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    Forms of DNA

    B - DNA: Watson-Crick Structure10 residues/turn

    A - DNA: wider helix - tilted basesno minor groove

    RNA-DNA hybrids

    Z - DNA: High salt form - left handedantibodies to Z - DNA exist

    in nature

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    Procaryotic DNA

    no structural proteins bound

    circular DNA

    supercoiling: conserves spaceeasier to unwind helix

    L = T + W

    3

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    Topoisomerase

    I

    L = 25 + 0

    Helicase

    L = 23 + 0

    DNA Gyrase

    L = 25 - 2

    4

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    Eucaryotic DNA

    50:50 DNA:Protein

    Histones: basic proteins

    octamer core = (H2A, H2B, H3, H4)2

    H1: phophorylated prior to mitosis

    Histones are highly conserved

    5

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    =

    H2A H2B

    H3 H4

    Histones are highly conserved

    2/102 base changes in H4 between pea and cow

    1% divergence every 600 million years

    (compare to 6 million for Hb)

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    H1

    P

    Nucleosome

    ~ 200 bp per nucleosome

    8

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    Eucaryotic Chromosome

    nucleosomes -

    like beads on string

    additional coiling &

    supercoiling makes

    compact chromatin.

    9

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    Genetic Information

    RNA

    Transcription

    Protei

    n

    Translation

    DNA

    DNA

    Replication

    Gene

    Expression

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    Replication

    dATP

    dGTP

    dTTP

    dCTP

    DNA

    DNA Polymerase

    DNA

    Substrates

    Template

    Enzyme

    Product

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    REPLICATION

    Connects dNTPs(3-5 Phosphodiester bond)

    Template instructions: parent cells DNA

    Single copy only: prior to cell division

    Error Rate = 1 in 1 x 109

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    Procaryotic DNA is circular

    Origin ofReplication = Replicon Site

    DNA Gyrase +

    Helicase +SSB Protein

    replication fork

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    DNA Polymerase

    Replication Fork

    Primer = ~ 10 base oligo RNAinserted by Primase

    5

    35

    The new strand is formed 5 3

    28

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    In Real System DNA Pol

    dimers act in tandem

    5 3

    5 3

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    Thymine has an equilibrium between

    keto & enol forms

    N

    NO

    O

    H CH3

    keto

    N

    NO

    HO

    CH3

    enol

    1

    10,000 or

    104

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    H2

    N

    NN

    N N

    Adenine (amino)

    Adenine (imino)

    NH

    NN

    N N

    H

    Imino form of Adenine pairs with C

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    Fidelity of

    Replication

    AUCAG

    GCT

    T A

    GTCCG

    AATGC

    T

    C

    A

    C

    GT AG

    CChance of error1 in 104

    33

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    DNA Polymerase Activities

    5-3 Polymerase: (normal)

    3-5 exonuclease: removes last basein growing strand if incorrect

    5-3 exonuclease: removes RNA primer(while 5-3 Polymerase fills in gap)

    34

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    Fidelity of

    Replication

    AUCAG

    GCT

    T A

    GTCCG

    AATGC

    T

    C

    A

    C

    GT AG

    C

    Chance of error

    1 in 104

    Chance of correctionfailure : 1 in 104

    error rate = 1 in 1 x 108/9

    35

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    continuous strand

    no additional primer

    Continuationof Replication

    38

    discontinuous (lagging) strand

    requires new primer

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    discontinuous ~2,000 bps

    continuous

    Okasaki Fragments

    39

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    E coli DNA Polymerases

    DNA Pol I: primer removal & gap filling

    DNA Pol II: DNA repair?

    Pol III: main polymerase

    ) 2 2 4 ( )2 compl

    5-3 polymerase

    proof reading

    dimerization

    processivity

    40

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    E coli DNA Polymerases

    DNA Pol I: primer removal & gap filling

    DOMAIN 1 323 5'-3' EXONUCLEASE.DOMAIN 324 517 3'-5' EXONUCLEASE.

    DOMAIN 521 928 POLYMERASE.

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    Primer Excision:

    5-3 Exonuclease Activity....

    53

    .... removes RNA primerwhile regular polymerase

    activity fills in gap

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    DNA Ligase seal nicks

    Nick: no bases missingone phosphate ester bond not formed

    S-P-S-P-S-P-S-P S-P-S-P-S-PT A G C A C A

    S-P-S-P-S-P-S-P-S-P-S-P-S-P

    TGTGCTA

    44

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    Mutations

    in DNA SequenceCan occur due to uncorrected

    enol/imino pairing during replication:rate = 1 in 1 x 109

    Can also occur due to DNA damageduring interphase

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    N

    N

    H2N

    O

    C

    N

    NO

    O

    HU

    Spontaneous

    deamination

    Why both T & U?

    The lack of CH3 on U in DNA marks as error

    and instigates removing and repair to restore C

    48

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    Restriction Endonucleases/Methylases

    Recognize and bind to specific DNA sequence

    Function in bacteria is to destroy foreign DNA

    1) if methylated ignore

    2) if methylated methylate other strand

    3) if unmethylated - cut

    Feature #2 allows for marking of self DNAFollowing replication/Cell Division

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    Enz Seq Enz SeqAluI AGCT

    TCGA

    HpaII CCGG

    GGCC

    BamHI GGATCC

    CCTAGG

    KpaI GGTACC

    CCATGG

    BglII AGATCT

    TCTAGA

    MboI GATC

    CTAG

    ClaI ATCGAT

    TAGCTA

    PstI CTGCAG

    GACGTC

    EcoRI GAATTC

    CTTAAG

    PvuI CGATGC

    GCTACG

    HaeII CGCCGCGG

    SalI GTGCACCACGTG

    HindII GTpPAC

    CAPpTG

    SmaI CCCGGG

    GGGCCC

    HindIII AAGCTT

    TTCGAA

    XmaI CCCGGG

    GGGCCC

    Restriction Endonucleases11

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    RFLPs

    1

    2

    3

    4

    5

    5

    4

    3

    2

    1

    1

    2

    3

    4

    5

    5

    4

    2

    3

    1

    12

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    DNA Sequencing

    Cut DNA into managable pieces

    Create DNA Library of pieces

    Select desired piece (Southern Blot)

    Clone & amplify desired piece - PCR

    Sequence cloned fragment

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    Southern Blot

    1

    2

    3

    4

    5

    5

    4

    3

    2

    1

    add 32 P 2 to

    nitrocellulose blot

    14

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    Polymerase Chain Reaction (PCR)

    Heat +

    15

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    0

    1

    2

    Long strands add while short strands double

    16

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    Generation Parent Long Short

    0 1 0 01 1 1 0

    2 1 2 1

    3 1 3 44 1 4 11

    5 1 5 26

    6 1 6 577 1 7 120

    8 1 8 247

    17

    = 1 l + 1 2s + l

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    |_O - P -

    |

    O_

    O

    O

    O

    A

    |

    O ||O - P -

    |O_

    O ||O - P -

    |O_

    O ||-O - P -

    |O_

    O

    OH

    O

    C

    |

    dideoxy nucleotides will terminate DNA

    chain because there is no 3 end to add to.5 end

    3 end19

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    3- XXXXACGTAGCTTACC - 5

    5-32 PXXXX

    Add ddTTP + dTTP, dATP, dCTP, dGTP

    Get : XXXXT

    XXXXTGCAT XXXXTGCATCGAAT

    piece length = Primer + 1,5,10

    20

    21

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    3- XXXXACGTAGCTTACC - 5

    5-32 PXXXX

    Fragment Sizes :

    1 2 3 4 5 6 7 8 9 10 11 12

    ddT : 1, 5, 10

    ddA : 4, 8, 9

    ddC : 3, 6

    ddG : 2, 7, 11, 12

    21

    22

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    12

    11

    10

    98

    7

    6

    5

    4

    3

    2

    1

    T A C G

    5-XXXXTGCATCGAATGG

    22

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    Hasil Agarose Elektroforesis

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