Deck 14: Dna and the Gene: Synthesis and Repair
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Deck 14: Dna and the Gene: Synthesis and Repair
1

Figure 14.1
Refer to Figure 14.1.What bases will be added to the primer as DNA replication proceeds? The bases should appear in the order that they will be added starting at the 5' end of the new strand.
A)C,A,G,C,A,G,A
B)T,C,T,G,C,T,G
C)A,G,A,C,G,A,C
D)G,T,C,G,T,C,T
C
2
Who performed the classic experiments that proved DNA was copied by semiconservative replication?
A)Watson and Crick
B)Meselson and Stahl
C)Hershey and Chase
D)Franklin and Wilkins
A)Watson and Crick
B)Meselson and Stahl
C)Hershey and Chase
D)Franklin and Wilkins
B
3
How does the simple primary and secondary structure of DNA hold the information needed to code for the many features of multicellular organisms?
A)The hydrogen bonding among backbone constituents carries coded information.
B)The base sequence of DNA carries all the information needed to code for proteins.
C)The width of the double helix changes at each gene due to differences in hydrogen bonds.
D)The amino acids that make up the DNA molecule contain the information needed to make cellular proteins.
A)The hydrogen bonding among backbone constituents carries coded information.
B)The base sequence of DNA carries all the information needed to code for proteins.
C)The width of the double helix changes at each gene due to differences in hydrogen bonds.
D)The amino acids that make up the DNA molecule contain the information needed to make cellular proteins.
B
4
What is a major difference between eukaryotic DNA replication and prokaryotic DNA replication?
A)Prokaryotic replication does not require a primer.
B)Prokaryotic chromosomes have a single origin of replication,while eukaryotic chromosomes have multiple origins of replication.
C)DNA polymerase III of eukaryotes has both endonuclease and exonuclease activity,while that of prokaryotes has only exonuclease activity.
D)DNA polymerases of prokaryotes can add nucleotides to both 3′ and 5′ ends of DNA strands,while those of eukaryotes function only in the 5′→3′ direction.
A)Prokaryotic replication does not require a primer.
B)Prokaryotic chromosomes have a single origin of replication,while eukaryotic chromosomes have multiple origins of replication.
C)DNA polymerase III of eukaryotes has both endonuclease and exonuclease activity,while that of prokaryotes has only exonuclease activity.
D)DNA polymerases of prokaryotes can add nucleotides to both 3′ and 5′ ends of DNA strands,while those of eukaryotes function only in the 5′→3′ direction.
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5

Figure 14.2
Identify the lagging strand during duplication of DNA starting from a double helix in Figure 14.2.
A)a
B)b
C)c
D)d
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6
What is the difference between the leading strand and the lagging strand in DNA replication?
A)The leading strand is synthesized in the 3′→5′ direction in a discontinuous fashion,while the lagging strand is synthesized in the 5′→3′ direction in a continuous fashion.
B)The leading strand requires an RNA primer,whereas the lagging strand does not.
C)The leading strand is synthesized continuously in the 5′→3′ direction,while the lagging strand is synthesized discontinuously in the 5′→3′ direction.
D)There are different DNA polymerases involved in elongation of the leading strand and the lagging strand.
A)The leading strand is synthesized in the 3′→5′ direction in a discontinuous fashion,while the lagging strand is synthesized in the 5′→3′ direction in a continuous fashion.
B)The leading strand requires an RNA primer,whereas the lagging strand does not.
C)The leading strand is synthesized continuously in the 5′→3′ direction,while the lagging strand is synthesized discontinuously in the 5′→3′ direction.
D)There are different DNA polymerases involved in elongation of the leading strand and the lagging strand.
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7
DNA is synthesized through a process known as _____.
A)semiconservative replication
B)conservative replication
C)translation
D)transcription
A)semiconservative replication
B)conservative replication
C)translation
D)transcription
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8
What can you infer from the information presented in this table?

A)DNA polymerase I and DNA polymerase III are the same enzyme found in different organisms.
B)DNA polymerase I and DNA polymerase III have different functions.
C)The sliding clamp molecule is a ribozyme.
D)Topoisomerase is involved in proofreading activity.

A)DNA polymerase I and DNA polymerase III are the same enzyme found in different organisms.
B)DNA polymerase I and DNA polymerase III have different functions.
C)The sliding clamp molecule is a ribozyme.
D)Topoisomerase is involved in proofreading activity.
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9
What provides the energy for the polymerization reactions in DNA synthesis?
A)ATP
B)DNA polymerase
C)breaking the hydrogen bonds between complementary DNA strands
D)the deoxyribonucleotide triphosphate substrates
A)ATP
B)DNA polymerase
C)breaking the hydrogen bonds between complementary DNA strands
D)the deoxyribonucleotide triphosphate substrates
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10
Semiconservative replication involves a template.What is the template?
A)single-stranded binding proteins
B)DNA polymerase contains the template needed.
C)one strand of the DNA molecule
D)an RNA molecule
A)single-stranded binding proteins
B)DNA polymerase contains the template needed.
C)one strand of the DNA molecule
D)an RNA molecule
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11
DNA contains the template needed to copy itself,but as you learned in Chapter 4,it has no catalytic activity.What catalyzes the formation of phosphodiester bonds between adjacent nucleotides in the DNA polymer being formed?
A)ribozymes
B)DNA polymerase
C)ATP
D)deoxyribonucleotide triphosphates
A)ribozymes
B)DNA polymerase
C)ATP
D)deoxyribonucleotide triphosphates
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12
Watson and Crick elucidated the structure of DNA in 1953.Their research built on and helped explain the findings of other scientists before them,including _____.
A)X-ray diffraction studies by Rosalind Franklin and Maurice Wilkins.
B)Replication studies by Meselson and Stahl.
C)Okazaki's work on lagging-strand DNA fragments.
D)All of the above were important considerations in the elucidation of the structure of DNA.
A)X-ray diffraction studies by Rosalind Franklin and Maurice Wilkins.
B)Replication studies by Meselson and Stahl.
C)Okazaki's work on lagging-strand DNA fragments.
D)All of the above were important considerations in the elucidation of the structure of DNA.
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13
The bacterial transduction experiments done by Hershey and Chase,and the bacterial transformation experiments done by Griffith,supported the same conclusion,which was _____.
A)pathogenic molecules affect the health of all living organisms
B)DNA is the molecular substance of genetic inheritance
C)RNA is the molecular substance of genetic inheritance
D)lateral gene transfer is not possible between bacteria
A)pathogenic molecules affect the health of all living organisms
B)DNA is the molecular substance of genetic inheritance
C)RNA is the molecular substance of genetic inheritance
D)lateral gene transfer is not possible between bacteria
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14
In the polymerization of DNA,a phosphodiester bond is formed between a phosphate group of the nucleotide being added and _____ of the last nucleotide in the polymer.
A)the 5' phosphate
B)C₆
C)the 3′ OH
D)a nitrogen from the nitrogen-containing base
A)the 5' phosphate
B)C₆
C)the 3′ OH
D)a nitrogen from the nitrogen-containing base
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15
Hershey and Chase set out to determine what molecule served as the unit of inheritance.They completed a series of transduction experiments in which E.coli was infected by a T2 virus.Which molecular component of the T2 virus actually ended up inside the cell?
A)protein
B)RNA
C)ribosome
D)DNA
A)protein
B)RNA
C)ribosome
D)DNA
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16
The fact that within a double-stranded DNA molecule,adenine forms two hydrogen bonds with thymine and cytosine forms three hydrogen bonds allows _____.
A)variable width of the double helix
B)complementary base pairing
C)secondary structure of a DNA molecule
D)constant width of the double helix
A)variable width of the double helix
B)complementary base pairing
C)secondary structure of a DNA molecule
D)constant width of the double helix
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17
Put the following steps of DNA replication in chronological order.
1)Single-stranded binding proteins attach to DNA strands.
2)Hydrogen bonds between base pairs of antiparallel strands are broken.
3)Primase binds to the site of origin.
4)DNA polymerase binds to the template strand.
5)An RNA primer is created.
A)1,2,3,4,5
B)2,1,3,5,4
C)3,2,1,5,4
D)3,1,2,4,5
1)Single-stranded binding proteins attach to DNA strands.
2)Hydrogen bonds between base pairs of antiparallel strands are broken.
3)Primase binds to the site of origin.
4)DNA polymerase binds to the template strand.
5)An RNA primer is created.
A)1,2,3,4,5
B)2,1,3,5,4
C)3,2,1,5,4
D)3,1,2,4,5
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18

Figure 14.2
In Figure 14.2,which is the template strand?
A)a
B)b
C)c
D)d
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19

Figure 14.3
Refer to Figure 14.3.Which structure is responsible for stabilizing DNA in its single-stranded form?
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20

Figure 14.3
Refer to Figure 14.3.Which of the structures in the figure breaks hydrogen bonds between complementary bases?
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21
In a healthy cell,the rate of DNA repair is equal to the rate of DNA mutation.When the rate of repair lags behind the rate of mutation,what is a possible fate of the cell?
A)The cell can be transformed to a cancerous cell.
B)RNA may be used instead of DNA as inheritance material.
C)It will become embryonic.
D)DNA synthesis will continue in an attempt to repair itself.
A)The cell can be transformed to a cancerous cell.
B)RNA may be used instead of DNA as inheritance material.
C)It will become embryonic.
D)DNA synthesis will continue in an attempt to repair itself.
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22
Telomere shortening is a problem in which types of cells?
A)prokaryotes
B)eukaryotes
C)both prokaryotes and eukaryotes
A)prokaryotes
B)eukaryotes
C)both prokaryotes and eukaryotes
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23
In the mismatch repair process,enzyme complexes replace bases that were incorrectly inserted into the newly synthesized DNA strand.To function,they must be able to distinguish between the parent DNA strand and the new strand.How is this accomplished?
A)The new strand contains the diphosphate bases.
B)The parent strand is methylated.
C)The new strand contains ribose sugars.
D)The parent strand is usually radiolabeled.
A)The new strand contains the diphosphate bases.
B)The parent strand is methylated.
C)The new strand contains ribose sugars.
D)The parent strand is usually radiolabeled.
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24
Which of the following cells do not have active telomerase activity?
A)most normal somatic cells
B)most normal germ cells
C)most cancer cells
A)most normal somatic cells
B)most normal germ cells
C)most cancer cells
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25
If DNA repair mechanisms fail,which of the following is not a direct result?
A)a mutation
B)a defective enzyme
C)cancer
D)DNA polymerase is slower at replicating DNA.
A)a mutation
B)a defective enzyme
C)cancer
D)DNA polymerase is slower at replicating DNA.
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26
Recent studies have shown that xeroderma pigmentosum (an error in the nucleotide excision repair process)can result from mutations in one of seven genes.What can you infer from this finding?
A)There are seven genes that produce the same protein.
B)These seven genes are the most easily damaged by ultraviolet light.
C)There are several enzymes involved in the nucleotide excision repair process.
D)These mutations have resulted from translocation of gene segments.
A)There are seven genes that produce the same protein.
B)These seven genes are the most easily damaged by ultraviolet light.
C)There are several enzymes involved in the nucleotide excision repair process.
D)These mutations have resulted from translocation of gene segments.
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27
What is a telomere?
A)the mechanism that holds two sister chromatids together
B)DNA replication during telophase
C)the site of origin of DNA replication
D)the ends of linear chromosomes
A)the mechanism that holds two sister chromatids together
B)DNA replication during telophase
C)the site of origin of DNA replication
D)the ends of linear chromosomes
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28
Researchers found E.coli that had mutation rates 100 times higher than normal.What is a possible explanation for these results?
A)The single-stranded binding proteins were malfunctioning.
B)There were one or more mismatches in the RNA primer.
C)The proofreading mechanism of DNA polymerase was not working properly.
D)The DNA polymerase was unable to add bases to the 3′ end of the growing nucleic acid chain.
A)The single-stranded binding proteins were malfunctioning.
B)There were one or more mismatches in the RNA primer.
C)The proofreading mechanism of DNA polymerase was not working properly.
D)The DNA polymerase was unable to add bases to the 3′ end of the growing nucleic acid chain.
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29
Which of the following would be least likely to cause DNA damage in an individual suffering from xeroderma pigmentosum?
A)natural lighting
B)low-level ultraviolet lights
C)incandescent lightbulbs
D)reflected sunlight
A)natural lighting
B)low-level ultraviolet lights
C)incandescent lightbulbs
D)reflected sunlight
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30
DNA replication is highly accurate.It results in about one mistake per billion nucleotides.For the human genome,how often would errors occur?
A)on average,once or twice in the lifetime of an individual
B)on average,6 times each time the entire genome of a cell is replicated
C)on average,once every 6 cell divisions
D)on average,once a lifetime in 10% of the population
A)on average,once or twice in the lifetime of an individual
B)on average,6 times each time the entire genome of a cell is replicated
C)on average,once every 6 cell divisions
D)on average,once a lifetime in 10% of the population
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31
Which of the following are important in reducing the errors in DNA replication in E.coli organisms?
A)proofreading activity of the epsilon subunit of DNA polymerase III
B)mismatch repair
C)nucleotide excision repair
D)All of the above minimize errors in DNA replication in E.coli.
A)proofreading activity of the epsilon subunit of DNA polymerase III
B)mismatch repair
C)nucleotide excision repair
D)All of the above minimize errors in DNA replication in E.coli.
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32
Hereditary nonpolyposis colorectal cancer (HNPCC)is an inherited disorder.The genetic defect identified is an error in the mismatch repair mechanism.Which of the following would be an expected result of this mutation?
A)increased rate of errors by wild-type DNA polymerase
B)increased rate of formation of pyrimidine dimers
C)increased rate of repair of pyrimidine dimers
D)decreased ability to repair certain DNA mutations
A)increased rate of errors by wild-type DNA polymerase
B)increased rate of formation of pyrimidine dimers
C)increased rate of repair of pyrimidine dimers
D)decreased ability to repair certain DNA mutations
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33
The epsilon (ε)subunit of DNA polymerase III of E.coli has exonuclease activity.How does it function in the proofreading process?
A)The epsilon subunit can remove a mismatched nucleotide.
B)The epsilon subunit excises a segment of DNA around the mismatched base.
C)The epsilon subunit can recognize which strand is the template or parent strand,and which is the new strand of DNA.
D)It adds nucleotide triphosphates to the 3′ end of the growing DNA strand.
A)The epsilon subunit can remove a mismatched nucleotide.
B)The epsilon subunit excises a segment of DNA around the mismatched base.
C)The epsilon subunit can recognize which strand is the template or parent strand,and which is the new strand of DNA.
D)It adds nucleotide triphosphates to the 3′ end of the growing DNA strand.
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34
Which of the following agents are unlikely to cause mutations in DNA?
A)aflatoxins that are found in moldy grains
B)free radicals that are formed as by-products of aerobic respiration
C)ultraviolet radiation from sunlight
D)radiation from an incandescent lightbulb
A)aflatoxins that are found in moldy grains
B)free radicals that are formed as by-products of aerobic respiration
C)ultraviolet radiation from sunlight
D)radiation from an incandescent lightbulb
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35
Bodnar et al (1998)used telomerase to extend the life span of normal human cells.Telomere shortening puts a limit on the number of times a cell can divide.How might adding telomerase affect cellular aging?
A)Telomerase will speed up the rate of cell proliferation.
B)Telomerase ensures that the ends of the chromosomes are accurately replicated and eliminates telomere shortening.
C)Telomerase shortens telomeres and thus delays cellular aging.
D)Telomerase would have no effect on cellular aging.
A)Telomerase will speed up the rate of cell proliferation.
B)Telomerase ensures that the ends of the chromosomes are accurately replicated and eliminates telomere shortening.
C)Telomerase shortens telomeres and thus delays cellular aging.
D)Telomerase would have no effect on cellular aging.
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36
In humans,xeroderma pigmentosum is a disorder of the nucleotide excision repair mechanism.These individuals are unable to repair DNA damage caused by ultraviolet light.Which of the following are the most prominent types of mutations in individuals suffering from xeroderma pigmentosum?
A)mismatch errors
B)telomere shortening
C)methylation of purines
D)pyrimidine dimers
A)mismatch errors
B)telomere shortening
C)methylation of purines
D)pyrimidine dimers
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37
What are pyrimidine dimers?
A)pyrimidines (C and T)that have gained an extra nitrogen-containing ring structure
B)pyrimidines on antiparallel DNA strands that form complementary base pairs
C)adjacent pyrimidines on the same DNA strand that join by covalent bonding
D)pyrimidines formed by demethylation of purines
A)pyrimidines (C and T)that have gained an extra nitrogen-containing ring structure
B)pyrimidines on antiparallel DNA strands that form complementary base pairs
C)adjacent pyrimidines on the same DNA strand that join by covalent bonding
D)pyrimidines formed by demethylation of purines
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