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Molecular Genetics 3 > Genetic Recombination > Flashcards

Genetic Recombination Flashcards

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1
Q

What are the five mechanism of Genetic Recombination?

A
  • Homologous Recombination
  • Non-homologous end joining
  • Transposition
  • Site-specific recombination
  • Independent assortments of chromosomes\

The first four affect chromosome integrity and thus genome stability

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2
Q

What is the basis of homologous recombination?

A

Present in all organisms and faithful pathway to repair DNA damage such as DNA ds breaks

  • Following DNA ds breakage, DNA ends are processed by nuclease (RecBCD complex)
  • Broken DNA invades an intact homologous DNA molecule due to recombinase (RecA)
  • Broken DNA repaired and the unbroken DNA is used as a template
  • Holliday Junctions are cleaved by resolvase (RuvABC)
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3
Q

What is non-homologous end joining?

A

Repairs ds breaks (not as faithful) and can be carried out without homologous sequences.

Important pathway for ds breaks occurring in most of the eukaryotic cell cycle

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4
Q

What is the basis of NHEJ?

A
  • Ends are recognised and protected by KU. Keeps em next to each other
  • Ends are first processed
    ○ Carried out by different enxyme
    ○ Nt added by Pol.
    ○ Nt removed by endonuclease
    ○ Nt modified by kinase phosphatase or phosphodesterase
    ○ Change original DNA sequence and thus create mutations
  • DNA Lig4 (along with other proteins) ligate the two ends together.
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5
Q

Use E.coli chromosome dimer resolution as an example of site-specific recombination

A

The second replicating molecule is used as a template for repair. There are two outcomes: non-crossover and cross-over.

Site-specific recombination at dif target sites is carried out by XerCD recombinases. Creates a new crossover which allows unlinking.

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6
Q

Use Bacteriophage λ integration and excision as an example of site-specific recombination

A
  • Lambda phage DNA recircularise after entry into the bacteria. There are now two circular molecules and the possibility of inserting at att site
  • AttP in lambda genome and attB in the E.coli chromosome
  • They use the Int (integrase) to integrate its chromosome. This is reversible because of the head-to-tail orientation after insertion of lambda genome
  • Composite site formed are called attL and attR
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7
Q

Use flagellar phase variation in Salmonella as an example of site-specific recombination

A

Site-specific recombination using an inversion system –> head-to-head orientation

fljB encodes for protein forming salmonella flagellum and fljA encodes for regulatory protein that repress the expression of fljC which encodes another protein to form a flagellum

Promoter of fljB is between two head-to-head target sites. When promoter is in front of fljA –> fljB protein
Hin gene in the promoter encodes a recombinase that can invert region by SSR
When sequence inverted, fljA and fljB are not produced so that fljC is no longer repressed

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8
Q

Use bacteriophage Mu tail fibre variation as an example of site-specific recombination

A

Constant part encodes by Sc and variable part encoded by Sv or Sv’. Second protein encoded by U or U’

Gin gene encodes recombinase that can invert region between the target sites to produce:
ScSv U or ScSv’ U’

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9
Q

What was the initial model of Holliday Junctions?

A

Recombination was initiated by nicks at identical locations in the DNA strands which generated a 4-way structure (i.e. Holliday Junction)

Cleavage of this structure, in either one direction or the other, would generate non-crossover or crossover products

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10
Q

What is the general principle of double-strand break repair?

A

It is initiated by RecBCD, which generates single-strand ends.

RecA will coat these single-stranded ends and will mediate synapsis and strand exchange leading to a structure will Holliday Junctions

Resolution of these junctions can cause crossover outcomes or non-crossover outcomes

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11
Q

What is the general principle for gap repair?

A

Single-strand ends are already present.

Therefore, RecFOR will help the RecA bind to the single-strand gaps.

Similarly to DSB repair, RecA will mediate synapsis and strand exchange leading to structure with Holliday Junctions. Resolution of these junctions will cause cross-over outcomes or non-crossover outcomes (if identical strands are cleaved)

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12
Q

What is the action of RecA

A

RecA binds to a single strand of DNA. It polymerises in a 5’ to 3’ direction and stretches the DNA out.

The RecA-coated ssDNA will find a homologous DNA sequence and mediates synapsis

Hydrolysis of ATP mediates the removal of RecA

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13
Q

What can be said about RecA-coated ssDNA vs its uncoated counterpart?

A

RecA-coated ssDNA has been stretched out with a contour length 50% longer than its counterpart.

RecA spirals around DNA and binds every 3 nt. Therefore, there are triplets of bases that have not been stretched out and these are the bases that find their homologous sequence

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14
Q

Explain the structure of RecBCD and how it relates to its function

A

3’ end enters the RecB, and the 5’ enters the RecD.

Both RecB and RecD begin simultaneously unwinding and cleaving their respective ends.
RecD works faster than RecB so there will be a small single strand loop generated ahead of RecB

RecB nuclease will digest the 3’ end frequently until it is trapped at the Chi sequence in RecC

At this point, RecD starts slowing down, but DNA can still go through. This generates a loop in B and B domain can no longer attack the 3’ end but frequently attacks 5’ end

Continues to unwind strand until the RecA protein can be loaded onto the 3’ strand

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15
Q

How does RuvABC work?

A

RuvA binds the Holliday Junctions in the open square planar structure

RuvB circles the two DNA strands on opposite side and pulls the strands to facilitate junction migration

RuvC is responsible for cleavage

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16
Q

Give an overview of the initiation of DNA replication in E.coli

A

Initiation starts at the OriC locus.

At this locus, it contains two sets of sequences: 13-mer DNA repeats and 9-mer DNA repeats

9-mer repeats are recognised by DnaA initiator protein. This only occurs when DnaA is loaded with ATP –> binding allows it to bind to 13-mer which causes an unwinding at the cluster of A-T

DnaB binds to unwound DNA. Once DnaB binds, SSBs prevent the refolding which allows the loading of the replisomes

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17
Q

Give an overview of the termination of DNA replication in E.coli

A

Terminus region allows binding of Tus proteins. It will bind to some of these sites and prevent the replication from going completely around on both sides.

RecQ and Topo III will help with finishing and getting rid of pieces of DNA that have not been removed

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18
Q

Explain how Replication Fork Collapses would be fixed.

A

Occur due to nicks/gaps in DNA

Dealt with:
- This is a replication problem that leads to a ds break.
- Ds is recognised and processed by RecBCD, and then RecA is loaded
- RecA mediates synapsis and genetic exchange
- Holliday junction resolved by RuvABC to form a three-way junction which allows PriA-dependent replication restart

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19
Q

What is PriA dependent restart?

A

PriA will recognise 3-way junctions and apply its intrinsic 3’-5’ helicase, which will open the DNA a little bit which allows loading of accessory proteins priB, DnaT and DnaC

Once DnaC has been loaded then DnaB (main helicase for replication) can be loaded and then the entire replisome is loaded and replication is restarted

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20
Q

How does replication fork reversal occur?

A

Occurs when the replication fork trying to replicate the chromosome encounters an obstacle

Obstacles can be:
- RNA pol/ which is trying to transcribe a gene in that region (going slowly), and thus fast-paced replisome collides with it
- Might be replication defects
- Very strong binding of proteins on DNA

When it reaches the stop, replication will reverse and use the newly synthesised strand as a template. Strands now complementary and will anneal

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21
Q

How is replication fork reversal fixed?

A

RuvAB will recognise the three-way structure produced and will simply branch-migrate it. Facilitates Reversal

If you have a four-way structure on which RuvAB has been loaded, RuvC will recognise and will cleave it and will create a DSB.

Then it will lead from the DSB using DNA repair machinery

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22
Q

Explain what will happen in the case DnaB(Ts) mutants during replication fork reversal?

A

Remember that lagging strand is synthesised in a discontinuous manner.

  • Therefore, when the replication fork hits an obstacle, there will be a small part of the DNA that is a single strand, and the RecA protein can be loaded
  • The annealing of recA between complementary strand will help reversal of blue and green strand because of the single strand part
  • RecA behaves in a manner where it is not involved in repair of replication fork but instead promoting the reversal to cause a DB
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23
Q

What are Transposable Elements?

A

Mobile DNA that is capable of copying itself around the genome.

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24
Q

What are the two types of TEs?

A
  • DNA transposons: “Cut and Paste elements” aka Class II
  • Retrotransposons: “Copy and Paste elements” aka Class I
25
Q

What determines the proportion of the genome made up of TEs?

A

The proportion of the genome made up of TEs is variable.

This is determined by:
- Transposition Rate
- Acquisition rate of new TEs
- Efficiency of selection in removing TEs

Variation in the proportion of TEs is determined by a balance between these factors.

26
Q

Why are TEs considered to be selfish genes?

A

Even if they cause harm, so long as the host’s reduction in fitness is less than the TE gains through copying itself, the TE will spread

27
Q

What is the general process used by retrotransposons?

A

TEs will be transcribed as RNA and a Reverse Transcriptase will be expressed.
Reverse transcriptase will copy cDNA back into another location in the genome

28
Q

What is the general process used by DNA transposons?

A

They are cut-and-paste transposons. However, they exploit different routes to help increase their probability of being inherited:

  • Cut self out and rely on the host to repair the damage. Could use other chromatid as template for synthesis
  • Once copied, cut themselves out and move ahead. Thus copied once and copied again
29
Q

What are the consequences of TEs? (4)

A
  • If they break the reading frame or promoter of a protein-coding gene, loss of protein
  • TEs carry regulatory sequences which can have an impact on neighbouring sequence
  • These effects are strong for new insertions , suggesting expression change is associated with insertion event
  • Movement of TEs implies near-identical sequences everywhere in the genome. These can be misrecognised as places where recombination can happen
30
Q

What does the conflict of interest refer to?

A

On average, TEs are bad for the genomes that carry them.

TEs are selected to spread at the cost of the genomes that bear them, and Hosts are selected to suppress TE activity.

Thus TEs prefer transposition in germline as it increases the frequency of TE in next gen. But host prefers transposition in soma as it is still bad for host but TE will not be passed down

31
Q

What are the benefits of TEs? (2)

A
  • TEs are mutagenic, and mutations can be good
  • They provide complex mutations that include ORFs and regulation. Complex more likely to be beneficial than single-base mutation

Though they can contribute useful things, this is different from net benefit

32
Q

What is the general structure of a DNA transposon?

A

They are generally 1kbp - 5kbp, and encode a single protein, which is the transposase that mediates excision and insertion

There are inverted terminal repeats at either ends of the transposable elements, required for transposition

33
Q

Explain Mariner Transposition (6)

A

DNA transposition

  1. Transposase expressed from sequence, will bind to terminal inverted repeat and the sequence is cut 2nt inside the transposon
  2. Dimer transposase bring together ends to form an upside-down U structure
  3. There will be a second cut so that there is a free loop of DNA bound together by dimerised transposase
  4. 3’OH mediates integration and attack at the target site resulting in a hairpin structure
  5. Transposase recognise target site and will cut on either side of a A-T
  6. Insert the TE and will leave short region to be repaired –> Target site duplication
34
Q

How do you recognise a new TE? (3)

A
  • Repetitive Sequence? Good sign of TE
  • Cross-species transfer? Move around among species unlike rest of genome
  • Recognisable sequence signature (TIR/ Transposase)
35
Q

Explain the transposition of Helitrons (with use of diagram)

A
  1. TE plus strand cleaved at the LTS and replicate
  2. Leading strand synthesis (5’ to 3’) reconstitutes the donor TE. Replication generates the ds DNA.
  3. Lagging strand synthesis results in a dsDNA circle
  4. The circle may serve as a transposon donor
  5. Plus strand-cleavage at the circle LTS: RTS junction and replicate
  6. Leading strand synthesis regenerates the circle
  7. Lagging strand synthesis makes a dsDNA transposon copy for integration, or forms a new circle
36
Q

Explain the transposition of Polintons

A
  1. Integrase-catalysed excision of a Polinton element from host DNA leads to an extrachromosomal ss Polinton that forms a racket-like structure
  2. Polinton pol B replicates the extrachromosomal Polinton
  3. Initiation of the replication requires the terminal protein (TP) that binds a free 5’ end of Polinton
  4. After ds Polinton is synthesised, the integrase molecule bind its termini and catalyse its integration in the host
37
Q

What are the two sub-types of retrotransposons?

A
  • LTRs (Long Terminal Repeat elements)
  • Non-LTRs, including:
    a. LINEs (Long Interspersed Nuclear Elements)
    b. Penelope-like elements
    c. DIRs
38
Q

Explain the process of Target-Primed Reverse Transcription (Non-LTR transposition)

A
  1. Cleavage of one strand on the target site by endonuclease
  2. non-LTR retrotransposon mRNA anneals to cleaved end of target site
  3. Minus strand synthesis using retrotransposon mRNA as a template. OH group on cleaved DNA target site provides target site for synthesis
  4. After minus strand synthesis, cleavage occurs by retrotransposon endonuclease either downstream (duplication) or upstream (deletion)
  5. Template jump of retrotransposon from mRNA to top strand of the target site
  6. Plus strand synthesis using the cleaved top strand as primer (OH as target)
  7. Completion of synthesis:
    a. Downstream cleavage: fill in of gaps leads to target site duplication
    b. Upstream cleavage: degradation of non-homologous flaps by nuclease
39
Q

What is the general structure of LTR retrotransposons?

A

Generally 5 - 7 kbp long and encode 2 or 3 open reading frames

Characterised by having long terminal repeats at each end that are a few hundred bases long

Pol gene encodes integrase, reverse transcriptase and RNAse H Domains

40
Q

Explain the process of retrotransposition by LTR elements

A

The TE is transcribed between the LTS, and the RNA is exported to cytoplasm, where proteins are expressed. Then:

  1. tRNA binds at a priming site near the 5’ end
  2. This primes the start of reverse-transcription (produces cDNA) which proceeds through a unique sequence and then a repeat to the 5’ end of the template
  3. The repeat sequence of the new DNA can now bind to the 3’ end of the RNA
  4. This primes continued reverse-transcription through the whole TE transcript
  5. The RNA template is degraded by RNAse H, leaving a fragment that primers second strand DNA synthesis
  6. Synthesis moves through the 3’ unique sequence, the repeat, and the 5’ unique sequence
  7. This fragment primes (from the 5’ end)
  8. Synthesis completes in both directions
  9. The resulting dsDNA enters the nucleus with a viral integrase that allows the 3’ OH strand to attack the target site at sites a few base pairs apart, giving a target site duplicatiob
41
Q

What are Gag and Env?

A

Gag protein associates with LTR element transcripts to form virus-like particles

Where the Env gene is present (e.g. Gypsy), it also contributes to forming virus-like particles and may contribute to cross-species transimission

42
Q

What is the difference between LTR elements and retroviruses?

A

Both retroviruses and LTRs contain the same genes in the same order and replicate the same way

However, LTR elements are not horizontally transmitted

43
Q

What does hyper-parasitism refer to in terms of TEs?

A

Given that there are an abundance of TEs making transposase that can still recognise the TIRs, TEs without own transposases can still be excised using these other ones.

Ex. SINEs

44
Q

What are Alu Elements?

A

They are SINEs that are 280 nt long. They are ancestrally derived from the RNA components of the signal recognition particle RNP complex

Non-autonomous elements with the necessary motifs necessary to be transposed by an autonomous element

45
Q

How do KRAB Domain Zinc Fingers suppress TEs?

A

These TFs bind DNA via tandem zinc finger domain, and form a stable complex with TRIM28 via the KRAB domain.

TRIM28 recruits other chromatin-related co-repressors, inducing inactive heterochromatin preventing its expression and thus transposition

46
Q

Describe the piRNA pathway and its suppression of TEs

A

In somatic fly cells:

  1. Long transcripts are transcribed from
    piRNA generating clusters which comprise neighbouring and non-functional TEs in heterochromatic
  2. piRNA intermediates with a 5’ uracil are generated by Zucchini from the cluster transcripts.
  3. They are then loaded into Piwi and the 3’ end is formed by another Zucchini cleavage - forming primary piRNA
  4. Following 3’ 2-O-methylation by Hen-1, the mature piRNA-Piwi complex enters the nucleus to mediate transcriptional silencing of TEs
47
Q

Explain Ping-Pong amplification of piRNA pathway

A

In fly germ cells, Aub-piRNA (cluster piRNA) binds and cleaves transcripts. This results in post-transcriptional silencing

Cleavage also creates the 5’ end of a new piRNA, which is loaded into Ago3. Ago3-piRNA binds and cleaves cluster transcripts to generate more antisense piRNAs

Maturation of Aub-piRNAs can be done by Nibbler or Zucchini cleavage

Zucchini products can associate with Piwi - which again enters the nucleus to mediate transcriptional silencing

48
Q

What is the regulon?

A

Group of genes/operators that are under the control of the same regulator (LexA).

49
Q

How does the position of the LexA box affect the SOS response?

A

The LexA box is located near or in RNA pol binding site, thus interferes with action of RNA pol. Therefore, LexA acts as a repressor of SOS genes

When the sequence of the LexA box becomes more similar to perfect consensus, it binds more strongly and thus affects expression of SOS genes

50
Q

Explain the SOS system in E. coli

A
  1. Intracellular or extracellular stress causes double-strand break
  2. The ends are processed by RecBCD and RecA is loaded onto the ends to form a filament
  3. Active RecA- ssDNA filament acts as a co-protease on LexA. This induces self-cleavage of LexA
  4. Once cleaved, LexA further degraded by cytoplasmic protease. As [LexA] decreases less LexA able to bind
  5. Absence of LexA allows RNA pol to bind and there will be an increase in transcription of SOS genes
  6. The longer RecA and thus DSB is present, the more SOS genes will be expressed.
51
Q

Explain the feedback loop of SOS response

A

Feedback loop is created through LexA and RecA control.

The Positive Feedback amplifies the response if problem has not been fixed (RecA)

The Negative Feedback inhibits the process if the problem has been solved (LexA)

52
Q

Describe the Timeline of SOS Response

A

Early Response - focus on trying to repair DNA damage while protecting its integrity
1. Polymerase (High-fidelity)
2. Homologous Recombination
3. Nucleotide Excision Repair, Other action on DNA and LexA (where LexA will stop SOS if fixed)
4. If not SOS not stopped, Cell Division Stopped and Toxic proteins to stop cell growth
5. Polymerase (Low-Fidelity)
6. Toxic

53
Q

What are the consequences of the SOS Response?

A
  • Homologous Recombination leads to genome rearrangement. Can lead to gene deletion, duplication and inversion
  • Increase mutation rate (bc of low fidelity polymerase). Can lead to acquirement of antimicrobial resistance
  • Stops Cell Cycle which is not in best interest as growth and cell division is required for environment colonisation and thus survival
54
Q

Describe the three early models for the mechanism of recombination (with diagram)

A
  • Break-Join Model: both ds DNA are broken and the wrong ends are ligated together. No DNA synthesis required to create new genetic material
  • Copy-choice Model: each DNA strand is replicated and during replication the templates are exchanged. New genetic material is totally newly synthesised
  • Break-Copy Model: on strand of DNA is broken. Replication uses the other molecule a template for repair. Parts of the new genetic material is newly synthesised
55
Q

Explain Meselson and Weigle’s experiment

A

Focus on genes: c (turbidity) and mi (plaque size)

A bacterium was co-infected with a wild-type bacteriophage (cultivated in presence of heavy isotopes) and mutant bacteriophage (normal molecules)

After co-infection progeny was collected and centrifuged in Cesium chloride. The infection phenotype in the different fractions were analysed

56
Q

What were the results of the Meselson and Weigle experiment?

A

When graphed, similar population densities were observed as the expected wild-type DNA after several rounds of replication.

This implies that recombination occurred.

57
Q

Explain Meselson (1995) Experiment

A

Focus on genes: c (turbidity)

A bacterium was co-infected with a wild-type bacteriophage (cultivated in presence of heavy isotopes) and mutant bacteriophage (normal molecules)

After co-infection progeny was collected and centrifuged in Cesium chloride. The infection phenotype in the different fractions were analysed

58
Q

What were the results of the Meselson (1995) experiment?

A

There was a new phenotype: mottled (clear plaques with turbid circles within). This indicates that the DNA in the phage had info for both clear and turbid.

Observed DNA molecules with 3/4 and 3/8 proportions of heavy atoms. Therefore, this proves that recombination leaves a ‘scar’ in DNA in the form of heteroduplex DNA at the site of cross-over

59
Q

What did Amati and Meselsons experiment reveal?

A

The coefficient of coincidence increases as the distance between the markers decreases –> negative interference.

This means that one recombination event increase the chance of another one happening

Molecular Genetics 3 (5 decks)

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