Essential Cell Biology 4th Edition by Bruce Alberts  – Test Bank

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Essential Cell Biology 4th Edition by Bruce Alberts  – Test Bank

 

Sample  Questions

 

ESSENTIAL CELL BIOLOGY, FOURTH EDITION

CHAPTER 6: DNA REPLICATION, REPAIR, AND

RECOMBINATION

© 2014 GARLAND SCIENCE PUBLISHING

 

DNA Replication

 

6-1      The     process           of         DNA    replication      requires         that     each    of         the       parental            DNA    strands           be

used    as        a          ___________________        to         produce          a          duplicate        of         the            opposing        strand.

(a)       catalyst

(b)       competitor

(c)

template

(d)       copy

 

6-2      DNA    replication      is         considered     semiconservative      because          ____________________________.

(a)       after    many   rounds            of         DNA    replication,     the       original           DNA    double            helix    is         still

intact.

(b)       each    daughter        DNA    molecule         consists          of         two      new     strands            copied from    the

parent DNA    molecule.

(c)       each    daughter        DNA    molecule         consists          of         one      strand from    the            parent DNA

molecule         and      one      new     strand.

(d)       new     DNA    strands           must    be        copied from    a          DNA    template.

 

6-3      The     classic experiments   conducted      by        Meselson        and      Stahl    demonstrated            that            DNA

replication      is         accomplished by        employing      a          ________________            mechanism.

(a)       continuous

(b)       semiconservative

(c)       dispersive

(d)       conservative

 

6-4      Initiator          proteins          bind    to         replication      origins            and      disrupt            hydrogen            bonds between

the       two      DNA    strands           being   copied.            Which of         the       factors            below            does    not      contribute      to

the       relative           ease    of         strand separation      by        initiator          proteins?

(a)       replication      origins            are      rich     in         A-T      base    pairs

(b)       the       reaction          can      occur  at         room   temperature

(c)       they     only     separate         a          few      base    pairs   at         a          time

(d)       once    opened,          other   proteins          of         the       DNA    replication      machinery            bind    to         the

origin

 

6-5      If          the       genome          of         the       bacterium       E.        coli      requires         about  20            minutes          to         replicate         itself,

how     can      the       genome          of         the       fruit     fly        Drosophila     be        replicated       in            only     3          minutes?

(a)       The     Drosophila     genome          is         smaller           than    the       E.        coli      genome.

(b)       Eukaryotic      DNA    polymerase    synthesizes    DNA    at         a          much   faster  rate            than

prokaryotic    DNA    polymerase.

 

Page    1          of         25

 

(c)       The     nuclear           membrane     keeps  the       Drosophila     DNA            concentrated in         one      place

in         the       cell,      which  increases        the       rate     of         polymerization.

(d)       Drosophila     DNA    contains          more   origins            of         replication            than    E.        coli      DNA.

6-6      Meselson        and      Stahl    grew   cells     in         media  that     contained       different            isotopes          of         nitrogen         (15N     and      14N)     so        that     the            DNA    molecules       produced       from    these   different         isotopes            could   be        distinguished by        mass.

A.         Explain           how     “light”  DNA    was     separated       from    “heavy”            DNA    in         the       Meselson

and      Stahl    experiments.

B.         Describe         the       three   existing           models            for       DNA            replication      when   these   studies

were   begun, and      explain            how     one      of         them   was     ruled            out      definitively     by        the

experiment    you      described       for       part     A.

C.         What   experimental result  eliminated      the       dispersive      model of            DNA

 

replication?

           
6-7      Indicate          whether          the       following        statements     are      true     or            false.   If          a          statement       is         false,   explain            why     it          is            false.

A.         When  DNA    is         being   replicated       inside  a          cell,      local    heating            occurs,            allowing          the

two      strands           to         separate.

B.         DNA    replication      origins            are      typically          rich     in         G-C            base    pairs.

C.         Meselson        and      Stahl    ruled   out      the       dispersive      model for            DNA    replication.     D.         DNA    replication      is         a          bidirectional            process           that     is         initiated          at         multiple

locations         along   chromosomes            in         eukaryotic      cells.

6-8      How    many   replication      forks   are      formed           when   an        origin  of            replication      is         opened?         (a)       1

(b)       2

(c)       3

(d)       4

6-9      Answer           the       following        questions       about  DNA    replication.

A.         On       a          DNA    strand that     is         being   synthesized,   which  end      is            growing—the            3′         end,

the       5′         end,     or        both    ends?  Explain           your    answer.

B.         On       a          DNA    strand that     is         being   used    as        a            template,        where is         the       copying

occurring       relative           to         the       replication      origin—3′       of            the       origin, 5′,        or        both?

6-10    How    does    the       total    number          of         replication      origins            in            bacterial         cells     compare         with     the       number          of         origins            in         human            cells?

(a)       1          versus 100

(b)       5          versus 500

(c)       10        versus 1000

(d)       1          versus 10,000

           
6-11    Which of         the       following        statements     correctly         explains          what    it            means for       DNA    replication      to         be        bidirectional?

 

Page    2          of         25

 

(a)       The     replication      fork     can      open   or        close,   depending      on        the       conditions.

(b)       The     DNA    replication      machinery      can      move   in         either  direction         on        the            template

strand.

(c)       Replication-fork        movement      can      switch directions       when   the       fork     converges

on        another          replication      fork.

(d)       The     replication      forks   formed           at         the       origin  move   in         opposite            directions.                 

6-12    The     chromatin      structure        in         eukaryotic      cells     is         much   more   complicated            than    that

observed        in         prokaryotic    cells.    This     is         thought           to         be        the       reason            that     DNA    replication

occurs much   faster  in         prokaryotes.  How    much   faster  is         it?

(a)       2×

(b)       5×

(c)       10×

(d)       100

×

 

6-13    DNA    polymerase    catalyzes         the       joining            of         a          nucleotide      to         a            growing          DNA    strand.

What   prevents         this      enzyme           from    catalyzing       the       reverse           reaction?

(a)       hydrolysis      of         pyrophosphate          (PPi)    to         inorganic        phosphate      (Pi)      +            Pi

(b)       release            of         PPi       from    the       nucleotide

(c)       hybridization of         the       new     strand to         the       template

(d)       loss     of         ATP     as        an        energy            source

 

6-14    Use      your    knowledge     of         how     a          new     strand of         DNA    is         synthesized    to            explain            why     DNA

replication      must    occur  in         the       5′-to-3′            direction.        In         other   words,            what    would be        the

consequences            of         3′–to-5′           strand elongation?

 

6-15    Figure Q6-15  shows a          replication      bubble.

 

 

 

Figure      Q6-15
A. On       the       figure, indicate           where the       origin  of         replication            was     located            (use     O).
B. Label   the       leading-strand           template         and      the       lagging-strand            template         of         the
right-hand      fork     [R]       as        X          and      Y,         respectively.
C. Indicate          by        arrows            the       direction         in         which  the            newly  made   DNA    strands
(indicated       by        dark    lines)   were   synthesized.
D. Number          the       Okazaki           fragments       on        each    strand as            1,         2,         and      3          in         the       order  in
which  they     were   synthesized.
E. Indicate          where the       most    recent DNA    synthesis        has            occurred        (use     S).
F. Indicate          the       direction         of         movement      of         the            replication      forks   with     arrows.

 

Page    3          of         25

 

           

           

Use     the      following       information  about a          series of        in         vitro    DNA   replication            experiments to        answer          questions     6-16    through         6-22.              

 

You     prepare          bacterial         cell      extracts          by        lysing  the       cells     and      removing            insoluble        debris via       centrifugation.           These  extracts          provide           the       proteins            required         for       DNA    replication.     Your    DNA    template         is         a          small,  double-stranded         circular           piece   of         DNA    (a         plasmid)         with     a          single  origin  of            replication      and      a          single  replication      termination    site.     The     termination    site      is            on        the       opposite         side     of         the       plasmid           from    the       origin.

 

 

6-16    In         addition          to         the       extracts          and      the       plasmid           DNA,   are      there   any            additional       materials

you      should            add      to         this      in         vitro     replication      system?          Explain            your    answer.

6-17    Which of         the       following        statements     is         true     with     respect           to         this      in            vitro     replication

system?

(a)       There  will      be        only     one      leading            strand and      one      lagging            strand            produced       using

this      template.

(b)       The     leading            and      lagging            strands           compose         one      half      of            each    newly  synthesized

DNA    strand.

(c)       The     DNA    replication      machinery      can      assemble        at         multiple          places on            this

plasmid.

(d)       One     daughter        DNA    molecule         will      be        slightly            shorter           than    the            other.

You     decide to         use      different         bacterial         strains            (each   having one      protein           of            the       replication      machinery      mutated)        in         order  to         examine          the       role     of            individual       proteins          in         the       normal            process           of         DNA    replication.

 

6-18    What   part     of         the       DNA    replication      process           would be        most    directly            affected          if          a          strain

of         bacteria          lacking            primase          were   used    to         make   the       cell      extracts?

(a)       initiation         of         DNA    synthesis

(b)       Okazaki           fragment        synthesis

(c)       leading-strand           elongation

(d)       lagging-strand           completion

 

6-19    What   part     of         the       DNA    replication      process           would be        most    directly            affected          if          a          strain

of         bacteria          lacking            the       exonuclease   activity            of         DNA    polymerase            were   used    to         make

the       cell      extracts?

 

Page    4          of         25

 

(a)       initiation         of         DNA    synthesis

(b)       Okazaki           fragment        synthesis

(c)       leading-strand           elongation

(d)       lagging-strand           completion

           
6-20    What   part     of         the       DNA    replication      process           would be        most            directly           affected          if          a          strain  of         bacteria          lacking            helicase          were   used    to         make   the       cell      extracts?

(a)       initiation         of         DNA    synthesis

(b)       Okazaki           fragment        synthesis

(c)       leading-strand           elongation

(d)       lagging-strand           completion

           
6-21    What   part     of         the       DNA    replication      process           would be        most            directly           affected          if          a          strain  of         the       cell

bacteria      lacking            single-strand  binding           protein           were   used    to        make

extracts?

(a)       initiation         of         DNA    synthesis

(b)       Okazaki           fragment        synthesis

(c)       leading-strand           elongation

(d)       lagging-strand           completion

6-22    What   part     of         the       DNA    replication      process           would be        most            directly           affected          if          a          strain  of         bacteria          lacking            DNA    ligase  were   used    to         make   the       cell      extracts?

(a)       initiation         of         DNA    synthesis

(b)       Okazaki           fragment        synthesis

(c)       leading-strand           elongation

(d)       lagging-strand           completion

           
6-23    Which of         the       following        statements     about  the       newly  synthesized            strand of         a          human            chromosome is         true?

(a)       It          was     synthesized    from    a          single  origin  solely  by            continuous     DNA    synthesis.       (b)       It          was     synthesized    from    a            single  origin  by        a          mixture           of         continuous     and

discontinuous            DNA    synthesis.

(c)       It          was     synthesized    from    multiple          origins            solely  by            discontinuous            DNA

synthesis.

(d)       It          was     synthesized    from    multiple          origins            by        a            mixture           of         continuous     and

discontinuous            DNA    synthesis.

6-24    You     have    discovered     an        “Exo–” mutant            form    of         DNA            polymerase    in         which  the       3′-to-5′            exonuclease   function            has      been    destroyed       but      the       ability  to         join      nucleotides            together         is         unchanged.    Which of         the       following            properties      do        you      expect the       mutant            polymerase    to            have?

(a)       It          will      polymerize     in         both    the       5′-to-3′            direction            and      the       3′-to-5′            direction.        (b)       It          will      polymerize            more   slowly than    the       normal            Exo+    polymerase.

(c)       It          will      fall       off       the       template         more   frequently      than            the       normal            Exo+    polymerase.   (d)       It          will      be        more            likely   to         generate         mismatched   base    pairs.

 

Page    5          of         25

 

 

6-25    A          molecule         of         bacterial         DNA    introduced     into     a          yeast   cell      is            imported        into     the       nucleus

but      fails     to         replicate         with     the       yeast   DNA.   Where            do        you      think   the            block   to         replication

arises?            Choose           the       protein           or        protein           complex          below  that     is            most    probably

responsible    for       the       failure to         replicate         bacterial         DNA.   Give     an            explanation    for       your

answer.

(a)       primase

(b)       helicase

(c)       DNA    polymerase

(d)       initiator          proteins

 

6-26    Most    cells     in         the       body   of         an        adult   human            lack     the       telomerase            enzyme           because          its

gene    is         turned            off       and      is         therefore        not      expressed.      An       important            step     in         the

cell,  growth

conversion     of         a          normal            cell      into     a          cancer which  circumvents   normal

control,           is         the       resumption    of         telomerase     expression.    Explain           why            telomerase     might

be        necessary       for       the       ability  of         cancer cells     to         divide  over    and      over            again.

 

6-27    Which diagram          accurately      represents     the       directionality of         DNA    strands           at            one      side

of         a          replication      fork?  

 

           

 

Page    6          of         25

 

6-28    Indicate          whether          the       following        statements     are      true     or            false.   If          a          statement       is         false,   explain            why     it          is            false.

A.         Primase          is         needed           to         initiate            DNA    replication            on        both    the       leading            strand and

the       lagging            strand.

B.         The     sliding clamp  is         loaded            once    on        each    DNA    strand,            where it          remains

associated      until    replication      is         complete.

C.         Telomerase    is         a          DNA    polymerase    that     carries            its            own     RNA    molecule         to         use      as

a          primer            at         the       end      of         the       lagging            strand.

D.         Primase          requires         a          proofreading function          that     ensures            there   are      no        errors in

the       RNA    primers          used    for       DNA    replication.

6-29    Because          all        DNA    polymerases  synthesize      DNA    in         the       5′-to-3′            direction,        and      the       parental          strands           are      antiparallel,            DNA    replication      is         accomplished with     the       use      of         two            whether          the

mechanisms:      continuous     and      discontinuous            replication.     Indicate

following        items   relate  to         (1)       continuous     replication,     (2)            discontinuous            replication,     or        (3)       both    modes of            replication.

______   primase

______   single-strand  binding           protein

______   sliding clamp

______   RNA    primers

______   leading            strand

______   lagging            strand

______   Okazaki           fragments

______   DNA    helicase

______   DNA    ligase

6-30    The     synthesis        of         DNA    in         living   systems          occurs in         the            5′-to-3′            direction.        However,        scientists        synthesize      short            DNA    sequences      needed           for       their    experiments   on        an            instrument     dedicated       to         this      task.

A.         The     chemical         synthesis        of         DNA    by        this      instrument            proceeds        in         the       3′-to-5′

direction.        Draw   a          diagram          to         show   how     this      is            possible          and      explain            the

process.

B.         Although        3′-to-5′            synthesis        of         DNA    is         chemically            possible,         it          does    not      occur  in

living   systems.          Why    not?

           
6-31 DNA    polymerases  are      processive,     which  means that     they     remain            tightly associated      with     the       template         strand while   moving            rapidly            and      adding            nucleotides    to         the       growing            daughter        strand.            Which piece   of         the       replication            machinery      accounts         for       this      characteristic?

(a)       helicase

(b)       sliding clamp

(c)       single-strand  binding           protein

(d)       primase

 

Page    7          of         25

 

6-32    Use      the       components   in         the       list       below  to         label    the       diagram          of         a            replication      fork     in

Figure Q6-32.

  1. DNA polymerase
  2. single-strand binding           protein
  3. Okazaki fragment
  4. primase
  5. sliding clamp
  6. RNA primer
  7. DNA helicase

 

 

 

Figure      Q6-32

 

6-33    Researchers   have    isolated           a          mutant            strain  of         E.        coli      that     carries            a          temperature-

sensitive         variant            of         the       enzyme           DNA    ligase.  At        the       permissive            temperature,  the

mutant            cells     grow   just      as        well     as        the       wild-type        cells.    At        the            nonpermissive

temperature,  all        of         the       cells     in         the       culture            tube    die       within 2            hours. DNA    from    mutant

cells     grown at         the       nonpermissive           temperature  for       30        minutes          is            compared       with     the

DNA    isolated           from    cells     grown at         the       permissive     temperature.  The     results            are

shown in         Figure Q6-33, where DNA    molecules       have    been    separated       by        size      by

means of         electrophoresis         (P,        permissive;    NP,      nonpermissive).        Explain           the

appearance    of         a          distinct            band   with     a          size      of         200     base    pairs            (bp)     in         the       sample

collected         at         the       nonpermissive           temperature.

 

 

 

Figure            Q6-33

 

Page    8          of         25

 

6-34    Indicate          whether          the       following        statements     are      true     or        false.   If          a            statement       is         false,

explain            why     it          is         false.

  1. The repair polymerase    is         the       enzyme           that     proofreads     the       newly            synthesized

strands           to         ensure            the       accuracy         of         DNA    replication.

  1. There is         a          single  enzyme           that     degrades        the       RNA    primers          and            lays     down  the

corresponding           DNA    sequence        behind            it.

  1. DNA ligase  is         required         to         seal     the       sugar–phosphate      backbone            between         all        the

DNA    fragments       on        the       lagging            strand.

  1. The repair polymerase    does    not      require           the       aid       of         the       sliding            clamp, because

it          is         only     synthesizing   DNA    over    very    short   stretches.

           

6-35    Which of         the       following        statements     about  sequence        proofreading during DNA

replication      is         false?

(a)       DNA

The     exonuclease   activity            is         in         a          different         domain           of         the            polymerase.

(b)       The     exonuclease   activity            cleaves            DNA    in         the       5′-to-3′            direction.

(c)       The     DNA    proofreading activity            occurs concomitantly            with     strand elongation.

(d)       If          an        incorrect        base    is         added, it          is         “unpaired”     before removal.

 

6-36    The     sliding clamp  complex          encircles         the       DNA    template         and      binds   to            DNA

polymerase.   This     helps   the       polymerase    synthesize      much   longer stretches        of            DNA

without           dissociating.   While  the       loading           of         the       clamp  only     occurs once    on            the       leading

strand,            it          must    happen           each    time     a          new     Okazaki           fragment        is            made   on        the       lagging

strand.            How    does    the       cell      expedite         this      process?

 

6-37    The     DNA    duplex            consists          of         two      long     covalent          polymers        wrapped            around           each    other

many   times   over    their    entire  length. The     separation      of         the       DNA    strands           for

replication      causes the       strands           to         be        “overwound” in         front   of         the            replication      fork.

How    does    the       cell      relieve            the       torsional         stress  created           along   the            DNA    duplex            during

replication?

(a)       Nothing          needs  to         be        done   because          the       two      strands           will      be            separated       after

replication      is         complete.

(b)       Topoisomerases       break  the       covalent          bonds of         the       backbone       allowing            the       local

unwinding      of         DNA    ahead  of         the       replication      fork.

(c)       Helicase          unwinds         the       DNA    and      rewinds          it          after    replication      is            complete.

(d)       DNA    repair enzymes         remove           torsional         stress  as        they     replace            incorrectly

paired bases.

           

6-38    Telomeres      serve   as        caps    at         the       ends    of         linear  chromosomes.           Which of            the       following

is         not       true     regarding       the       replication      of         telomeric        sequences?

(a)       The     lagging-strand           telomeres       are      not      completely     replicated       by            DNA

polymerase.

(b)       Telomeres      are      made   of         repeating        sequences.

(c)       Additional      repeated         sequences      are      added to         the       template         strand.

(d)       The     leading            strand doubles          back    on        itself    to         form    a          primer            for       the       lagging

strand.

 

Page    9          of         25

 

 

DNA Repair

 

6-39    Sickle-cell       anemia            is         an        example          of         an        inherited        disease.            Individuals     with     this

disorder         have    misshapen      (sickle-shaped)          red      blood  cells     caused            by        a            change            in         the

sequence        of         the       β-globin          gene.   What   is         the       nature of         the       change?

(a)       chromosome loss

(b)       base-pair        change

(c)       gene    duplication

(d)       base-pair        insertion

 

6-40    Even    though            DNA    polymerase    has      a          proofreading function,         it          still            introduces      errors

in         the                   what    degree

newly   synthesized    strand at         a          rate     of         1          per      107nucleotides.          To

does    the       mismatch        repair system            decrease         the       error   rate     arising            from    DNA

replication?

(a)       2-fold

(b)       5-fold

(c)       10-fold

(d)       100-fold

 

6-41    Which of         the       choices           below  represents     the       correct            way     to         repair the            mismatch

shown in         Figure Q6-41?

 

 

Figure      Q6-41

 

 

           

6-42    A          mismatched   base    pair     causes a          distortion       in         the       DNA    backbone.       If            this      were   the

only     indication       of         an        error   in         replication,     the       overall            rate     of            mutation         would be        much

higher.            Explain           why.

 

Page    10        of         25

 

 

6-43    Beside the       distortion       in         the       DNA    backbone       caused            by        a            mismatched   base    pair,    what

additional       mark   is         there   on        eukaryotic      DNA    to         indicate           which  strand            needs  to         be

repaired?

(a)       a          nick     in         the       template         strand

(b)       a          chemical         modification   of         the       new     strand

(c)       a          nick     in         the       new     strand

(d)       a          sequence        gap      in         the       new     strand

 

6-44    A          pregnant        mouse is         exposed          to         high     levels  of         a          chemical.            Many   of         the       mice    in         her

litter    are      deformed,      but      when   they     are      interbred        with     each    other,  all            their    offspring

are      normal.           Which two      of         the       following        statements     could   explain            these   results?

(a)       In         the       deformed       mice,   somatic           cells     but      not      germ   cells     were            mutated.

(b)

The     original           mouse’s          germ   cells     were   mutated.

(c)       In         the       deformed       mice,   germ   cells     but      not      somatic           cells     were            mutated.

(d)       The     toxic    chemical         affects development  but      is         not      mutagenic.

 

6-45    The     repair of         mismatched   base    pairs   or        damaged        nucleotides    in         a            DNA    strand

requires         a          multistep        process.          Which choice below  describes        the       known            sequence        of

events in         this      process?

(a)       DNA    damage           is         recognized,    the       newly  synthesized    strand is         identified            by        an

existing           nick     in         the       backbone,       a          segment          of         the       new            strand is         removed         by

repair proteins,         the       gap      is         filled    by        DNA    polymerase,   and      the            strand is         sealed

by        DNA    ligase.

(b)       DNA    repair polymerase    simultaneously          removes         bases  ahead  of         it          and

polymerizes   the       correct            sequence        behind            it          as        it          moves            along   the       template.

DNA    ligase  seals    the       nicks   in         the       repaired         strand.

(c)       DNA    damage           is         recognized,    the       newly  synthesized    strand is         identified            by        an

existing           nick     in         the       backbone,       a          segment          of         the       new            strand is         removed         by        an

exonuclease,  and      the       gap      is         repaired         by        DNA    ligase.

(d)       A          nick     in         the       DNA    is         recognized,    DNA    repair proteins          switch out            the       wrong

base    and      insert  the       correct            base,   and      DNA    ligase  seals    the       nick.

 

6-46    Human            beings with     the       inherited        disease           xeroderma     pigmentosum            have    serious

problems        with     lesions            on        their    skin     and      often   develop          skin     cancer            with     repeated

exposure        to         sunlight.          What   type     of         DNA    damage           is         not      being            recognized     in         the       cells

of         these   individuals?

(a)       chemical         damage

(b)       X-ray   irradiation      damage

(c)       mismatched   bases

(d)       ultraviolet      irradiation      damage

 

 

6-47    You     are      examining      the       DNA    sequences      that     code    for       the       enzyme

phosphofructokinase           in         skinks and      Komodo         dragons.         You     notice  that     the            coding

sequence        that     actually           directs            the       sequence        of         amino acids   in         the            enzyme           is         very

 

Page    11        of         25

 

similar            in         the       two      organisms      but      that     the       surrounding            sequences      vary    quite   a          bit.       What   is         the       most    likely            explanation    for       this?

(a)       Coding            sequences      are      repaired         more   efficiently.

(b)       Coding            sequences      are      replicated       more   accurately.

(c)       Coding            sequences      are      packaged        more   tightly in         the            chromosomes            to         protect

them   from    DNA    damage.

(d)       Mutations       in         coding sequences      are      more   likely   to         be            deleterious     to         the

organism        than    mutations       in         noncoding      sequences.

6-48    In         somatic           cells,    if          a          base    is         mismatched   in         one            new     daughter        strand during DNA    replication,     and      is         not            repaired,        what    fraction           of         the       DNA    duplexes         will            have    a          permanent     change            in         the       DNA    sequence            after    the       second            round of         DNA    replication?    (a)       1/2

(b)       1/4

(c)       1/8

(d)       1/16

           
6-49    Sometimes,     chemical         damage           to         DNA    can      occur  just      before            DNA    replication      begins,            not      giving  the       repair system            enough           time     to         correct            the       error   before the       DNA    is            duplicated.     This     gives   rise      to         mutation.        If          the       cytosine            in         the       sequence        TCAT  is         deaminated   and      not            repaired,        which  of         the       following        is         the       point            mutation         you      would observe          after    this      segment          has            undergone     two      rounds            of         DNA    replication?    (a)       TTAT

(b)       TUAT

(c)       TGAT

(d)       TAAT

6-50 During            DNA    replication      in         a          bacterium,      a          C          is            accidentally    incorporated instead            of         an        A          into     one            newly  synthesized    DNA    strand.            Imagine          that     this      error            was     not      corrected       and      that     it          has      no        effect   on            the       ability  of         the       progeny          to         grow   and      reproduce.     A.            After   this      original           bacterium       has      divided           once,   what            proportion     of         its

progeny          would you      expect to         contain           the       mutation?

B.         What   proportion     of         its        progeny          would you      expect to            contain           the       mutation

after    three   more   rounds            of         DNA    replication      and      cell            division?

6-51 Sometimes,     chemical         damage           to         DNA    can      occur  just      before            DNA    replication      begins,            not      giving  the       repair system            enough           time     to         correct            the       error   before the       DNA    is            duplicated.     This     gives   rise      to         mutation.        If          the            adenosine      in         the       sequence        TCAT  is         depurinated   and            not      repaired,        which  of         the       following        is         the       point            mutation         you      would observe          after    this      segment          has            undergone     two      rounds            of         DNA    replication?    (a)       TCGT

(b)       TAT

(c)       TCT

(d)       TGTT

           

 

Page    12        of         25

 

6-52    Which of         the       following        statements     is         not       an        accurate         statement            about  thymine

dimers?

(a)       Thymine         dimers            can      cause  the       DNA    replication      machinery      to            stall.

(b)       Thymine         dimers            are      covalent          links    between         thymidines     on            opposite         DNA

strands.

(c)       Prolonged      exposure        to         sunlight          causes thymine          dimers            to            form.

(d)       Repair proteins          recognize       thymine          dimers            as        a          distortion       in            the       DNA

backbone.

 

6-53    Indicate          whether          the       following        statements     are      true     or        false.   If          a            statement       is         false,

explain            why     it          is         false.

  1. Ionizing radiation        and      oxidative        damage           can      cause  DNA    double-strand

breaks.

  1. After backbone       are

damaged           DNA    has      been    repaired,        nicks   in         the       phosphate

maintained     as        a          way     to         identify           the       strand that     was     repaired.

  1. Depurination of DNA    is         a          rare     event   that     is         caused            by            ultraviolet      irradiation.
  2. Nonhomologous end      joining            is         a          mechanism     that     ensures          that            DNA    double-

strand breaks            are      repaired         with     a          high     degree            of         fidelity            to         the       original           DNA

sequence.

 

6-54    Several           members        of         the       same   family  were   diagnosed      with     the       same            kind     of         cancer

when   they     were   unusually       young. Which one      of         the       following        is         the            most    likely

explanation    for       this      phenomenon?           It          is         possible          that     the       individuals            with     the       cancer

have    _______________________.

(a)       inherited        a          cancer-causing          gene    that     suffered          a          mutation         in            an        ancestor’s

somatic           cells.

(b)       inherited        a          mutation         in         a          gene    required         for       DNA    synthesis.

(c)       inherited        a          mutation         in         a          gene    required         for       mismatch            repair.

(d)       inherited        a          mutation         in         a          gene    required         for       the       synthesis            of         purine

nucleotides.

 

6-55    You     have    made   a          collection        of         mutant            fruit     flies     that     are      defective            in         various           aspects

of         DNA    repair. You     test      each    mutant            for       its        hypersensitivity         to            three   DNA-damaging

agents:            sunlight,          nitrous            acid     (which            causes deamination   of         cytosine),            and      formic

acid     (which            causes depurination).           The     results            are      summarized   in            Table  Q6-55,

where a          “yes”   indicates         that     the       mutant            is         more   sensitive         than    a            normal            fly,       and

blanks indicate           normal            sensitivity.

 

Page    13        of         25

 

 

Table       Q6-55

 

  1. polymerase?

Which mutant            is         most    likely   to         be        defective         in         the       DNA            repair

  1. What aspect of         repair is         most    likely   to         be        affected          in         the            other   mutants?

6-56    The     deamination   of         cytosine          generates       a          uracil  base.   This     is         a            naturally         occurring

nucleic            acid     base,   and      so        does    not      represent       a          DNA    lesion  caused            by        damage           due      to

chemicals       or        irradiation.     Why    is         this      base    recognized     as        “foreign”         and            why     is         it

important       for       cells     to         have    a          mechanism     to         recognize       and      remove            uracil  when   it          is

found  in         the       DNA    duplex?

 

Homologous Recombination

 

6-57    Select  the       option that     best     completes       the       following        statement:      Nonhomologous            end

joining            is         a          process           by        which  a          double-stranded       DNA    end      is            joined ___________________.

(a)       to         a          similar            stretch            of         sequence        on        the       complementary            chromosome.

(b)       after    repairing        any      mismatches.

(c)       to         the       nearest           available         double-stranded       DNA    end.

(d)       after    filling   in         any      lost      nucleotides,    helping           to         maintain         the            integrity          of         the

DNA    sequence.

           

6-58    Nonhomologous       end      joining            can      result  in         all        but      which  of         the            following?

(a)       the       recovery         of         lost      nucleotides    on        a          damaged        DNA    strand

(b)       the       interruption   of         gene    expression

(c)       loss     of         nucleotides    at         the       site      of         repair

(d)       translocations            of         DNA    fragments       to         an        entirely           different            chromosome            

6-59    Homologous  recombination           is         an        important       mechanism     in         which  organisms            use      a

“backup”         copy    of         the       DNA    as        a          template         to         fix        double-strand            breaks            without           loss     of

genetic            information.   Which of         the       following        is         not       necessary       for            homologous

recombination           to         occur?

(a)       3′         DNA    strand overhangs

(b)       5′         DNA    strand overhangs

(c)       a          long     stretch            of         sequence        similarity

 

Page    14        of         25

 

(d)       nucleases

 

6-60    In         addition          to         the       repair of         DNA    double-strand            breaks,            homologous   recombination           is

a          mechanism     for       generating      genetic            diversity         by        swapping        segments            of         parental

chromosomes.           During            which  process           does    swapping        occur?

(a)       DNA    replication

(b)       DNA    repair

(c)       meiosis

(d)       transposition

           

6-61    Recombination          has      occurred        between         the       chromosome segments        shown in            Figure

Q6-61. The     genes  A         and      B,         and      the       recessive        alleles a          and      b,         are            used    as        markers          on

the       maternal         and      paternal          chromosomes,           respectively.   After   alignment       and

,        b          have

homologous   recombination,          the       specific           arrangements            of         A,         B,            aand

changed.

 

 

 

Figure      Q6-61
Which of         the       choices           below  correctly         indicates         the       gene            combination   from    the       replication      products         of         the       maternal            chromosome?

(a)       AB       and      aB

(b)       ab        and      Ab

(c)       AB       and      Ab

(d)       aB       and      Ab

6-62    The     events listed   below  are      all        necessary       for       homologous            recombination           to         occur  properly:

A.         Holliday          junction          cut       and      ligated

B.         strand invasion

C.         DNA    synthesis

D.         DNA    ligation

E.         double-strand            break

F.         nucleases       create uneven           strands

 

Page    15        of         25

 

Which of         the       following        is         the       correct            order  of         events during            homologous

recombination?

(a)       E,         B,         F,         D,         C,         A

(b)       B,         E,         F,         D,         C,         A

(c)       C,         E,         F,         B,         D,         A

(d)       E,         F,         B,         C,         D,         A

 

6-63    Homologous  recombination           is         initiated          by        double-strand            breaks            (DSBs)            in         a

chromosome. DSBs   arise    from    DNA    damage           caused            by        harmful          chemicals            or        by

radiation        (for     example,         X-rays).           During            meiosis,          the       specialized     cell            division           that

produces        gametes          (sperm            and      eggs)   for       sexual reproduction,            the            cells     intentionally

cause  DSBs   so        as        to         stimulate        crossover       homologous   recombination.          If            there   is         not

at         least    one      occurrence     of         crossing-over            within each    pair     of            homologous

chronot          segregate

mosomes            during meiosis,          those   noncrossover            chromosomes            will

properly.

 

Figure      Q6-43

 

  1. Consider the       copy    of         Chromosome 3          that     you      received         from            your    mother.           Is         it

identical          to         the       Chromosome 3          that     she      received         from    her            mother           (her

maternal         chromosome)            or        identical          to         the       Chromosome 3          she            received         from

her      father  (her     paternal          chromosome),           or        neither?          Explain.

  1. Starting with     the       representation          in         Figure Q6-43  of         the       double-stranded

maternal         and      paternal          chromosomes            found  in         your    mother,            draw   two

possible          chromosomes            you      may     have    received         from    your    mother.

  1. What does    this      indicate           about  your    resemblance  to         your    grandfather   and

grandmother?

 

6-64    Indicate          whether          the       following        statements     are      true     or        false.   If          a            statement       is         false,

explain            why     it          is         false.

  1. Homologous recombination           cannot            occur  in         prokaryotic    cells,    because            they

are      haploid,          and      therefore        have    no        extra   copy    of         the            chromosome to         use      as        a

template         for       repair.

  1. The first     step     in         repair requires         a          nuclease         to         remove           a            stretch            of         base    pairs

from    the       5′         end      of         each    strand at         the       site      of         the       break.

  1. The 3′         overhang        “invades”        the       homologous   DNA    duplex,            which  can            be        used

as        a          primer            for       the       repair DNA    polymerase.

 

Page    16        of         25

 

  1. The DNA    template         used    to         repair the       broken            strand is         the            homologous

chromosome inherited        from    the       other   parent.

 

 

 

 

Page    17        of         25

 

ANSWERS

 

6-1      (c)

 

6-2      Choice (c)       is         the       correct            answer.           Choices           (a)       and      (b)       are            false.   Although        choice (d)       is

a          correct            statement,      it          is         not      the       reason            that     DNA    replication            is         called

semiconservative.

 

6-3      (b)

 

6-4      (b)

 

6-5      Choice (d)       of         and

is   the       correct            answer.           Bacteria          have    one      origin  replication,

Drosophila     has      many.  Choice (a)       is         incorrect        because          the       Drosophila            genome          is

bigger than    the       E.        coli      genome.          Choice (b)       is         incorrect,        because            eukaryotic

polymerases  are      not      faster  than    prokaryotic    polymerases.

 

6-6      A.         The     DNA    samples          collected         were   placed into     centrifuge       tubes   containing

cesium            chloride.         After   high-speed     centrifugation            for       2          days,   the            heavy  and

light     DNA    products         were   separated       by        density.

  1. The three   models            were   conservative, semiconservative,     and      dispersive.      The

conservative  model suggested       a          mechanism     by        which  the       original            parental

strands           stayed together         after    replication      and      the       daughter        duplex            was     made

entirely           of         newly  synthesized    DNA.   The     semiconservative      model proposed

that     the       two      DNA    duplexes         produced       during replication      were   hybrid

molecules,      each    having one      of         the       parental          strands           and      one      of            the       newly

synthesized    strands.          The     dispersive      model predicted        that     the       new            DNA

duplexes         each    contained       segments        of         parental          and      daughter            strands           all        along

the       molecule.        The     conservative  model was     ruled   out      by        the       density-gradient

experiments.

  1. The dispersive      model was     ruled   out      by        using   heat     to         denature        the            DNA

duplexes         and      then    comparing      the       densities         of         the       single-stranded            DNA.   If          the

dispersive      model had      been    correct,           individual       strands           should            have    had      an

intermediate  density.           However,        this      was     not      the       case;    only     heavy            strands           and

light     strands           were   observed,       which  convincingly   supported      the

semiconservative      model for       DNA    replication.

 

6-7      A.         False.  The     two      strands           do        need    to         separate         for       replication      to            occur,  but      this      is

accomplished by        the       binding           of         initiator          proteins          at         the            origin  of         replication.

  1. False. DNA    replication      origins            are      typically          rich     in         A-T      base            pairs,   which  are

held     together         by        only     two      hydrogen       bonds (instead          of         three   for            C-G      base

pairs), making            it          easier  to         separate         the       strands           at         these            sites.

  1. True.
  2. True.

 

 

Page    18        of         25

 

6-8      (b)

 

6-9      A.         The     3′         end.     DNA    polymerase    can      add      nucleotides    only     to         the       3′-OH       end      of         a

nucleic            acid     chain.

  1. Both, as        a          result  of         the       bidirectional  nature of         chromosomal            replication.

 

6-10    (d)

           

6-11    (d)

           

6-12    (c)

 

6-13    (a)

 

6-14    There  would be        several            detrimental    consequences            to         3′–to-5′           strand            elongation.

One     of         those   most    directly           linked  to         the       processes       of         DNA    replication            involves

synthesis        of         the       lagging            strand.            After   the       RNA    primers          are            degraded,       the       DNA

segments        remaining       will      have    5′         ends    with     a          single  phosphate      group. The            incoming

nucleotide      will      have    a          3′-OH   group. Without          the       energy            provided        by            the       release            of

PPi       from    the       5′         end,     the       process           of         elongation      would no        longer be            energetically

favorable.

 

6-15    See      Figure A6-15.

 

 

Figure      A6-15

 

6-16    You     will      probably        add      exogenous     nucleoside      triphosphates            to         serve   as        the       building          blocks needed           to         make            new     strands           of         DNA.   Although        these   monomers     will            be        present           in         the       extracts,          they     will      be        present            at         lower  concentrations          than    are      normally         found  inside            the       cell.      They   may     also     be        subject            to         hydrolysis,            and      the       nucleoside      diphosphates that     are      the       products            of         this      hydrolysis      are      not      usable substrates      for       DNA            replication.     For      both    of         these   reasons,          it          is            important       to         add      excess nucleotides    to         the       reaction            mixture           for       efficient          DNA    replication      to         occur.
           
6-17    (b)       Leading          and      lagging            strands           are      synthesized            bidirectionally           from    the       replication      origin, and      are      joined            together         by        DNA    ligase  where the       two      replication      forks            meet    at         the       termination    site.     Choice (a)       is         not      correct,            because          this      answer           implies            that     the       replication            fork     is         not      bidirectional  and      that     replication      continues            around           the       plasmid           until    the       process           makes it            back    to         the       origin  of         replication.     Choice (c)       is

 

Page    19        of         25

 

incorrect        because          the       origin  is         a          specialized     sequence            where initiator          proteins          bind    and      open   the       DNA    so            that     the       DNA    replication      machinery      can      assemble.       Choice            (d)       is         incorrect        because          the       daughter        DNA            molecules       will      be        same   size      as        the       original           plasmid            (and    each    other).
           
6-18    Choice (a)       is         the       best     answer           because          DNA    synthesis            cannot            begin   without           the       initial   primers.          Choice (b)       is            a          good   answer           because          lagging-strand           synthesis            requires         continual        use      of         RNA    primers          for            discontinuous            replication      to         occur.
           
6-19    (d)
           
6-20    (a)       Because          helicase          unwinds         the       two      DNA    template            strands,          replication      of         both    strands           depends         upon            the       activity            of         helicase          at         the       time     of            initiation.
           
6-21    (b)
           
6-22    (d)
           
6-23    (d)       Each    newly  synthesized    strand in         a          daughter        duplex            was     synthesized    by        a          mixture           of         continuous     and            discontinuous            DNA    synthesis        from    multiple          origins.            Consider         a          single  replication      origin. The     fork     moving           in            one      direction         synthesizes    a          daughter        strand continuously  as            part     of         leading-strand           synthesis;       the       fork     moving           in            the       opposite         direction         synthesizes    a          portion           of            the       same   daughter        strand

discontinuously         as        part     of         lagging-strand           synthesis.

6-24    (d)
6-25    Choice (d)       is         the       correct            answer.           DNA    from    all            organisms      is         chemically      identical          except for       the            sequence        of         nucleotides.    The     proteins          listed   in         choices            (a)       to         (c)       can      act       on        any      DNA    regardless      of            its        sequence.       In         contrast,         the       initiator          proteins            recognize       specific           DNA    sequences      at         the       origins            of            replication.     These  sequences      differ   between         bacteria          and            yeast.
6-26    In         the       absence          of         telomerase,    the       life-span         of         a            cell      and      its        progeny          cells     is         limited.            With    each            round of         DNA    replication,     the       length of         telomeric        DNA            will      shrink,            until    finally  all        the       telomeric        DNA    has            disappeared.  Without          telomeres       capping           the       chromosome            ends,   the       ends    might  be        treated            like      breaks            arising            from    DNA    damage,          or        crucial genetic            information    might            be        lost.     Cells    whose DNA    lacks    telomeres       will      stop     dividing            or        die.      However,        if          telomerase     is         provided        to            cells,    they     may     be        able     to         divide  indefinitely     because            their    telomeres       will      remain            a          constant         length despite            repeated         rounds            of         DNA    replication.
6-27    (d)
6-28    A.         True.

 

Page    20        of         25

 

  1. False. Although        the       sliding clamp  is         only     loaded            once    on        the            leading            strand,

the       lagging            strand needs  to         unload            the       clamp  once    the            polymerase    reaches

the       RNA    primer            from    the       previous         segment          and      then    reload it            where a          new

primer            has      been    synthesized.

  1. True.
  2. False. Primase          does    not      have    a          proofreading function,         nor      does    it            need    one

because          the       RNA    primers          are      not      a          permanent     part     of         the            DNA.   The     primers

are      removed,        and      a          DNA    polymerase    that     does    have    a            proofreading function

fills      in         the       remaining       gaps.

 

6-29    ___3___ primase

___2___ single-strand  binding           protein

___3___ sliding clamp

___3___ RNA

primers

___1___ leading            strand

___2___ lagging            strand

___2___ Okazaki           fragments

___3___ DNA    helicase

___2___ DNA    ligase

 

6-30    A.         The     actual  chemical         reaction          in         DNA    synthesis        is         the       same            regardless      of

whether          going   in         the       5′-to-3′            or        in         the       3′-to-5′            direction.            The     most    important

distinction      between         these   two      options           is         that     if          DNA    is            synthesized    in         the       3′-

to-5′ direction,           the       5′         end      of         the       elongating      strand,            rather            than    the       3′         end,

will      have    a          nucleoside      triphosphate.

 

  1. DNA synthesis        from    3′         to         5′         does    not      allow   proofreading.            If            the       last

nucleotide      added is         mispaired       and      is         removed,        the       last      nucleotide            on        the

growing          strand is         a          nucleoside      monophosphate        and      the       nucleotide            coming            in

only     has      a          hydroxyl         group  on        the       3′         end.     Thus,   there   is         no            favorable

hydrolysis      reaction          to         drive   the       addition          of         new     nucleotides.

 

6-31    (b)

           

 

Page    21        of         25

 

6-32                See      Figure A6-32.

 

 

Figure      A6-32

 

6-33    DNA    ligase  has      an        important       role     in         DNA    replication.     After            Okazaki           fragments       are      synthesized,   they     must    be        ligated            (covalently     connected)     to         each    other   so        that     they     finally            form    one      continuous     strand.            At        the       nonpermissive            temperature  this      does    not      happen,          and      although         there            may     be        a          range  of         fragments,      the       notable           band   at            200     base    pairs   is         the       typical size      of         an        individual            Okazaki           fragment.
6-34    A.         False.  The     repair polymerase    is         used    to         fill        in         the            spaces left       vacant after    the

RNA    primers          are      degraded.

B.         False.  This     is         a          two-step         process           that     requires            two      different         enzymes.        First,   a

nuclease         removes         the       RNA    primers.          Then,  the       repair            polymerase    fills      in         the

complementary         DNA    sequence.

C.         True.

D.         True.

           
6-35    (b)
6-36    The     cell      employs          an        additional       protein           in         order  to            make   the       constant         reloading        of         the       sliding clamp  on            the       lagging            strand much   more   efficient.          The     protein,            called  the       clamp  loader,            harnesses       energy            from    ATP            hydrolysis      to         lock     a          sliding clamp  complex          around            the       DNA    for       every  successive      round of         DNA    synthesis.
6-37    (b)
           
6-38    (d)
6-39    (b)
6-40    (d)
6-41    (a)

 

Page    22        of         25

 

6-42    The     distortion       in         the       DNA    backbone       is         insufficient            information    for       the       mismatch        repair system            to         identify            which  base    is         incorrect        and      which  was     originally        part     of            the       chromosome when   replication      began. Without          additional            marks that     identify           the       difference       between         the       newly            synthesized    strand and      the       template         strand,            the       repair            would be        corrected       only     50%    of         the       time     by        random            chance.           The     error   rate     (and    therefore        the       mutation            rate)    would still      be        less      than    in         a          system            that            lacked the       mismatch        repair enzymes         (1        mistake           per            107      base    pairs), but      greater           than    the       error   rate     in         a            system            that     accurately      identifies        the       newly  synthesized            strand (1        mistake           per      109      base    pairs).
6-43 (c)
6-44 Choices           (a)       or        (d)       are      correct.           Choice (b)       cannot            account           for       these   results            because          a          mutation            the       fetuses            she

in    the       original           mouse’s          germ   cells     would have    no        effect   on

was     already           carrying.         Neither           can      choice (c),       because            mutations       in         the       germ   cells     of         the       fetuses            while   in            utero   would have    had      no        effect   on        their    development, but            they     might  have    led       to         mutant            mice    among            their            offspring.

6-45 (a)
6-46 (d)
6-47 (d)
6-48 (b)
6-49 (a)
6-50 A.         One-half,         or        50%.   DNA    replication      in         the       original            bacterium       will      create two

new     DNA    molecules,      one      of         which  will      now     carry   a            mismatched   C-T      base    pair.

So        one      daughter        cell      of         that     cell      division           will            carry   a          completely     normal            DNA

molecule;        the       other   cell      will      have    the       molecule         with            the       mutation         mispaired

to         a          correct            nucleotide.

B.         One-quarter,  or        25%.   At        the       next     round of         DNA            replication      and      cell      division,

the       bacterium       carrying          the       mismatched   C-T      will      produce            and      pass    on        one

normal            DNA    molecule         from    the       undamaged    strand            containing      the       T          and      one

mutant            DNA    molecule         with     a          fully     mutant            C-G            base    pair.    So        at         this      stage,  one

out      of         the       four     progeny          of         the       original            bacterium       is         mutant.           Subsequent    cell

divisions         of         these   mutant            bacteria          will      give     rise            only     to         mutant            bacteria,

whereas         the       other   bacteria          will      give     rise      to         normal            bacteria.         The     proportion

of         progeny          containing      the       mutation         will      therefore            remain            at         25%.

6-51 (c)
6-52 (b)

 

Page    23        of         25

 

6-53    A.         True.

  1. False. It          is         believed          that     the       nicks   are      generated       during DNA            replication      as        a

means of         easy    identification  of         the       newly  synthesized    strand but      are            sealed

by        DNA    ligase  shortly            after    replication      is         completed.

  1. False. Depurination occurs constantly      in         our      cells     through          spontaneous

hydrolysis      of         the       bond   linking the       DNA    base    to         the       deoxyribose            sugar.

  1. False. Homologous  recombination           can      repair double-strand            breaks            without

any      change            in         DNA    sequence,       but      nonhomologous        end      joining            always

involves          a          loss     of         genetic            information    because          the       ends    are            degraded        by

nucleases       before they     can      be        ligated back    together.

 

6-54    Choice (c)       is         the       correct            answer.           In         fact,     affected          individuals     in            some   families           with     a

history            of         early-onset     colon   cancer have    been    found  to         carry   mutations       in            mismatch

repair choice (a)       is

genes.  Mutations       arising            in         somatic           cells     are      not      inherited,        so

incorrect.        A          defect  in         DNA    synthesis        or        nucleotide      biosynthesis   would            probably        be

lethal,  so        choices           (b)       and      (d)       are      incorrect.

 

6-55    A.         Mr       Self-destruct  is         more   likely   than    the       other   mutants          to         be            defective         in         the

DNA    repair polymerase    because          Mr       Self-destruct  is         defective         in         the            repair of

all        three   kinds   of         DNA    damage.          The     repair pathways        for       all            three   kinds   of

damage           are      similar            in         the       later    steps,  including        a            requirement  for       the       DNA

repair polymerase.

  1. The other   mutants          are      specific           for       a          particular       type     of            damage.          Thus,   the

mutations       are      likely   to         be        in         genes  required         for       the       first            stage   of         repair—the

recognition    and      excision          of         the       damaged        bases.  Dracula           and            Mole    are      likely   to

be        defective         in         the       recognition    or        excision          of         thymidine            dimers;           Faust   is         likely

to         be        defective         in         the       recognition    or        excision          of         U-G            mismatched   base    pairs;

and      Marguerite     is         likely   to         be        defective         in         the       recognition    or            excision          of

abasic sites.

 

6-56    Uracil  is         an        RNA    base    and      it          is         recognized     as        a          mutational            lesion  because,         as        it          is

formed           from    the       deamination   of         cytosine,         it          will      be        paired with     a            guanine          in         the

context           of         the       DNA    duplex.            Uracil  pairs   by        forming          two      hydrogen            bonds, similar            to

thymine,         and      is         thus     a          poor    partner           for       guanine,          which  forms            three   hydrogen       bonds

with     cytosine.         The     mismatch        causes a          distortion       of         the       DNA    backbone,            allowing          the

repair machinery      to         recognize       the       uracil  as        a          lesion. Because          uracil            pairs   preferably

with     adenine          (its       partner           in         double-stranded       RNA),  the       deamination   of            cytosine          to

uracil  is         highly  mutagenic.      If          unrepaired,    it          can      result  in         the       transition            of         a          C-G      base

pair     to         a          T-A      base    pair.

 

6-57    (c)

           

6-58    (a)

 

6-59    (b)

 

 

Page    24        of         25

 

6-60    (c)
6-61    (d)      
6-62    (d)
6-63    A.         Neither.          The     copy    of         Chromosome 3          you      received            from    your    mother           is         a

hybrid of         the       ones    she      received         from    her      mother            and      her      father.

B.         See      Figure A6-43. The     correct            answers          include            any            chromosome in         which  a

portion           matches          the       information    from    the       paternal            chromosome and      the

remainder      matches          the       information    from    the       maternal            chromosome.

 

Figure      A6-43

 

C. As        a          result  of         extensive        crossing-over,            you            resemble        both    your    grandmother
and      your    grandfather.   If          there   were   no        crossing-over,            you      might  have    a          much
stronger         resemblance  to         one      than    the       other.
6-64 A. False.  Homologous  recombination           also     occurs in         prokaryotic            cells,    and
typically          occurs very    shortly            after    DNA    replication,     when            the       newly
replicated       duplexes         are      in         close   proximity.
B. True.
C. True.
D. False.  Although        it          is         called  homologous   recombination,            this      is         not      a          process
that     depends         on        the       proximity       of         parental            homologs.      When  used    as        a
mechanism     for       DNA    repair, homologous   recombination           uses            the       sister
chromatids     in         an        undamaged,   newly  replicated       (homologous)            DNA    helix    as        a
template.

 

Page    25        of         25

ESSENTIAL CELL BIOLOGY, FOURTH EDITION

CHAPTER 7: FROM DNA TO PROTEIN

© 2014 GARLAND SCIENCE PUBLISHING

7-1 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
The instructions specified by the DNA will ultimately specify the sequence of proteins. This process involves DNA, made up of ____ different nucleotides, which gets _________________ into RNA, which is then _________________ into proteins, made up of _____ different amino acids. In eukaryotic cells, DNA gets made into RNA in the

 

_________________,while proteins are produced from RNA in the

_________________. The segment of DNA called a _________________ is the portion that is copied into RNA; this process is catalyzed by RNA _________________.

4   gene   proteasome

20   Golgi   replisome

109   kinase   sugar-phosphate

128   nuclear pore  transcribed

cytoplasm  nucleus  transferase

exported  polymerase  translated

7-2 Use the numbers in the choices below to indicate where in the schematic diagram of a eukaryotic cell (Figure Q7-2) those processes take place.
 

Figure Q7-2

 

 

 

1. transcription
2. translation
3. RNA splicing

 

Page 1 of 29

 

  1. polyadenylation
  2. RNA capping

 

From DNA to RNA

 

7-3 Consider two genes that are next to each other on a chromosome, as arranged in Figure

Q7-3.

 

 

Figure Q7-3

Which of

the following statements is true?

(a) The two genes must be transcribed into RNA using the same strand of DNA.

(b) If gene A is transcribed in a cell, gene B cannot be transcribed.

(c) Gene A and gene B can be transcribed at different rates, producing different

amounts of RNA within the same cell.

(d) If gene A is transcribed in a cell, gene B must be transcribed.

 

7-4 RNA in cells differs from DNA in that ___________________.

(a) it contains the base uracil, which pairs with cytosine.

(b) it is single-stranded and cannot form base pairs.

(c) it is single-stranded and can fold up into a variety of structures.

(d) the sugar ribose contains fewer oxygen atoms than does deoxyribose.

7-5 Transcription is similar to DNA replication in that ___________________.

(a) an RNA transcript is synthesized discontinuously and the pieces are then joined

together.

(b) it uses the same enzyme as that used to synthesize RNA primers during DNA

replication.

(c) the newly synthesized RNA remains paired to the template DNA.

(d) nucleotide polymerization occurs only in the 5′-to-3′ direction.

 

Figure Q7-6 is to be used with Questions 7-6, 7-7, and 7-8. These three questions can be used separately or together.

 

Figure Q7-6

 

Page 2 of 29

 

7-6 Figure Q7-6 shows a ribose sugar. RNA bases are added to the part of the ribose sugar pointed to by arrow _____.

(a) 3.

(b) 4.

(c) 5.

(d) 6.

7-7 Figure Q7-6 shows a ribose sugar. The part of the ribose sugar that is different from the deoxyribose sugar used in DNA is pointed to by arrow ____.

(a) 1.

(b) 4.

(c) 5.

(d) 6.

7-8 Figure Q7-6 shows a ribose sugar. The part of the ribose sugar where a new

ribonucleotide will attach in an RNA molecule is pointed to by arrow ____.

(a) 1.

(b) 3.

(c) 4.

(d) 5.

7-9 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
For a cell’s genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________________. Various kinds of RNA are produced, each with different functions. __________________ molecules code for proteins, __________________ molecules act as adaptors for protein synthesis, __________________ molecules are integral components of the ribosome, and __________________ molecules are important in the splicing of RNA transcripts.
incorporation  rRNA   translation

mRNA   snRNA  transmembrane

pRNA   transcription  tRNA

proteins

7-10 Match the following structures with their names.

 

Page 3 of 29

 

 

 

Figure Q7-10

 

7-11 Which of the following statements is false?

(a) A new RNA molecule can begin to be synthesized from a gene before the

previous RNA molecule’s synthesis is completed.

(b) If two genes are to be expressed in a cell, these two genes can be transcribed with

different efficiencies.

(c) RNA polymerase is responsible for both unwinding the DNA helix and catalyzing

the formation of the phosphodiester bonds between nucleotides.

(d) Unlike DNA, RNA uses a uracil base and a deoxyribose sugar.

7-12 Unlike DNA, which typically forms a helical structure, different molecules of RNA can fold into a variety of three-dimensional shapes. This is largely because

___________________.

(a) RNA contains uracil and uses ribose as the sugar.

(b) RNA bases cannot form hydrogen bonds with each other.

(c) RNA nucleotides use a different chemical linkage between nucleotides compared

to DNA.

(d) RNA is single-stranded.

7-13 Which of the following molecules of RNA would you predict to be the most likely to fold into a specific structure as a result of intramolecular base-pairing?

(a) 5′-CCCUAAAAAAAAAAAAAAAAUUUUUUUUUUUUUUUUAGGG-3′

(b) 5′-UGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUG-3′

(c) 5′-AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA-3′

(d) 5′-GGAAAAGGAGAUGGGCAAGGGGAAAAGGAGAUGGGCAAGG-3′

7-14 Which one of the following is the main reason that a typical eukaryotic gene is able to respond to a far greater variety of regulatory signals than a typical prokaryotic gene or operon?

(a) Eukaryotes have three types of RNA polymerase.

(b) Eukaryotic RNA polymerases require general transcription factors.

 

Page 4 of 29

 

(c) The transcription of a eukaryotic gene can be influenced by proteins that bind far

from the promoter.

(d) Prokaryotic genes are packaged into nucleosomes.

 

7-15 Match the following types of RNA with the main polymerase that transcribes them.

 

7-16 List three

ways in which the process of eukaryotic transcription differs from the process

of bacterial transcription.

 

7-17 For each of the following sentences, fill in the blanks with the best word or phrase

selected from the list below. Not all words or phrases will be used; each word or phrase

should be used only once.

 

In eukaryotic cells, general transcription factors are required for the activity of all

promoters transcribed by RNA polymerase II. The assembly of the general transcription

factors begins with the binding of the factor __________________ to DNA, causing a

marked local distortion in the DNA. This factor binds at the DNA sequence called the

__________________ box, which is typically located 25 nucleotides upstream from the

transcription start site. Once RNA polymerase II has been brought to the promoter DNA,

it must be released to begin making transcripts. This release process is facilitated by the

addition of phosphate groups to the tail of RNA polymerase by the factor

__________________. It must be remembered that the general transcription factors and

RNA polymerase are not sufficient to initiate transcription in the cell and are affected by

proteins bound thousands of nucleotides away from the promoter. Proteins that link the

distantly bound transcription regulators to RNA polymerase and the general transcription

factors include the large complex of proteins called the__________________. The

packing of DNA into chromatin also affects transcriptional initiation, and histone

__________________ is an enzyme that can render the DNA less accessible to the

general transcription factors.

 

activator  lac   TFIIA

CAP   ligase   TFIID

deacetylase  Mediator  TFIIH

enhancer  TATA

 

7-18 You have a piece of DNA that includes the following sequence:

 

5′-ATAGGCATTCGATCCGGATAGCAT-3′

3′-TATCCGTAAGCTAGGCCTATCGTA-5′

 

Page 5 of 29

 

Which of the following RNA molecules could be transcribed from this piece of DNA? (a) 5′-UAUCCGUAAGCUAGGCCUAUGCUA-3′

(b) 5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′

(c) 5′-UACGAUAGGCCUAGCUUACGGAUA-3′

(d) none of the above

7-19 The following segment of DNA is from a transcribed region of a chromosome. You know that RNA polymerase moves from left to right along this piece of DNA, that the promoter for this gene is to the left of the DNA shown, and that this entire region of DNA is made into RNA.
5′-GGCATGGCAATATTGTAGTA-3′

 

3′-CCGTACCGTTATAACATCAT-5′

Given this information, a student claims that the RNA produced from this DNA is:
3′-GGCATGGCAATATTGTAGTA-5′
Give two reasons why this answer is incorrect.
7-20 You have a segment of DNA that contains the following sequence:
5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′

3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′

You know that the RNA transcribed from this segment contains the following sequence:
5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA–3′
Which of the following choices best describes how transcription occurs?

(a) the top strand is the template strand; RNA polymerase moves along this strand

from 5′ to 3′

(b) the top strand is the template strand; RNA polymerase moves along this strand

from 3′ to 5′

(c) the bottom strand is the template strand; RNA polymerase moves along this strand

from 5′ to 3′

(d) the bottom strand is the template strand; RNA polymerase moves along this strand

from 3′ to 5′

7-21 Imagine that an RNA polymerase is transcribing a segment of DNA that contains the following sequence:
5′-AGTCTAGGCACTGA-3′

3′-TCAGATCCGTGACT-5′

 

Page 6 of 29

 

A. If the polymerase is transcribing from this segment of DNA from left to right,

which strand (top or bottom) is the template?

B. What will be the sequence of that RNA (be sure to label the 5′ and 3′ ends of your

RNA molecule)?

 
7-22 The sigma subunit of bacterial RNA polymerase ___________________.

(a) contains the catalytic activity of the polymerase.

(b) remains part of the polymerase throughout transcription.

(c) recognizes promoter sites in the DNA.

(d) recognizes transcription termination sites in the DNA.

7-23 Which of the following might decrease the transcription of only one specific gene in a bacterial cell?

(a) a

decrease in the amount of sigma factor

(b) a decrease in the amount of RNA polymerase

(c) a mutation that introduced a stop codon into the DNA that precedes the gene’s

coding sequence

(d) a mutation that introduced extensive sequence changes into the DNA that

precedes the gene’s transcription start site

7-24 There are several reasons why the primase used to make the RNA primer for DNA replication is not suitable for gene transcription. Which of the statements below is not one of those reasons?

(a) Primase initiates RNA synthesis on a single-stranded DNA template.

(b) Primase can initiate RNA synthesis without the need for a base-paired primer. (c) Primase synthesizes only RNAs of about 5–20 nucleotides in length.

(d) The RNA synthesized by primase remains base-paired to the DNA template.

7-25 You have a bacterial strain with a mutation that removes the transcription termination signal from the Abd operon. Which of the following statements describes the most likely effect of this mutation on Abd transcription?

(a) The Abd RNA will not be produced in the mutant strain.

(b) The Abd RNA from the mutant strain will be longer than normal.

(c) Sigma factor will not dissociate from RNA polymerase when the Abd operon is

being transcribed in the mutant strain.

(d) RNA polymerase will move in a backward fashion at the Abd operon in the

mutant strain.

7-26 Transcription in bacteria differs from transcription in a eukaryotic cell because __________________________.

(a) RNA polymerase (along with its sigma subunit) can initiate transcription on its

own.

(b) RNA polymerase (along with its sigma subunit) requires the general transcription

factors to assemble at the promoter before polymerase can begin transcription. (c) the sigma subunit must associate with the appropriate type of RNA polymerase to

produce mRNAs.

 

Page 7 of 29

 

(d) RNA polymerase must be phosphorylated at its C-terminal tail for transcription to

proceed.

7-27 Which of the following does not occur before a eukaryotic mRNA is exported from the nucleus?

(a) The ribosome binds to the mRNA.

(b) The mRNA is polyadenylated at its 3′ end.

(c) 7-methylguanosine is added in a 5′-to-5′ linkage to the mRNA.

(d) RNA polymerase dissociates.

7-28 Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads to which the polynucleotide 5′-TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached. After a short incubation, the beads are then extracted from the mixture. When to beads, which

you analyze the cellular nucleic acids that have stuck the

of the following is most abundant?

(a) DNA

(b) tRNA

(c) rRNA

(d) mRNA

7-29 Name three covalent modifications that can be made to an RNA molecule in eukaryotic cells before the RNA molecule becomes a mature mRNA.
7-30 Which of the following statements about RNA splicing is false?

(a) Conventional introns are not found in bacterial genes.

(b) For a gene to function properly, every exon must be removed from the primary

transcript in the same fashion on every mRNA molecule produced from the same

gene.

(c) Small RNA molecules in the nucleus perform the splicing reactions necessary for

the removal of introns.

(d) Splicing occurs after the 5′ cap has been added to the end of the primary

transcript.

7-31 The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this difference?
7-32 Why is the old dogma “one gene—one protein” not always true for eukaryotic genes?
7-33 Genes in eukaryotic cells often have intronic sequences coded for within the DNA. These sequences are ultimately not translated into proteins. Why?

(a) Intronic sequences are removed from RNA molecules by the spliceosome, which

works in the nucleus.

(b) Introns are not transcribed by RNA polymerase.

(c) Introns are removed by catalytic RNAs in the cytoplasm.

 

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(d) The ribosome will skip over intron sequences when translating RNA into protein.
7-34 snRNAs ___________________.

(a) are translated into snRNPs.

(b) are important for producing mature mRNA transcripts in bacteria.

(c) are removed by the spliceosome during RNA splicing.

(d) can bind to specific sequences at intron–exon boundaries through complementary

base-pairing.

7-35 Is this statement true or false? Explain your answer.
“Since introns do not contain protein-coding information, they do not have to be removed precisely (meaning, a nucleotide here and there should not matter) from the primary transcript

during RNA splicing.”

7-36 You have discovered a gene (Figure Q7-36A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure Q7-36B. The lines connecting the exons that are included in the mRNA indicate the splicing. From your experiments, you know that protein translation begins in exon 1. For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10. Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of the following statements about exons 2 and 3 is the most accurate? Explain your answer.
 

Figure Q7-36

 

 

 

(a) Exons 2 and 3 must have the same number of nucleotides.
(b) Exons 2 and 3 must contain an integral number of codons (that is, the number of
nucleotides divided by 3 must be an integer).
(c) Exons 2 and 3 must contain a number of nucleotides that when divided by 3,
leaves the same remainder (that is, 0, 1, or 2).
(d) Exons 2 and 3 must have different numbers of nucleotides.

 

Page 9 of 29

 

 

 

From RNA to Protein

 

7-37 Which of the following statements about the genetic code is correct?

(a) All codons specify more than one amino acid.

(b) The genetic code is redundant.

(c) All amino acids are specified by more than one codon.

(d) All codons specify an amino acid.

 

NOTE: The following codon table is to be used for Problems Q7-38 to Q7-49.

 

7-38 The piece of RNA below includes the region that codes for the binding site for the initiator tRNA needed in translation.
5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′
Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome?

(a) methionine

(b) arginine

(c) cysteine

(d) valine

7-39 The following DNA sequence includes the beginning of a sequence coding for a protein. What would be the result of a mutation that changed the C marked by an asterisk to an A?
5′-AGGCTATGAATGGACACTGCGAGCCC…

*

7-40 Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to carry? (a) lysine

 

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(b) glutamic acid

(d) leucine

(d) phenylalanine

7-41 Which of the following pairs of codons might you expect to be read by the same tRNA as a result of wobble?

(a) CUU and UUU

(b) GAU and GAA

(c) CAC and CAU

(d) AAU and AGU

7-42 Below is a segment of RNA from the middle of an mRNA.
 

5′-UAGUCUAGGCACUGA-3′

If you were told that this segment of RNA was part of the coding region of an mRNA for a large protein, give the amino acid sequence for the protein that is encoded by this segment of mRNA. Write your answer using the one-letter amino acid code.
7-43 Below is the sequence from the 3′ end of an mRNA.
5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′
If you were told that this sequence contains the stop codon for the protein encoded by this mRNA, what is the anticodon on the tRNA in the P site of the ribosome when release factor binds to the A site?

(a) 5′-CCA-3′

(b) 5′-CCG-3′

(c) 5′-UGG-3′

(d) 5′-UUA-3′

7-44 One strand of a section of DNA isolated from the bacterium E. coli reads:
5′-GTAGCCTACCCATAGG-3′
A. Suppose that an mRNA is transcribed from this DNA using the complementary

strand as a template. What will be the sequence of the mRNA in this region (make

sure you label the 5′ and 3′ ends of the mRNA)?

B. How many different peptides could potentially be made from this sequence of

RNA, assuming that translation initiates upstream of this sequence?

C. What are these peptides? (Give your answer using the one-letter amino acid

code.)

7-45 A strain of yeast translates mRNA into protein inaccurately. Individual molecules of a particular protein isolated from this yeast have variations in the first 11 amino acids compared with the sequence of the same protein isolated from normal yeast cells, as

 

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listed in Figure Q7-45. What is the most likely cause of this variation in protein
sequence?
Figure Q7-45

 

 

 

(a) a mutation in the DNA coding for the protein

(b) a

mutation in the anticodon of the isoleucine-tRNA (tRNAIle)

(c) a mutation in the isoleucyl-tRNA synthetase that decreases its ability to

distinguish between different amino acids

(d) a mutation in the isoleucyl-tRNA synthetase that decreases its ability to

distinguish between different tRNA molecules

7-46 A mutation in the tRNA for the amino acid lysine results in the anticodon sequence 5′-

UAU-3′ (instead of 5′-UUU-3′). Which of the following aberrations in protein

synthesis might this tRNA cause?

(a) read-through of stop codons

(b) substitution of lysine for isoleucine

(c) substitution of lysine for tyrosine

(d) substitution of lysine for phenylalanine

7-47 After treating cells with a mutagen, you isolate two mutants. One carries alanine and the

other carries methionine at a site in the protein that normally contains valine. After

treating these two mutants again with mutagen, you isolate mutants from each that now

carry threonine at the site of the original valine (see Figure Q7-47). Assuming that all

mutations caused by the mutagen are due to single nucleotide changes, deduce the codons

that are used for valine, alanine, methionine, and threonine at the affected site.

 

Figure Q7-47

 

 

 

7-48 What do you predict would happen if you created a tRNA with an anticodon of 5′-CAA-

3′ that is charged with methionine, and added this modified tRNA to a cell-free

translation system that has all the normal components required for translating RNAs?

 

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(a) methionine would be incorporated into proteins at some positions where

glutamine should be

(b) methionine would be incorporated into proteins at some positions where leucine

should be

(c) methionine would be incorporated into proteins at some positions where valine

should be

(d) translation would no longer be able to initiate

7-49 In a diploid organism, the DNA encoding one of the tRNAs for the amino acid tyrosine is mutated such that the sequence of the anticodon is now 5′-CTA-3′ instead of 5′-GTA-3′. What kind of aberration in protein synthesis will this tRNA cause? Explain your answer.
7-50 The ribosome Which of the

is important for catalyzing the formation of peptide bonds.

following statements is true?

(a) The number of rRNA molecules that make up a ribosome greatly exceeds the

number of protein molecules found in the ribosome.

(b) The large subunit of the ribosome is important for binding to the mRNA.  (c) The catalytic site for peptide bond formation is formed primarily from an rRNA. (d) Once the large and small subunits of the ribosome assemble, they will not

separate from each other until degraded by the proteasome.

7-51 Which of the following statements is true?

(a) Ribosomes are large RNA structures composed solely of rRNA.

(b) Ribosomes are synthesized entirely in the cytoplasm.

(c) rRNA contains the catalytic activity that joins amino acids together.

(d) A ribosome binds one tRNA at a time.

7-52 Figure Q7-52A shows the stage in translation when an incoming aminoacyl-tRNA has bound to the A site on the ribosome. Using the components shown in Figure Q7-52A as a guide, show on Figures Q7-52B and Q7-52C what happens in the next two stages to complete the addition of the new amino acid to the growing polypeptide chain.

 

Page 13 of 29

 

Figure Q7-52

 

7-53 A poison added to an in vitro translation mixture containing mRNA molecules with the sequence 5′-AUGAAAAAAAAAAAAUAA-3′ has the following effect: the only product made is a Met-Lys dipeptide that remains attached to the ribosome. What is the most likely way in which the poison acts to inhibit protein synthesis?

(a) It inhibits peptidyl transferase activity.

(b) It inhibits movement of the small subunit relative to the large subunit. (c) It inhibits release factor.

(d) It mimics release factor.

7-54 In eukaryotes, but not in prokaryotes, ribosomes find the start site of translation by ____________________________.

(a) binding directly to a ribosome-binding site preceding the initiation codon. (b) scanning along the mRNA from the 5′ end.

(c) recognizing an AUG codon as the start of translation.

(d) binding an initiator tRNA.

7-55 Which of the following statements about prokaryotic mRNA molecules is false? (a) A single prokaryotic mRNA molecule can be translated into several proteins. (b) Ribosomes must bind to the 5′ cap before initiating translation.

(c) mRNAs are not polyadenylated.

(d) Ribosomes can start translating an mRNA molecule before transcription is

complete.

 

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7-56 Figure Q7-56 shows an mRNA molecule.
 

Figure Q7-56

 

 

 

A. Match the labels given in the list below with the label lines in Figure Q7-56. (a)

ribosome-binding site

(b) initiator codon

(c) stop codon

(d) untranslated 3′ region

(e) untranslated 5′ region

(f) protein-coding region

B. Is the mRNA shown prokaryotic or eukaryotic? Explain your answer.

7-57 You have discovered a protein that inhibits translation. When you add this inhibitor to a mixture capable of translating human mRNA and centrifuge the mixture to separate polyribosomes and single ribosomes, you obtain the results shown in Figure Q7-57. Which of the following interpretations is consistent with these observations?
 

Figure Q7-57

 

 

 

(a) The protein binds to the small ribosomal subunit and increases the rate of initiation of translation.

 

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(b) The protein binds to sequences in the 5′ region of the mRNA and inhibits the rate

of initiation of translation.

(c) The protein binds to the large ribosomal subunit and slows down elongation of the

polypeptide chain.

(d) The protein binds to sequences in the 3′ region of the mRNA and prevents

termination of translation.

7-58 The concentration of a particular protein, X, in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed. You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that following

levels of X mRNA in the mutant cells are normal. Which of the

mutations in the gene for X could explain these results?

(a) the introduction of a stop codon that truncates protein X at the fourth amino acid (b) a change of the first ATG codon to CCA

(c) the deletion of a sequence that encodes sites at which ubiquitin can be attached to

the protein

(d) a change at a splice site that prevents splicing of the RNA

7-59 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the __________________ subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the __________________ subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the __________________ site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the __________________ site by forming base pairs with the exposed codon in the mRNA. The __________________ enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the __________________ called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the __________________.
A   medium   protein

central   P    RNA

DNA   peptidyl transferase  small

E   polymerase   T

large   proteasome    ubiquitin

 

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7-60 Which of the following methods is not used by cells to regulate the amount of a protein in

the cell?

(a) Genes can be transcribed into mRNA with different efficiencies.

(b) Many ribosomes can bind to a single mRNA molecule.

(c) Proteins can be tagged with ubiquitin, marking them for degradation.

(d) Nuclear pore complexes can regulate the speed at which newly synthesized

proteins are exported from the nucleus into the cytoplasm.

 

7-61 Which of the following statements about the proteasome is false?

(a) Ubiquitin is a small protein that is covalently attached to proteins to mark them

for delivery to the proteasome.

(b) Proteases reside in the central cylinder of a proteasome.

(c) Misfolded proteins are delivered to the proteasome, where they are sequestered

 

from the cytoplasm and can attempt to refold.

(d) The protein stoppers that surround the central cylinder of the proteasome use the

energy from ATP hydrolysis to move proteins into the proteasome inner chamber.

RNA and the Origins of Life

 

7-62 Which of the following molecules is thought to have arisen first during evolution?

(a) protein

(b) DNA

(c) RNA

(d) all came to be at the same time

 

7-63 According to current thinking, the minimum requirement for life to have originated on

Earth was the formation of a _______________.

(a) molecule that could provide a template for the production of a complementary

molecule.

(b) double-stranded DNA helix.

(c) molecule that could direct protein synthesis.

(d) molecule that could catalyze its own replication.

 

7-64 Ribozymes catalyze which of the following reactions?

(a) DNA synthesis

(b) transcription

(c) RNA splicing

(d) protein hydrolysis

 

7-65 You are studying a disease that is caused by a virus, but when you purify the virus

particles and analyze them you find they contain no trace of DNA. Which of the

following molecules are likely to contain the genetic information of the virus?

(a) high-energy phosphate groups

(b) RNA

(c) lipids

(d) carbohydrates

 

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7-66 Give a reason why DNA makes a better material than RNA for the storage of genetic

information, and explain your answer.

 

How We Know: Cracking the Genetic Code

 

7-67 When using a repeating trinucleotide sequence (such as 5′-AAC-3′) in a cell-free

translation system, you will obtain:

(a) three different types of peptides, each made up of a single amino acid

(b) peptides made up of three different amino acids in random order

(c) peptides made up of three different amino acids, each alternating with each other

in a repetitive fashion

(d) polyasparagine, as the codon for asparagine is AAC

 

7-68 You have discovered an alien life-form that surprisingly uses DNA as its genetic

material, makes RNA from DNA, and reads the information from RNA to make protein

using ribosomes and tRNAs, which read triplet codons. Because it is your job to decipher

the genetic code for this alien, you synthesize some artificial RNA molecules and

examine the protein products produced from these RNA molecules in a cell-free

translation system using purified alien tRNAs and ribosomes. You obtain the results

shown in Table Q7-68.

 

 

Table Q7-68
From this information, which of the following peptides can be produced from poly UAUC?

(a) Ile-Phe-Val-Tyr

(b) Tyr-Ser-Phe-Ala

(c) Ile-Lys-His-Tyr

(d) Cys-Pro-Lys-Ala

7-69 An extraterrestrial organism (ET) is discovered whose basic cell biology seems pretty much the same as that of terrestrial organisms except that it uses a different genetic code

 

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to translate RNA into protein. You set out to break the code by translation experiments using RNAs of known sequence and cell-free extracts of ET cells to supply the necessary protein-synthesizing machinery. In experiments using the RNAs below, the following results were obtained when the 20 possible amino acids were added either singly or in different combinations of two or three:
RNA 1: 5′-GCGCGCGCGCGCGCGCGCGCGCGCGCGC-3′

RNA 2: 5′-GCCGCCGCCGCCGCCGCCGCCGCCGCCGCC-3′

Using RNA 1, a polypeptide was produced only if alanine and valine were added to the reaction mixture. Using RNA 2, a polypeptide was produced only if leucine and serine and cysteine were added to the reaction mixture. Assuming that protein synthesis can start anywhere on the template, that the ET genetic code is nonoverlapping and linear, and that the many

each codon is the same length (like terrestrial triplet code), how

nucleotides does an ET codon contain?

(a) 2

(b) 3

(c) 4

(d) 5

(e) 6

7-70 NASA has discovered an alien life-form. You are called in to help NASA scientists to deduce the genetic code for this alien. Surprisingly, this alien life-form shares many similarities with life on Earth: this alien uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs. Even more amazing, this alien uses the same 20 amino acids, like the organisms found on Earth, and also codes for each amino acid by a triplet codon. However, the scientists at NASA have found that the genetic code used by the alien life-form differs from that used by life on Earth. NASA scientists drew this conclusion after creating a cell-free protein-synthesis system from alien cells and adding an mRNA made entirely of uracil (poly U). They found that poly U directs the synthesis of a peptide containing only glycine. NASA scientists have synthesized a poly AU mRNA and observe that it codes for a polypeptide of alternating serine and proline amino acids. From these experiments, can you determine which codons code for serine and proline? Explain.
Bonus question. Can you propose a mechanism for how the alien’s physiology is altered so that it uses a different genetic code from life on Earth, despite all the similarities?

 

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ANSWERS
7-1 The instructions specified by the DNA will ultimately specify the sequence of proteins.

This process involves DNA, made up of 4 different nucleotides, which gets transcribed

into RNA, which is then translated into proteins, made up of 20 different amino acids. In

eukaryotic cells, DNA gets made into RNA in the nucleus, while proteins are produced

from RNA in the cytoplasm. The segment of DNA called a gene is the portion that is

copied into RNA; this process is catalyzed by RNA polymerase.

7-2 See Figure A7-2.

 

Figure A7-2

 

7-3 (c)
7-4 Choice (c) is correct. Choice (a) is untrue because although RNA contains uracil, uracil pairs with adenine, not cytosine. Choice (b) is false because RNA can form base pairs with a complementary RNA or DNA sequence. Choice (d) is false because ribose contains one more oxygen atom than deoxyribose.
7-5 Choice (d) is correct. Choice (a) is incorrect because an RNA transcript is made by a single polymerase molecule that proceeds from the start site to the termination site without falling off. The enzyme used to make primers during DNA synthesis is indeed an RNA polymerase, but it is a special enzyme, primase, and not the enzyme that is used for transcription, which is why choice (b) is incorrect. Choice (c) is false.
7-6 (a)
7-7 (c)

 

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7-8 (c)
7-9 For a cell’s genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called transcription. Various kinds of RNA are produced, each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors for protein synthesis, rRNA molecules are integral components of the ribosome, and snRNA molecules are important in the splicing of RNA transcripts.
7-10 A—4; B—2; C—1; D—3
7-11 Choice (d) is false. RNA nucleotides contain the sugar ribose.
7-12 Choice (d) why RNA

is correct. Choice (a) is true, but is not the main reason different molecules can form different three-dimensional structures (although ribose does increase potential hydrogen-bonding potentials compared to deoxyribose). Choices (b) and (c) are untrue.

7-13 Choice (a) is correct. Choices (b) and (c) do not have any opportunity for intramolecular base-pairing and thus a specific structure is unlikely. Although there is some opportunity for intramolecular base-pairing in choice (d), choice (a) has much more intrastrand complementarity and is a better choice.
7-14 (c)
7-15 A—1; B—3; C—3; D—2; E – 2
7-16 Any three of the following are acceptable.

1. Bacterial cells contain a single RNA polymerase, whereas eukaryotic cells have

three.

2. Bacterial RNA polymerase can initiate transcription without the help of additional

proteins, whereas eukaryotic RNA polymerases need general transcription factors. 3. In eukaryotic cells, transcription regulators can influence transcriptional initiation

thousands of nucleotides away from the promoter, whereas bacterial regulatory

sequences are very close to the promoter.

4. Eukaryotic transcription is affected by chromatin structure and nucleosomes,

whereas bacteria lack nucleosomes.

7-17 In eukaryotic cells, general transcription factors are required for the activity of all promoters transcribed by RNA polymerase II. The assembly of the general transcription factors begins with the binding of the factor TFIID to DNA, causing a marked local distortion in the DNA. This factor binds at the DNA sequence called the TATA box, which is typically located 25 nucleotides upstream from the transcription start site. Once RNA polymerase II has been brought to the promoter DNA, it must be released to begin making transcripts. This release process is facilitated by the addition of phosphate groups to the tail of RNA polymerase by the factor TFIIH. It must be remembered that the

 

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general transcription factors and RNA polymerase are not sufficient to initiate

transcription in the cell and are affected by proteins bound thousands of nucleotides away from the promoter. Proteins that link the distantly bound transcription regulators to RNA polymerase and the general transcription factors include the large complex of proteins called the Mediator. The packing of DNA into chromatin also affects transcriptional initiation, and histone deacetylase is an enzyme that can render the DNA less accessible to the general transcription factors.

7-18 Choice (b) is correct. The molecules listed in choices (a) and (c) have incorrect polarity.
7-19 First, the RNA molecule should have uracil instead of thymine bases. Second, the polarity of the molecule is incorrectly labeled. The correct RNA molecule produced, using the bottom strand of the DNA duplex as a template, would be:

 

5′-GGCAUGGCAAUAUUGUAGUA-3′

7-20 (d) The bottom strand can hybridize with the RNA molecule and thus is the template strand. The polymerase moves along the DNA in a 3′-to-5′ direction, because the RNA nucleotides are joined in a 5′-to-3′ polarity.
7-21 A.         The     bottom            strand.

B. 5′-AGUCUAGGCACUGA-3′

7-22 (c)
7-23 (d) Such changes would probably destroy the function of the promoter, making RNA polymerase unable to bind to it. Decreasing the amount of sigma factor or RNA polymerase [choices (a) or (b)] would affect the transcription of most of the genes in the cell, not just one specific gene. Introducing a stop codon before the coding sequence [choice (c)] would have no effect on transcription of the gene, because the transcription machinery does not recognize translational stops.
7-24 Choice (b) is true for both primase and RNA polymerase, so it does not describe why primase cannot be used for gene transcription.
7-25 (b) Without the termination signal, the polymerase will not halt and release from the DNA template at the normal location when transcribing the Abd operon. Most probably, the polymerase will continue to transcribe RNA until it reaches a sequence in the DNA that can serve as a termination sequence, either from the next downstream operon or in the intervening sequence between the Abd operon and the next operon. Dissociation of sigma factor occurs once an approximately 10-nucleotide length of RNA has been synthesized by RNA polymerase and should not be affected by the lack of a termination signal [choice (c)].
7-26 Choice (a) is correct. Eukaryotic cells, but not bacteria, require general transcription factors [choice (b)]. There is only a single type of RNA polymerase in bacterial cells

 

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[choice (c)]. The general transcription factor TFIIH phosphorylates the C-terminal tail of RNA polymerase in eukaryotic cells but not in bacteria [choice (d)].
7-27 (a) Ribosomes are in the cytosol and will bind to the mRNA once it has been exported from the nucleus.
7-28 (d) mRNA is the only type of RNA that is polyadenylated, and its poly-A tail would be able to base-pair with the strands of poly T on the beads and thus stick to them. DNA would not be found in the sample, because the poly-A tail is not encoded in the DNA and long runs of T are rare in DNA.
7-29 1. A poly-A tail is added.

2. A 5′ cap is added.

3.

Introns can be spliced out.

7-30 (b) The primary transcript of a gene can sometimes be spliced differently so that different exons can be stitched together to produce distinct proteins in a process called alternative splicing.
7-31 The gene contains one or more introns.
7-32 The transcripts from some genes can be spliced in more than one way to give mRNAs containing different sequences, thus encoding different proteins. A single eukaryotic gene may therefore encode more than one protein.
7-33 (a)
7-34 (d) snRNAs are part of the snRNPs, which include proteins and RNA molecules. The proteins within the snRNPs are encoded by their own genes and not translated from snRNPs, which is why choice (a) is incorrect. Bacteria do not have introns, which is why choice (b) is incorrect.
7-35 False. Although it is true that the sequences within the introns are mostly dispensable, the introns must still be removed precisely because an error of one or two nucleotides would shift the reading frame of the resulting mRNA molecule and change the protein it encodes.
7-36 Choice (c) is the only answer that must be true for exons 2 and 3. Although choices (a), (b), and (d) could be true, they do not have to be. Because the protein sequence is the same in segments of the mRNA corresponding to exons 1 and 10, the choice of either exon 2 or exon 3 would not alter the reading frame. To maintain the normal reading frame, whatever it is, the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder.

 

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7-37 (b) Most amino acids can be specified by more than one codon. Each codon specifies only one amino acid [choice (a)]. Tryptophan and methionine are encoded by only one codon [choice (c)]. Some codons specify translational stop signals [choice (d)].
7-38 Choice (b) is correct. The initiator methionine is underlined on the RNA molecule below.
5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′
The first tRNA to bind at the A site is the second codon of the protein, because the initiator tRNA is already bound to the P site when translation begins. The codon that follows the binding site for the initiator tRNA is CGU, which codes for arginine.
7-39 The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the (the of the coding

protein-coding sequence and in the correct reading frame beginning

sequence is indicated by the ATG). Thus, translation would terminate after only four amino acids had been joined together, and the complete protein would not be made.

7-40 (a) As is conventional for nucleotide sequences, the anticodon is given reading from 5′ to 3′. The complementary base-pairing occurs between antiparallel nucleic acid sequences, and the codon recognized by this anticodon will therefore be 5′-AAG-3′.
7-41 Choice (c) is the correct answer. These two codons differ only in the third position and also encode the same amino acid, which is the definition of wobble. Although the codons GAU and GAA [choice (b)] also differ only in the third position, they are unlikely in normal circumstances to be read by the same tRNA, because they encode different amino acids.
7-42 SLGT is the correct answer. (Reading frame two is the only reading frame that does not contain a stop codon.)
7-43 (a) The stop codon (UAA) is underlined in the mRNA sequence below; this is the only stop codon on this piece of mRNA. The codon (UGG) preceding the stop codon will be binding to a tRNA in the P site of the ribosome when release factor binds to the A site. The anticodon of the tRNA will bind to the codon UGG and will be CCA.
5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′
7-44 A.         5′-GUAGCCUACCCAUAGG-3′

B. Two. (There are three potential reading frames for each RNA. In this case, they

are

GUA GCC UAC CCA UAG …

UAG CCU ACC CAU AGG …

AGC CUA CCC AUA GG? …

The center one cannot be used in this case, because UAG is a stop codon.)

 

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C. VAYP

SLPIG

Note: PTHR will not be a peptide because it is preceded by a stop codon.

7-45 (c) A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between amino acids would allow an assortment of amino acids to be attached to the tRNAIle. These assorted aminoacyl-tRNAs would then base-pair with the isoleucine codon and cause a variety of substitutions at positions normally occupied by isoleucine. A mutation in the gene encoding the protein would cause only a single variant protein to be made [choice (a)]. A mutation in the anticodon loop of tRNAIle

[choice (b)] or a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules [choice (d)] would cause the substitution of isoleucine for some other amino acid (which is the opposite of what is observed).

 

(b) The mutant tRNALys will be able to pair with the codon 5-AUA-3, which codes for

7-46 ′′

isoleucine.

7-47 Given that there are only single nucleotide changes, the only codons consistent with the changes are GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine.
7-48 (b) The 5′-CAA-3′ anticodon binds to the 5′-UUG-3′ codon, which normally codes for leucine.
7-49 If the DNA sequence specifying the anticodon is changed from 5′-GTA-3′ to 5′-CTA-3′, this tRNA will now pair with the 5′-UAG-3′ codon (instead of 5′ -UAC-3′). The UAG codon normally serves as a stop codon. Thus, this change will result in the amino acid tyrosine being incorrectly incorporated where there is a stop codon, resulting in the addition of amino acids at the end of proteins that normally would come to a stop because of the UAG codon in the mRNA. (Note that the tyrosine codons will NOT cause premature termination of translation, as tyrosine should continue to be incorporated into proteins, as there are additional tyrosine-tRNA genes in the cell that will provide a normal supply of tyrosine-tRNAs.)
7-50 Choice (c) is correct. A ribosome is built from many more proteins than rRNA molecules, although the ribosome is about two-thirds RNA and one-third protein by weight. Thus, (a) is incorrect. The small subunit binds to mRNA, so (b) is incorrect. Choice (d) is incorrect, as the assembly and disassembly of the small subunit with the large subunit occurs every time a protein is produced from an mRNA. When release factor binds to an mRNA, the ribosome will release the mRNA and dissociate into its two subunits, to be recycled during another round of protein synthesis.
7-51 Choice (c) is correct. Ribosomes contain proteins as well as rRNA [choice (a)]. rRNA is synthesized in the nucleus, and ribosomes are partly assembled in the nucleus [choice (b)]. A ribosome must be able to bind two tRNAs at any one time [choice (d)].
7-52 See Figure A7-52.

 

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Figure A7-52

 

7-53 Choice (b) is correct. Choice (a) would prevent all peptide bond formation. Choice (c) would have no effect on translation until the stop codon was reached. Choice (d) would be likely to result in a mixture of polypeptides of various lengths; a poison mimicking a release factor could conceivably cause only Met-Lys to be made, but this dipeptide would not remain bound to the ribosome.
7-54 Choice (b) is correct. Choice (a) is true only for prokaryotes. Choices (c) and (d) are true for both prokaryotes and eukaryotes.
7-55 (b) Bacterial mRNAs do not have 5′ caps. Instead, ribosome-binding sites upstream of the start codon tell the ribosome where to begin searching for the start of translation.
7-56 A.         (a)—3;            (b)—2;            (c)—4;            (d)—6;(e)—1;           (f)—5

B. The mRNA is prokaryotic. It contains coding regions for more than one protein,

as shown by the multiple initiation codons, each preceded by a ribosome-binding

site. It contains an unmodified 5′ end, as shown by the three phosphate groups,

and an unmodified 3′ end, as shown by the absence of a poly-A tail.

7-57 (b) The results in Figure Q7-57 show a marked decrease in the number of polyribosomes formed relative to normal. Polyribosomes form because the initiation of translation is fairly rapid: ribosomes can bind successively to the free 5′ end of an mRNA molecule and start translation before the first ribosome has had a chance to finish translating the message. Therefore, inhibition of the rate of initiation will tend to decrease the number of

 

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ribosomes in the polyribosome, and in the extreme case there will be only one ribosome per mRNA. Conversely, increasing the rate of initiation or slowing the rate of elongation would result in an increased number of ribosomes per polyribosome (up to a maximum point), making choices (a) and (c) false. Choice (d) is incorrect, because preventing termination would prevent release of the ribosomes at the end of the coding sequence and would be expected to “freeze” the assembled polyribosomes, so that the ratio of polyribosomes to ribosomes would be much as normal.
7-58 (c) The decrease in the level of protein X in the normal cell is most probably due to protein degradation, because levels of mRNA remain constant. The inability of the mutant cell to divide could be due to a mutation that inhibits protein degradation. This would be achieved by the removal of sites for attachment of ubiquitin, which targets proteins for destruction. Choices (a), (b), and (d) would probably not produce the results

mutant cells described, because without the production of a functional protein X,the

could not grow in size.

7-59 Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the large subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the small subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the P site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the A site by forming base pairs with the exposed codon in the mRNA. The peptidyl transferase enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the protein called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the proteasome.
7-60 (d) Proteins are synthesized in the cytoplasm and therefore newly synthesized proteins would not be exported from the nucleus into the cytoplasm.
7-61 (c) Once proteins are sent to the proteasome, proteases degrade them. Chaperone proteins provide a place for misfolded proteins to attempt to refold.
7-62 (c) Because RNA is known to catalyze reactions within the cell, because the components of RNA are thought to be more readily formed in the conditions on primitive Earth, and because RNA can contain genetic information, it is the most likely of the three molecules to have arisen first in evolution.
7-63 Choice (d) is correct. Choice (a) is incorrect in that, although this may have been a step in self-replication, it would not by itself be sufficient. Choices (b) and (c) are incorrect, as these stages in the evolution of the cell must have succeeded the formation of the first self-replicating molecules.
7-64 (c)

 

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7-65 (b)

 

7-66 Three possible answers are:

  1. The deoxyribose sugar of DNA makes the molecule much less susceptible than

RNA to breakage, because of the lack of the hydroxyl group on carbon 2 of the

ribose sugar.

  1. DNA is double-stranded and therefore the complementary strand provides a

template from which damage can be repaired accurately.

  1. The use of “T” in DNA instead of “U” (as in RNA) protects against the effect of

deamination, a common form of damage. Deamination of T produces an aberrant

base (methyl C), whereas deamination of U generates C, a normal base. The

presence of an abnormal base eases the cell’s job of recognizing the damaged

 

strand.

 

7-67 (a) An mRNA composed of a trinucleotide repeat of AAC can be “read” in three different

frames: AAC, ACA, and CAA. Thus, this mRNA will yield polyasparagine (codon = AAC),

polythreonine (codon = ACA), and polyglutamine (codon = CAA).

 

7-68 (d) All other answers are not possible, because poly UAUC cannot code for Tyr. Tyr must

be encoded by AUA, because both poly AUA and poly UA lead to the synthesis of Tyr (see

Table A7-68).

 

 

Table A7-68
7-69 (d) An organism having codons with an even number of nucleotides (such as 2, 4, or 6) could read 5′-GCGCGCGCGC-3′ (RNA 1) in either of two ways, namely “GC GC GC GC …” or “CG CG CG CG …” Either of the two amino acids alone could have supported protein synthesis, so you would not need them in combination [thus eliminating choices (a), (c), and (e)]. An organism having three bases per codon could read the sequence 5′-GCCGCCGCCGCCGCC-3′ (RNA 2) in one of three ways, namely “GCC GCC GCC GCC …,” “CCG CCG CCG CCG …,” or “CGC CGC CGC

 

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CGC …,” and so again, any one of the three amino acids could have supported synthesis of a polypeptide, and you would not need to add all three amino acids to produce a polypeptide chain, thus eliminating choice (b). Only a five-nucleotide code gives you two different consecutive codons for RNA 1 and three different consecutive codons for RNA 2.
7-70 No, you cannot definitively determine the codons that code for serine or proline, because it could be either UAU or AUA.
Bonus. The alien aminoacyl-tRNA synthetases could adapt a different amino acid to each tRNA, thus matching an amino acid with a different codon compared with the codons used by life on Earth.

 

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