genetics hw

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genetics hw
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genetics homework for ch12-14
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  1. WHAT IS SEMICONSERVATIVE REPLICATION?
    • In semiconservative
    • replication, the original two strands of the double helix serve as
    • templates for new strands of DNA. When replication is complete, two doublestranded

    • DNA molecules will be present. Each will consist of one original
    • template

    • strand and one newly synthesized strand that is complementary to
    • the template.
  2. HOW DID MESELSON AND STAHL DEMONSTRATE THAT
    REPLICATION IN E. COLI TAKES PLACE IN A SEMICONSERVATIVE MANNER?
    • Meselson and Stahl grew E.
    • coli cells in a medium containing the heavy isotope of
    • nitrogen (15N) for several generations. The 15N was incorporated in the DNA of the E. coli cells. The E. coli cells were then switched to a medium containing the common form of nitrogen (14N) and allowed to proceed through a few cycles
    • of cellular generations. Samples of the bacteria were removed at each cellular generation. Using equilibrium density gradient centrifugation, Meselson and Stahl were able to distinguish DNAs that contained only 15N from DNAs that contained only 14N or a
    • mixture of 15N and 14N because DNAs containing the 15N isotope are “heavier.” The more 15N a DNA molecule contains, the further it will sediment during equilibrium density gradient centrifugation. DNA from cells grown in the 15N medium produced

    • only a single band at the expected position during centrifugation.
    • After one round of

    • replication in the 14N medium, one band was present following
    • centrifugation, but the band was located at a position intermediate to that of a DNA band containing only 15N and a DNA band containing only 14N. After two rounds of replication, two
    • bands of DNA were present. One band was located at a position
    • intermediate to that of a DNA band containing only 15N and a DNA band containing only 14N, while the other band was at a position expected for DNA containing only 14N. These results
    • were consistent with the predictions of semiconservative replication and incompatible with the predictions of conservative and dispersive replication.
  3. DRAW A
    MOLECULE OF DNA UNDERGOING THETA REPLICATION. ON YOUR DRAWING, IDENTIFY (1)
    ORIGIN, (2) POLARITY (5' & 3' ENDS) OF ALL TEMPLATE STRANDS AND NEWLY
    SYNTHESIZED STRANDS, (3) LEADING AND LAGGING STRANDS, (4) KAZAKI FRAGMENTS, AND
    (5) LOCATION OF PRIMERS.
    
  4. DRAW A MOLECULE OF DNA UNDERGOING ROLLING-CIRCLE
    REPLICATION. ON YOUR DRAWING, IDENTIFY (1) ORIGIN, (2) POLARITY (5' &
    3' ENDS) OF ALL TEMPLATE STRANDS AND NEWLY SYNTHESIZED STRANDS, (3)
    LEADING AND LAGGING STRANDS, (4) KAZAKI FRAGMENTS, AND (5) LOCATION OF
    PRIMERS.
  5. DRAW A MOLECULE OF DNA UNDERGOING EUKARYOTIC
    LINEAR REPLICATION. ON YOUR DRAWING, IDENTIFY (1) ORIGIN, (2) POLARITY (5'
    & 3' ENDS) OF ALL TEMPLATE STRANDS AND NEWLY SYNTHESIZED STRANDS, (3)
    LEADING AND LAGGING STRANDS, (4) KAZAKI FRAGMENTS, AND (5) LOCATION OF
    PRIMERS.
  6. WHAT SUBSTRATES ARE USED IN THE DNA SYNTHESIS
    REACTION?
  7. The substrates for DNA synthesis are the four types of deoxyribonucleoside triphosphates: deoxyadenosine triphosphate, deoxyguanosine triphosphate, deoxycytosine triphosphate, and deoxythymidine triphosphate. The substrates for DNA synthesis are the four types of deoxyribonucleoside triphosphates: deoxyadenosine triphosphate, deoxyguanosine
    triphosphate, deoxycytosine triphosphate, and deoxythymidine triphosphate.
  8. LIST THE DIFFERENT PROTEINS AND ENZYMES TAKING
    PART IN BACTERIAL REPLICATION. GIVE THE FUNCTION OF EACH IN THE
    REPLICATION PROCESS.
    • DNA polymerase III is the primary replication polymerase. It
    • elongates a new

    nucleotide strand from the 3¢–OH of the primer.

    • DNA polymerase I removes the RNA nucleotides of the primers and
    • replaces them

    with DNA nucleotides.

    • DNA ligase connects Okazaki fragments by sealing nicks in the sugar
    • phosphate

    backbone.

    • DNA primase synthesizes the RNA primers that provide the 3¢–OH
    • group needed

    for DNA polymerase III to initiate DNA synthesis.

    • DNA helicase unwinds the double helix by breaking the hydrogen
    • bonding between

    the two strands at the replication fork.

    • DNA gyrase reduces DNA supercoiling and torsional strain that is
    • created ahead of

    • the replication fork by making double-stranded breaks in the DNA
    • and passing

    • another segment of the helix through the break before resealing it.
    • Gyrase is also

    called topoisomerase II.

    • Initiator proteins bind to the replication origin and unwind short
    • regions of DNA.

    • Single-stranded binding protein (SSB protein) stabilizes
    • single-stranded DNA prior

    • to replication by binding to it, thus preventing the DNA from
    • pairing with complementary sequences.
  9. WHAT SIMILARITIES AND DIFFERENCES EXIST IN THE
    ENZYMATIC ACTIVITIES OF DNA POLYMERASES I, II, AND III: WHAT IS THE
    FUNCTION OF EACH TYPE OF DNA POLYMERASE IN BACTERIAL CELLS?
    • Each of the three DNA
    • polymerases has a 5¢ to 3¢ polymerase activity. They differ in their exonuclease activities.
    • DNA polymerase I has a 3¢ to 5¢ as well as a 5¢ to 3¢ exonuclease activity.
    • DNA polymerase II and DNA polymerase III have only a 3' to 5' exonuclease activity.

    (1) DNA polymerase I carries out proofreading. It also removes and replaces the RNA primers used to initiate DNA synthesis.

    • (2) DNA polymerase II functions as a DNA repair polymerase. It
    • restarts replication after DNA damage has halted replication. It has proofreading activity.

    (3) DNA polymerase III is the primary replication enzyme and also has a proofreading function in replication.
  10. WHY IS PRIMASE REQUIRED FOR REPLICATION?
    • Primase is a DNA-dependent
    • RNA polymerase. Primase synthesizes the short RNA
    • molecules, or primers, that have a free 3'–OH to which DNA
    • polymerase can attach deoxyribonucleotides in replication initiation. The DNA polymerases require a free 3'–OH to which they add nucleotides, and therefore they cannot initiate replication.

    Primase does not have this requirement.
  11. WHAT 3 MECHANISMS ENSURE THE ACCURACY OF
    REPLICATION IN BACTERIA?
    (1) Highly accurate nucleotide selection by the DNA polymerases when pairing bases.

    (2) The proofreading function of DNA polymerase, which removes incorrectly inserted bases.

    (3) A mismatch repair apparatus that repairs mistakes after replication is complete.
  12. IN WHAT WAYS IS EUKARYOTIC REPLICATION SIMILAR
    TO BACTERIAL REPLICATION, AND IN WHAT WAYS IS IT DIFFERENT?
    • Eukaryotic and bacterial replication of DNA replication share some basic principles:
    • (1) Semiconservative replication.
    • (2) Replication origins serve as starting points for replication.
    • (3) Short segments of RNA called primers provide a 3¢–OH for DNA polymerases to begin synthesis of the new strands.
    • (4) Synthesis occurs in a 5' to 3' direction.
    • (5) The template strand is read in a 3' to 5' direction.
    • (6) Deoxyribonucleoside triphosphates are the substrates.
    • (7) Replication is continuous on the leading strand and
    • discontinuous on the lagging strand.
    • Eukaryotic DNA replication differs from bacterial replication in
    • that:
    • (1) It has multiple origins of replications per chromosome.
    • (2) It has several different DNA polymerases with different
    • functions.
    • (3) Immediately following DNA replication, assembly of nucleosomes takes place.
  13. OUTLINE IN WORDS AND PICTURES HOW TELOMERES AT
    THE ENDS OF EUKARYOTIC CHROMOSOMES ARE REPLICATED
    • For DNA polymerases to work, they need the presence of a 3' OH group to which to add a nucleotide. At the ends of the chromosomes when the RNA primer is removed, there is no adjacent 3' OH group to which to add a nucleotide, thus no
    • nucleotides are added leaving a gap at the end of the chromosome.
    • Telomerase can extend the single stranded protruding end by pairing with the overhanging 3' end of the DNA and adding a repeated sequence of nucleotides. In the absence of telomerase, DNA polymerase will be unable to add nucleotides to the
    • end of the strand. After multiple rounds of replication without a functional telomerase the chromosome will become progressively shorter.
  14. What are some of the enzymes taking part in
    recombination in E. coli, and what roles do they play?
    (1) RecBCD protein unwinds double-stranded DNA and can cleave polynucleotide strands.

    (2) RecA protein allows a single strand to invade a double-stranded DNA.

    • (3) RuvA and RuvB proteins promote branch migration during
    • homologous recombination.

    • (4) RuvC protein is resolvase, a protein that resolves the Holliday
    • structure by cleavage of the DNA.

    • (5) DNA ligase repairs nicks or cuts in the DNA generated during
    • recombination
  15. Phosphorous is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. coli in a medium containing nonradioactive phosphorous for many generations. A sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32P).
    Samples of the bacteria are removed immediately after the transfer
    and after one and two rounds of replication. What will be the distribution of radioactivity in the DNA of the bacteria in each sample? Will radioactivity be detected in neither, one, or both
    strands of the DNA?
    • In the initial sample removed immediately after transfer, no 32P
    • should be incorporated into the DNA because replication in the medium containing 32P has not yet occurred. After one round of replication in the 32P containing medium, one strand of each newly synthesized DNA molecule will contain 32P, while the
    • other strand will contain only nonradioactive phosphorous. After two rounds of replication in the 32P containing medium, 50% of the DNA molecules will have 32P in both strands, while the remaining 50% will contain 32P in one strand and nonradioactive
    • phosphorous in the other strand.
  16. A circular molecule of DNA contains 1 million base pairs. If DNA synthesis at a replication fork occurs at a rate of 100,000 nucleotides per minute, how long will theta replication require to completely replicate the molecule, assuming that theta replication is bidirectional? How long will replication of this circular chromosome take by rolling-circle replication? Ignore replication of the displaced strand in rolling circle replication.
    • In bidirectional replication there are two replication forks, each
    • proceeding at a rate of 100,000 nucleotides per minute. Therefore, it would require 5 minutes for the circular DNA molecule to be replicated by bidirectional replication because
    • each fork could synthesize 500,000 nucleotides (5 minutes × 100,000 nucleotides per minute) within the time period. Because rolling-circle replication is unidirectional and thus has only one replication fork, 10 minutes will be required to replicate the entire circular molecule.
  17. A bacterium synthesizes DNA at each replication
    fork at a rate of 1000 nucleotides per second. If this bacterium completely replicates its circular chromosome by theta replication in 30 minutes, how many base pairs of DNA will its chromosome
    contain?
    Each replication complex is synthesizing DNA at each fork at a rateof 1000 nucleotides per second. So for each second, 2000 nucleotides arebeing synthesizedby both forks (1000 nucleotides / second × 2 forks = 2000nucleotides / second) or120,000 nucleotides per minute. If the bacterium requires 30minutes to replicate its chromosome, then the size of the chromosome is 3,600,00 nucleotides(120,000nucleotides / minute × 30 minutes = 3,600,000). How would DNA replication be affected in a cellthat is lacking topoisomerase?Topoisomerase II or gyrase reduces the positive supercoiling ortorsional strain that develops ahead of the replication fork due to the unwinding of the double helix. If the topoisomerase activity was lacking, then the torsional strainwould continue to increase, making it more difficult to unwind the double helix.Ultimately, the increasing strain would lead to an inhibition of the replicationfork movement.
  18. Draw an RNA nucleotide and a DNA nucleotide,highlighting the differences. Howis the structure of RNA similar to that of DNA? How is it different?
    • RNA and DNA are polymers of nucleotides that are held together by phosphodiester bonds. An RNA nucleotide contains a ribose sugar, whereas a DNA nucleotide contains a deoxyribose sugar. Also, the pyrimidine base uracil is found in RNA butthymine is not. DNA, however, contains thymine but not uracil.Finally, an RNA polynucleotide is typically single-stranded even though RNA molecules can pair with other complementary sequences. DNA molecules are almost always double-stranded.
  19. What are the major classes of cellular RNA?
    Where would you expect to find each class of RNA within eukaryotic cells?
    • Cellular RNA molecules are made up of six classes:
    • (1) Ribosomal RNA, or rRNA, is found in the cytoplasm.
    • (2) Transfer RNA, or tRNA, is found in the cytoplasm.
    • (3) Messenger RNA is found in the cytoplasm (however, pre-mRNA is found only in the nucleus).
    • (4) Small nuclear RNA, or snRNA, is found in the nucleus as part of riboproteins called snrps.
    • (5) Small nucleolar RNA, snoRNA, is found in the nucleus.
    • (6) Small cytoplasmic RNA, or scRNA, is found in the cytoplasm.
  20. Why is DNA more stable than RNA?
    The presence of the free 2' OH in the ribose sugar makes RNA more susceptible to degradation under alkaline conditions. DNA molecules contain thesugar deoxyriboseand lack the 2' OH found in ribose sugars, so DNA is more stable.
  21. What parts of DNA make up a transcription unit?
    Draw and label a typicaltranscription unit in a bacterial cell.
  22. What is the substrate for RNA synthesis?
    How is this substrate modified and joined together to produce an RNA molecule?

    Four ribonucleoside triphosphates serve as the substrate for RNA
    synthesis:

    adenosine triphosphate, guanosine triphosphate, cytosine
    triphosphate, and uridine

    monophosphate. The enzyme RNA polymerase uses a DNA polynucleotide
    strand as

    a template to synthesize a complementary RNA polynucleotide strand.
    The

    nucleotides are added to the RNA polynucleotide strand, one at
    time, at the 3'–OH of

    the RNA molecule. As each nucleoside triphosphate is added to the
    growing

    polynucleotide chain, two phosphates are removed from the 5' end of
    the nucleotide.

    The remaining phosphate is linked to the 3'–OH of the RNA molecule
    to form the

    phosphodiester bond.


    Describe the structure of bacterial RNA
    polymerase.


    Bacterial RNA polymerase consists of several polypeptides. The RNA
    polymerase

    core enzyme is composed of four polypeptide subunits: two copies of
    the alpha

    subunit, the beta subunit, and the beta prime subunit. The addition
    of a sigma factor to

    the core enzyme forms the RNA polymerase holoenzyme.


    Give the names of the three RNA polymerases
    found in eukaryotic cells and the types


    of RNA they transcribe.

    (1) RNA polymerase I transcribes rRNA.

    (2) RNA polymerase II transcribes pre-mRNA and some snRNAs.

    (3) RNA polymerase III transcribes small RNA molecules such as 5S
    rRNA,

    tRNAs, and some small nuclear RNAs.




    What are the three basic stages of
    transcription? Describe what happens at each stage.


    (1) Initiation: Transcription proteins assemble at the promoter to
    form the basal

    transcription apparatus and begin synthesis of RNA.

    (2) Elongation: RNA polymerase moves along the DNA template in a 3'
    to 5'

    direction unwinding the DNA and synthesizing RNA in a 5' to 3'
    direction.

    (3) Termination: Synthesis of RNA is terminated, and the RNA
    molecule separates

    from the DNA template.


    Draw and label a typical bacterial promoter.
    Include any common consensus


    sequences.
    Four ribonucleoside triphosphates serve as the substrate for RNA synthesis: adenosine triphosphate, guanosine triphosphate, cytosinetriphosphate, and uridinemonophosphate. The enzyme RNA polymerase uses a DNA polynucleotide strand as a template to synthesize a complementary RNA polynucleotide strand.The nucleotides are added to the RNA polynucleotide strand, one attime, at the 3'–OH ofthe RNA molecule. As each nucleoside triphosphate is added to thegrowingpolynucleotide chain, two phosphates are removed from the 5' end ofthe nucleotide.The remaining phosphate is linked to the 3'–OH of the RNA moleculeto form the phosphodiester bond.
  23. Describe the structure of bacterial RNA polymerase.
    Bacterial RNA polymerase consists of several polypeptides. The RNApolymerase core enzyme is composed of four polypeptide subunits: two copies ofthe alphasubunit, the beta subunit, and the beta prime subunit. The addition of a sigma factor to the core enzyme forms the RNA polymerase holoenzyme.
  24. Give the names of the three RNA polymerasesfound in eukaryotic cells and the types of RNA they transcribe.
    • (1) RNA polymerase I transcribes rRNA.
    • (2) RNA polymerase II transcribes pre-mRNA and some snRNAs.
    • (3) RNA polymerase III transcribes small RNA molecules such as 5SrRNA,tRNAs, and some small nuclear RNAs.
  25. What are the three basic stages oftranscription? Describe what happens at each stage.
    • (1) Initiation: Transcription proteins assemble at the promoter to form the basal transcription apparatus and begin synthesis of RNA.
    • (2) Elongation: RNA polymerase moves along the DNA template in a 3'to 5'direction unwinding the DNA and synthesizing RNA in a 5' to 3'direction.
    • (3) Termination: Synthesis of RNA is terminated, and the RNAmolecule separates from the DNA template
  26. Draw and label a typical bacterial promoter.Include any common consensus sequences
    • The typical bacterial promoter consists of the –35 and –10
    • consensus sequences.

    • An upstream element rich in AT sequences is found only in some
    • bacterial promoters

    and is located upstream of the –35 consensus sequence
  27. What are the two basic types of terminators
    found in bacterial cells? Describe the structure of each.
    • The two basic types of terminators in bacterial cells arerho-independent and rhodependent terminators.
    • Rho-independent terminators consist of inverted repeats that can form a hairpin structure. Immediately following the inverted repeats is a string of six adenine nucleotides.
    • Rho-dependent terminators require the interaction of the protein rho with RNA polymerase. Two features are typical forrho-dependenttermination:
    • (1) variable DNA sequences that cause RNA polymeraseto pause during transcription and
    • (2) upstream from variable region lies DNAsequence that encodes RNA devoid of secondary structure, which also serves as the rhobinding site.
  28. How is the process of transcription in eukaryotic cells different from that in bacterial cells?
    Eukaryotic transcription requires the action of three RNA polymerases. Each type of polymerase recognizes and transcribes from different types of promoters. Binding to the promoter and initiation from the promoter requires the action of many protein factors; different promoters require different sets of protein factors. The RNA molecule produced by transcription in eukaryotic cells usuallyrequires extensiveprocessing, such as the addition of a 5' cap, a 3' poly(A) tail,and the removal ofintrons prior to becoming functional. Bacterial promoters tend tobe more uniform in composition, and only one RNA polymerase does transcription. Bacterial RNAs are typically functional once transcription has taken place.
  29. Compare the roles of general transcriptionfactors and transcriptional activator proteins.
    General transcription factors form the basal transcription apparatus together with RNA polymerase and are needed to initiate minimal levels of transcription.Transcriptional activator proteins bring about higher levels of transcription by stimulating the assembly of the basal transcriptional apparatus at the start site.
  30. How are the processes of transcription in archaeans and eukaryotes different? Howare they similar?
    A comparison of transcription between eukaryotes and archaea shows that transcription in eukaryotes shares more similarities with archaeal transcription than they do with transcription in bacteria. Organisms in domain archaeause a single RNA polymerase for transcription while eukaryotes have three RNA polymerases for transcribing nuclear genes. Archaea lack a nucleus so transcriptionoccurs within the cytoplasm. The RNA polymerase from archaea is similar to RNApolymerases of eukaryotes. Archaea possess a TATA-binding protein, which is also found in eukaryotes and is critical for all three nuclear RNA polymerases.The TATA-bindingprotein in archaea binds the TATA box with the assistance of TFIIB,which is found in eukaryotes as well.
  31. An RNA molecule has the following percentages of bases: A = 23%, U = 42%,C = 21%, G = 14%.
    a. Is this RNA single-stranded or double-stranded? How can you tell?
    b. What would be the percentages of bases in the template strand ofthe DNA thatcontains the gene for this RNA?
    • a. The RNA molecule is likely to be single-stranded. If the moleculewas doublestranded,we would expect nearly equal percentages of adenine and uracil, aswellas equal percentages of guanine and cytosine. In this RNA molecule,the percentages of these potential base pairs are not equal, so the molecule is single stranded .
    • b. Because the DNA template strand is complementary to the RNAmolecule, we would expect equal percentages for bases in the DNA complementaryto the RNA bases. Therefore, in the DNA we would expect A = 42%, T = 23%, C =14%,and G = 21%.
  32. The following diagram represents DNA that is part of the RNA-coding sequence of atranscription unit. The bottom strand is the template strand. Givethe sequence foundon the RNA molecule transcribed from this DNA and label the 5' and3' ends of the RNA.
    • 
    • The RNA
    • molecule would be complementary to the template strand, contain uracil,

    • and be
    • synthesized in an antiparallel fashion. The sequence would be:

    • 5'–A U A
    • G G C G A U G C C A–3'.

    • The RNA
    • strand contains the same sequence as the nontemplate DNA strand except

    • that the
    • RNA strand contains uracil in place of thymine.
  33. 19. Write the consensus sequence for the following set of
    nucleotide sequences:
    A G G A G T T

    A G C T A T T

    T G C A A T A
    A C G A A A A
    T C C T A A T
    T G C A A T T
    • The consensus sequence is identified by determining which nucleotide is used most frequently at each position. For the two nucleotides that occur at an equal frequency at the first position, both are listed at that position in the sequence and identified by a slash mark:
    • T/A G C A A T T
  34. RNA molecules have three phosphates at their 5'end, but DNA molecules never do.Explain this difference.
    During initiation of DNA replication, DNA nucleotide triphosphatesmust be attached to a 3'–OH of a RNA molecule by DNA polymerase. This process removes the terminal two phosphates of the nucleotides. If the RNA molecule is subsequently removed, then a single phosphate would remain at the 5' end of theDNA molecule.RNA polymerase does not require the 3'–OH to initiate synthesis of RNA molecules.Therefore, the 5' end of a RNA molecule will retain all three of the phosphates from the original nucleotide triphosphate substrate.
  35. Write out a hypothetical sequence of bases that might be found in the first 20 nucleotides of a promoter of a bacterial gene. Include both strands of DNA and label the 5' and 3' ends of both strands. Be sure to include the start site for transcription and any consensus sequences found in the promoter.
    • The –10
    • region, or Pribnow Box, has the consensus sequence of TATAAT.

    • However,
    • few bacterial promoters actually contain the exact consensus sequence. A

    • common
    • sequence at the transcription start site is 5'–CAT–3' with transcription

    • beginning
    • at the “A.”
  36. What would be the most likely effect of a mutation at the following locations in an E. coli gene?
    a. –8
    b. -35
    c,-20
    d. start site
    • a. A mutation at the –8 position would probably affect the –10 consensus sequence(TATAAT), which is centered on position –10. This consensus sequence is necessary for binding of RNA polymerase. A mutation in there would most likely decrease transcription.
    • b. A mutation in the –35 region could affect the binding of the sigmafactor to the promoter. Deviations away from the consensus typically reduce transcription, so transcription is likely to be reduced or inhibited.
    • c.The –20 region is located between the consensus sequences of an E.coli promoter. Although the holoenzyme may cover the site, it is unlikely that amutation will have any effect on transcription.
    • d. A mutation in the start site would have little effect on transcription.
  37. The following diagram represents a transcription unit on a DNA molecule.
    a. Assume that this DNA molecule is from a bacterial cell. Draw in the approximate location of the promoter and terminator for this transcription unit.
    b. Assume that this DNA molecule is from a eukaryotic cell. Draw in the approximate location of an RNA polymerase II promoter.
    • a.
    • b.
  38. The following DNA nucleotides are found near the
    end of a bacterial transcription unit. Find the terminator in this sequence:
    3'–AGCATACAGCAGACCGTTGGTCTGAAAAAAGCATACA–5'
    a. Mark the point at which transcription will terminate.
    b. Is this terminator intrinsic or rho-dependent?
    c. Draw a diagram of the RNA that will be transcribed from this DNA, including its nucleotide sequence and any secondary structures that form.
    • a.
    • b. Based on the potential hairpin structure that can form and the run of U’s that willbesynthesized in the RNA, this terminator is an intrinsic terminator.
    • c.
  39. Computer programmers, working with molecular geneticists, have developed computer programs that can identify genes within long stretches of DNA sequences. Imagine that you are working with a computer programmer on such a project. Based on what you know about the process of transcription, what sequences might you suggest the computer program look for to identify the beginning and end of a gene?
    The sequences recommended to the programmer will depend on the type of organism whose DNA is being studied. However, both bacteria and eukaryotic organisms have both promoter and termination consensus DNA sequences that could be recognized by the programmer’s computer program as potential sites of interest for genes. The program should be able to recognize the consensus sequences as well as deviations from the consensus since the consensus sequence may only rarel be present. The identification by the program of both promoter sequences, terminator sequences and/or potential cleavage sites in a region of the DNA should provide evidence of a gene’s presence in that region. For bacterial promoters, the –5 sequences and –10 sequences would be good starting points. At the end of bacterial genes areterminators. The program should be able to find inverted repeats and strings of A’s,which are indicative of Rho-independent termination. For rho-dependent termination sequences may prove more difficult to identify. The presence ofboth the promotersequences and a termination sequence in a region of DNA would beevidence of abacterial gene. Similar promoter and termination sequences as wel las 3' cleavage site sequences in eukaryotic cells could also be used. For RNApolymerase II genes,identifying sequences of the core promoter such as the TFIIB recognition element(BRE) sequences, the TATA box, the initiator element sequence, and the downstreamcore promoter element sequences might indicate the beginning of a gene. Although not every RNA polymerase II promoter contains each sequence, the core promoter will contain at least one of these sequences. Transcription terminators at the ends of RNA polymerase II transcribed genes could prove difficult to recognize. However,another potential sequence for identification would be the cleavage site consensus sequences at the 3' regions of the gene.
  40. Elaborate repair mechanisms are associated with replication to prevent permanent mutations in DNA, yet no similar repair is associated with transcription. Can you think of a reason for these differences in replication andtranscription? (Hint: Thinkabout the relative effects of a permanent mutation in a DNAmolecule compared withone in an RNA molecule.)
    RNA molecules are constantly being synthesized and subsequently degraded.Typically, RNA is produced regularly by new transcription. A defective RNA molecule would soon be replaced and would only briefly affect a cell. Furthermore,the defective RNA would not be passed to offspring of that cell. DNA mutations,however, are permanent. All RNAs transcribed from that gene would be affected by a change in the DNA sequence. The mutation in the DNA molecule would be passed to progeny cells and propagated. Finally, replication occurs only once during the cell cycle. Once the mutation occurs, there is no subsequent replication that could produce the correct sequence.
  41. What are some characteristics of introns?
    Eukaryotic genes commonly contain introns. However, introns are rare in bacterial genes.The number of introns found in an organism’s genome is typically related to complexity—more complex organisms possess more introns. Introns typically do not encode proteins and are usually larger in size than exons.
  42. What are the three principal elements in mRNAsequences in bacterial cells?
    • (1) The 5' untranslated region, which contains the Shine-Dalgarnosequence
    • (2) The protein-encoding region
    • (3) The 3' untranslated region
  43. a. What is the 5' cap?
    b. How is the 5' cap added to eukaryotic pre-mRNA?
    c. What is the function of the 5' cap?
    • a.The 5' end of eukaryotic mRNA is modified by the addition of the 5'cap. The capconsists of an extra guanine nucleotide linked 5' to 5' to the mRNAmolecule. Thisnucleotide is methylated at position 7 of the base. The ribosesugars of adjacent basesmay be methylated at the 2' –OH
    • b. Initially, the terminal phosphate of the three 5' phosphates linkedto the end of themRNA molecule is removed. Subsequently, a guanine nucleotide isattached to the 5'end of the mRNA using a 5' to 5' phosphate linkage. Next, a methylgroup is attachedto position 7 of the guanine base. Ribose sugars of adjacentnucleotides may also bemethylated, but at the 2'–OH
    • c. CAP binding proteins recognize the 5' cap and stimulate binding of the ribosome to the 5' cap and to the mRNA molecule. The 5' cap may also increase mRNA stability in the cytoplasm. Finally, the 5' cap is needed for efficient splicing of the intron that is nearest the 5' end of the pre-mRNA molecule.
  44. How is the poly(A) tail added to pre-mRNA? Whatis the purpose of the poly(A) tail?
    Initially, a complex consisting of several proteins forms on the 3'UTR of the pre-mRNAmolecule. Cleavage and polyadenylation specificity factor (CPSF) bind to the AAUAAAconsensus sequence, which is located upstream of the 3' cleavagesite. Another protein,cleavage stimulation factor (CsTF), binds downstream of the cleavage site. Two cleavage factors (CFI and CFII) and polyadenylate polymerase (PAP) also become part of the complex. Once the complex has formed, the pre-mRNA is cleaved. CsTF and the two cleavage factors leave the complex. PAP adds approximately 10adenine nucleotides tothe 3' end of the pre-mRNA molecule. The addition of the short poly(A) tail allows for the binding of the poly(A) binding protein (PABII) to the tail.PABII increases the rate of polyadenylation, which subsequently allows for more PABII protein to bind the tail. The presence of the poly(A) tail increases the stability of themRNA molecule through the interaction of proteins at the poly(A) tail.
  45. What makes up the spliceosome? What is the function of the spliceosome?
    The spliceosome consists of five small ribonucleoproteins (snRNPs).Each snRNP is composed of multiple proteins and a single small nuclear RNAmolecule or snRNA. These nRNPs are identified by which snRNA (U1, U2, U3, U4, U5, or U6)each contains. Splicing of pre-mRNA nuclear introns takes place within the spliceosome.
  46. Explain the process of pre-mRNA splicing innuclear genes.
    Removal of an intron from the pre-mRNA requires the assembly of the spliceosome complex on the pre-mRNA, cleavage at both the 5' and 3'splice sites of the intron, and two transesterification reactions ultimately leading to the joining of the two exons.Initially, snRNP U1 binds to the 5' splice site through complementary base pairing of the U1 snRNA. Next, snRNP U2 binds to the branch point within the intron. The U5 andU4–U6 complex joins the spliceosome, resulting in the looping of the intron so that the branch point and 5' splice site of the intron are now adjacent to each other. U1 and U4 now disassociate from the spliceosome, and the spliceosome is activated. The pre-mRNA is then cleaved at the 5' splice site, producing an exon with a 3'–OH. The 5' end of the intron folds back and forms 5'–2' phosphodiester linkage through the first transesterification reaction with the adenine nucleotide at the branch point of the intron.This looped structure is called the lariat. Next, the 3' splicesite is cleaved and then immediately ligated to the 3'–OH of the first exon through the second transesterification reaction. Thus, the exons are now joined and the intron has been excised.
  47. Summarize the different types of processing that can take place in pre-mRNA.
    • Several modifications to pre-mRNA take place to produce maturemRNA.
    • (1) Addition of the 5' cap to the 5' end of the pre-mRNA.
    • (2) Cleavage of the 3' end of a site downstream of the AAUAAAconsensus sequence ofthe last exon.
    • (3) Addition of the poly(A) tail to the 3' end of the mRNA immediately following cleavage.
    • (4) Removal of the introns (splicing).
  48. At the beginning of the chapter, we considered Duchenne muscular dystrophy and the dystrophin gene. We learned that the gene causing Duchenne musculardystrophyen compasses more than 2 million nucleotides, but less than 1% of the gene encodes the protein dystrophin. On the basis of what you now know about gene structure and RNA processing in eukaryotic cells, provide a possible explanation for the large size of the dystrophin gene.
    The large size of the dystrophin gene is likely due to the presence of many intervening sequences or introns within the coding region of the gene. Excision of the introns through RNA splicing yields the mature mRNA that encodes the dystrophinprotein.
  49. How do the mRNA of bacterial cells and the pre-mRNA of eukaryotic cells differ? the mature mRNAs of bacterial and eukaryotic cells differ?
    Bacterial mRNA is translated immediately upon being transcribed.Eukaryotic pre-mRNA must be processed. Bacterial mRNA and eukaryotic pre-mRNA have similarities in structure. Each has a 5' untranslated region as well as a 3'untranslated region. Both also have protein-coding regions. However, the protein-coding region of the pre-mRNA is disrupted by introns. The eukaryotic pre-mRNA must be processed to produce the mature mRNA. Eukaryotic mRNA has a 5' cap and a poly(A) tail, unlike bacterial mRNAs.Bacterial mRNA also contains the Shine-Dalgarno consensus sequence.Eukaryotic mRNA does not have the equivalent.
  50. Draw a typical eukaryotic gene and the pre-mRNAand mRNA derived from it. Assume that the gene contains three exons. Identify the following items and, for each item, give a brief description of its function
    a. 5' untranslated region
    b. Promoter
    c. AAUAAA consensus sequence
    d. Transcription start site.
    e. 3' untranslated region.
    f. Introns
    g. Exons
    h. Poly(A) tail
    • a. The 5' untranslated region lies upstream of the translation startsite. In bacteria, theribosome binding site or Shine-Dalgarno sequence is found withinthe 5' untranslatedregion. However, eukaryotic mRNA does not have the equivalentsequence, and a eukaryotic ribosome binds at the 5' cap of the mRNA molecule.
    • b. The promoter is the DNA sequence that the transcription apparatus recognizes and binds to initiate transcription.
    • c. The AAUAAA consensus sequence lies downstream of the coding regionof thegene. It determines the location of the 3' cleavage site in thepre-mRNA molecule.
    • d. The transcription start site begins the coding region of the geneand is located 25 to 30nucleotides downstream of the TATA box
    • e. The 3' untranslated region is a sequence of nucleotides at the 3'end of the mRNA thatis not translated into proteins. However, it does affect thetranslation of the mRNAmolecule as well as the stability of the mRNA
    • f. Introns are noncoding sequences of DNA that intervene within codingregions of a gene
    • g. Exons are transcribed regions that are not removed in intronprocessing. They includethe 5'UTR, coding regions that are translated into amino acidsequences, and the3'UTR
    • h. A poly(A) tail is added to the 3' end of the pre-mRNA. It affects mRNA stability.i. 5' capThe 5' cap functions in the initiation of translation and mRNAstability.
  51. How would the deletion of the Shine-Dalgarno
    sequence affect a bacterial mRNA?
    In bacteria, the small ribosomal subunit binds to the Shine-Dalgarno sequence to initiate translation. If the Shine-Dalgarno sequence is deleted, then translation initiation cannot take place, preventing protein synthesis.
  52. How would the deletion of the following sequences or features most likely affect a eukaryotic pre-mRNA?
    a. AAUAAA consensus sequence
    b. 5' cap
    c. Poly(A) tail
    • a. The deletion of the AAUAA consensus sequence would prevent bindingof thecleavage and polydenylation factor (CPSF), thus resulting in nocleavage orpolyadenylation of the pre-mRNA. This would affect the stabilityand translation ofthe mRNA.
    • b. The deletion of the 5' cap would most likely prevent splicing ofthe intron that isnearest to the 5' cap. Ultimately, elimination of the cap will affect the stability of the pre-mRNA as well as its ability to be translated.
    • c. Polyadenylation increases the stability of the mRNA. If eliminated from the premRNA,then the mRNA would be degraded quickly by nucleases in the cytoplasm.

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