DNA Repair Lecture 2a PDF

VDJ recombination and antibody diversity Antigens (about 1 million antigens) Variable light region chain Constant heavy...

VDJ recombination and antibody diversity Antigens (about 1 million antigens) Variable light region chain Constant heavy region chain Antibody (about 1 million types) VDJ recombination Antibody differentiate synthesis Stem B Cell Cell Each B cell produces a single species of antibody, each with a unique antigen-binding site. Each B cell has unique antibody genes encoding light and heavy chain. However, DNA sequences of all the stem cells are same. The mammalian immune system has evolved unique genetic mechanisms that enable it to generate an almost unlimited number of different light and heavy chain encoding genes during B cell development VDJ recombination Stem Variable region Constant region Cell VH(150) DH(9) JH(4) VD recombination DJ recombination B Cell Transcription and splicing mRNA Number of VDJ combination = 150x9x4=5400 VDJ recombination and NHEJ Recombinase:RAG1+RAG2 1. Form DSB with hairpin end Ku70/Ku80 + DNA-PK Artemis 2. Removal of hairpin end XRCC4/Ligase IV 3. Join DSB ends Model for Homologous Recombination Repair template Model of Homologous Recombination Double strand break DNA resection: Produce 3’-ssDNA for homologous sequence search DNA synthesis: copy homologous sequence from template DNA to repair damaged sequence Ligation: form covalent bond between two DNA fragments. It results in double holiday junctions. Resolution: to separate two joint DNA molecules Non-crossover crossover DNA replication of the leading strand DNA replication is one direction: from 5’ to 3’. The paring of ssDNA with the template DNA is required for its further extension. Step1: DNA End Resection The first step of homologous combination is to produce 3’-ssDNA tail from a DSB Formation of 3’-ssDNA requires nucleolytic degradation of the 5′-terminated strands, a process referred to as 5′– 3′ end resection 5′–3′ end resection 3’ 3’ 5′–3′ end resection Nucleolytic degradation: carried out by deoxyribonuclease (DNase) Endonuclease: cleave the phosphodiester bond within a DNA chain 5’ 3’ 5’ 5’ 3’ 3’ Exonuclease: cleave nucleotide one at a time from one end of a DNA chain. Digestion polarity: 5’ 5’ to 3’ 3’ 5’ 3’ to 5’ 3’ MRE11 complex initialises the resection MRE11 interacts with RAD50 and NBS1 to form a MRN complex The Mre11p complex is required for homologous recombination Mre11p binds DSB sites at the early stage of repair Mre11p has a 3’ to 5’ exonuclease activity Mre11p also has an endonuclease activity Both nuclease activities are required for resection How does Mre11 complex produce 3’-ssDNA using both ENDO- and 3’ to 5’ EXO- nuclease activities ? DNA digestion by Mre11 complex How does Mre11 complex produce 3’-ssDNA using both ENDO- and EXO- nuclease activities ? Both endo- and exo- nuclease activities of Mre11 are required for DNA resection in a sequential manner. Question: How do cells regulate these activities? This regulation needs a protein called Sae2 (in yeast). Its human orthologue is called CtIP or Ctp1. Effect of Sae2 on Mre11’s nuclease activity  50 bp DNA substrate 5’ 3’ 3’ 5’ P 3’3’ P 5’ 5’ P 18-35 nt 3’ P 5’ Endonuclease 3’ P 5’ 3’ P 5’ 1nt Exonuclease Sae2 stimulates Mre11’s endonuclease Nature 514, 122–125 activity, (2014) but has no effect on its exonuclease activity. How does Mre11 complex initialise the DNA resection? Why do cells use such complicated pathway to initialise the DNA resection, not simply start DNA resection by 5’ to 3’ digestion? This is because that 5’ to 3’ resection requires an accessible 5’-end to an exonuclease. But it is often that this accessibility is blocked. By covalently-bound proteins at 5-end TopII Spo11 in meiosis By the secondary structure at the end of DSB Hairpin end However, DNA end resection by Mre11 complex is limited to only 100-300 nt. It is not efficient for homologous repair In fact, the resection can be processed up to 3.5 kb in human cells. This suggests that there are other resection pathways, which are more efficient. Two redundant resection pathways take over the resection: Exo1 Sgs1/Dna2 Extended resection I: Exo1 pathway Exo1 is a 5’ to 3’ exonuclease The resection carried out by Exo1 is faster than Mre11 complex The dsDNA with a short 3-ssDNA generated by Mre11 is a preferential substrate for Exo1 Exo1 Exo1 Exo1 preferentially resects dsDNA with 3’-ssDNA tail Mimicking the product after Mre11-mediated resection Extended resection II: Sgs1/Dna2 1. Sgs1 is a DNA helicase, which unwinds dsDNA to produce 5’-flap ssDNA as Dna2’s substrate 2. Dna2 is a 5'-flap endonuclease, which cleaves within single- stranded regions of substrates that form flaps. Two-step DNA resection Mre11 complex binds DSB Sae2 stimulates its endonuclease activity to nick DNA Mre11 resects DNA by 3’ to 5’ exonuclease activity DNA is further resected by Exo1 or Sgs1/Dna2  Questions to think about 1. What is the role of NHEJ in antibody diversity? 2. What is homologue recombination and its process? 3. What is DNA resection? 4. What is the role of DNA resection in DSB repair? 5. What are the biochemical properties of Mre11 complex? 6. How does Mre11 complex initialise DNA resection? 7. How does Exo1 carry on the extended DNA resection? 8. How does Sgs1/Dna2 carry on the extended DNA resection?