposition

2. The cellular response to double-strand breaks results in the localization of the break sites to repair foci

3. The first step in repairing double-strand breaks through homologous recombination is recognition of the damage by recruitment of ATM (ataxia telangeictasia mutated)

a. ATM is a serine-threonine kinase

b. ATM acts to phosphorylate proteins involved in double-strand break repair and cell cycle control (including p53)

4. In addition, Rad52 is also recruited to the site of the double-strand break (note Rad stands for radiation)

5. Next, the MRN1 complex is recruited to the double-strand break site and initiates the repair

a. Mre11

b. Rad50

c. Nbs1

6. The DNA is first processed by the 3’ 5’ exonuclease activity of Mre11 to form single-stranded tails

7. Rad52 recognizes the single stranded tails

8. Rad51 mediates strand invasion of the 3’ tails with intact homologous sequences

9. Other proteins involved involved in homologous recombination are below

a. Rad54

b. Rad55

c. Rad57

d. BRCA1

e. BRCA2

10. Strand exchange generates a joint molecule between damaged and undamaged duplex DNA

11. After strand exchange occurs, the sequence that is missing at the double strand break site is restored by DNA synthesis (by either DNA polymerase δ/ε

12. The repair synthesis results in formation of a Holliday Junction (A recombination intermediate)

13. The interlinked double stranded molecules are first resolved by branch migration

14. Next, the Holliday Junctions are resolved by the Resolvasome

a. Resolvasome is an enzyme complex

b. Consists of Rad51C and XRCC3

15. Finally, repair ends with ligation of the repaired DNA strands

16. Figure 7.13: Cellular Responses to DNA Damage: Double Strand Break Repair (Homologous Recombination)

O. Cellular Responses to DNA Damage: Double Strand Break Repair (Homologous Recombination) and Hereditary Breast Cancer

1. Mutations of major cancer predisposition genes account for anywhere between 3-10% of all cases of breast cancer

2. The overall prevalence of hereditary breast cancer is anywhere from 1 in 300 to 1 in 800

3. Hereditary (or familial) breast cancer only accounts for about 10% of all breast cancer cases

4. The two genes implicated in hereditary breast cancer are BRCA1 and BRCA2 which are involved in double strand break repair

a. BRCA1 mutation prevalence in North America is between 1 in 500 to 1 in 1000

b. BRCA2 mutation prevalence in North America is between 1 in 250-1 in 500

5. Mutations in the BRCA1 and BRCA2 genes account for approximately 80% of all familial breast cancers, but account for only a small portion of all breast cancer cases

6. BRCA1 and BRCA2 are tumor suppressor genes that encode ubiquitously expressed proteins involved in double strand break repair

7. Loss of BRCA 1 or BRCA2 function permits accumulation of other mutations

8. The accumulation of other mutations are responsible for neoplasia and subsequent breast cancer development

9. Tumor formation in carriers of BRCA1 or BRCA2 mutations follows the two-hit hypothesis

a. Individuals are born with one mutant allele

b. In order for the disease to occur, both alleles must be mutant

10. Current recommendations for women with germline mutations of BRCA1 or BRCA2 include frequent breast and ovarian exams

a. Self-exams

b. Regular mammograms starting at an early age

11. Familial breast cancer can also affect men

a. Regular breast exams

b. Frequent prostate exams

12. In families with known germline mutations, molecular analysis can focus surveillance or prophylaxis on members carrying a BRCA1 or BRCA2 mutation

13. Total bilateral mastectomy may reduce the risk of breast cancer by more than 90%

14. Please note that bilateral mastectomy does not eliminate risk due to some breast tissue being left

P. Cellular Responses to DNA Damage: Double Strand Break Repair (Non-Homologous End Joining)

1. In mammals, double-strand breaks are primarily repaired through non-homologous end joining

2. This method of repair in mammals is essential to maintaining the integrity of the genome

3. However, the repair process can lead to mutation, thus it comes at a cost

a. Can result in insertions or deletions at the break site

b. Two broken ends can be ligated together regardless of whether they come from the same chromosome

4. In vitro studies have led to the delineation of a mechanism for Non-Homologous End Joining

5. In order to initiate non-homologous end joining, the double-strand break must first be recognized by the Ku70/Ku80 heterodimer

6. The Ku70/Ku80 heterodimer forms a scaffold that holds the broken ends of the double-strand break in close proximity to one another

7. The Ku70/Ku80 complex in turn recruits a series of other proteins/protein complexes to mediate repair

a. The nuclease complex (Artemis/DNA-PKCS)

b. DNA polymerase μ and λ

c. The ligase complex (XRCC4-DNA ligase IV)

8. The Artemis/DNA-PKCS trims excess or damaged DNA at the break site

9. DNA polymerases μ or λ mediate any fill in synthesis or extension of 5’ or 3’ overhangs

10. The rejoining of the ends is then carried out by XRCC4/DNA ligase IV complex