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Advanced Molecular Bio > Intro to Recombination > Flashcards

Intro to Recombination Flashcards

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Biology 101 flashcards
1
Q

Phases of the cell cycle (2)

A
  1. Interphase- G1, S, G2
  2. Mitosis
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2
Q

Gap 1 phase (G1)

A

During this phase, the cell grows to double its original size and functions normally. There is a high degree of protein synthesis, high degree of function (metabolism), and mitochondrial division

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3
Q

Synthesis (S) phase

A

DNA replication occurs during this phase. Only during and after S phase will we have duplicated copies of the chromosomes

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4
Q

Gap 2 (G2) phase

A

During this phase, there is continued cell growth in preparation of division. The cell continues to function mostly normally

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5
Q

Mitosis

A

This is the phase where there is active cell division. Eventually, with cytokinesis, the cell divides in two. There is also partitioning of replicated chromosomes

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6
Q

Homologous recombination

A

This occurs during and after the S phase (S and G2). This is because it requires two copies of the chromosome, which is only possible during and after the S phase. Also occurs during meiosis. There is no loss of nucleotides, so it is more efficient than nonhomologous recombination. The non damaged chromosome is used as a template to repair the other, damaged chromosome. There is homology between the sister chromatids because they are exactly the same, which promotes strand crossover and the use of sister chromatids as a DNA template for the lesion.

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7
Q

Where does homologous recombination occur?

A

It occurs between two DNA helices that have regions of sequence similarity. With replicated chromosomes, these sequences should be identical. Base pairing is essential. The 2 DNA duplexes “sample” each other’s sequences via extensive base pairing between the single strands from each duplex

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8
Q

Hybridization

A

When there is extensive base pairing between sister chromatids during homologous recombination. Once the DNA strands have hybridized, it is referred to as heteroduplex DNA

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9
Q

Homologous recombination mechanism (4)

A
  1. DNA is damaged by a double stranded break
    1 DNA strand must be freed from the double helix for the process to occur.
  2. Exonuclease degrades the 5’ end of the double stranded break, chewing away part of the single stranded DNA (to make 3’ overhangs).
  3. The overhangs of ssDNA at the 3’ end then invades the homologous DNA duplex. During strand invasion, damaged DNA hybridizes with the undamaged DNA. Several specialized proteins direct this invasion
  4. DNA polymerase heals the break. There is still a small piece of the top strand that’s missing, but the bottom strand acts as a template to completely heal the break
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10
Q

Which specialized proteins direct ssDNA invasion? (4)

A
  1. SSBs
  2. RecA (bacteria) or Rad51(eukaryotes)
  3. Rad52
  4. Specialized helicases
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11
Q

SSBs

A

Used initially to stabilize the ssDNA during homologous recombination

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12
Q

RecA/Rad51

A

RecA in bactria, Rad 51 in eukaryotes. Binds tightly in long, cooperative clusters to ssDNA, wrapping around it and forming a filament. This wrapping promotes heteroduplex formation. These proteins have more than 1 DNA binding site, which allows it to hold ssDNA and the double helix together and promote hybridization

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13
Q

Rad52

A

Displaces SSBs and allows for binding of Rad51 during homologous recombination. It promotes annealing of complementary single strands, so it also helps to promote hybridization

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14
Q

Specialized helicases

A

Use ATP to move DNA and help to form a Holliday junction. This structure is formed during strand invasion

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15
Q

RecA/Rad51 mechanism (3)

A
  1. Intertwines invading ssDNA & duplex in a sequence-independent manner. It clamps down with the undamaged double helix of the undamaged chromosome, clamping together the whole heteroduplex region
  2. Helps ssDNA “search” for homologous sequences. This is not well-understood, may involve transient base pairs
  3. When the homologous sequence is located- strand invasion occurs. ssDNA displaces 1 strand of duplex = heteroduplex
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16
Q

How does Rad52 promote annealing of complementary strands? (2)

A
  1. ssDNA binds a deep groove in the protein, running along the Rad52 surface. Therefore, the protein binds ssDNA and mediates its hybridization
  2. Bases of DNA are exposed which helps mediate annealing of 2 complementary strands
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17
Q

Pros and cons of non-homologous recombination

A

Cons: loss of nucleotides
Pros: can occur at any time in the cell cycle

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18
Q

Pros and cons of homologous recombination

A

Pro: not losing nucleotides
Cons: only occurs at specific times in the cell cycle

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19
Q

Phases of homologous recombination (5)

A

Occurs in a programmed manner during meiosis. There’s no damaged DNA in this case, but we still need a double stranded break
1. Spo11 breaks double strand, Mre11 nuclease complex “chews” Spo11-bound strands leaving 3’ overhangs in the DNA
2. RecA/Rad51 family proteins direct DNA contact and strand invasion
3. Formation of Holliday junction- transient structure of cross-strand exchange
4. The break in DNA is healed specifically for genetic exchange. Bottom strand is repaired and used as a template
5. Resolution.

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20
Q

What does resolution of homologous recombination involve?

A

Involves cleavage of the Holliday junction. How the junction is cleaved determines how much genetic material is swapped- sections of chromosomes from mom and dad are exchanged. Results in hybrid chromosomes that contain genetic info from maternal & paternal homologs. Can result in crossover (extensive swapping of genetic information) & gene conversion

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21
Q

Purpose of meiotic recombination

A

Results in a new arrangement of paternal and maternal alleles on the same chromosome. This is a unique combination of parental alleles that makes you, you

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22
Q

Genetic crossover during recombination

A

Crossover is when Holliday junctions are cleaved oppositely (on non-crossing strands). As a result, DNA upstream and downstream of the Holliday junctions is swapped. 10% of recombination in humans are crossovers

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23
Q

Crossover and non-crossover in meiosis

A
  1. 2 Holliday junctions form
  2. If crossing strands of the junctions are cleaved, this is non-crossover. DNA is unaltered except for the regions between junctions
  3. If cleavage forms at non-crossing and crossing strands. Upstream DNA is altered in addition to the DNA between the Holliday junctions. This is crossover
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24
Q

What occurs during meiotic recombination?

A

Whether we have crossover or non-crossover, meiotic recombination results in a heteroduplex region where the paternal homolog is base-paired with the maternal homolog. Some parts of the gene are from mom, some parts from dad. Non-crossover happens around 90% of the time, in contrast to crossover which occurs 10% of the time. These are potential sites of gene conversion

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25
Q

Meiosis

A

During normal meiosis, a diploid cell gives rise to 4 haploid cells. There is an equal distribution of alleles from mom & dad

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26
Q

Gene conversion

A

When there is a mismatch between the DNA inherited from mom and the DNA. If dad’s DNA is removed due to a mismatch and is replaced with mom’s DNA, this would be gene conversion. Dad’s gene would be converted into mom’s gene. Result is 3 alleles from mom, 1 from dad

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27
Q

Gene rearrangement during homologous recombination

A

Rearrangements occur b/w gene segments but order of genes on interacting chromosomes remains the same. There are no “vast changes”

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28
Q

Gene rearrangement during non-homologous recombination

A

Does not require any DNA homology. This process can cause alterations to gene order along a chromosome & cause unusual mutations that add new info to genome (similar to evolution)

29
Q

Transposition (site-specific recombination)

A

This occurs when there is movement of mobile genetic elements, called “jumping genes”. These genes randomly move around the genome and randomly insert themselves into the genome. Causes spontaneous mutations and therefore is essential for evolution. Carries unique set of genes with 1 often encoding enzyme that catalyzes movement

30
Q

Spontaneous mutations

A

The vast majority of these mutations are deleterious. Some result in increased fitness for survival, which are essential for evolution

31
Q

Transposons

A

Mobile genetic elements that are required for transposition, called “jumping genes”. These genes randomly move around the genome and randomly insert themselves into the genome. Causes spontaneous mutations and therefore is essential for evolution.

32
Q

3 classes of transposons

A

Classified based on their structure and transposition mechanisms.
1. DNA-only transposons
2. Retroviral-like retrotransposons
3. Nonretroviral retrotransposons

33
Q

Transposase

A

An enzyme that acts on specific DNA sequences at each end of the transposon, causing it to insert into a new target DNA site. Catalyzes the gene’s insertion into a random area of the genome. No homology is necessary. Transposons move rarely, throughout evolutionary time.

34
Q

DNA-only transposons

A

Transposons that predominate in bacteria and are responsible for the spread of antibiotic resistance. They are also essential for VDJ recombination in developing vertebrates (which produces antibody and T cell receptor diversity). They exist only as DNA during their movement. Can relocate by two mechanisms: cut-and-paste or replicative transposition

35
Q

DNA-only transposons cut and paste mechanism (4)

A
  1. Transposase monomers bind sequences flanking the transposon
  2. 2 monomers come together (dimerize) forming a functional transpososome
  3. Transposase catalyzes direct attack of the element’s flanking sequences on target DNA. The transposon has been removed from its original site and now there is an attack of a completely different site in the gene. The transposon inserts
  4. Gaps in target DNA from staggered cuts healed by DNA Pol & ligase
36
Q

Replicative transposition

A

Transposon is replicated during transposition- the original remains at original site. Its replicated version is inserted at a new location. Aside from replication of the transposon, most of the mechanism follows cut-and-paste

37
Q

Retroviral-like retrotransposons

A

Found in retroviruses, move as RNA. Integrate as DNA due to reverse transcription. These transposons move around as RNA and integrate as DNA due to the reverse transcriptase activity

38
Q

Retroviral enzymes (2)

A
  1. Reverse transcriptase is an enzyme unique to these viruses
  2. Integrase- helps viral DNA to integrate into host DNA
39
Q

Retroviral-like retrotransposons mechanism

A
  1. The retrotransposon is transcribed to RNA
  2. Part of the RNA is translated to protein. Some of the protein will be reverse transcriptase and some will be integrase
  3. Reverse transcriptase binds to RNA and reverse transcribes it to DNA
  4. We now have a single strand of DNA. DNA polymerase comes in to make the second strand
  5. Insertion of retrotransposon DNA into another site in the genome. Integrase helps with this
40
Q

Integrase mechanism (3)

A
  1. Integrase binds to the ends of the retrotransposon. It creates 3’ overhangs
  2. The 3’ overhangs bind to the DNA where it is going to be inserted. The overhangs promote insertion via hybridization
  3. Gaps are filled by DNA repair
41
Q

Nonretroviral retrotransposons

A

Move as RNA. Involves a reverse transcriptase and endonuclease complex. There is no integrase here. Reverse transcriptase & RNA have much more direct role than in retroviral-like retrotransposons, although the mechanism is similar

42
Q

Nonretroviral retrotransposons mechanism (6)

A
  1. The DNA transposon is transcribed into RNA
  2. The RNA is translated into protein, which includes reverse transcriptase
  3. Reverse transcriptase binds transposon RNA
  4. Cleavage of the top strand of the destination DNA
  5. The reverse transcriptase uses the transposon RNA to which it is bound, and reverse transcribes it to DNA
  6. The reverse transcribed DNA inserts in the other part of the genome
43
Q

Transposons and evolution

A

DNA-only transposons were very active well before divergence of humans and Old World monkeys (25-35 million years ago). They have been dormant in human lineage since this time. Most retroviral-like retrotransposons are also dormant. Only 1 family transposed in human genome since divergence of humans & chimpanzees (6 million years ago).

44
Q

Nonretroviral transposons and evolution

A

Some nonretroviral retrotransposons are still active today
Movement of Alu element seen once every 100-200 human births. Responsible for a small but significant fraction of new human mutations- accounts for one third of the human genome

45
Q

Conservative site-specific recombination

A

Breakage & joining of DNA occur at 2 special sites- 1 on each participating DNA molecule. Different from transposons, which did not require special sequences. They contain recognition sites for the recombinase that catalyzes the reaction. The special site is only required on the transposon for transposition, not on mobile element & target DNA.

46
Q

Results of conservative site-specific recombination (3)

A

Depending on positions and orientations of the 2 sites, 3 things may occur:
1. DNA integration into a new site of the genome
2. DNA excision- needed for DNA to actually move
3. DNA inversion- flipping around of the gene

47
Q

Bacteriophage lambda and site-specific recombination

A

Bacteriophage lambda utilizes site-specific recombination. It is the
first mobile element to be understood in biochemical detail. Same type of recombination enables lambda DNA to exit integration site & enter lytic pathway. This machinery is exploited in creating bacterial knockouts in the lab

48
Q

Lambda integrase

A

A enzyme found in bacteriophage lambda, which mediates joining of viral DNA to bacterial chromosome

49
Q

Central dogma of biology

A

Genetic information is stored as DNA, transcribed to short lived RNA, then translated to proteins (the functional product of the gene)

50
Q

RNA polymerase

A

A large enzyme responsible for RNA synthesis. It is a 10 subunit protein in eukaryotes; 5 subunits in bacteria. RNA is synthesized in a 5’ to 3’ direction like DNA, using single stranded DNA as a template. Within the polymerase, there is a transient interaction between RNA and DNA during RNA synthesis. We transcribe certain genes rather than entire chromosomes at a time, so many RNA polymerases may be active to transcribe many different genes at once

51
Q

Transcription mechanism (4)

A
  1. ssDNA used as template
  2. Transient RNA-DNA interaction (~ 9 nucleotides)
  3. Short region of DNA transcribed. Genes – transcript
  4. Many RNA Pols may be active at 1 time. Diff genes transcribed at same time
52
Q

RNA polymerase mechanism

A
  1. RNA polymerase binds to and unwinds DNA. It has its own helicase activity
  2. RNA polymerase moves along the gene, reading
  3. As it reads, RNA polymerase adds complementary ribonucleotides through an uptake channel
  4. RNA nucleotides pair with complementary DNA nucleotides in the active site. The DNA nucleotides are on the single stranded DNA
  5. This transcription results in a short DNA/RNA helix (around 9 base pairs)
  6. DNA/RNA interaction is transient, replaced by DNA/DNA
  7. When RNA polymerase reaches the end of the gene, transcription is completed, and the mRNA transcript is released
  8. As RNA is released from DNA, it exits the polymerase through a 5’ channel
53
Q

Core RNA polymerase structure in prokaryotes

A

There is a 2.5 nm deep cleft between the β and β’ subunits (looks like a crab claw). This cleft binds to the DNA template. It has 5 subunits: αI, αII, β, β’, and ω. Lined with positive charges due to its amino acid content. Catalytic Mg2+ is located deep in the cleft (forming an active site, which is buried when the cleft is closed). Has an overall negative charge and interacts with DNA

54
Q

Core RNA polymerase mechanism (4)

A
  1. Active site is buried when the cleft is closed
  2. 2’ site on β’ allows entry of nucleotide triphosphates (NTPs) to the active site
  3. Allows binding/entry of cleavage factors (Gre)
  4. Cleavage factors GreA and GreB induce transcript cleavage when polymerase stalls
55
Q

Different functions of RNA in the cell

A

mRNA codes for proteins, rRNA makes up ribosomes, tRNAs transfer nucleotides during translation

56
Q

Sigma (σ) factor

A

Transcription factor in bacteria. Associates with RNA Pol, forming the RNA polymerase holoenzyme. It binds to the promoter region at the -35 and -10 sequences (usually T-A repeats). The holoenzyme binds strongly to the promoter region but very weakly to other DNA sites. Induces opening of a short stretch of the DNA helix. This process does not require ATP, DNA and Pol (σ2) change into a reversible conformation more energetically favorable than initial binding

57
Q

Consensus promoter sequences

A

An average of the nucleotides that show up most frequently in the bacterial promoter sequences. They are usually A-T rich.

58
Q

RNA polymerase holoenzyme binding

A

Sigma (σ) factor has 4 subunits:
σ4 binds -35, while σ2 binds -10 (closed complex). It is domain σ2 binding to -10 that initiates DNA unwinding.
Template DNA is fed through the active site for transcription (open complex), the polymerase uses its own helicase activity

59
Q

Abortive initiation

A

Several cycles of “False starts” in transcription. There is transcription of short RNA oligomers until RNA polymerase gets to its most favorable binding conformation with the DNA template. Once this occurs, productive initiation can begin

60
Q

Productive initiation

A

When more than 10 nucleotides are synthesized

61
Q

Phases of transcription (3)

A
  1. Initiation
  2. Elongation
  3. Termination
62
Q

Elongation

A

After productive initiation, the core enzyme (polymerase) breaks interactions with promoter & σ factor (which is only needed for the first step). RNA Polymerase synthesizes RNA at rate of ~ 50 nucleotides/sec

63
Q

Termination of transcription in bacteria

A

Occurs when RNA Pol reaches the terminator sequence. This is usually A-T repeats. The terminator sequence
may be preceded by a DNA seq that causes RNA to fold into a hairpin structure. This is one way for the transcript to be released from RNA polymerase.
Releases RNA chain & DNA, Pol is now free to bind σ factor & start all over again

64
Q

Elongation and termination in bacteria (4)

A
  1. After productive initiation, RNA polymerase dissociates from the promoter and Sigma (σ)
  2. RNA polymerase enters the elongation phase, transcribing at a high rate
  3. When the terminator is reached, the hairpin (sometimes) forms in RNA, pulling out the transcript
  4. Alternatively, Rho helicase dissociates the transcript
65
Q

RNA hairpin

A

A loop made of complementary base pairs. Has high GC content, creating strong hydrogen bonds. There is a string of 6-8 uracils on the 3’ end that are necessary for termination. The DNA:RNA A-U base pairing at the 3’ region is weak, making this region unstable. The combination of weak DNA:RNA interactions and strong RNA:RNA interactions “yanks” RNA off of the DNA and out of polymerase

66
Q

Rho-dependent termination (bacteria)

A

Requires termination factor (Rho helicase), which is an ATP-dependent RNA:DNA helicase. Rho is a spherical protein with a central pore. The pore is made of very specific dimensions and can only accommodate single stranded DNA. Binds rho utilization sites (rut) in RNA behind RNA Pol, which has a high G-C content. Pulls apart RNA:DNA in ATP-dependent manner & feeds ssRNA through pore

67
Q

Rho helicase structure

A

A spherical hexamer with a central pore.

68
Q

Alternate cleavage of Holliday junctions (2 methods)

A
  1. The junctions can be cleaved on crossing strands- there is less swapping of genetic information
  2. If they are cleaved on non-crossing strands, it results in more extensive genetic swapping. This is called crossover
69
Q

Recombinase

A

Catalyzes site-specific recombination. Similar function to transposes/integrases

Advanced Molecular Bio (13 decks)

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