Conventional DSBR - Homologous Recombination

Question 1

What is the advantage of HR?

Answer 1

It is a high fidelity, error-free repair pathway.

Question 2

Where in the cell cycle does HR occur?

Answer 2

S and G2 phases.

Question 3

Step 1 - HR

Answer 3

MRN complex:BRCA1:CtIP, whose formation depends on CDK activity and where CtIP mediates the transition from sensing to processing that causes checkpoint activation, resect the break generating highly unstable 3’ ssDNA overhangs.

Question 4

Step 2 - HR

Answer 4

RPA and RAD52 initially bind overhangs.

Question 5

Step 3 - HR

Answer 5

RAD51 displaces RPA with the help of BRCA1/2 and the RAD51 parlous.

Question 6

Step 4 - HR

Answer 6

RAD51 forms a nucleoprotein filament that makes a strand exchange with the sister chromatid with the help of RAD51 paralogs and the search for sequence homology on the template strand of the sister chromatid in the displacement loop starts.

Question 7

Step 5 - HR

Answer 7

DNA polymerase extends the 3’ end of the invading filament where it either finish off repair by synthesis dependent strand annealing (SDSA) or Second End Capture (SEC)

Question 8

What happens in SDSA

Answer 8

The newly synthesised strand is displaced from the sister chromatid strand and acts as a template for the second 3’ end.

Question 9

What happens in SEC?

Answer 9

The second 3’ overhang can anneal to the other strand of the system chromatid and initiate synthesis resulting in Holliday Junctions that ca either lead to dissolution where no cross over of the two strands occur and integrity of them both is maintained or resolution where the cross overs between the two strands occur and cause loss of heterozygosity opening up opportunity for mutations.