Chapter 11 - Homologous Recombination at the Molecular Level Flashcards

1
Q

What are the key steps in homologous recombination?

A
  1. Alignment of two homologous DNA molecules.
  2. Introduction of breaks in the DNA.
  3. Strand invasion.
  4. Formation fo the Holliday junction.
  5. Resolution of the Holliday junction.
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2
Q

What does the strand-exchange protein do?

A

Catalyses strand invasion.

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3
Q

When do you generate a crossover product?

A

When the cut sites occur in the two DNA strands that are composed entirely of DNA from one of the two parental DNA molecules.

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4
Q

When do you generate non-crossover products or patch products?

A

When the cut sites occur the two DNA strands that contain regions of sequence from both parental molecules.

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5
Q

Describe the double-strand break-repair pathway.

A

Starts with introduction a DSB in one of two homologous duplex DNA molecules. The other DNA duplex remains intact. The asymmetric initial breakage of the two DNA molecules in the DSB-repair model necessitates the later stages in the recombination process are also asymmetric.

DNA-cleaving enzyme degrades the broken DNA molecules to generate regions of single-stranded DNA (ssDNA tail). These ssDNA tails terminate with 3’ ends. The ssDNA tails then invade the unbroken homologous DNA duplex. Elongation from these DNA ends serves to generates the region of DNA that were destroyed during the processing of the strands at the break site.

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6
Q
What is the function of the following proteins: 
RecA
RecBCD helicase/nuclease
RecBCD and RecFOR
RuvAB complex
RuvC
A

RecA - Pariring homologous DNAs and strand invasion.

RecBCD helicase/nuclease - processing DNA breaks to generate single strands for invasion.

RecBCD and RecFOR - Assembly of strand-exchange proteins.
RuvAB complex - Holliday junction recognition and branch migration.

RuvC - resolution of Holiday junctions.

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7
Q

What is the functions of the RecBCD enzyme?

A

To process broken DNA molecules to generate with ssDNA extensions or tails.

RecBCD also helps load the RecA strand-exchange protein onto these ssDNA ends.

Provides means for the cell to “determine” whether to recombine with or destroy DNA molecules that enter the cell.

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8
Q

Which subunits is the RecBCD enzyme composed of and what are their function? Which element controls the activity of RecBCD?

A

Subunits: recB, recC, recD. The enzyme has both helices (B & C) and nuclease activities. The complex binds to DNA molecules at site of the DSBand tracks along DNA using the energy of ATP hydrolysis.

RecB subunit contains a 3’-to-5’ helices and has also a multifunctional nuclease domain that digest DNA as it moves along. RecD is a 5’-to-3’ helices and RecC functions to recognise Chi sites.

The activity of the enzyme is controlled by DNA sequences known as Chi sites.

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9
Q

What happens the the RecBCD complex encounters Chi sites?

A

Chi sites within the DNA act as molecular throttle to regulate the activities of the helices and therefore the speed of DNA.

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10
Q

How does the RecBCD complex move along the DNA?

A

RecB and RecD helices “motors” move independently along opposite strands of the DNA duplex and at different speeds. Together, the are capable of driving the complex along the DNA at rates of > 1000 bp per second.

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11
Q

Explain the steps of DNA processing by RecBCD.

A

RecBCD enter the DNA at the site of the DSB and moves along the DNA, unwinding the strand. During the initial phase the two motors of the complex are not moving at equal speed - the RecD subunit rush faster than the RecB and therefore leads the complex. As RecB tries to keep up a loop of ssDNA from the 3’ end bulges out ahead of the complex. Upon encountering the Chi sequence, the complex pauses for a few seconds, then continues at about one-half the initial rate.
During the pause tree events occur:
1. The looped-out ssDNA is pulled back in the RecB subunit, and RecB becomes the primary motor leading the complex.
2. Conformational change occurs that results in uncoupling of the RecD subunit.
3. The nuclease activity of the RecBCD complex is altered.

As the complex moves into the sequence beyond the Chi sites, the nuclease no longer cleaves the DNA strand with 3’–> 5’ polarity. The opposite DNA strand is cleaved more frequently than it was before the Chi site was encountered. As a result the DNA duplex now has a 3’ single-strand extension terminating with the Chi sequence at 3’end. This structure is ideal for assembly of RecA and initiation of strand exchange.

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12
Q

Smart future Jasmine, write this question.

A

DNA encounters a pin protruding from he RecC subunit that splits the duplex and guides the two individual strands of DNA to the two motors within the enzyme. The RecC subunit channels the 3’ strand tot he RecB motor and the 5’ strand to the RecD motor. RecC contributes to the overall efficiency of the helices activity. The 3’ DNA tail is fed along the groove that emerges at the nuclease active site on the RecB subunit. As a result, before the enzyme encounters the chi sites, the strand is efficiently and processively degraded. The 5’ DNA tail also moves past the nuclease active site upon leasing the RecD motor, but it is digested less frequently than the 3’ tail. However, upon encountering a Chi site, the situation changes. RecC recognises and bind tighly to this DNA site, and once this 3’ end I bound, it is prevented from entering the nuclease. This binding therefore both prevents further digestion of the 3’ tail and promotes digestion fo the 5’ tail, by removing its competitor.

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13
Q

Why is recombination stimulated specifically only on one side of the Chi site?

A

The DNA between DSB and the Chi site is cut into small pieces by the enzyme and is therefore not available for recombination. In contrast, DNA sequences met by RecBCD after its encounter with Chi are preserved in a recombinogenic, single-strand form and are specifically loaded with RecA.

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14
Q

What do strand-exchange proteins do and which enzyme is its founding member?

A

RecA. Proteins that catalyse the pairing of homolog DNA molecules.

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15
Q

What are the important features of DNA molecules for pairing of homologous DNA molecules?

A
  1. DNA sequence complementarity between the two partner molecules.
  2. A region of ssDNA on at least one molecule to allow RecA assembly.
  3. The presence of a DNA end within the region of complementarity, enabling the DNA strands in the newly formed duplex to intertwine.
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16
Q

Explain how RecA filament formation occurs.

A

RecA bind cooperatively to DNA. RecA binding and assembly are much more rapid on ssDNA than on double stranded DNA. The filament grows by the addition of RecA subunits in the 5’-to-3’ direction, such that a DNA strand that terminates in 3’ ends in most likely to be coated by RecA.

17
Q

Name the different RecA-catalyzed reaction stages in strand exchange.

A

RecA filament must assemble on one of the participating DNA molecules. RecA must look foe base-pair complementarity between the DNA within the filament and a new DNA molecule. The structure of RecA with ssDNA and with dsDNA reveal varied DNA stretching and imply a mechanism for strand exchange.

This homology search is promoted by RecA because the filament structure has two distinct DNA-binding sites: a primary site (bound by the first DNA molecule) and a secondary. The secondary DNA site can be occupied by double-stranded DNA. Binding to this site is rapid, weak and transient, and independent of DNA sequence. In this way RecA filament can bind and rapidly “sample” huge stretches of DNA for sequence homology.

18
Q

How does ssDNA and ATP bind to RecA to form a helical “presynaptic filament”?

A

The ssDNA within the filament becomes underfund and stretched, a change that likely allows for more optimal Watson-Crick base pairing with dsDNA. to form a synaptic filament that samples for homology between the ssDNA and dsDNA. Within the presynaptic filament, ssDNA binds with a stoichiometry of 3 nucleotides per RecA, arranged in a B DNA-like conformation. After each 3 nucleotides there is a large “step” responsible for the extended helix compared to naked DNA.

The DNA is the secondary binding site is transiently opened and tested for complementarity with the ssDNA in the primary sites. Once a region of past-pair complementary is located, RecA promotes the formation of stable complex with Watson-Crick hydrogen bonding between these two DNA molecules. The repeating unit is now a triple of stacked base pairs that is quite similar to B-form (called a joint molecule).

19
Q

What are RuvA proteins? What are their function?

A

Holiday junction-specific DNA-binding protein that recognises the structure of the DNA junction. Therefore, they recruit the RuvB proteins to the site.

20
Q

What are RuvB proteins? What are their function?

A

Hexameric ATPase, similar to the hexametric helicases involved in DNA replication. The RuvB ATPase provides the energy to drive the exchange of base pairs that moves the DNA branch. This energy is needed to move the branch rapidly and in one direction.

21
Q

What is the function of RuvC?

A

Resolution by RuvC occurs when it recognises the holiday junction and specifically nicks tow of the homologs DNA strands that have the same polarity. This cleavage results in DNA ends that terminates with 5’-phosphate and 3’-OH groups that can be directly joined with ligase.