Molecular Biology Exam 4 Study Guide

Which two cellular events indicate to the cell that there is extensive DNA damage?

1. Accumulation of single-stranded DNA (e.g. due to replication fork stalling or lesions in the DNA) 2. Double-stranded breaks (e.g. due to nick in the DNA downstream during DNA synthesis by ionizing radiation)

Homologous recombination, homologous repair, and gene conversion begin in a similar way. Describe the similar features of the initiation of these processes.

1. All begin with a double-stranded break followed by 5' resection, leading to a 3' single strand that is bound by either RecA (E. coli) or Rad51/Dmc1 (eukaryotes) 2. The 3' OH of the single strand undergoes strand invasion and invades undamaged homologous duplex DNA (homologous or sister) to from a D-loop. 3. The 3' OH of the invading strand serves as a primer for its own synthesis using the undamaged homologous strand as a template.

How are mismatch repair and nucleotide excision repair similar?

1. Both operate on mutations in single-stranded DNA (mismatch repair: single bp mutation or slipped-strand misreplication, nucleotide excision repair: bulky lesions) 2. Both involve the removal of the damaged DNA, followed by repair of the gap by repair polymerases + ligation (mismatch repair: degradation by exonuclease, nucleotide excision repair: excision by endonuclease)

How does DNA repair differ from DNA replication?

1. DNA repair and DNA replication processes each have specialized polymerases 2. In DNA repair, the damaged strand uses its 3' OH to carry out strand invasion and formation of a D-loop, which does not occur in DNA replication. 3. In DNA repair, the damaged strand uses an undamaged strand from a homologous DNA duplex as a template for its own synthesis; the end result is conserved replication. In DNA replication, each parental strand serves as a template for DNA synthesis through semiconservative replication (i.e. no recapture is involved).

What happens if a depurinated base is not repaired before replication?

A depurinated base that is not repaired before replication results in the creation of an abasic site. With respect to replication, an abasic site can lead to stalled replication forks; if bypassed by translesion DNA synthesis, it can lead to point mutations.

How does Ecoli alkyltransferase Ada repair methylated bases?

Ada is an alkyltransferase that recognizes methylated bases and removes the methyl group from the base and covalently attaches it to the active site of the enzyme. Once Ada itself is methylated, it adopts a conformation that allows it to be a sequence-specific DNA-binding protein. Ada binds to the DNA that encodes Ada, and activates its own expression, thereby ensuring its replacement.

How is the collapse of replication forks repaired in E. coli?

Collapsed replication forks are repaired in a mechanism similar to homology-directed repair. The exposed 3' OH undergoes strand invasion into a homologous duplex and forms a D-loop. The 3' invading strand then serves as a primer for synthesis of DNA using the undamaged homologous DNA as a template. The displaced strand from the D loop then base pairs with the lone single strand (non-invading strand), where lagging strand synthesis takes place (requires replication restart and repeated assembly of the replisome) and a Holliday junction is formed. Once synthesis is completed, the Holliday junction is resolved.

Correct the false statement: Bacteriophage recombinases such as lambda integrase and Cre recombinase require host proteins for site specific recombination at target sites.

Cre recombinase does not require additional proteins for CSSR at target sites, but lambda integrase does.

How is Cre recombinase activity distinct from topoisomerase activity?

Cre recombinase is a sequence-specific topoisomerase whereas general topoisomerases do not have this sequence specificity. Also, Cre recombinase tetramerizes to carry out site-specific recombination whereas topoisomerase nicks a strand to relieve DNA supercoiling.

What is the role of the resolvase RuvC in homologous recombination in E. coli?

During homologous recombination, branch migration is facilitated by RuvA and RuvB. RuvA tetramerizes to hold the Holliday junction in open conformation while RuvB translocates the junction through ATP hydrolysis. RuvAB recruits RuvC, which then resolves the junction by cleavage of the intercrossed strands. The direction of RuvC-mediated resolution decides whether the resulting DNA duplexes are recombinant or non-recombinant. If both Holliday junctions are cleaved in the same direction (e.g. both vertical or both horizontal), then the non-recombinant product is formed. If the Holliday junctions are cleaved in different directions, then the recombinant product is formed.

Describe the process used by Ecoli to repair mismatches detected following replication. Be sure to identify key proteins involved.

First MutS recognizes and binds to the site of mismatch by identifying altered DNA structure. MutS recruits MutH and MutL to form the MutSLH complex. MutH binds to hemimethylated DNA at GATC sites and uses its endonuclease to nick the unmethylated (new) DNA strand. MutL recruits UvrD (helicase), which unwinds and displaces the nicked DNA. An exonuclease (either 3' to 5' or 5' to 3') then degrades the displaced DNA strand that contains the mismatch. The gap is then filled in by repair polymerases, followed by ligation to seal the backbone.

Trypanosomes evade the immune system of infected hosts by switching gene expression of variant surface glycoprotein (VSG).. What process is used for changing the identity of the expressed VSG?

Gene conversion

How does a DNA transposon with intact terminal repeats but mutant/inactive transposase transpose?

Since the TIRs are still present, the transposon element itself can still transpose as long as the necessary protein (i.e. transposase) is available. In this case, the transposon is nonautonomous and does not encode its own functional transposase; thus, it must rely on a nearby autonomous element to encode transposase.

Compare and contrast Cre recombinase with type I topoisomerases.

-Both create single-stranded nicks in the DNA -Cre is used to carry out conservative site-specific recombination between two specific lox sites (which Cre nicks followed by facilitation of end joining), whereas type I topoisomerases nick single-stranded DNA to relieve torsional stress introduced by DNA supercoiling.

How are thymidine dimers repaired in E. coli at night?

In the absence of light, photoreactivation will not work. Thus nucleotide excision repair is needed to repair thymidine dimers.

Compare and contrast global genomic repair with transcription coupled repair.

Nucleotide excision repair in which damage is recognized anywhere in the genome is referred to as global genomic repair. However, repair is preferentially targeted to genes that are being actively transcribed since damage to these genes could disrupt gene expression and lead to widespread detrimental effects; this kind of repair is termed transcription coupled repair (TCR). TCR involves additional recognition proteins and is initiated by recognition of stalled RNA polymerase molecules due to encountering DNA damage. Repair is then specifically targeted to the template DNA strand (the one that is being transcribed into RNA). The stalled polymerase is then removed along with the damaged DNA.

Correct the false statement: P elements are the major transposon in the human genome.

P elements are the major transposon in drosophila

Describe the process used in B cells to generate the VDJ exon.

RAG1 and RAG2 bind to and cleave a single strand of the RSS12/RSS23 (heptamer and nonamer separated by 12 or 23 base pairs) that flank the V, D, and J segments upon forming a complex. The 3' OH of one strand then carries out an intramolecular attack on the other strand, forming hairpins in both segments. NHEJ subsequently occurs, where Ku70/Ku80 bind to the DSBs and recruit DNA-PKcs. DNA-PKcs autophosphorylate, become activated, and recruit and phosphorylate Artemis. Artemis then cleaves the hairpins, and the ends are ligated by a complex comprised of DNA ligase IV, XRCC4, and XLF/Cerunnos. Note: A D segment is first joined to a J-C segment. Next, a V segment recombines with the DJC combined segment.

How does RPA initiate DNA damage response?

RPA binds to single-stranded DNA and recruits and activates ATR. Once activated, ATR leads to cell cycle control, replication fork stabilization, and replication origin control.

Correct the false statement: RecA is the eukaryotic equivalent of MutH.

RecA is the bacterial equivalent of Rad51

How are heteroduplexes formed in homology-directed repair?

The 3' OH of the free strand undergoes strand invasion to form a D-loop. In HDR, there is no extensive formation of heteroduplex during resynthesis using the 3' end as a primer and the homologous strand as a template.

What are the major steps in repair of double strand breaks when sister chromatid is not available?

If no sister chromatid is present, non-homologous end joining is used to repair double-stranded breaks. Ku70 and Ku80 form a heterodimer that bind to the ends of double-stranded breaks after resection. Ku70/Ku80 then recruit DNA-PKcs, which then become autophosphorylated. Once activated DNA-PKcs recruit and phosphorylate Artemis. Artemis trims single-stranded tails on each DSB, and the DNA ends are ligated by a complex involving DNA ligase IV, XRCC4, and XLF/Cerunnos. In addition, homology-directed repair can also occur in the absence of sister chromatids by using homologous chromosomes instead. The 3' OH of the damaged strand invades homologous duplex DNA. The 3' end serves as a primer for DNA synthesis without the formation of extensive heteroduplex DNA. The newly synthesized DNA is then recaptured by the other 3' overhang. Now replication can take place on the other strand, followed by ligation.

Compare and contrast the mechanism DNA only cut-and-paste transposition and LTR retroelements transposition.

In DNA-only transposition, an encoded transposase binds to the TIRs of the transposon and oligomerize to form a transpososome. End cleavage occurs in trans where the transposase at one end cleaves the DNA at the other end and vice versa, thereby assuring transposon ends are together. The transpososome then binds to target DNA and the 3' ends are joined to the 5' ends of target DNA at staggered positions. ssDNA is then filled in by host repair, which results in the duplication of DNA flanking the transposon. Cut-and-paste transposition leaves a gap in the donor DNA, which can be fixed by NHEJ or HDR. In LTR retrotransposition, replication and transposition of the donor site through an RNA intermediate is used. A tRNA is primed adjacent to the U5 coding segment of the RNA, and its 3' OH is used as a primer for reverse transcription. The cDNA synthesized is then transported to the nucleus and uses its 3' OH to attack to attack target DNA at staggered positions, leading to directly repeated sequences flanking the transposed element. In both cases, the transposition is autonomous; DNA-only transposons encode their own transposon, and LTR retrotransposons encode their own reverse transcriptase.

Contrast homology directed repair and homologous recombination. What are the major differences in these two processes?

In HDR, the 3' OH of the DSB invades the duplex of undamaged homologous duplex DNA (can be sister chromatid or homologous chromosome) to form a D-loop. The 3' end serves as a primer for its own synthesis using the strand from the undamaged duplex as template (note: during DNA synthesis, there is little heteroduplex formation). When long enough, the newly synthesized strand is recaptured by the other 3' single-strand, and becomes the template for the resynthesis of its complementary strand (conservative replication). In homologous recombination, the 3' OH invades homologous duplex DNA and forms a D-loop. However, after resynthesis of the invading strand through extensive heteroduplex formation, there is no recapture. Instead, the undamaged strand displaced by the D-loop base pairs with the other single strand and carries out its resynthesis by serving as its template. This leads to the formation of a Holliday junction, which subsequently becomes resolved by a resolvase enzyme.

How does MRN initiate the DNA damage response?

In eukaryotes, MRN binds to double-stranded breaks and recruits ATM. ATM becomes activated and phosphorylates histone variant H2AX. H2AX then serves as a binding site for MDC1, which amplifies the amount of MRN and ATM. H2AX and MDC1 both serve as mediators that recruit effector proteins including DNA repair proteins, p53 binding protein, and ubiquitin ligases.

Contrast gene conversion and homology-directed repair. Indicate one significant difference in these two processes.

In gene conversion, a special endonuclease (e.g. HO endonuclease in the MAT locus) is used to form a DSB in facilitating resynthesis of the DNA. Since a non-sister chromatid is used, re-synthesis results in the damaged strand acquiring the sequence present in the template strand (gene conversion). In contrast, HDR utilizes a sister chromatid during repair, so resynthesis of the damaged strand does not acquire new sequences (no gene conversion).

Deamination of cytosine yields uracil and a U-G base pair. How is this repaired in somatic cells?

In somatic cells, this is repaired through base excision repair since the damage is small and localized to only one nucleotide. DNA glycosylase recognizes the incorrect base pair and flips the damaged base out of DNA. The flipped out base is then excised via hydrolysis by the glycosylase, leaving an abasic site. AP endonuclease recognizes the abasic site and cleaves the phosphodiester backbone on the sides of the abasic site. The gap is then filled in by repair polymerases using the undamaged strand as a template. Finally, DNA ligase seals the phosphodiester backbone.

What is the role of the Ku70/Ku80 dimer?

Ku70/Ku80 dimer plays a major role in NHEJ. Ku70/Ku80 form a heterodimer that contains a loop into which the broken ends of DNA can fit. This aligns the strands and protects them from nucleases. Ku70/Ku80 recruits and activates DNA-PKcs, which phosphorylates and activates downstream events in NHEJ.

Compare and contrast the transposition of LTR retroelements (e.g. yeast Ty element) with the transposition of nonLTR retroelements (e.g. human L1). Be sure to address the following: 1. Template for transposition, 2. cellular location for DNA synthesis, 3. primer/initiation of cDNA synthesis, 4. mechanism of integration

LTR retroelements: -Contains long terminal repeats flanking the element -Use encoded reverse transcriptase -Use 3' OH of tRNA adjacent to U5 segment as a primer for reverse transcription in the cytoplasm -The 3' OH of synthesized cDNA carries out an intermolecular attack at staggered positions in the target DNA upon entry into the nucleus -The flanking gaps are filled in by repair polymerases, resulting in short direct repeat sequences surrounding the transposed element -Viruses contain ENV which allow for survival outside of the cell during infection of other cells Non-LTR retroelements: -Does not contain LTRs flanking the element -Uses encoded reverse transcriptase only in LINEs but not SINEs (non-autonomous) -A 3' poly(A) tail is added to non-LTR RNA -LINE RNA is exported to the cytoplasm where it associates with ORF1 and ORF2 and returns to the nucleus -ORF2-encoded reverse transcriptase/endonuclease nicks T-rich region of target DNA -Poly(A) tail of LINE RNA base-pairs with the T-rich region and uses its 3' end as a primer for cDNA synthesis -Reverse transcription in which cDNA is formed and integrates into target site takes place in the nucleus at the target site in situ

Correct the false statement: All DNA damage results from the action of chemical or physical mutagen.

Not all DNA damage results from chemical or physical mutagens; DNA damage can also come from replication errors.

Insertion of the transposon results in a direct repeat of the target site. How does this happen?

The 3' OH of the transposon attacks the target DNA strand at the 5' ends in staggered positions, leaving gaps between the 5' end of the transposon and the 3' end of the target DNA flanking the transposon. These gaps are filled in by host repair polymerases, leading to flanking direct repeats.

Lox sites and att sites contain spacer regions (O). What is the role of the spacer region in the site specific recombination event?

The role of the spacer region is to orient the lox sites or att sites for recombination in the correct direction (provide directionality).

Why is it that SINEs in the human genome are more likely to cause mutation by unequal homologous recombination than by transposition?

This is because SINEs are nonautonomous (do not encode their own proteins needed to transpose themselves) and therefore cannot undergo transposition unless an autonomous element (e.g. LINE) is also present. Thus, SINEs rely on the use of the transposition proteins (reverse transcriptase, endonucleases) encoded by the autonomous element to transpose. SINEs can, however, undergo unequal homologous recombination, which is not restricted and is therefore more likely to occur.

How are thymidine dimers repaired in E. coli in daylight?

Thymidine dimers are repaired through photoreactivation by using DNA photolyase

Correct the false statement: Base excision in humans involves XP genes, homologs of E. coli Uvr genes.

XP genes are similar in function to Uvr genes, but are NOT homologs