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.