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C. del Barrio, L. Embún, B. García, G. Miró, R. Pinacho, D. S. Rivera DNA POLYMERASES: A Structural Approach.

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Presentación del tema: "C. del Barrio, L. Embún, B. García, G. Miró, R. Pinacho, D. S. Rivera DNA POLYMERASES: A Structural Approach."— Transcripción de la presentación:

1 C. del Barrio, L. Embún, B. García, G. Miró, R. Pinacho, D. S. Rivera DNA POLYMERASES: A Structural Approach


3 Process of copying a double-stranded DNA molecule Important in all known life forms Each DNA strand holds the same information, so both strands can serve as templates for the reproduction of the opposite strand Template strand is preserved and new strand is assembled from nucleotides: semiconservative replication Resulting double-stranded DNA molecules are identical Proofreading and error-checking mechanisms ensure fidelity DNA replication is also performed in the laboratory: PCR Prokaryotes Eukaryotes between cell divisions during S phase preceding mitosis or meiosis I DNA Replication

4 Synthesis of DNA proceeds from a replication fork where the strands of the parent DNA are separated Both strands serve as templates for replication during which new DNA strands are 5-3 synthesized One strand is synthesized as a continuous chain The other strand (which uses 5-3 parent strand as template) is made as a series of short DNA molecules: Okazaki fragments Okazaki fragments require an RNA primer at 5 ends to initiate DNA synthesis started by DNA pol III adds nucleotides in 5-3 until it encounters RNA primer of previous Okazaki fragment removes the ribonucleotide primer (5-3 nuclease activity) continues DNA synthesis by filling the gaps with deoxyribonucleotides (polymerase activity) DNA pol I DNA Replication 5 Parental DNA duplex Direction of fork movement Daughter duplex Leading strand Short RNA primer Okazaki fragment Lagging strand Point of joining

5 DNA Replication Helicase Separates the double-helical configuration Topoisomerase Catalyzes and guides unknotting of DNA (topological unlinking of the 2 strands) SSB / SSBP Bind to single-strands, keeping them separated and allowing DNA replication machinery to perform its function RNA primase Performs new RNA primer synthesis No known DNA pol can initiate the synthesis of a DNA strand without initial RNA primers DNA ligase Its nick sealing joins new Okazaki fragment to the growing chain OTHER IMPORTANT ENZYMES INVOLVED IN DNA REPLICATION

6 DNA polymerase families FamilyMain membersFeatures Are they crystallized? A Prokaryotic DNA pol I mitochondrial pol γ phage pols T3, T5 and T7 Found primarily in organisms related to prokaryotes Yes [92 entries] B Phage pols T4 and T6 herpes virus pol archeal pol Vent mammalian pols α, δ and ε Present in phages, viruses, archea and eukaryotes Many of these pol function replicate the host genome Yes [17 entries] X Mammalian pols β, λ and μ Function during DNA repair Yes [117 entries] RT RTs from retroviruses eukaryotic telomerases Use of a RNA template to synthesize the DNA strand Yes [158 entries] most MMLV and HIV Pol III Bacterial DNA pols Replicate the majority of bacterial genomes Yes [135 entries] UmuC/DinB Pols η, ι and κ Rev1 (terminal deoxycytidyl transferase) TLS pols, which have low fidelity on undamaged templates and replicate through damaged DNA Yes [17 entries]

7 DNA polymerase I lineage Class Multi-domain proteins (alpha & beta) Folds consisting of 2 or more domains belonging to different classes FoldDNA/RNA polymerases 3 morphological domains (palm, thumb and fingers) All members conserve the palm domain SuperfamilyDNA/RNA polymerases Palm domain has a ferrodoxin-like fold, related to that of an adenylyl cyclase domain 6 members FamilyDNA polymerase I Protein domain DNA polymerase I (Klenow fragment) Species Escherichia coli Termophilus aquaticus

8 DNA polymerase I domains polymerase3-5 exonuclease5-3 exonuclease Klenow fragment primer DNA template N C residue number distance between polymerase active site and exonuclease binding site = 30 Å

9 DNA polymerase I functions POLYMERASE REACTION T 3 5 primer strand -OH T T TT A 5 template strand AAA A A A A 3 A T TTT -OH 5 3 C 3 AAAAA A A A 5 A It catalyzes stepwise addition of a deoxyribonucleotide to 3-OH end of the primer strand that is paired to a second template strand The new strand grows in 5-3 direction Each incoming deoxyribonucleoside triphosphate must pair with the template strand to be recognized by pol this strand determines which deoxyribonucleotide will be added

10 DNA polymerase I functions 3-5 EXONUCLEASE ACTIVITY A 53 AAAAAAAAAA hydrolysis site 3 T 5 3 G T TT TT TT T One domain catalyzes hydrolysis of nucleotides at the 3 end of DNA chains To be removed, a nucleotide must have a free 3-OH terminus and must not be part of a double helix 5-3 EXONUCLEASE ACTIVITY 53 AAAAAAAAAAA hydrolysis site G TT T TT TT TT The second domain hydrolyzes DNA starting from the 5 end of DNA It can occur at the 5 terminus or at a bond several residues away from it The cleaved bond must be in a double-helical region

11 Taq DNA polymerase I: structure and domains

12 Taq polymerase I (Klentaq fragment) [3KTQ] E. coli polymerase I (Klenow fragment) [1D8YA] Why do we use Taq polymerase I instead of the one from E. coli? RMS: 1.75

13 Taq polymerase I N-term C-term PDB: 1CMW

14 Taq polymerase Linker region (links Klentaq to N-term fragment) 5-3 exonuclease domain Klentaq fragment PDB: 1CMW

15 5-3 exonuclease domain N-term resolvase-like domain C-term SAM fold PDB: 1CMW

16 Klentaq fragment Thumb Palm 3-5 exonuclease-like domain Fingers Linker region (links to N-term fragment) PDB: 3KTQ

17 Klentaq fragment Thumb Palm 3-5 exonuclease-like domain Fingers PDB: 3KTQ

18 DNA polymerase I Klenow fragment LARGE DOMAIN C N α+β Type with 6-stranded antiparallel β sheet Connection between β strands 9 and 12 makes a long excursion that builds up one side of the DNA-binding cleft It contains helices L-Q as well as the antiparallel hairpin of β strands 11 and 12

19 DNA polymerase I Klentaq fragment LARGE DOMAIN PDB: 3KTQ

20 DNA pol Klenow fragment SMALL DOMAIN C nucleotide binding site α/β Type with 1 antiparallel βstrand 5-stranded βsheet with 2 connecting helices and 1 helix (C) at the carboxy terminus (E.coli) UNIQUE TOPOLOGY FOR A MIXED β SHEET

21 Highly conserved regions

22 Region 1 Region 2 Motif A Motif B Region 6 Motif C

23 Motif A Hydrophobic residues Asp610 DYSQIELR

24 Motif B Arg659 Lys663 Gly668 Tyr671 RRxhKhhNFGhhY

25 Motif C His784 Asp785 HDE


27 E+TPE-TP dNTP E-TP-dNTP PPi E-TP n+1 -PPi E-TP n+1 Pathway of DNA Synthesis 1 Polymerase (E) binds with template-primer (TP) 2 Appropriate dNTP binds with polymerase-DNA complex 3 Nucleophilic attack results in phosphodiester bond formation 4 Pyrophosphate (PPi) is released dynamic interactions between polymerase with its nucleic acid and dNTP substrates rate limiting step phosphodiester bond formation conformational change preceding nucleotide incorporation

28 Pathway of DNA Synthesis Polymerases undergo 4 significant conformational changes: During DNA binding step Subsequent to dNTP binding step and immediately preceding chemical catalysis Subsequent to nucleotide incorporation during PPi release During translocation towards new primer 3-OH terminus E+TPE-TP dNTP E-TP-dNTP PPi E-TP n+1 -PPi E-TP n+1

29 First step: polymerase binds to template Region 1 Template PDB: 2KTQ

30 Open conformation 3.21 Å 2.40 Å PDB: 2KTQ REGION 1 DNA 1

31 90º 2.40Å 3 2 Low dNTPs concentration OPEN CONFORMATION Tyr671 1st base DCT PDB: 2KTQ 1st base Stacking interaction


33 When dNTPs concentration increases... CLOSED CONFORMATION Open conformation (2KTQ) Closed conformation (3KTQ) RMS: 0.75

34 O-helix Open conformation (2ktq.pdb) Closed conformation (3ktq.pdb) 40º Å

35 Closed conformation Asp610 O-helix Arg587 Tyr671 Asp785 Gln613 Lys663 Arg659 His639 PDB: 3KTQ

36 Tyr671 Hydrophobic pocket Tyr671 Phe Å PDB: 3KTQ

37 Hydrogen bonding Hydrogen bonds PDB: 1QSS dNTP ~3Å

38 Pentose Base Triphosphate β γ α Structure dNTP PDB: 1QSS

39 Electrostatic Interaction α- phosphate Arg587 PDB: 1QSS + - 5,6Å

40 Electrostatic Interaction Gln613 β phosphate PDB: 1QSS 4,07Å + -

41 Electrostatic Interaction PDB: 1QSS Lys663 5,02Å + - Β-Β- β phosphate

42 Electrostatic Interaction His639 γ phosphate + - PDB: 1QSS ~5Å

43 Arg659 2,45 Å PDB: 1QSS Electrostatic Interaction + - γ phosphate

44 Stacking interactions Tyr671 Base Phe667 PDB: 1QSS 4,23Å 3,19Å

45 Metal-mediated interactions Asp785 Catalytic site triphosphate Mg Asp610 PDB: 3KTQ

46 Taq polymerase: Active site GLU 786 ASP 785 ASP 610 Catalytic Triad: 3 active site carboxylates Motif AAsp610 Motif CAsp 785 Glu 786 Equivalence in Escherichia coli: Asp 705Glu 710 Asp 882Glu viable triads!! PDB: 3KTQ

47 Taq polymerase: Active site Mg A Mg B Catalytic Triad: PDB: 3KTQ

48 Taq polymerase: Active site PαPα PβPβ PγPγ Coordination of the Mg B Asp 610 Asp 785 Tyr 611 PDB: 3KTQ

49 Taq pol : Active site PO 4 α Tyr 611 Asp 610 Asp 785 PO 4 γ PO 4 β Coordination of the Mg B PDB: 3KTQ

50 Taq pol : Active site Coordination of the Mg A PDB: 3KTQ O H2O O-O- O-O- Asp 785 PO 4 α O H2O Mg A Asp 610 O-O- 3OH -

51 Asp 610Asp 785Tyr 611 P P P MgA Asp ,773,25,29 2,862,09 Asp 785 3,043,142,85 4,382,12 Tyr 611 4,613,24 2,862,24 P 3,4 3,532,39 P 3,542,214 P 2,228 MgA Asp 610Asp 785HOH3003HOH3125 P 3'OH MgB Asp 610 3,853,43,733,2 ?2,37 Asp 785 4,843, ?2,48 HOH3003 3,653,02 ?2,54 HOH3125 4,312 ?2,2 P ?2,24 3'OH ? MgB 3,78 MgA Distance between atoms

52 Taq pol : Active site Coordination of the Mg A: Distances PDB: 3KTQ 3.8 A 2.2 A

53 Structural Superposition PDB: 3KTQ, 1D8Y,1XLW, 1T7P Cluster: (1T7P_phage & 3KTQ_Taq 1D8Y_KF 1XWL_Bst) Sc 7.04 RMS 2.28 Polymerases I from Thermus aquaticus, E. Coli, B. Stearothermofilus and Bateriophage T7

54 Structural Superposition Score = 7,75 RMSD=1,86 *alignfit PDB: 3KTQ, 1D8Y,1XLW

55 Evolution within the same genus

56 Evolution among eukaryotic organisms

57 Conclusions These enzimes have a characteristic structure, similar to a hand with 3 differentiated domains: palm, fingers and thumb. Distinct conformational changes lead to the appropriate DNA sinthesis pathway. The active site is the most conserved part of the sequence and its mechanism is based on metal coordination complexes. In other enzimes with polymerase activity, like RT, the 3D- disposition and sequence are different, but similar mechanisms have been kept. Structure and sequence are highly and significantly conserved among some polymerases, determining its importance in different organisms.

58 Preguntas tipo PEM Señale la respuesta FALSA, referente a las familias de DNA polimerasas: a)La DNA polimerasa I procariota forma parte de la familia UmuC/DinB b) La familia RT contiene las retrotranscriptasas de los retrovirus c) Los miembros de la familia X funcionan durante la reparación del DNA d) La familia Pol III contiene la mayoría de DNA polimerasas bacterianas e) Existen 6 familias de DNA polimerasas Pregunta 1

59 Preguntas tipo PEM ¿Qué dominio/s de la DNA polimerasa I de E. coli constituye/n el fragmento Klenow? a)Dominio polimerasa b) Dominio 3-5 exonucleasa c) Los dos anteriores d) Dominio 5-3 exonucleasa e) Todos los anteriores Pregunta 2

60 Preguntas tipo PEM Señale la respuesta VERDADERA, en relación a la topología de los dominios del fragmento Klenow: a)El dominio grande es del tipo α+β b) El dominio pequeño es del tipo α/β c) Las dos anteriores d) La lámina β mixta del dominio pequeño constituye una topología única e) Todas las anteriores Pregunta 3

61 Preguntas tipo PEM Respecto a la vía de síntesis de DNA, ¿qué paso/s es/son limitante/s en la reacción? 1.Unión de la polimerasa al cebador molde (template primer) de DNA 2. Cambio conformacional de la polimerasa, previo a la incorporación de dNTPs 3. Liberación de pirofosfato (PPi) 4. Formación del enlace fosfodiéster a)1, 2 y 3 b)1 y 3 c)2 y 4 d)4 e)1, 2, 3 y 4 Pregunta 4

62 Preguntas tipo PEM Señale la respuesta FALSA, referente al fragmento klenow de la DNA polimerasa I: a) El fragmento klenow se encuentra en la región N terminal de la DNA polimerasa I. b) El fragmento klenow contiene el dominio exonucleasa 3-5 y el dominio con actividad polimerasa. c) El dominio con actividad polimerasa tiene una morfología similar a una mano derecha con tres dominios bien diferenciados: palma, pulgar y dedos. d) Cuando hablamos de la Taq polimerasa I, el fragmento klenow pasa a denominarse fragmento klentaq. e) El dominio palma es el más conservado. Pregunta 5

63 Sobre la estructura de la DNA polimerasa I, señala la respuesta VERDADERA: a)El dominio 5-3 exonucleasa tiene una estructura típica de barril beta y se encuentra englobado en el dominio pulgar. b) El dominio 5-3 exonucleasa se compone de dos dominios: un dominio N terminal tipo resolvasa, y otro dominio C terminal con un plegamiento tipo SAM. c) Los dominios palma, pulgar y dedos no son relevantes en la interacción con el DNA y por tanto no participan en la reacción de formación del enlace fosfodiéster. d) El dominio 3-5 exonucleasa del fragmento klenow no está formado por ninguna lámina beta. e) El dominio pulgar del fragmento klenow presenta un plegamiento tipo sándwich de dos capas. Preguntas tipo PEM Pregunta 6

64 Sobre las regiones conservadas de la DNA pol, señala la respuesta VERDADERA: a)Las regiones más conservadas son los dominios A, B y C. b) El dominio A contiene la Asp610, que juega un papel importante en la interacción con el DNA. c) Las dos anteriores son ciertas. d) Algunos aminoácidos del motivo B y del motivo C tienen un papel muy importante en la estabilización del dNTP entrante. e) Todas las anteriores son ciertas. Preguntas tipo PEM Pregunta 7

65 Respecto a la conformación abierta de la DNA polimerasa señala la respuesta FALSA: a)La Tyr671 del dominio dedos ocupa el lugar que correspondería a la primera base del DNA. b)La Tyr671 establece puentes de hidrógeno con la primera base del DNA. c)La primera base del DNA experimenta un giro de 90º respecto al eje de la hélice. d)El dNTP se apila sobre el primer par de bases adyacente a la Tyr671. Preguntas tipo PEM Pregunta 8

66 Preguntas tipo PEM Pregunta 9 ¿En qué mecanismo se basa el centro activo de una DNA- polimerasa? a)En un complejo de coordinación octahédrico en torno a iones metálicos divalentes. b)En una coordinación octagonal entre residuos de la polimerasa y dos átomos de Mg++. c)En la unión covalente con el sustrato, en este caso el DNA. d)En una interacción forzada con el sustrato mediada por puentes salinos. e)Ninguna de las anteriores.

67 Preguntas tipo PEM Pregunta 10 ¿Por qué se caracterizan evolutivamente las DNA-polimerasas? a)Por el grado de conservación estructural de los dominios A, B y C del centro activo y por su mecanismo de acción. b)Por la elevada identidad de secuencia y conservación estructural de los dominios A, B y C del centro activo y por su mecanismo de acción. c)Por el grado de conservación de secuencia de los dominios A, B y C del centro activo y por su mecanismo de acción. d)Por adoptar una conformación espacial fija, gracias a la presencia de residuos hidrofóbicos en el bolsillo catalítico. e)No se han hallado evidencias de que estos enzimas estén relacionados entre sí evolutivamente.

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