BCH3033 Biochemistry 1 Chapter 25 PDF

BCH3033: Biochemistry 1 Chapter 25 04.05.2024 Donella Beckwith, Ph.D. [email protected] 1 The Functions of DNA gene = a segment of a DNA molecule that contains the information required for the synthesis of a functional biological product, whether protein or RNA only known functions of DNA: – storage...

BCH3033: Biochemistry 1 Chapter 25 04.05.2024 Donella Beckwith, Ph.D. [email protected] 1 The Functions of DNA gene = a segment of a DNA molecule that contains the information required for the synthesis of a functional biological product, whether protein or RNA only known functions of DNA: – storage of biological information – transmission of that information to the next generation 2 Gene Naming: Bacterial typically named using three italicized lowercase letters reflecting a function – examples: dna, uvr, and rec dna: affects DNA replication uvr: genes that remove UV-induced lesions from DNA rec: involved in recombination capital letters added to abbreviation reflect order of discovery, not enzymatic order – examples: dnaA, dnaB, and dnaQ (establish replication fork machinery; DNA polymerase III) 3 Bacterial Protein Naming often named after their genes using nonitalicized, roman type with the first letter capitalized – examples: DnaA (encoded by dnaA) and RecA (encoded by recA) DnaA: activates initiation of DNA replication in bacteria RecA: activates a mobile element-encoded mutagenic DNA polymerase 4 Eukaryotic Gene Naming no single convention exists for all eukaryotic systems in Saccharomyces cerevisiae, gene names are three italicized uppercase letters followed by an italicized number – example: COX1 (brewer’s yeast, cytochrome c oxidase subunit 1) 5 Eukaryotic Protein Naming complex and variable in yeast, some proteins have long common names – example: cytochrome oxidase other yeast proteins have the same name as the gene, with one uppercase and two lowercase letters in roman type, followed by a number and the letter “p” – example: Rad51p (encoded by RAD51) 6 Question Which answer choice shows a correctly named bacterial gene? A. B. C. D. dnaA dnaA DnaA DnaA 7 DNA as a Template template = a structure that would allow molecules to be lined up in a specific order and joined to create a macromolecule with a unique sequence and function the structure of DNA revealed one strand is the complement of the other – each strand provides the template for a new strand 8 DNA Replication Follows a Set of Fundamental Rules basic principles that apply to DNA synthesis in every organism: – DNA replication is semiconservative – replication begins at an origin and usually proceeds bidirectionally – DNA synthesis proceeds in a 5′→3′ direction and is semidiscontinuous 9 DNA Replication Is Semiconservative semiconservative replication = each DNA strand serves as a template for the synthesis of a new strand – produces two new DNA molecules, each with one new strand and one old strand – every new DNA double helix would be a hybrid that consisted of 1 strand of old DNA bound to 1 strand of newly synthesized DNA 10 Replication Begins at an Origin and Usually Proceeds Bidirectionally replication forks = dynamic points where parent DNA is being unwound and separated strands replicated both DNA strands are replicated simultaneously 11 DNA Synthesis Proceeds in a 5′→3′ Direction and Is Semidiscontinuous new DNA strands are always synthesized in the 5′ → 3′ direction – the free 3′-OH serves as the point of elongation Okazaki fragments = short DNA fragments that are synthesized in the replication of one of the new DNA strands – ~150 to 200 nucleotides long in eukaryotes – ~1,000 to 2,000 nucleotides long in bacteria – spliced together by DNA ligase 12 Leading and Lagging Strands leading strand = 5′ → 3′ synthesis proceeds in the same direction that the replication fork moves – synthesized continuously lagging strand = 5′ → 3′ synthesis proceeds in the opposite direction that the replication fork moves – synthesized discontinuously through the synthesis of Okazaki fragments 13 Take Away About DNA Strands at the Replication Fork new DNA strands are always synthesized in the 5′→3′ direction the template is read in the 3′→5′ direction 14 Question What type of bond must be made between Okazaki fragments in order to make a complete DNA strand? A. B. C. D. E. hydrogen disulfide ester phosphodiester ionic 15 DNA Is Degraded by Nucleases nucleases = enzymes that degrade nucleic acids – DNases = specifically degrade DNA exonucleases = degrade nucleic acids from one end of the molecule – many operate in only one direction (either 5′→3′ or 3′→5′) endonucleases = degrade nucleic acids at specific internal sites in the molecule 16 DNA Is Synthesized by DNA Polymerases DNA polymerase: complex enzymes that can synthesize DNA – many have additional activities – E. coli cells have at least 5 DNA polymerases, including DNA polymerase I Reaction is a phosphoryl group transfer The 3′-OH of the nucleotide at the 3′ end of the strand (the nucleophile) attacks the α phosphorus of the incoming dNTP One Mg2+ ion helps deprotonate the 3′-OH group One Mg2+ ion binds the incoming dNTP and facilitates departure of PPi 17 The General DNA Polymerase Reaction the general reaction is (dNMP)n + dNTP → (dNMP)n+1 + PPi DNA (25-1) lengthened DNA where dNMP and dNTP are a deoxynucleoside 5′monophosphate and 5′-triphosphate, respectively 18 Question Which statement correctly describes the reaction catalyzed by DNA polymerase I? A. The nucleophile is the α phosphate of the incoming deoxynucleoside 5′-triphosphate. B. The nucleophile is the 3′-OH group of the nucleotide at the 3′ end of the growing strand. C. The nucleophile is the 3′-OH group of the nucleotide at the 5′ end of the growing strand. D. The nucleophile is the γ phosphate of the incoming deoxynucleoside 5′-triphosphate. 19 DNA Polymerases Require a Template polymerization is guided by a template DNA strand according to Watson and Crick base-pairing rules insertion site = the portion of the active site where the incoming nucleotide binds postinsertion site = the portion of the active site where the new base pair resides when the polymerase slides forward 20 DNA Polymerases Require a Primer primer = a strand segment with a free 3′-OH group to which a nucleotide can be added – must be complementary to the template – many are RNA oligonucleotides primer terminus = the free 3′ end of the primer 21 Question Which statement does NOT describe a fundamental feature of DNA replication? A. B. C. D. It requires a template. It is semiconservative. It begins at a DNA location known as an origin. It involves polymerizing deoxyribonucleotides only. 22 The Processivity of DNA Polymerases processivity = the average number of nucleotides added before a polymerase dissociates DNA polymerases vary greatly in processivity – ranges from a few nucleotides to many thousand 23 Error Correction by DNA Polymerase I translocation of the enzyme is inhibited when an incorrect nucleotide is added many DNA polymerases have intrinsic 3′→5′ exonuclease proofreading activity – permits the enzyme to remove a newly added nucleotide – requires the expenditure of 3 high-energy bonds (energetically expensive) 24 Question Processivity is defined as: A. how many errors are incorporated into the DNA per 1,000 bases. B. the average number of nucleotides added before a polymerase dissociates. C. the rate at which nucleotides are added to the DNA. D. the number of DNA polymerases bound to one DNA molecule. E. which DNA polymerase is performing the elongation. 25 Question DNA polymerases: A. B. C. D. all have a 5′→3′ exonuclease activity for proofreading. only polymerize in a 5′→3′ direction. are all highly processive. are only found at replication forks. 26 DNA Replication Requires Many Enzymes and Protein Factors DNA replicase system (replisome) = the entire complex of enzymes and proteins required for replication in E. coli – consists of 20+ different enzymes and proteins Main Classes of Replication Enzymes helicases = enzymes that move along the DNA and separate the strands – requires chemical energy from ATP topoisomerases = relieve topological stress created by strand separation DNA-binding proteins = stabilize separated strands primases = synthesize short segments of RNA to serve as primers 27 Main Classes of Replication Enzymes, Continued DNA polymerase I = removes and replaces RNA primers RNase H1 = a specialized nuclease that degrades RNA in RNA-DNA hybrids – can remove RNA primers DNA ligases = seals nicks in the DNA backbone following removal and replacement of an RNA primer 29 Question Which protein is NOT necessary to initiate DNA replication in E. coli? A. B. C. D. E. helicases primases topoisomerases DNA-binding proteins All of them are necessary 30 Question The lagging strand in DNA replication: A. requires its own replisome. B. is synthesized after the leading strand. C. is an inescapable consequence of replicating both strands of template DNA at a single replication fork. D. causes the formation of Okazaki fragments in the leading strand. 31 Question A research project using E. coli has come to a halt because DNA replication is blocked. It seems that the DNA strands are not separating. What enzyme is likely faulty? A. B. C. D. E. helicase (DnaB protein) topoisomerase I DNA ligase one of the DNA primases DNA polymerase I 32 Eukaryotes Have Several Types of DNA Polymerases DNA polymerase ε = synthesizes the leading strand – highly processive – has 3′→5′ proofreading exonuclease activity DNA polymerase δ = synthesizes the lagging strand – has 3′→5′ proofreading exonuclease activity DNA polymerase α = a DNA polymerase/primase that synthesizes RNA primers and also extends them to initiate synthesis of each Okazaki fragment – does not have 3′→5′ proofreading exonuclease activity 33 Eukaryotes Have Several Types of DNA Polymerases 34 Mutations mutation = a permanent change in the nucleotide sequence – substitution mutation = replacement of one base pair with another – insertion mutation = the addition of 1+ base pairs – deletion mutation = the deletion of 1+ base pairs silent mutation = a mutation that affects nonessential DNA or has a negligible effect on gene function mutations rarely confer some biological advantage most nonsilent mutations are neutral or deleterious 35 Question DNA repair is necessary because: A. any mutation in DNA will affect the function of the product of the gene in which it occurs. B. while some mutation is necessary for evolution, most mutations do not confer a biological advantage and therefore represent more danger than they do opportunity. C. DNA replication results in approximately one mistake made per 106 nucleotides polymerized. D. DNA is chemically unstable. 36 Mutations Can Lead to Cancer Strong correlation between accumulation of mutation and cancer Bruce Ames invented the Ames Test: measures the potential of a given chemical compound to promote certain easily detected mutations in a specialized bacterial strain Few of the chemicals we encounter in daily life score as mutagens in this test However, of the compounds known to be carcinogenic from extensive animal trials, more than 90% are found to be mutagenic The strong correlation between mutagenesis and carcinogenesis, is why 37 the Ames test is still widely used Question Which statement is true about the Ames test? A. B. C. D. It can be used to screen for potential human carcinogens. It requires a histidine-rich growth medium. It requires a strain of antibiotic-resistant bacteria. It involves sacrificing lab animals and evaluating them for tumors. 38 All Cells Have Multiple DNA Repair Systems nearly 200 genes in the human genome encode proteins dedicated to DNA repair many DNA repair processes seem to be extraordinarily inefficient energetically – irrelevant because the integrity of the genetic information is more important 39 Mismatch Repair mismatches are corrected to reflect template strand information – distinguished from the newly synthesized strand by the presence of methyl group tags on the template DNA in E. coli, Dam methylase methylates DNA at the N6 position of adenines within (5′)GATC sequences (strand discrimination) 40 -Methylation and Mismatch Repair- 41 Mismatch Repair on the 5′ Side of the Cleavage Site the unmethylated strand is unwound and degraded in the 3′→5′ direction from the cleavage site through the mismatch this segment is replaced with new DNA by: – SSB (single strand DNA binding protein) – exonuclease I, exonuclease X, or exonuclease VII – DNA polymerase III – DNA ligase 42 Mismatch Repair on the 3′ Side of the Cleavage Site repair is similar to mismatches on the 5′ side the exonuclease is either exonuclease VII or RecJ nuclease 43 Question Which statement regarding DNA mismatch repair in E. coli is false? A. If both strands appear to be parental (“correct”), the mismatch repair system can still repair the strand with the mistake (through a process still poorly understood). B. The parental (“correct”) strand can be distinguished by its methylation. C. A single mismatch may require replacing hundreds of nucleotides on the unmethylated strand. D. It involves exonucleases. 44 Eukaryotic Mismatch Repair Systems several proteins are structurally and functionally analogous to the bacterial MutS and MutL proteins many details of eukaryotic mismatch repair are unknown identification of newly synthesized DNA strands does not involve GATC sequences 45 Base-Excision Repair DNA glycosylases = recognize common DNA lesions and remove the affected base by cleaving the N-glycosyl bond in the process of base-excision repair – generally specific for one lesion type AP site or abasic site = an apurinic or apyrimidinic site in the DNA resulting from removal of a base by DNA glycosylase 46 Uracil DNA Glycosylases uracil DNA glycosylases = specifically remove from DNA the uracil that results from spontaneous deamination of cytosine – does not remove uracil residues from RNA or thymine residues from DNA – deamination is 100-fold faster in ssDNA 47 Other DNA Glycosylases other DNA glycosylases recognize: – formamidopyrimidine (from purine oxidation) – 8-hydroxyguanine (from purine oxidation) – hypoxanthine (from adenine deamination) – alkylated bases – pyrimidine dimers 48 Repair at AP Sites in Bacteria the deoxyribose 5′-phosphate left behind is removed and replaced with a new nucleotide AP endonucleases = cut the DNA strand containing the AP site DNA polymerase I replaces the DNA DNA ligase seals the remaining nick 49 Question In the base-excision repair pathway, a DNA glycosylase creates: A. B. C. D. an AP site. an alkylated base. a pyrimidine dimer. a gap of 12 to 13 nucleotides. 50 Nucleotide-Excision Repair nucleotide-excision system = repairs DNA lesions that cause large distortions in the DNA helical structure excinuclease = a multisubunit enzyme that hydrolyzes two phosphodiester bonds, one on either side of the distortion DNA polymerase I (E. coli) or DNA polymerase ε (humans) fills the gap DNA ligase seals the nick 51 Eukaryotic Excinuclease similar mechanism to that of the bacterial enzyme requires 16+ polypeptides with no similarity to the E. coli excinuclease subunits hydrolyzes the sixth phosphodiester bond on the 3′ side and the twenty-second phosphodiester bond on the 5′ side – produces a fragment of 27 to 29 nucleotides 52 Question Nucleotide-excision repair: A. is required when the DNA lesion to be repaired does not cause a detectable distortion in the DNA structure. B. in both bacteria and eukaryotes requires removal of 60–70 nucleotides. C. involves DNA polymerase II in E. coli. D. occurs in all free-living organisms. 53 Direct Repair pyrimidine dimers result from a UV-induced reaction DNA photolyases = promote direct photoreactivation of cyclobutane pyrimidine dimers – use energy from absorbed light – generally contain two cofactors: FADH2 and folate 54 Formation of Nucleotides with Alkylation Damage O6-methylguanine = a modified nucleotide that forms in the presence of alkylating agents – common and highly mutagenic lesion – tends to pair with thymine rather than cytosine 55 Repair of Nucleotides with Alkylation Damage O6-methylguanine-DNA methyltransferase = catalyzes transfer of the methyl group of O6methylguanine to one of its own Cys residues – a single methyl transfer event permanently methylates the protein, inactivating it 56 Question DNA must be replicated with extremely high fidelity, and maintained free of structural defects in order to propagate a species. Which structural change in DNA, considered damage, CANNOT be repaired? A. pyrimidine dimers B. mismatched bases C. methylation of bases, such as 1-methylguanine or 3-methycytosine D. uracil incorporated into DNA E. All of these can be repaired. 57 The Interaction of Replication Forks with DNA Damage Can Lead to Error-Prone Translesion DNA Synthesis double-strand breaks and lesions in ssDNA arise from: – a replication fork encountering an unrepaired DNA lesion – ionizing radiation – oxidative reactions 58