Biochemical Identification Nucleic Acid Structure, DNA ...flywing.co.kr/zokbo/file/KAIST/Common/BS120/ch11.pdf · 2015-04-06 1 Chapter 11 Nucleic Acid Structure, DNA Replication,

2015-04-06

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Chapter 11Nucleic Acid Structure, DNA Replication,

and Chromosome Structure

n Biochemical Identification of the Genetic Material

n Nucleic Acid Structure

n An Overview of DNA Replication

n Molecular Mechanism of DNA Replication

n Molecular Structure of Eukaryotic Chromosomes

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Key Concepts:

n What is the genetic material?n Four criteria necessary for genetic material:

1. Information2. Replication3. Transmission4. Variation

n Late 1800s – biochemical basis of heredity postulated

n Researchers became convinced that chromosomescarry the genetic information

n 1920s to 1940s – scientists expected the protein portion of chromosomes would turn out to be the genetic material 2

Biochemical Identification of the Genetic Material

Griffith’s bacterial transformationn Late 1920s – Frederick Griffith was working with

Streptococcus pneumoniae bacteria

n Two strains of S. pneumoniae:q Strains that secrete capsules look smooth (S)

and infections are fatal in miceq Strains that do not secrete capsules look rough (R) and

infections are not fatal in mice

n The capsule shields the bacteria from the immune system, so they survive in the blood

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

1 Type S cellsare virulent.

Control:Injected livingtype R bacteriainto mouse.

2 Type R cellsare benign.

Control:Injected livingtype S bacteriainto mouse.

n Smooth strains (S) with capsule are fatal; rough strains (R) without capsule are not

n If mice are injected with heat-killed type S, they survive (because bacteria are dead)

n However, mixing live R with heat-killed S kills the mouseq Blood is found to contain living type S bacteriaq Known as transformation

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5


Control:Injected livingtype R bacteriainto mouse.

Control:Injected heat-killed type S bacteria into mouse.

Virulent type Sstrain in deadmouse’s blood

4

3

2

1

Living typeR cells havebeentransformedinto virulenttype S cellsby asubstancefrom theheat-killedtype S cells.

Treatment Result Conclusion

Type S cellsare virulent.

Type R cellsare benign.

Heat-killedtype S cellsare benign.

Control:Injected livingtype S bacteriainto mouse.

Injected livingtype R andheat-killedtype S bacteriainto mouse.

n How is this possible?

n Genetic material had been transferred from the heat-killed type S bacteria to the living type R bacteria

n This gave them the capsule-secreting trait and was passed on to their offspring

n What was the biochemical basis of this transforming principle? At the time there was no way to know

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Avery, MacLeod, and McCarty Used Purification Methods to Reveal That

DNA is the Genetic Materialn 1940s – interest in finding biochemical basis of bacterial

transformation

n Only purified DNA from type S could transform type R

n But, purified DNA might still contain traces of contamination that may be the transforming principle

n Added DNase, RNase and proteasesn RNase and protease had no effect

n When DNase was added, no transformation took place

n Surprising conclusion: DNA is the genetic material


1

Control

A B C D E

A B C D E

A B C D E

Experimental level Conceptual level

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HYPOTHESIS A purified macromolecule from type S bacteria, which functions as the genetic material, will be able to convert typeR bacteria into type S.

KEY MATERIALS Type R and type S strains of Streptococcus pneumoniae.

Purify DNA from a type S strain.This involves breaking open cellsand separating the DNA away fromother components bycentrifugation.

Mix the DNA extract with type Rbacteria. Allow time for the DNAto be taken up by the type R cells,converting a few of them to type S.Also, carry out the same steps butadd the enzymes DNase, RNase, orprotease to the DNA extract, whichdigest DNA, RNA, and proteins,respectively. As a control, don’t addany DNA extract to some type R cells.

Add an antibody, a protein made bythe immune system of mammals,that specifically recognizes type Rcells that haven’t beentransformed. The binding of theantibody causes the type R cells toaggregate.

Addantibody

+ DNA + DNA+ DNase

+ DNA+ RNase

+ DNA+ Protease

± DNase

± RNase± Protease+ Type R cells

A B C D E

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Centrifuge

AB

CD

E

Control

4

5

Remove type R cells bycentrifugation. Plate the remainingbacteria (if any) that are in thesupernatant onto petri plates.Incubate overnight.

THE DATA

DNA extract

DNA extract + DNaseDNA extract + RNase

DNA extract + protease

Type S cellsin supernatant

Type R cellsin pellet

CONCLUSION DNA is responsible for transforming type R cells into type S cells.

SOURCE Avery, O.T., MacLeod, C.M., and McCarty, M. 1944. Studies on the Chemical Nature of the Substance Inducing Transformation of Pneumococcal Types. Journal of Experimental Medicine 79:137–156.

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Hershey and Chasen 1952 – studied a T2 virus that infects Escherichia coli

q Bacterial virus is known as bacteriophage or phage

n Phage coat made entirely of proteinn DNA found inside capsid

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DNA

Protein

(a) Schematic drawing ofT2 bacteriophage

DNA

Phage head(capsid)

Sheath

Tail fiber

T2 geneticmaterial beinginjected intoE. coli

E. coli cell

Base plate

(b) An electron micrograph of T2 bacteriophageinfecting E. coli

50 nm

© Eye of Science/Photo Researchers, Inc.

n What does the phage inject into the bacteria –DNA or protein?

n Shear force from a blender separates the phage coat from surface of the bacteria

n Tag each component with a radioactive labelq 35S labels proteinsq 32P labels DNA

n Conclusion: DNA is the genetic material

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Experiment 1 Experiment 2

1

2

Bacterial cell Bacterial cell

Sheared empty phage

Tota

l iso

tope

in s

uper

nata

nt (%

)

Agitation time in blender (min)

100

80

60

40

20

00 1 2 3 4 5 6 7 8

3

4 THE DATA

E. coli cells wereinfected with35S-labeled phageand subjected toblender treatment.

Phage DNA

35S-labeled shearedempty phage

E. coli cells wereinfected with32P-labeled phageand subjected toblender treatment.

32P-labeledphage DNA

Using a Geiger counter,determine the amount ofradioactivity in the supernatant.

Geiger (radioisotope)counter

Transfer to tubeand centrifuge.

Supernatanthas 35S-labeledempty phage.

Transfer to tubeand centrifuge.

Supernatant hasunlabeled emptyphage.

Pellet hasE. coli cellsinfected with32P-labeledphage DNA.

Extracellular 35S

Extracellular 32P

Blending removes 80%of 35S from cells.

Most of the 32P (65%)remains with intact cells.Pellet has

E. coli cellsinfected withunlabeledphage DNA.

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Levels of DNA Structure:1. Nucleotides – the building blocks of DNA and RNA

2. Strand – a linear polymer strand of DNA or RNA

3. Double helix – the two strands of DNA

4. Chromosomes – DNA associated with an array of different proteins into a complex structure

5. Genome – the complete complement of genetic material in an organism

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Nucleic Acid Structure

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Single strand

Nucleotides

Double helix

DNA associates withproteins to form achromosome.

DNAn Formed from nucleotides (A, G, C, T)

n Nucleotides composed ofthree componentsq Phosphate group

q Pentose sugarn Deoxyribosen DNA = Deoxyribonucleic Acid

q Nitrogenous basen Purines – Adenine (A), Guanine (G)n Pyrimidines – Cytosine (C), Thymine (T)

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Base

Phosphate

Deoxyribose

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DNA nucleotidesCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

(a) DNA nucleotide

Phosphate

Deoxyribose

Base

Thymine (T)Adenine (A)

Cytosine (C)Guanine (G)

NH2

O

H

H

N

N

NH2

NH H

H H

NH

N

N

O

NH2

HN

N

NH

N

H

HOH

HH

OOO–

CH2

O–

PO

H

H

O

N

O

NH

Purines(double ring)

Pyrimidines(single ring)

CH3

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RNAn Formed from nucleotides (A, G, C, U)

n Nucleotides composed ofthree componentsq Phosphate group

q Pentose sugarn Ribosen RNA = Ribonucleic Acid

q Nitrogenous basen Purines – Adenine (A), Guanine (G)n Pyrimidines – Cytosine (C), Uracil (U)

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Base

Phosphate

Ribose

H

OH

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RNA nucleotidesCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

(b) RNA nucleotide

RiboseUracil (U)Adenine (A)

Guanine (G)

H

Cytosine (C)

NH2

O

H

H

N

N

H

H H

H

H

H

NH2

N

NH

N

N

O

NH2

HN

N

NH

N

Phosphate

Base

H

OHOH

HH

OO

O–

CH2

O–

PO

H

n Sugar carbons are 1’ to 5’n Base attached to 1’ carbon on sugarn Phosphate attached to 5’ carbon on sugar

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Nucleotide numbering system


HH

HOH

HH

OO

O–

O–5′

4′ 1′

2′3′

3

21

6

54

PO

O

H

H

N

O

N

ThymineCH3

CH2

Phosphate

Deoxyribose

Strandsn Nucleotides are

covalently bondedn Phosphodiester bond

– phosphate group links two sugars

n Backbone – formed from phosphates and sugars

n Bases project away from backbone

n Written 5’ to 3’n ex: 5’ – TACG – 3’

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O

H

H

N

O

O–

N

N

H

N

N

HH

H

3′

2′3′

N

Thymine (T)

Adenine (A)

BasesBackbone

CH3

Sugar (deoxyribose)OH

H

Phosphate

NH2

H H

OO P

O–

OO

O

5′

4′

3′

P

5′ CH2

CH2

H

H

H

O

1′

2′

O

O

N

N

HH

H

HH

OOO P

O–

HN

N

N

H

N

HH

OOO

O

P

5′

4′ 1′

2′3′

5′4′ 1′

Cytosine (C)

Guanine (G)

Phosphodiesterlinkage

Singlenucleotide

CH2

NH2

NH2

CH2

H

H

H

H

H

O

1′

2′3′

O

H HO–

5′

4′

O–

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Solving the structure of DNAn 1953, James Watson and Francis Crick,

proposed the structure of the DNA double helix

n Watson and Crick used Linus Pauling’s method of working out protein structures using simple ball-and-stick models

n Rosalind Franklin’sX-ray diffraction results were crucial evidence, suggesting a helical structurewith uniform diameter

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X-rays diffracted by DNAonto photographic platePattern represents theatomic array in wet fibers

Wet DNA fibers

X-ray beam

n Erwin Chargoff analyzed base composition of DNA from many different species

n Results consistently showed

amount of adenine (A) = amount of thymine (T)

amount of cytosine (C) = amount of guanine (G)

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Base-pairing

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Watson and Crickn Put together these pieces of information

n Found ball-and-stick model consistent with dataq Double-stranded helixq Base-pairing: A with T and G with C

n James Watson, Francis Crick, and Maurice Wilkins awarded Nobel Prize in 1962

n Rosalind Franklin had died and the Nobel Prize is not awarded posthumously

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n Double stranded

n Antiparallel strands

n Right-handed helix

n Sugar-phosphatebackbone

n Bases on the inside

n Stabilized by H-bonding

n Specific base-pairing

n ~10 nts per helical turn26

Features of DNACopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Bases

Hydrogen bond

2 nm

(a) Double helix

5′ end

3′ end

Sugar-phosphatebackbone

One nucleotide0.34 nm

3′ end5′ end

Complete turnof the helix3.4 nm


HH

H

H

OOO P

N

O H

H

O

CH3

CH2

O

H N

N

N

H

N

H H

H

HH

OOO PCH2

O

O

HN

N

NH

N

H H

H

HH

OOO PCH2

H N

NH

N

N

H H

H

HH

OOO PCH2

HO

Cytosine

Cytosine

Guanine

Guanine

Thymine

Adenine

O

HH

N

N

HH

H

HH

OO

O

P CH2

O–

O

O

HH

N

N

HH

HOH

HH

OO

O

P CH2 O

O

5′ phosphate

3′ hydroxyl

(b) Base pairing

Key Features• Two strands of DNA form a double helix.• The bases in opposite strands hydrogen-

bond according to the AT/GC rule.• The 2 strands are antiparallel.• There are ~10 nucleotides in each

strand per complete turn of the helix.

5′ end

O–

O–

O–

3′ end

HN

H

H

HN

HN

H

H

HN

NH H

Hydrogenbond

O–

O–

O–

O–

3′ end

5′ end

H

n Chargoff’s ruleq A pairs with Tq G pairs with Cq Keeps width consistent

n Complementary DNA strandsq 5’ – GCGGATTT – 3’q 3’ – CGCCTAAA – 5’

n Antiparallel strandsq One strand 5’ to 3’q Other stand 3’ to 5’

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n Grooves are revealed in the space-filling model

n Major grooveq Proteins bind to affect gene

expression

n Minor grooveq Narrower

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Major groove

Minor groove

Major groove

Minor groove

n Late 1950s – three different models were proposed for DNA replicationq Semiconservative Modelq Conservative Modelq Dispersive Model

n Newly-made strands are “daughter strands”

n Original strands are “parental strands”

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An Overview of DNA Replication

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Semiconservative MechanismCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Second roundof replication

First roundof replication

Originaldouble helix

Parental strandDaughter strand

(a) Semiconservative mechanism. DNA replication producesDNA molecules with 1 parental strand and 1 newly madedaughter strand.

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Conservative MechanismCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.




(b) Conservative mechanism. DNA replication produces 1 doublehelix with both parental strands and the other with 2 newdaughter strands.

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Dispersive MechanismCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.




(c) Dispersive mechanism. DNA replication produces DNAstrands in which segments of new DNA are interspersedwith the parental DNA.

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Parental strandDaughter strand

(a) Semiconservative mechanism. DNA replication producesDNA molecules with 1 parental strand and 1 newly madedaughter strand.

(b) Conservative mechanism. DNA replication produces 1 doublehelix with both parental strands and the other with 2 newdaughter strands.

(c) Dispersive mechanism. DNA replication produces DNAstrands in which segments of new DNA are interspersedwith the parental DNA.

n In 1958, Matthew Meselson and Franklin Stahl devised an experiment to differentiate among the three proposed DNA replication mechanisms

n Nitrogen comes in a common light form (14N) and a rare heavy form (15N)

n Grew E. coli in medium with 15N to label, then switched to medium with 14N, collecting samples after each generation

n Original parental strands would be 15N while newly made strands would be 14N

n Conclusion: Semiconservative DNA replication35

Meselson and Stahl experiment

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© Meselson, M., Stahl, F., (1958) “The replication of DNA in Escherichia coli,”PNAS, 44(7):671–82, Fig. 4a

5

Approximate generations after transfer to 14N medium.

Light

Half-heavy

Heavy

< 1.0 3.01.0 2.0

THE DATA

1

3

4

2Grow bacteria in 15Nmedia.

Transfer to 14N media andcontinue growth for <1,1.0, 2.0, or 3 generations.

14N medium(light)

15N medium(heavy)

Isolate DNA after each generation. TransferDNA to CsCl gradient, and centrifuge.

DNA

CsCl gradient

Centrifuge

Observe DNA under UV light.

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Semiconservative replication

n The two parental strands separate and serve as template strands

n New nucleotides must obey the AT/GC rule

n End result: two new double helices with same base sequence as original

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T A

T A

GG

AC

C

T A

G CC G

T A

T A

C G

C G

A

A T

A T

T A

T A

T A

C G

C G

G CG C

G CG C

C G

A T

A T

A T

T A

T A

T A

C G

C G

G CG C

G CG C

C G

A T

A T

A T

T A

T A

T A

C G

C G

G CG C

G CG C

C G

A T

Incomingnucleotides

Original(template)strand

Newlysynthesizeddaughter strand

Original(template)strand 5′ 5′3′3′

(b) The products of replication(a) The mechanism of DNA replication

3′

5′ 3′ 5′ 3′

5′3′

3′5′

3′5′

A TC G

C G

C G

A T

C GT A

5′

T A

A T

A

CT

TG

ReplicationforkA T

CG

n Origin of replication provides an opening called a replication bubble that forms two replication forks

n DNA replication proceeds outward from forks

n Bacteria have single origin of replication

n Eukaryotes have multiple origins of replication

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Molecular Mechanismof DNA Replication

40


2

1

3

2

DNA strands unwind.

DNA replication begins outwardfrom two replication forks.

DNA replicationcontinues in bothdirections.

2

Replicationforks

Replicationfork

Replicationfork

DNA replicationis completed.

Site whereDNA replicationendsDNA strands unwind,

and DNA replicationbegins.

DNA strands unwind,and DNA replicationbegins at multipleorigins of replication.

DNA replicationis completed.

Kinetochore proteinsat the centromere

(c) Multiple origins of replication in eukaryotes(b) Single origin of replication in bacteria(a) Bidirectional replication

Circularbacterialchromosome

Origin ofreplication Origin of

replicationOrigin ofreplication

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n DNA helicaseq Binds to DNA and travels 5’ to 3’ using

ATP to separate strand and move fork forward

n DNA topoisomeraseq Relives additional coiling ahead of

replication fork

n Single-strand binding proteinsq Keep parental strands open to act as

templates

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5′

3′

3′

DNA topoisomerasetravels slightly aheadof the replication forkand alleviates coilingcaused by the actionof helicase.

5′

DNA helicase travelsalong one DNA strandin the 5′ to 3′ directionand separates the DNAstrands.

Direction of replication fork

Single-strand binding proteinscoat the DNA strands to preventthem from re-forming a double helix.

n DNA polymeraseq Covalently links nucleotidesq Deoxynucleoside triphosphates

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(a) Action of DNA polymerase

5′

3′Incomingdeoxynucleosidetriphosphates

DNA polymerasecatalytic site

Deoxynucleoside triphosphatesq Free nucleotides with three phosphate groupsq Breaking covalent bond to release pyrophosphate (two phosphates)

provides energy to connect nucleotidesCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

HH

H

HH

OOO

O–

P N

O

H

O

CH3

O–

O

N

HN

N

NH

N

H H

H

HH

OOO

O–

PCH2

O

OH

N

N

NH

N

H H

H

HH

OOO

O–

PCH2

O–

N

HH

H

H N

NH

H

H

H

N

N

H H

H

HH

OOO

O–

PCH2

+

O

HH

NN

HH

H

HH

OO

O

P

O–

NO

OH

O

HH

H

HH

OOO

O–

P N

O H

H

O

CH3

OH

N

N

NH

N

H H

H

HH

OOO

O–

P

O

OH

N

N

NH

N

H H

H

HHO

OOO–

PCH2

O–

H N

NH

N

N

H H

H

HH

OOO

O–

PCH2

O

HH

NN

HH

H

HH

OO

O

P CH2

O–O

O

HH

NN

HH

H

HH

OOOP CH2

O–O

O

H

T A

C

C

G

G

T

C

G

C

G

A

T

C

G

C

G

A

N

H

H

N

H

H

NH

H

H

HN

N

H

H

H

HN

O O––O

O O

P P–O O–

5′ end

CH2

CH2

3′ end

HO

5′ end

O–

CH2

Templatestrand

3′ end 3′ end

5′ end

Phosphate

3′ end Pyrophosphate

New phosphoesterbond

5′ end

CH2

OH +

An incoming nucleotide(a deoxynucleoside triphosphate)(b) Chemistry of DNA replication

HO

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Features of DNA polymerase

1. DNA polymerase cannot begin synthesis on a bare template strand

q Requires a primer to get startedq DNA primase makes the primer from RNAq The RNA primer is removed and replaced with

DNA later

2. DNA polymerase only works 5’ to 3’

45 46

5′ 3′

3′ 5′

DNA polymerase can linknucleotides only in the 5′ to 3′ direction.

DNA polymerase is able tocovalently link nucleotidestogether from a primer, whichis made by DNA primase.

(b) 5′ to 3′ direction ofDNA synthesis

(a) Need for a primer

RNAprimer


n Leading strandq DNA synthesized in as one long moleculeq DNA primase makes a single RNA primerq DNA polymerase adds nucleotides in a 5’ to 3’

direction as it slides forward

n Lagging strandq DNA synthesized 5’ to 3’ but as Okazaki fragmentsq Okazaki fragments consist of RNA primers plus DNA

n In both strandsq RNA primers are removed by DNA polymerase and

replaced with DNAq DNA ligase joins adjacent DNA fragments

47 48


1

2

3

5′5′

5′

3′

5′

3′

3′

3′

5′

5′

3′

3′

3′

DNA strands separate at anorigin of replication, creating2 replication forks.

Replicationforks

RNA primerLeadingstrand

Primers are needed to initiateDNA synthesis. The synthesisof the leading strand begins inthe direction of the replicationfork. In the lagging strand, thefirst Okazaki fragment is madein the opposite direction.

The leading strand elongates,and a second Okazaki fragmentis made.

5′

3′

5′

Primer

First Okazaki fragmentof the lagging strand

Direction ofreplication fork

FirstOkazakifragment

SecondOkazakifragment

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4

3′

5′

5′

3′

5′

5′3′

3′

The leading strand continuesto elongate. A third Okazakifragment is made, and the firstand second are connectedtogether.

First and second Okazakifragments have beenconnected to each other.

ThirdOkazakifragment

50


5′

3′ 5′

3′

Leadingstrand

Laggingstrand

Replicationfork

Replicationfork

(b) Replication from an origin

Origin of replication

Leadingstrand

Laggingstrand


2

3′

5′

5′

3′5′

3′

5′

3′

5′

1DNAprimaseDNA primase makes RNA primers to begin

the replication process.

DNA polymerase III makes DNA from theRNA primers. DNA primase hops back tothe opening of the fork and makes a secondRNA primer for the lagging strand.

Direction of replication forkDNApolymerase III

SecondprimerDNAprimase

FirstRNA primer

DNApolymerase III

Leadingstrand

Clampprotein

RNAprimer

Lagging strand(Okazakifragment)

3′

5′

3′

5′

3′

5′

3′

5′

3′

5′

3′

5′

4

3 DNA polymerase III continues to elongatethe leading strand. In the lagging strand,DNA polymerase III synthesizes DNAfrom the second primer. DNA polymeraseI removes the first primer and replaces itwith DNA.

Thirdprimer

Secondprimer

Missingcovalent bond

DNA ligase

Thirdprimer

In the lagging strand, DNA ligase forms acovalent bond between the first and secondOkazaki fragments. A third Okazakifragment is made. The leading strandcontinues to elongate.

DNApolymerase I

3′

5′ DNA replication is very accurate

n Three mechanisms for accuracy1. Hydrogen bonding between A and T,

and between G and C is more stable than mismatched combinations

2. Active site of DNA polymerase is unlikely to form bonds if pairs mismatched

3. DNA polymerase can proofread to remove mismatched pairsn DNA polymerase backs up and digests linkagesn Other DNA repair enzymes as well

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DNA Polymerases Are a Family of Enzymes With Specialized Functionsn Important issues for DNA polymerase are speed, fidelity,

and completeness

n Nearly all living species have more than one type of DNA polymerase

n Genomes of most species have several DNA polymerase genes due to gene duplication

n Independent genetic changes produce enzymes with specialized functions

n E. coli has 5 DNA polymerasesq DNA polymerase III – multiple subunits,

responsible for majority of replicationq DNA polymerase I – a single subunit, rapidly

removes RNA primers and fills in DNAq DNA polymerases II, IV and V – DNA repair and

can replicate damaged DNAn DNA polymerases I and III stall at DNA damagen DNA polymerases II, IV and V don’t stall but go slower

and make sure replication is complete

n Humans have 12 or more DNA polymerasesq Designated with Greek letters

q DNA polymerase α – its own built in primase subunitq DNA polymerase δ and ε – extend DNA at a faster rateq DNA polymerase γ – replicates mitochondrial DNA

q When DNA polymerases α, δ or ε encounter abnormalities they may be unable to replicate

q Lesion-replicating polymerases may be able to synthesize complementary strands to the damaged area

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Telomeres

n Series of short nucleotide sequences repeated at the ends of chromosomes in eukaryotes

n Specialized form of DNA replication only in eukaryotes in the telomeres

n Telomere at 3’ does not have a complementary strand and is called a 3’ overhang

57 58


G G G T T AG G G T T A G G G T T A G G G T T A G G GC C C A A T C C C A A T C C C A A T

T T A

3′

5′

Telomere repeat sequences

3′ overhang

n DNA polymerase cannot copy the tip of the strand with a 3’ endq No place for upstream primer to be made

n If this replication problem were not solved, linear chromosomes would become progressively shorter

n Telomerase enzyme attaches many copies of DNA repeat sequence to the ends of chromosomes

59

n Shortening of telomeres is correlated with cellular senescence

n Telomerase function is reduced as an organism ages

n 99% of all types of human cancers have high levels of telomerase

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UC C AAC

5′

3′

3′

3′

5′

TelomereEukaryoticchromosome

Telomere

RNA intelomerase

TelomeraseTelomerase synthesizesa 6-nucleotide repeatsequence.

Telomerase binds to aDNA repeat sequence.

1

2

T A G G GA T C C C A A T C C C A A T C C CA A U C C C

T T A G G G T T A G G G T T AA A UT T AG G G

C C CT T AG G G

A A AAU UC C CA G G G T T A G G G T T A GG G T T A T T A TG G GG G GT

A T C C C A A T

5′

3′5′

3′

Primase makes an RNA primer near the end ofthe telomere, and DNA polymerase synthesizesa complementary strand in the 5′ to 3′ direction.The RNA primer is eventually removed.

Telomerase moves 6 nucleotidesto the right and begins to makeanother repeat.

RNA primer that iseventually removed

3

4

T A G G G T T A G G G T T A T T AA A U

G G GA T C C C A A T

5′

5′

3′

Repeat sequence

A G G G T T A G G G T T A T TT AG G G GG GA A U C C C AAU

TA T C C C A A T

n Typical eukaryotic chromosome may be hundreds of millions of base pairs longq Length would be 1 meterq But must fit in cell 10-100µm

n Chromosomeq Discrete unit of genetic material

n Chromosomes composed of chromatinq DNA-protein complex

62

Molecular Structure of Eukaryotic Chromosomes

Three levels of DNA compaction

1. DNA wrapping q DNA wrapped around histones to form nucleosomeq Shortens length of DNA molecule 7-fold

2. 30-nm fiberq Current model suggests asymmetric, 3D zigzag of

nucleosomesq Shortens length another 7-fold

63 64


Nucleosome:8 histone proteins + 146 or 147 nucleotidebase pairs of DNA

DNA

Linkerregion

Aminoterminaltail ofhistoneprotein

H2BH2B

H4H4

H2A

H3

H111 nm

30 nm

(a) Micrograph of a 30-nm fiber

(b) Three-dimensional zigzag modela: Photo courtesy of Dr. Barbara HamkaloZ


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3. Radial loop domainsq Interaction between

30-nm fibers and nuclear matrix

q Each chromosome located in discrete territory

n Level of compaction is not uniform

q Heterochromatinq Euchromatin

65


Protein fiber inside the nucleus

30-nm fiber

Radial loopdomain

Protein that attaches the baseof a DNA loop to a protein fiber

Cell division

n When cells prepare to divide, chromosomes become even more compacted

q Euchromatin not as compact

q Hetrochromatin much more compact

n Metaphase chromosomes highly compacted

66

67


2 nm

30 nm

1

2

11 nm

Histone H1

Wrapping of DNA aroundhistone proteins

Formation of a 3-dimensionalzigzag structure via histoneH1 and other DNA-bindingproteins

Histones

Nucleosome

DNA double helix

(a) DNA double helix

(b) Nucleosomes (“beads on a string”)

(c) 30-nm fiber

a: © Dr. Gopal Murti/Visuals Unlimited; b: © Ada L. Olins and Donald E. Olins/Biological Photo Service; c: Courtesy Dr. Jerome B. Rattner, Cell Biology and Anatomy, University of Calgary

68


4

3

5

Anchoring of radial loopdomains to the nuclear matrix

Further compaction of radialloops to form heterochromatin

Metaphase chromosome with2 copies of the DNA

1,400 nm

700 nm

300 nm

(d) Radial loop domains

(e) Heterochromatin

(f) Metaphase chromosomed: Courtesy of Paulson, J.R. & Laemmli, U.K. James R. Paulson, U.K. Laemmli, “The structure of histonedepleted

metaphase chromosomes,” Cell, 12:817–28, Copyright Elsevier 1977; e-f: © Peter Engelhardt/Department of Virology, Haartman Institute

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69


2 nm

30 nm

1

4

3

2

5

11 nm

Histone H1

Wrapping of DNA aroundhistone proteins

Formation of a 3-dimensionalzigzag structure via histoneH1 and other DNA-bindingproteins

Anchoring of radial loopdomains to the nuclear matrix

Further compaction of radialloops to form heterochromatin

Metaphase chromosome with2 copies of the DNA

1,400 nm

700 nm

300 nm

Histones

Nucleosome

DNA double helix

(a) DNA double helix

(b) Nucleosomes (“beads on a string”)

(c) 30-nm fiber

(d) Radial loop domains

(e) Heterochromatin

(f) Metaphase chromosomea: © Dr. Gopal Murti/Visuals Unlimited; b: © Ada L. Olins and Donald E. Olins/Biological Photo Service; c: Courtesy Dr. Jerome B. Rattner, Cell Biology and

Anatomy, University of Calgary; d: Courtesy of Paulson, J.R. & Laemmli, U.K. James R. Paulson, U.K. Laemmli, “The structure of histonedepleted metaphase chromosomes,” Cell, 12:817–28, Copyright Elsevier 1977; e-f: © Peter Engelhardt/Department of Virology, Haartman Institute

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Biochemical Identification Nucleic Acid Structure, DNA ...flywing.co.kr/zokbo/file/KAIST/Common/BS120/ch11.pdf · 2015-04-06 1 Chapter 11 Nucleic Acid Structure, DNA Replication,