Download Chapter 25 DNA Metabolism

yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Chapter 25 DNA Metabolism
Multiple Choice Questions
1. DNA replication
Page: 977 Difficulty: 2 Ans: C
The Meselson-Stahl experiment established that:
DNA polymerase has a crucial role in DNA synthesis.
DNA synthesis in E. coli proceeds by a conservative mechanism.
DNA synthesis in E. coli proceeds by a semiconservative mechanism.
DNA synthesis requires dATP, dCTP, dGTP, and dTTP.
newly synthesized DNA in E. coli has a different base composition than the preexisting
2. DNA replication
Page: 978 Difficulty: 2 Ans: D
When a DNA molecule is described as replicating bidirectionally, that means that it has two:
independently replicating segment.
replication forks.
termination points.
3. DNA replication
Page: 979 Difficulty: 2 Ans: D
An Okazaki fragment is a:
fragment of DNA resulting from endonuclease action.
fragment of RNA that is a subunit of the 30S ribosome.
piece of DNA that is synthesized in the 3'  5' direction.
segment of DNA that is an intermediate in the synthesis of the lagging strand.
segment of mRNA synthesized by RNA polymerase.
4. DNA replication
Pages: 979-984 Difficulty: 2 Ans: C
Which one of the following statements about enzymes that interact with DNA is true?
A) E. coli DNA polymerase I is unusual in that it possesses only a 5'  3' exonucleolytic
B) Endonucleases degrade circular but not linear DNA molecules.
C) Exonucleases degrade DNA at a free end.
D) Many DNA polymerases have a proofreading 5'  3' exonuclease.
E) Primases synthesize a short stretch of DNA to prime further synthesis.
5. DNA replication
Page: 982 Difficulty: 2 Ans: C
E. coli DNA polymerase III:
can initiate replication without a primer.
is efficient at nick translation.
is the principal DNA polymerase in chromosomal DNA replication.
represents over 90% of the DNA polymerase activity in E. coli cells.
requires a free 5'-hydroxyl group as a primer.
6. DNA replication
Page: 982 Difficulty: 2 Ans: D
The 5'  3' exonuclease activity of E. coli DNA polymerase I is involved in:
formation of a nick at the DNA replication origin.
formation of Okazaki fragments.
proofreading of the replication process.
removal of RNA primers by nick translation.
sealing of nicks by ligase action.
7. DNA replication
Pages: 982-983 Difficulty: 2 Ans: C
Prokaryotic DNA polymerase III:
A) contains a 5'  3' proofreading activity to improve the fidelity of replication.
B) does not require a primer molecule to initiate replication.
C) has a  subunit that acts as a circular clamp to improve the processivity of DNA
D) synthesizes DNA in the 3'  5' direction.
E) synthesizes only the leading strand; DNA polymerase I synthesizes the lagging strand.
8. DNA replication
Page: 988 Difficulty: 2 Ans: E
At replication forks in E. coli:
DNA helicases make endonucleolytic cuts in DNA.
DNA primers are degraded by exonucleases.
DNA topoisomerases make endonucleolytic cuts in DNA.
RNA primers are removed by primase.
RNA primers are synthesized by primase.
9. DNA recombination
Page: 1012
Difficulty: 3 Ans: C
The bacteriophage  can lysogenize after infecting a bacterium, i.e. integrate into the host
bacterial chromosome by site-specific recombination, and may reside there for many
generations before an excision event regenerates the viral genome in an infective form.
Which one of the following is not a component of these events?
Excision requires two host proteins and two virally-encoded proteins.
Integration requires a viral-specific protein, called integrase.
RecA protein is required to catalyze the insertional recombination event.
The excision event relies on different sequences than the integration event.
The virus and the host DNAs share a 15 bp “core” region of perfect homology.
Short Answer Questions
10. DNA replication
Pages: 977-978 Difficulty: 2
Describe briefly how equilibrium density gradient centrifugation was used to demonstrate
that DNA replication in E. coli is semiconservative.
Ans: Equilibrium density gradient centrifugation separates DNA molecules of slightly
different buoyant density. For example, molecules containing 15N-labeled (“heavy”) DNA are
separable from identical molecules containing 14N (“light”) DNA. Meselson and Stahl grew
E. coli for many generations in a medium containing 15N, producing cells in which all DNA
was heavy. These cells were transferred to a medium containing 14N, and the buoyant density
of their DNA was determined (by equilibrium density gradient centrifugation) after 1, 2, 3,
etc., generations. After one generation, all DNA was of a density intermediate between fully
heavy and fully light, indicating that each double-stranded DNA molecule had one heavy
(parental) and one light (newly synthesized) strand; replication was semiconservative. (See
Fig. 25-2, p. 977.)
11. DNA replication
Page: 979 Difficulty: 2
The DNA below is replicated from left to right. Label the templates for leading strand and
lagging strand synthesis.
Ans: The polarity of the strands indicates that the top strand is the template for lagging strand
synthesis, and the bottom strand is the template for leading strand synthesis. (See Fig. 25-4,
p. 979.)
12. DNA replication
Page: 979 Difficulty: 2
All known DNA polymerases catalyze synthesis only in the 5'  3' direction. Nevertheless,
during semiconservative DNA replication in the cell, they are able to catalyze the synthesis
of both daughter chains, which would appear to require synthesis in the 3'  5' direction.
Explain the process that occurs in the cell that allows for synthesis of both daughter chains by
DNA polymerase.
Ans: During DNA replication, one strand is synthesized continuously and the other is
synthesized by a discontinuous mechanism. The daughter chain, which appears to be
growing in the 3'  5' direction (the “lagging strand”), is actually being synthesized by
continual initiation of new chains and their elongation in the 5'  3' direction.
13. DNA replication
Pages: 979, 987 Difficulty: 2
What is an Okazaki fragment? What enzyme(s) is (are) required for its formation in E. coli?
Ans: An Okazaki fragment is an intermediate in DNA replication in E. coli. It is a short
fragment of newly synthesized DNA, attached to the 3' end of a short RNA primer. Such
fragments are produced by the combined action of primase (part of the primosome) and DNA
polymerase III during replication of the lagging strand. (See Fig. 25-13, p. 987.)
14. DNA replication
Pages: 980-981 Difficulty: 2
A suitable substrate for DNA polymerase is shown below. Label the primer and template,
and indicate which end of each strand must be 3' or 5'.
To observe DNA synthesis on this substrate in vitro, what additional reaction components
must be added?
Ans: The top strand (the primer) has its 5' end to the left; the bottom (template) strand has
the opposite polarity. For DNA synthesis with this substrate in vitro, one would have to add
DNA polymerase, the four deoxynucleoside triphosphates, Mg2+, and a suitable buffer.
15. DNA replication
Pages: 981, 984 Difficulty: 2
All known DNA polymerases can only elongate a preexisting DNA chain (i.e., require a
primer), but cannot initiate a new DNA chain. Nevertheless, during semiconservative DNA
replication in the cell, entirely new daughter DNA chains are synthesized. Explain the
process that occurs in the cell that allows for the synthesis of daughter chains by DNA
Ans: In the cell, initiation of DNA chains occurs via the synthesis of an RNA primer by an
RNA polymerase type of enzyme (primase). This primer is elongated by DNA polymerase
to produce the daughter DNA chain. The RNA is removed by 5' exonucleolytic hydrolysis
before replication is completed.
16. DNA replication
Pages: 985-991 Difficulty: 2
DNA replication in E. coli begins at a site in the DNA called the (a) ___________. At the
replication fork the (b) ___________ strand is synthesized continuously while the (c)
_________ strand is synthesized discontinuously. On the strand synthesized discontinuously,
the short pieces are called (d) ____________ fragments. An RNA primer for each of the
fragments is synthesized by an enzyme called (e) __________, and this RNA primer is
removed after the fragment is synthesized by the enzyme (f) ___________, using its (g)
_____________ activity. The nicks left behind in this process are sealed by the enzyme (h)
Ans: (a) origin; (b) leading; (c) lagging; (d) Okazaki; (e) primase; (f) DNA pol I; (g) 5'  3'
exonuclease; (h) DNA ligase
17. DNA repair
Page: 993-994
Difficulty: 2
The high fidelity of DNA replication is due primarily to immediate error correction by the 3'
—> 5' exonuclease (proofreading) activity of the DNA polymerase. Some incorrectly paired
bases escape this proofreading, and further errors can arise from challenges to the chemical
integrity of the DNA. List the four classes of repair mechanisms that the cell can use to help
correct such errors.
Ans: The four classes are listed in Table 25-5 (p. 994), and consist of (1) mismatch repair,
(2) base-excision repair, (3) nucleotide-excision repair, and (4) direct repair.
18. DNA Recombination
Pages: 1004-1006 Difficulty: 2
Outline the four key features of the current model for homologous recombination during
meiosis in a eukaryotic cell.
Ans: (1) Homologous chromosomes are aligned. (2) A double-strand break is enlarged by an
exonuclease, leaving a single-strand extension with a free 3' hydroxyl end. (3) The exposed
3' ends invade the homologous intact duplex DNA, followed by branch migration to create a
Holliday junction. (4) Cleavage of the two crossover products creates the two recombinant
products. (See Fig. 25-33, p. 1006.)
19. DNA recombination
Pages: 1013-1014 Difficulty: 2
What distinguishes the simple from the complex class of bacterial transposon?
Ans: The simple class called insertion sequences contains only the information needed for
transposition and the genes for proteins (transposases) that carry out the process. Those in
the class of complex transposons carry additional genes, such as those for antibiotic
resistance, a property they confer upon any host bacterium that harbors them.
20. DNA recombination
Pages: 1014-1016 Difficulty: 2
Briefly describe the role of recombination in the generation of antibody (immunoglobin)
Ans: The genes for immunoglobin polypeptide chains are divided into segments, with
multiple versions of each segment (which code for slightly different amino acid sequences).
Recombination results in the joining of individual versions of each segment to generate a
complete gene. Antibody diversity results from the very large number of different
combinations that are possible. (See Fig. 25-46, p. 1015.)
Related documents