Question

Homologous Recombination

Endonuclease activity of RecBCD is greater in the 3'-5' strand, rather than the 5'-3' strand. Describe how this affects strand invasion, branch migration and resolution

Answer 1

RecBcD is unusual among helicases as it hastwo helicases that travel with different rates and because it can recognize and be altered by Chi DNA sequence. RecBCD avidly binds an end of linear dsDNA.Rec D helicase travel on strand with 5'end at which enzyme initiates unwinding and Rec B on strandwith 3'end.Rec B is slower then Rec D so that ss DNA loop accumulate ahead of Rec B .This produce DNA structures with two ss tails , a shorter 3'ended tail and longer 5'ended tail and one ss loop.

During unwinding nuclease in RecB can act in different ways depending on reaction conditions notably the ratio of Mg2+ ions and ATP.

1. If ATP is in excess enzyme simply nicks strand with Chi ie. strand with initial 3'end.Unwinding continues produce 3'ss tail with chi near terminus which can be bound by Rec A protein ehich promote strand exchange.when RecBCD reaches end of DNA all 3 subunits disaasemble .RecBCD acted at chidoes not attack another DNA molecule.

2.iF Mg+2 is excess,RecBCD cleaves both DNA strands Eendonucleatically although 5' tail is cleaved less often..When Rec BCD encounter chi site on 3'end unwinding pauses. and digestion of 3'tail is reduced.

under both reaction 3'end remain intact. recA protein isactively loaded on 3'tail .Rec BCD dissociate from DNA.rEC a initiates exchange of DNA. The joint DNA molecule is resolved by replication or by cleavage of D loop or formation of HOLIDAY JUNCTION whichcan be resolved by RuvABC COMPLEX.Each of these events generate intact DNA with new combination of genetic marker by which parental DNA differ.This process of HOMOLOGOUS RECOMBINATION completes repair of dsDNA break.

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