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Essential Cell Biology 4th Edition by Bruce Alberts – Test Bank
ESSENTIAL CELL BIOLOGY, FOURTH EDITION
CHAPTER 6: DNA REPLICATION, REPAIR, AND
© 2014 GARLAND SCIENCE PUBLISHING
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.
6-2 DNA replication is considered semiconservative because ____________________________.
(a) after many rounds of DNA replication, the original DNA double helix is still
(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.
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
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.
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|(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
|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
|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?|
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(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
(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?
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.
|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.|
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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
(a) There will be only one leading strand and one lagging strand produced using
(b) The leading and lagging strands compose one half of each newly synthesized
(c) The DNA replication machinery can assemble at multiple places on this
(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?
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|(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
(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
(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.
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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
(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
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?
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|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.
______ 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
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?
(b) sliding clamp
(c) single-strand binding protein
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6-32 Use the components in the list below to label the diagram of a replication fork in
- DNA polymerase
- single-strand binding protein
- Okazaki fragment
- sliding clamp
- RNA primer
- DNA helicase
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.
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6-34 Indicate whether the following statements are true or false. If a statement is false,
explain why it is false.
- The repair polymerase is the enzyme that proofreads the newly synthesized
strands to ensure the accuracy of DNA replication.
- There is a single enzyme that degrades the RNA primers and lays down the
corresponding DNA sequence behind it.
- DNA ligase is required to seal the sugar–phosphate backbone between all the
DNA fragments on the lagging strand.
- 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?
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
(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
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
(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
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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
6-41 Which of the choices below represents the correct way to repair the mismatch
shown in 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.
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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
(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.
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
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|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
|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
|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
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6-52 Which of the following statements is not an accurate statement about thymine
(a) Thymine dimers can cause the DNA replication machinery to stall.
(b) Thymine dimers are covalent links between thymidines on opposite DNA
(c) Prolonged exposure to sunlight causes thymine dimers to form.
(d) Repair proteins recognize thymine dimers as a distortion in the DNA
6-53 Indicate whether the following statements are true or false. If a statement is false,
explain why it is false.
- Ionizing radiation and oxidative damage can cause DNA double-strand
- After backbone are
damaged DNA has been repaired, nicks in the phosphate
maintained as a way to identify the strand that was repaired.
- Depurination of DNA is a rare event that is caused by ultraviolet irradiation.
- 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
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
(a) inherited a cancer-causing gene that suffered a mutation in an ancestor’s
(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
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.
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Which mutant is most likely to be defective in the DNA repair
- 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?
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
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
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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
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
|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
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Which of the following is the correct order of events during homologous
(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
mosomes during meiosis, those noncrossover chromosomes will
- 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.
- 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.
- What does this indicate about your resemblance to your grandfather and
6-64 Indicate whether the following statements are true or false. If a statement is false,
explain why it is false.
- 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.
- 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.
- The 3′ overhang “invades” the homologous DNA duplex, which can be used
as a primer for the repair DNA polymerase.
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- The DNA template used to repair the broken strand is the homologous
chromosome inherited from the other parent.
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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
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.
- 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
- 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.
- 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.
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6-9 A. The 3′ end. DNA polymerase can add nucleotides only to the 3′-OH end of a
nucleic acid chain.
- Both, as a result of the bidirectional nature of chromosomal replication.
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
6-15 See 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|
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|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.|
|6-20||(a) Because helicase unwinds the two DNA template strands, replication of both strands depends upon the activity of helicase at the time of initiation.|
|6-23||(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.
|6-25||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.|
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- 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.
- 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
___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.
- 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.
Page 21 of 25
6-32 See 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.
|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.|
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).|
|6-44||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.
|6-50||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%.
Page 23 of 25
6-53 A. True.
- 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.
- False. Depurination occurs constantly in our cells through spontaneous
hydrolysis of the bond linking the DNA base to the deoxyribose sugar.
- 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
- 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
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.
Page 24 of 25
|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.
|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.|
|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|
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.|
Page 1 of 29
- RNA capping
From DNA to RNA
7-3 Consider two genes that are next to each other on a chromosome, as arranged in Figure
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
(b) it uses the same enzyme as that used to synthesize RNA primers during DNA
(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.
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 _____.
|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 ____.
|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 ____.
|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
|7-10||Match the following structures with their names.|
Page 3 of 29
|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
(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
(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?
|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
7-18 You have a piece of DNA that includes the following sequence:
Page 5 of 29
|Which of the following RNA molecules could be transcribed from this piece of DNA? (a) 5′-UAUCCGUAAGCUAGGCCUAUGCUA-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.|
|Given this information, a student claims that the RNA produced from this DNA is:|
|Give two reasons why this answer is incorrect.|
|7-20||You have a segment of DNA that contains the following sequence:|
|You know that the RNA transcribed from this segment contains the following sequence:|
|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:|
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
|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?
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
(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
|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
(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
Page 7 of 29
|(d) RNA polymerase must be phosphorylated at its C-terminal tail for transcription to
|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?
|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
(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
|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.
Page 8 of 29
|(d) The ribosome will skip over intron sequences when translating RNA into protein.|
(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
|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.|
|(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.|
|Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome?
|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?|
|7-40||Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to carry? (a) lysine|
Page 10 of 29
|(b) glutamic acid
|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.|
|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.|
|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?
|7-44||One strand of a section of DNA isolated from the bacterium E. coli reads:|
|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
|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|
Page 11 of 29
|listed in Figure Q7-45. What is the most likely cause of this variation in protein|
|(a) a mutation in the DNA coding for the protein
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.
|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?
Page 12 of 29
|(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
(c) methionine would be incorporated into proteins at some positions where valine
(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
|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
Page 14 of 29
|7-56||Figure Q7-56 shows an mRNA molecule.|
|A. Match the labels given in the list below with the label lines in Figure Q7-56. (a)
(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?|
|(a)||The protein binds to the small ribosomal subunit and increases the rate of initiation of translation.|
Page 15 of 29
|(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
(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
(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
Page 16 of 29
7-60 Which of the following methods is not used by cells to regulate the amount of a protein in
(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?
(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
(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
(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
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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.
|From this information, which of the following peptides can be produced from poly UAUC?
|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?
|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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|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.|
|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.|
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|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.|
|7-15||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
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:
|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.
|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.
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.|
|7-34||(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.|
|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.|
B. Two. (There are three potential reading frames for each RNA. In this case, they
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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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
|7-47||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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|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.|
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7-66 Three possible answers are:
- 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
- DNA is double-stranded and therefore the complementary strand provides a
template from which damage can be repaired accurately.
- 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
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
|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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