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DNA POLYMERASES: A Structural Approach. C. del Barrio, L. Embún, B. García, G. Miró, R. Pinacho, D. S. Rivera. Table of contents. INTRODUCTION STRUCTURAL DESCRIPTION DNA SYNTHESIS PATHWAY DNA-ENZYME INTERACTIONS ACTIVE SITE EVOLUTIVE FEATURES CONCLUSIONS. DNA Replication. - PowerPoint PPT Presentation
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C. del Barrio, L. Embún, B. García, G. Miró, R. Pinacho, D. S. Rivera
DNA POLYMERASES:
A Structural Approach
Table of contents
• INTRODUCTION
• STRUCTURAL DESCRIPTION
• DNA SYNTHESIS PATHWAY
• DNA-ENZYME INTERACTIONS
• ACTIVE SITE
• EVOLUTIVE FEATURES
• CONCLUSIONS
• 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
• 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
Daugh
ter d
uple
x
Leading strand
Short RNA primer
Okazaki fragment
Lagging strand
Point of joining
5’
5’
3’
3’
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
DNA polymerase families
Family Main members FeaturesAre they
crystallized?
AProkaryotic 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 T6herpes 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 μFunction during DNA
repair Yes
[117 entries]
RTRTs from retroviruses
eukaryotic telomerasesUse of a RNA template to synthesize the DNA strand
Yes[158 entries]
most MMLV and HIV
Pol III Bacterial DNA polsReplicate the majority of
bacterial genomesYes
[135 entries]
UmuC/DinBPols 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]
DNA polymerase I lineage
ClassMulti-domain proteins
(alpha & beta)
Folds consisting of 2 or more domains belonging to different
classes
Fold DNA/RNA polymerases
3 morphological domains (“palm”, “thumb” and “fingers”)
All members conserve the “palm” domain
Superfamily DNA/RNA polymerases
“Palm” domain has a ferrodoxin-like fold, related to that of an
adenylyl cyclase domain
6 members
Family DNA polymerase I
Protein domain
DNA polymerase I (Klenow fragment)
SpeciesEscherichia coliTermophilus aquaticus
DNA polymerase I domains
polymerase 3’-5’ exonuclease 5’-3’ exonuclease
Klenow fragment
primer
DNA template
NC
3’
3’
5’
5’
residue number928 518 324 1
distance between polymerase active site and exonuclease binding site = 30 Å
DNA polymerase I functions
POLYMERASE REACTION
T3’5’
primer strand
-OHTT TT
A5’
template strand
A A A AA AA3’
AT T T T
-OH
5’3’
C
3’A A A AA A AA
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
DNA polymerase I functions
3’-5’ EXONUCLEASE ACTIVITY
A5’3’
A A A A A A A A A A
hydrolysis site3’
T5’ 3’GT T T T T T T 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
5’3’A A A A A A A A A A A
hydrolysis site5’
5’ 3’GT T T T T T T 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
Taq DNA polymerase I: structure and domains
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
Taq polymerase I
N-term C-term
PDB: 1CMW
Taq polymerase
Linker region (links Klentaq to N-term
fragment)
5’-3’ exonuclease domain
Klentaq fragment
PDB: 1CMW
5’-3’ exonuclease domain
N-term resolvase-like domain
C-term SAM fold
PDB: 1CMW
Klentaq fragment
Thumb
Palm
3’-5’ exonuclease-like
domain
Fingers
Linker region (links to N-term
fragment)
PDB: 3KTQ
Klentaq fragment
Thumb
Palm
3’-5’ exonuclease-like
domain
Fingers
PDB: 3KTQ
DNA polymerase I Klenow fragment
LARGE DOMAIN
C N
14 9 13 12 8 7
• 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
DNA polymerase I Klentaq fragment
LARGE DOMAIN
PDB: 3KTQ
DNA pol Klenow fragment
SMALL DOMAIN
C
nucleotide binding
site
5 2 3 4 1
• 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
Highly conserved regions
Highly conserved regions
Region 1
Region 2
Motif A
Motif B
Region 6
Motif C
Motif A
Hydrophobic residues
Asp610
“DYSQIELR”
Motif B
Arg659
Lys663Gly668
Tyr671
“RRxhKhhNFGhhY”
Motif C
His784
Asp785“HDE”
PATHWAY OF DNA SYNTESIS
Part 1: DNA binding
E+TP E-TP
dNTP
1 2 3 4
E-TP-dNTP
PPi
E-TPn+1-PPi E-TPn+1
Pathway of DNA Synthesis
1 Polymerase (E) binds with template-primer (TP)2 Appropriate dNTP binds with polymerase-DNA complex3 Nucleophilic attack results in phosphodiester bond formation4 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
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+TP E-TP
dNTP
1 2 3 4
E-TP-dNTP
PPi
E-TPn+1-PPi E-TPn+1
First step: polymerase binds to template
Region 1
Template
PDB: 2KTQ
Open conformation
3.21 Å
2.40 Å
PDB: 2KTQ
REGION 1
DNA
1
90º
2.40Å
3
2 Low dNTPs concentration
OPEN CONFORMATIONTyr671
1st base
DCT
PDB: 2KTQ1st base
Stacking interaction
PDB: 2KTQ
DCT
When dNTPs concentration increases...
CLOSED CONFORMATION
Open conformation (2KTQ)
Closed conformation (3KTQ)
RMS: 0.75
O-helix
Open conformation (2ktq.pdb)
Closed conformation (3ktq.pdb)
40º11-12 Å
Closed conformation
Asp610
O-helix
Arg587
Tyr671
Asp785
Gln613
Lys663
Arg659
His639
PDB: 3KTQ
Tyr671
Hydrophobic pocket
Tyr671
Phe667 4.11Å
PDB: 3KTQ
Hydrogen bonding
Hydrogen bonds
PDB: 1QSS
dNTP
~3Å
Pentose Base
Triphosphate
β
γ α
Structure dNTP
PDB: 1QSS
Electrostatic Interaction
α- phosphate
Arg587
PDB: 1QSS
+
-5,6Å
Electrostatic Interaction
Gln613
β phosphate
PDB: 1QSS
4,07Å +-
Electrostatic Interaction
PDB: 1QSS
Lys663
5,02Å+- Β-β phosphate
Electrostatic Interaction
His639
γ phosphate
+-
PDB: 1QSS
~5Å
Arg6592,45 Å
PDB: 1QSS
Electrostatic Interaction
+-
γ phosphate
Stacking interactions
Tyr671
Base
Phe667
PDB: 1QSS
4,23Å3,19Å
Metal-mediated interactions
Asp785
Catalytic site
triphosphate
Mg
Asp610
PDB: 3KTQ
Taq polymerase: Active site
GLU 786
ASP 785
ASP 610
Catalytic Triad:3 active site carboxylates
Motif AAsp610
Motif CAsp 785
Glu 786Equivalence in Escherichia coli:
Asp 705 Glu 710
Asp 882 Glu 883
2 viable triads!!PDB: 3KTQ
Taq polymerase: Active site
Mg A
Mg B
Catalytic Triad:
PDB: 3KTQ
Taq polymerase: Active site
Pα
Pβ
Pγ
Coordination of the Mg B
Asp 610
Asp 785Tyr 611
PDB: 3KTQ
Taq pol : Active site
PO4α
Tyr 611
Asp 610
Asp 785
PO4γ
PO4β
Coordination of the Mg B
PDB: 3KTQ
Taq pol : Active site
Coordination of the Mg A
PDB: 3KTQ
OH2O
O-
O-
Asp 785PO4 α
OH2O
MgA
Asp 610
O-3’OH-
Asp 610 Asp 785 Tyr 611 P P P MgA
Asp 610 3 2,77 3,2 5,29 2,86 2,09
Asp 785 3,04 3,14 2,85 4,38 2,12
Tyr 611 4,61 3,24 2,86 2,24
P 3,4 3,53 2,39
P 3,54 2,214
P 2,228
MgA
Asp 610 Asp 785 HOH3003 HOH3125 P 3'OH MgB
Asp 610 3,85 3,4 3,73 3,2 ? 2,37
Asp 785 4,84 3,62 2.81 ? 2,48
HOH3003 3,65 3,02 ? 2,54
HOH3125 4,312 ? 2,2
P ? 2,24
3'OH ?
MgB 3,78 MgA
Distance between atoms
Taq pol : Active site
Coordination of the Mg A: Distances
PDB: 3KTQ
3.8 A
2.2 A
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
Structural Superposition
Score = 7,75 RMSD=1,86 *alignfit
PDB: 3KTQ, 1D8Y,1XLW
Evolution within the same genus
Evolution among eukaryotic organisms
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.
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
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
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
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
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
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
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
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
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.
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.