DNA Replication and Recombination Lecture Notes PDF

Benjamin A. Pierce GENETICS A Conceptual Approach FIFTH EDITION CHAPTER 12 DNA Replication and Recombination © 2014 W. H. Freeman and Company  12.1 Genetic Information Must Be Accura...

Benjamin A. Pierce GENETICS A Conceptual Approach FIFTH EDITION CHAPTER 12 DNA Replication and Recombination © 2014 W. H. Freeman and Company  12.1 Genetic Information Must Be Accurately Copied Every Time a Cell Divides Replication has to be extremely accurate: – One error/million bp leads to 6400 mistakes every time a cell divides, which would be catastrophic. Replication also takes place at high speed: – E. coli replicates its DNA at a rate of 1000 nucleotides/second. 12.2 All DNA Replication Takes Place in a Semiconservative Manner Proposed DNA Replication Models: – Conservative replication model – Dispersive replication model – Semiconservative replication Blue = original DNA Red = new DNA 12.2 All DNA Replication Takes Place in a Semiconservative Manner Meselson and Stahl’s Experiment: – Two isotopes of nitrogen (nucleotides have nitrogenous bases): 14N common form; 15N rare heavy form E.coli were grown in a 15N media first, then transferred to 14N media Cultured E.coli were subjected to equilibrium density gradient centrifugation  Meselson and Stahl’s Experiment equilibrium density gradient centrifugation N 15 =heavy = bottom  Meselson and Stahl’s Experiment 15 N +14N - Conservative replication model: predicts one heavy band and one light band after one round of replication. 12.2 All DNA Replication Takes Place in a Semiconservative Manner Modes of Replication – Replicons: Units of replication. Replication origin (one in bacteria, many in eukaryotic chromosomes) – Theta replication: circular DNA, E. coli; single origin of replication forming a replication fork, and it is usually a bidirectional replication. – Rolling-circle replication: virus, F factor of E.coli; single origin of replication. Theta replication – circular DNA (bacteria) Bidirectional push semiconserv  Rolling-circle replication – viruses and F factor Inside is template and is replicated Outside is displaced F-factor of E. coli. Hybrid dna 12.2 All DNA Replication Takes Place in a Semiconservative Manner Linear eukaryotic replication – Eukaryotic cells – Thousands of origins – A typical replicon: ~ 200,000-300,000 bp in length. – big dna – Fig. 12.6 Linear eukaryotic replication  Characteristics of Replication of Origins among different organisms Characteristics of theta, rolling-circle, and linear eukaryotic replication 12.2 All DNA Replication Takes Place in a Semiconservative Manner Linear eukaryotic replication: Requirements of replication – A template strand – Raw material: nucleotides – Enzymes and other proteins  Synthesis of new DNA – lecture 10 H bonds need to be broken Tri phosphate needed to help rxn for phosphodiester 5’ 3’ *Look at Animation 12.2 All DNA Replication Takes Place in a Semiconservative Manner Linear eukaryotic replication: Direction of Replication: – DNA polymerase add nucleotides only to the 3 end of growing strand. – The replication can only go from 5  3 – Continuous and discontinuous replication Figs. 12.8 and 12.9  Direction of Replication Rep forks push out as dna is being replicated Pushed to the right DNA polymerases are enzymes that synthesize DNA. Add nucleotides only to the 3′ end of the growing strand (not the 5′ end), so new DNA strands always elongate in This is ex the same (5′→3′) direction. cannot sy how do y - Cont vs The antiparallel nature of the two DNA strands means that one template is exposed in the 5′→3′ direction and the other template is exposed in the 3′→5′ direction. So how can synthesis take place simultaneously on both strands at the fork? Continuous and Discontinuous Replication Keeps going  Direction of Synthesis in Different Models of Replication Continuous vs discontinuous = 2forks  Concept Check 2 Discontinuous replication is a result of which property of DNA? a. Complementary bases b. Antiparallel nucleotide strands c. A charged phosphate group d. Five-carbon sugar  12.3 Bacterial Replication Requires a Large Number of Enzymes and Proteins Bacterial DNA Replication (much more simple) – Initiation proteins: unwind dna and make rep bubble 245 bp in the oriC. (single origin replicon) an initiation protein (DnaA in E.coli) – Unwinding: Initiator protein DNA helicase Single-strand-binding proteins (SSBs) DNA gyrase (topoisomerase)  Initiation of DNA replication DnaA for E. Coli – stretch looks like rep bubble, causes it to open Holds open bubble Unwinding of DNA replication topisomerase 5’ 3’ 5’ 3’  12.3 Bacterial Replication Requires a Large Number of Enzymes and Proteins Elongation: 1st step is synthesis of primers – Primers: an existing group of RNA nucleotides with a 3-OH group to which a new nucleotide can be added. It is usually 10-12 nucleotides long. – DNA Primase: RNA polymerase Makes DNA primer  Elongation: synthesis of primers Replication of DNA requires a nucleotide with a 3′-OH group to which a new nucleotide can be added. How does DNA synthesis begin if there is no 3′-OH group? Rna and dna (not liked, nee *Look at Animation  12.3 Bacterial Replication Requires a Large Number of Enzymes and Proteins Elongation: carried out by DNA polymerase III Removing RNA primer: DNA polymerase I ‒ Connecting nicks after RNA primers are removed: DNA ligase ‒ Removes rna and places dna  DNA polymerases DNA polymerase III: a large multiprotein complex that acts as the main workhorse of replication. - Its 5′→3′ polymerase activity allows it to add new nucleotides in the 5′→3′ direction. - Its 3′→5′ exonuclease activity allows it to remove nucleotides in the 3′→5′ direction, enabling it to correct errors. Its 3′→5′ exonuclease activity is used to back up and remove incorrect nucleotides (then resumes its 5′→3′ polymerase activity). DNA polymerase I: primarily for RNA primer removal and replacement - 5′→3′ polymerase and 3′→5′ exonuclease activities. - 5′→3′ exonuclease (not in DNA pIII) activity, which is used to remove the primers laid down by primase and to replace them with DNA nucleotides by synthesizing in a 5′→3′ direction. - Takes out rna nucleotide and puts in dna nucleotide  DNA polymerase I and DNA ligase 5′→3′ exonuclease activity removes the primers (rna) and replaces them with only 1 phosphate @ nick DNA nucleotides by - Dna ligase gives energy f synthesizing in a 5′→3′ direction. Dnas do not form phosphodiest catalyzes the formation of a phosphodiester bond All that is left is dna without adding another nucleotide to the strand *Look at Animation  12.3 Bacterial Replication Requires a Large Number of Enzymes and Proteins The fidelity of DNA Replication (error rate = less than one mistake per billion nucleotides) Proofreading: DNA polymerase I: 3  5 exonuclease activity removes the incorrectly paired nucleotide. Mismatch repair: corrects errors (secondary structure deformities) after replication is complete. Requires the ability to distinguish the old and the new strands of DNA, because the enzymes need some way of determining which of the two incorrectly paired bases to remove. Termination: when replication fork meets or by termination protein. *Play Animation  look at other animations  12.4 Eukaryotic DNA Replication Is Similar to Bacterial Replication but Differs in Several Aspects Eukaryotic DNA Replication – Autonomously replicating sequences (ARSs) 100–120 bps (ie, Origin of replication) – Origin-recognition complex (ORC) binds to ARSs to initiate DNA replication. – The licensing (approval) of DNA replication by the replication licensing factor MCM: Minichromosome maintenance – binds DNA and initiates replication as a helicase on all origins. – Eukaryotic DNA polymerase  Creates translesion = templates with abnormal bases, distorted structures, and bulky lesions.  12.4 Eukaryotic DNA Replication Is Similar to Bacterial Replication but Differs in Several Aspects Eukaryotic DNA complexed to histone proteins in nucleosomes Nucleosomes reassembled quickly following replication Newly synthesized DNA being covered by nucleosomes (dots) Creation of nucleosomes requires: 1. Disruption of original nucleosomes on the parental DNA 2. Redistribution of pre-existing histones on the new DNA 3. The addition of newly synthesized histones to complete the formation of new nucleosomes Is it random? Do ne old come together Hybrid bands mean DNA is synthesized RANDOM  12.4 Eukaryotic DNA Replication Is Similar to Bacterial Replication but Differs in Several Aspects The location of DNA replication within the nucleus: - DNA polymerase is fixed in location and template RNA is threaded through it. Replication at the ends of chromosomes: ‒ Telomeres and telomerase. ‒ Fig. 12.18 ‒ Fig. 12.19 Replication at the Ends of Chromosomes Poly Noth Rna If this were the case, the chromsomes would shorten after each replication due to the inability to include this gap. Telomerase replicates the ends of chromosomes G-overhang Telomerase has Polymerase sees 3 protein and RNA OH and extends components the telomere End rep problem - always gaps on th Extends but the actual dna of chromosome is not chopped up Unclear how this strand is synthesized? *Play Animations  Concept Check 4 What would be the result if an organism’s telomerase were mutated and nonfunctional? a. No DNA replication would take place. b. The DNA polymerase enzyme would stall the telomerase. c. Chromosomes would shorten each generation. d. RNA primers could not be removed.  12.5 Recombination Takes Place Through the Breakage, Alignment, and Repair of DNA Strands Homologous recombination: exchange is between homologous DNA molecules during crossing over. Holliday junction and single-strand DNA break The double-strand DNA break model of recombination Holliday junction model Heteroduplex DNA Double-strand break model *Play Animation 12.5  12.5 Recombination Takes Place Through the Breakage, Alignment, and Repair of DNA Strands Gene conversion: – process of nonreciprocal genetic exchange – produces abnormal ratios of gametes Gene conversion arises through heteroduplex formation  Gene conversion Example: an individual organism with genotype Aa is expected to produce ½ A gametes and ½ a gametes. Sometimes, however, meiosis in an Aa individual produces ¾ A and ¼ a or ¼ A and ¾ a.