Deck 7: Dna: Structure and Replication
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Deck 7: Dna: Structure and Replication
1
Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus)and bacterial cells.In this experiment,how is transformation (altered phenotype)displayed?
A)The bacteria are transformed (by phage DNA)into virus synthesizing cells.
B)The bacteria are transformed from live to dead by the virus.
C)The bacteria are transformed from rough to smooth by the virus.
D)The bacteria are transformed into a virus.
E)Transformation is not exhibited by a virus-infected bacterial cell.
A)The bacteria are transformed (by phage DNA)into virus synthesizing cells.
B)The bacteria are transformed from live to dead by the virus.
C)The bacteria are transformed from rough to smooth by the virus.
D)The bacteria are transformed into a virus.
E)Transformation is not exhibited by a virus-infected bacterial cell.
A
2
Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus)and bacterial cells.In these experiments,they used radioactivity to label nucleic acids (in this case DNA)and proteins.What do radioactive sulfur and phosphate specifically label?
A)Sulfur labels DNA,and phosphate labels protein.
B)Sulfur labels both protein and DNA.
C)Phosphate labels both protein and DNA.
D)Sulfur labels protein,and phosphate labels DNA.
E)None of the answer options are correct.
A)Sulfur labels DNA,and phosphate labels protein.
B)Sulfur labels both protein and DNA.
C)Phosphate labels both protein and DNA.
D)Sulfur labels protein,and phosphate labels DNA.
E)None of the answer options are correct.
D
3
Chargaff's rules do NOT hold for which of the following genome types?
A)yeast
B)bacteria
C)single-stranded DNA virus
D)invertebrates
E)Archaeobacteria
A)yeast
B)bacteria
C)single-stranded DNA virus
D)invertebrates
E)Archaeobacteria
C
4
Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus)and bacterial cells.If phage are labeled with radioactive sulfur and allowed to infect bacterial cells,the radioactive sulfur will be localized to:
A)the inside of infected cells (in phage DNA).
B)the outside of infected cells (in phage ghosts).
C)in newly synthesized phage viruses in the host bacterial cell.
D)No radioactivity will remain after infection.
E)The sulfur will be metabolically consumed,and therefore the radioactivity will be destroyed.
A)the inside of infected cells (in phage DNA).
B)the outside of infected cells (in phage ghosts).
C)in newly synthesized phage viruses in the host bacterial cell.
D)No radioactivity will remain after infection.
E)The sulfur will be metabolically consumed,and therefore the radioactivity will be destroyed.
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5
Based on its association with chromosomes,this was once considered the hereditary molecule before the discovery of DNA.
A)phospholipids
B)proteins
C)glycogen
D)cholesterol
E)None of the answer options are correct.
A)phospholipids
B)proteins
C)glycogen
D)cholesterol
E)None of the answer options are correct.
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6
Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus)and bacterial cells.If phage are labeled with radioactive phosphate and allowed to infect bacterial cells,the radioactive phosphate will be localized to:
A)the inside of infected cells (in phage DNA).
B)the outside of infected cells (in phage ghosts).
C)newly synthesized phage viruses in the host bacterial cell.
D)No radioactivity will remain after infection.
E)The phosphate will be metabolically consumed,and therefore the radioactivity will be destroyed.
A)the inside of infected cells (in phage DNA).
B)the outside of infected cells (in phage ghosts).
C)newly synthesized phage viruses in the host bacterial cell.
D)No radioactivity will remain after infection.
E)The phosphate will be metabolically consumed,and therefore the radioactivity will be destroyed.
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7
Which statement below best describes the situation between nucleotides on opposite strands in a DNA molecule?
A)A-T and C-G bonding between opposite strands involves two hydrogen bonds for either pair.
B)A-T and C-G bonding between opposite strands involves three hydrogen bonds for either pair.
C)A-T bonding between opposite strands involves two hydrogen bonds,whereas G-C bonding between strands involves three hydrogen bonds.
D)A-T bonding between opposite strands involves three hydrogen bonds,whereas G-C bonding between strands involves two hydrogen bonds.
A)A-T and C-G bonding between opposite strands involves two hydrogen bonds for either pair.
B)A-T and C-G bonding between opposite strands involves three hydrogen bonds for either pair.
C)A-T bonding between opposite strands involves two hydrogen bonds,whereas G-C bonding between strands involves three hydrogen bonds.
D)A-T bonding between opposite strands involves three hydrogen bonds,whereas G-C bonding between strands involves two hydrogen bonds.
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8
Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus)and bacterial cells.Sulfur was used to label cellular components in this experiment because of its presence in the amino acid(s):
A)histidine.
B)methionine and cysteine.
C)tyrosine.
D)glycine.
E)serine.
A)histidine.
B)methionine and cysteine.
C)tyrosine.
D)glycine.
E)serine.
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9
Evidence of transformation of R strain cells by biomolecules from S strain cells includes:
A)the R strain cells begin to produce high levels of neurotoxin.
B)the transformed R strain cells gain the ability to kill mice.
C)isolation of smooth-looking cells from animals infected with R strain cells that had been incubated with heat-killed S strain cells.
D)the R strain cells produce neurotoxin and gain the ability to kill mice
E)the R strain cells gain the ability to kill mice,and smooth-looking cells were isolated from animals infected with R strain cells previously incubated with heat-killed S strain cells.
A)the R strain cells begin to produce high levels of neurotoxin.
B)the transformed R strain cells gain the ability to kill mice.
C)isolation of smooth-looking cells from animals infected with R strain cells that had been incubated with heat-killed S strain cells.
D)the R strain cells produce neurotoxin and gain the ability to kill mice
E)the R strain cells gain the ability to kill mice,and smooth-looking cells were isolated from animals infected with R strain cells previously incubated with heat-killed S strain cells.
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10
Of the three key building blocks of DNA,which type(s)of building block stabilizes the structure via weak van der Waals interactions (through stacking)and regular hydrogen bonds?
A)phosphate
B)deoxyribose
C)nitrogenous bases
D)both phosphate and nitrogenous bases
E)All of the answer options are correct.
A)phosphate
B)deoxyribose
C)nitrogenous bases
D)both phosphate and nitrogenous bases
E)All of the answer options are correct.
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11
Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design.The key difference was:
A)purifying S strain DNA and RNA from other biomolecules.
B)systematically eliminating the impact of classes of S strain biomolecules using enzymatic digestion before mixing with R strain live cells.
C)injecting S strain DNA directly into R strain cells and observing the change in phenotype.
D)Avery used Escherichia coli rather than Streptococcus as it is easier to transform.
E)All of the answer options are correct.
A)purifying S strain DNA and RNA from other biomolecules.
B)systematically eliminating the impact of classes of S strain biomolecules using enzymatic digestion before mixing with R strain live cells.
C)injecting S strain DNA directly into R strain cells and observing the change in phenotype.
D)Avery used Escherichia coli rather than Streptococcus as it is easier to transform.
E)All of the answer options are correct.
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12
Mendel identified evidence for the location of genes on separate structures (chromosomes)through his:
A)electrophoresis experiments.
B)purification of chromosomes from pea plants.
C)identification of dominant and recessive alleles of genes.
D)identification of independent assortment of distinct genes during meiosis.
E)use of high resolution microscopic techniques.
A)electrophoresis experiments.
B)purification of chromosomes from pea plants.
C)identification of dominant and recessive alleles of genes.
D)identification of independent assortment of distinct genes during meiosis.
E)use of high resolution microscopic techniques.
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13
Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design.They found that the enzyme ______________ was effective at destroying the transforming capacity of S strain biomolecules.
A)protease (protein destruction)
B)RNase (RNA destruction)
C)DNase (DNA destruction)
D)polysaccharide-destroying enzymes
E)lipase (lipid destruction)
A)protease (protein destruction)
B)RNase (RNA destruction)
C)DNase (DNA destruction)
D)polysaccharide-destroying enzymes
E)lipase (lipid destruction)
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14
Smooth (S)and rough (R)strains of Streptococcus pneumonia are distinct because of their ability to cause illness (death in rodents).This trait is controlled by genes that regulate:
A)the production of a protective polysaccharide coat (capsule)around each cell.
B)the production of lethal toxins.
C)the speed of bacterial growth.
D)the metabolic rate of microbial glycolysis.
E)the bacterial ability to penetrate through the membrane of animal cells.
A)the production of a protective polysaccharide coat (capsule)around each cell.
B)the production of lethal toxins.
C)the speed of bacterial growth.
D)the metabolic rate of microbial glycolysis.
E)the bacterial ability to penetrate through the membrane of animal cells.
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15
What structural feature of DNA suggests a possible method of its replication?
A)The nucleotides between nucleotide strands are complementarily paired.
B)The hydrogen bonds between A-T and G-C are the same on all DNA molecules.
C)The DNA strands run anti-parallel.
D)The phosphate bonds in the backbone of all DNA molecules.
A)The nucleotides between nucleotide strands are complementarily paired.
B)The hydrogen bonds between A-T and G-C are the same on all DNA molecules.
C)The DNA strands run anti-parallel.
D)The phosphate bonds in the backbone of all DNA molecules.
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16
Of the three key building blocks of DNA,which type(s)of building block is/are structurally different in RNA molecules?
A)phosphate
B)deoxyribose
C)nitrogenous bases
D)both deoxyribose and nitrogenous bases
E)All of the answer options are correct.
A)phosphate
B)deoxyribose
C)nitrogenous bases
D)both deoxyribose and nitrogenous bases
E)All of the answer options are correct.
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17
Which statement best describes the arrangement of components in a DNA molecule?
A)Nucleotides are located toward the outside of the strands and phosphates toward the interior.
B)Nucleotides are located toward the inside of the strands and the phosphates toward the outside.
C)Nucleotides,as well as the phosphates,are located toward the inside of the strands.
D)Nucleotides,as well as the phosphates,are located toward the outside of the strands.
A)Nucleotides are located toward the outside of the strands and phosphates toward the interior.
B)Nucleotides are located toward the inside of the strands and the phosphates toward the outside.
C)Nucleotides,as well as the phosphates,are located toward the inside of the strands.
D)Nucleotides,as well as the phosphates,are located toward the outside of the strands.
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18
In the classic experiment by Griffith,evidence of the action of a hereditary biomolecule was identified by:
A)growth of the bacterial cells.
B)transformation (phenotypic change)of the R strain by S strain biomolecules.
C)rapid death of mice following injection with bacteria.
D)isolation of pure DNA from the S strain cells.
E)the ability of heat to destroy deadly biomolecules.
A)growth of the bacterial cells.
B)transformation (phenotypic change)of the R strain by S strain biomolecules.
C)rapid death of mice following injection with bacteria.
D)isolation of pure DNA from the S strain cells.
E)the ability of heat to destroy deadly biomolecules.
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19
Of the three key building blocks of DNA,which type(s)of building block is/are negatively charged and oriented on the outside of the double helical structure?
A)phosphate
B)deoxyribose
C)nitrogenous bases
D)both phosphate and nitrogenous bases
E)All of the answer options are correct.
A)phosphate
B)deoxyribose
C)nitrogenous bases
D)both phosphate and nitrogenous bases
E)All of the answer options are correct.
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20
Which of the following is/are NOT a key structural property displayed by the heritable (transforming)material?
A)The structure must contain complex information (genetic code).
B)The structure must be composed of rare/unusual elements.
C)The structure must allow or tolerate small changes that facilitate nonlethal variation.
D)The structure must be enzymatically active,able to mediate cellular metabolism.
E)The structure must be composed of rare/unusual elements and must be enzymatically active,able to mediate cellular metabolism.
A)The structure must contain complex information (genetic code).
B)The structure must be composed of rare/unusual elements.
C)The structure must allow or tolerate small changes that facilitate nonlethal variation.
D)The structure must be enzymatically active,able to mediate cellular metabolism.
E)The structure must be composed of rare/unusual elements and must be enzymatically active,able to mediate cellular metabolism.
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21
When comparing the three key models of DNA replication,the model that included the separation of the two strands of the original DNA (template)and using those strands as templates to synthesize two new DNA strands is called:
A)dispersive replication.
B)conservative replication.
C)semiconservative replication.
D)liberal replication.
E)None of the answer options are correct.
A)dispersive replication.
B)conservative replication.
C)semiconservative replication.
D)liberal replication.
E)None of the answer options are correct.
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22
If Escherichia coli,grown for a period of time in 15N,is transferred to 14N for one generation of DNA replication,the resulting DNA should have 14N added to all "new" DNA.If conservative replication is occurring,the 14N-containing "new" DNA will compose:
A)one strand of all bacterial chromosomes.
B)both strands of DNA in half of all bacterial chromosomes.
C)regions dispersed throughout all bacterial DNA.
D)none of the new DNA;it will only be found in the old DNA.
E)All of the answer options are correct.
A)one strand of all bacterial chromosomes.
B)both strands of DNA in half of all bacterial chromosomes.
C)regions dispersed throughout all bacterial DNA.
D)none of the new DNA;it will only be found in the old DNA.
E)All of the answer options are correct.
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23
Initiation of replication occurs at an "origin of replication" site that typically includes an AT-rich region.Initiation benefits from these AT-rich regions because:
A)adenine-thymine pairs are held together by two H-bonds,making them easier to separate during unwinding.
B)the AT-rich region recruits DNA polymerase to begin the process of DNA replication.
C)the AT-rich region recruits topoisomerase to begin the process of DNA replication.
D)GC-rich regions are impossible to replicate because of their strong H-bonds.
E)None of the answers is correct.
A)adenine-thymine pairs are held together by two H-bonds,making them easier to separate during unwinding.
B)the AT-rich region recruits DNA polymerase to begin the process of DNA replication.
C)the AT-rich region recruits topoisomerase to begin the process of DNA replication.
D)GC-rich regions are impossible to replicate because of their strong H-bonds.
E)None of the answers is correct.
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24
The following data were obtained from three organisms : an RNA virus,a DNA virus,and a wombat (an Australian mammal).Which is likely the sample from the RNA virus? A.(a)
B)(b)
C)(c)
D)None of the answer options are correct.
B)(b)
C)(c)
D)None of the answer options are correct.
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25
In eukaryotic DNA replication,re-association of histones with newly formed DNA is accomplished by:
A)spontaneous association of histones with DNA.
B)chromatin assembly factor (CAF-1)attaching to the primase enzyme.
C)histones only associating with new DNA after the chromosome is completely replicated.
D)histones not releasing DNA during replication.
E)chromatin assembly factor 1 (CAF-1)and histones binding to the sliding clamp structure.
A)spontaneous association of histones with DNA.
B)chromatin assembly factor (CAF-1)attaching to the primase enzyme.
C)histones only associating with new DNA after the chromosome is completely replicated.
D)histones not releasing DNA during replication.
E)chromatin assembly factor 1 (CAF-1)and histones binding to the sliding clamp structure.
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26
Cell-cycle progress enables the initiation of genome replication by:
A)activating DNA polymerase III activity directly.
B)providing Cdc6 and Cdt1,which aid assembly of initiation components at the origin of replication.
C)activating histone degradation near the origin of replication.
D)activating expression of the ORC protein and helicase proteins.
E)All of the answer options are correct.
A)activating DNA polymerase III activity directly.
B)providing Cdc6 and Cdt1,which aid assembly of initiation components at the origin of replication.
C)activating histone degradation near the origin of replication.
D)activating expression of the ORC protein and helicase proteins.
E)All of the answer options are correct.
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27
Topoisomerase and helicase have distinct functions that include which of the following?
A)Topoisomerase is responsible for unwinding the double helix (separating strands).
B)Helicase is responsible for unwinding the double helix (separating strands).
C)Helicase relieves supercoiling that occurs in front of the replication fork.
D)The enzymes have nearly identical activities but are located at different sites during DNA replication.
E)These enzymes are nearly identical but have been given distinct names.
A)Topoisomerase is responsible for unwinding the double helix (separating strands).
B)Helicase is responsible for unwinding the double helix (separating strands).
C)Helicase relieves supercoiling that occurs in front of the replication fork.
D)The enzymes have nearly identical activities but are located at different sites during DNA replication.
E)These enzymes are nearly identical but have been given distinct names.
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28
Experiments on chromosome structure and function have shown that in eukaryotic chromatids:
A)there are at least 50 DNA molecules per chromatid.
B)there is more than one DNA replicating point per chromatid.
C)DNA molecules duplicate conservatively rather than semiconservatively.
D)DNA segregation to daughter chromatids occurs in a dispersive pattern.
E)most DNA synthesis occurs during the M (mitosis)phase of the cell cycle.
A)there are at least 50 DNA molecules per chromatid.
B)there is more than one DNA replicating point per chromatid.
C)DNA molecules duplicate conservatively rather than semiconservatively.
D)DNA segregation to daughter chromatids occurs in a dispersive pattern.
E)most DNA synthesis occurs during the M (mitosis)phase of the cell cycle.
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29
In 1959,Arthur Kornberg isolated DNA polymerase activity from Escherichia coli cells,revealing its ability to copy DNA by mixing ___________________ in a tube and measuring synthesis of a DNA strand.
A)dATP,dGTP,dUTP,dCTP,template DNA,and pure DNA polymerase enzyme
B)ATP,GTP,TTP,CTP,template DNA,and pure DNA polymerase enzyme
C)dATP,dGTP,dTTP,dCTP,template RNA,and pure RNA polymerase enzyme
D)dATP,dGTP,dTTP,dCTP,template DNA,and pure DNA polymerase enzyme
E)dATP,dGTP,dTTP,dCTP,template DNA,and cell enzymatic extract
A)dATP,dGTP,dUTP,dCTP,template DNA,and pure DNA polymerase enzyme
B)ATP,GTP,TTP,CTP,template DNA,and pure DNA polymerase enzyme
C)dATP,dGTP,dTTP,dCTP,template RNA,and pure RNA polymerase enzyme
D)dATP,dGTP,dTTP,dCTP,template DNA,and pure DNA polymerase enzyme
E)dATP,dGTP,dTTP,dCTP,template DNA,and cell enzymatic extract
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30
Linus Pauling proposed at one time that the DNA molecule was,in fact,a triple helix.If so,how would this situation affect cellular division?
A)The triple-helix DNA could be replicated and passed on to daughter cells as it normally would.
B)The triple-helix DNA could be replicated normally in mitosis but not meiosis.
C)Sperm and eggs produced by triple-helix DNA would be more fertile.
D)The fact that the DNA had three nucleotide strands would make it difficult to replicate the DNA and reassemble the copies.
A)The triple-helix DNA could be replicated and passed on to daughter cells as it normally would.
B)The triple-helix DNA could be replicated normally in mitosis but not meiosis.
C)Sperm and eggs produced by triple-helix DNA would be more fertile.
D)The fact that the DNA had three nucleotide strands would make it difficult to replicate the DNA and reassemble the copies.
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31
The complexity of lagging strand replication is necessary because:
A)there is room for only a single DNA polymerase III enzyme in the replisome.
B)the helicase can only unwind double-helical DNA slowly.
C)as polymerization occurs only in the 5′ to 3′ direction,the lagging strand must be synthesized in consecutive small fragments.
D)the RNA primase works better on the leading strand DNA.
E)DNA polymerase I is more often associated with the lagging strand.
A)there is room for only a single DNA polymerase III enzyme in the replisome.
B)the helicase can only unwind double-helical DNA slowly.
C)as polymerization occurs only in the 5′ to 3′ direction,the lagging strand must be synthesized in consecutive small fragments.
D)the RNA primase works better on the leading strand DNA.
E)DNA polymerase I is more often associated with the lagging strand.
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32
The spontaneous (though infrequent)change of a nitrogenous base to a slightly distinct form is called:
A)ketomerization.
B)tautomerization.
C)methylation.
D)mutagenesis.
E)proton exchange.
A)ketomerization.
B)tautomerization.
C)methylation.
D)mutagenesis.
E)proton exchange.
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33
When replicating the end of a chromosome,the lagging strand cannot copy the last ~10 nucleotides at the end of the chromosome.As a result,chromosomes contain telomere sequences at their ends,which are defined as:
A)special DNA sequences that do not require priming for replication.
B)noncoding,repetitive sequences that can be copied independent of the replisome.
C)noncoding,repetitive sequences that are not replicated during DNA replication.
D)protein-based structures at the ends of chromosomes that protect the chromosome end.
E)All of the answer options are correct.
A)special DNA sequences that do not require priming for replication.
B)noncoding,repetitive sequences that can be copied independent of the replisome.
C)noncoding,repetitive sequences that are not replicated during DNA replication.
D)protein-based structures at the ends of chromosomes that protect the chromosome end.
E)All of the answer options are correct.
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34
Telomerase activity relies upon ________________ for appropriate priming.
A)RNA primase
B)a short,telomeric RNA sequence that is carried within its structure
C)a primosome subunit associated with the telomere structure
D)a unique protein-based primer
E)DNA polymerase III holoenzyme
A)RNA primase
B)a short,telomeric RNA sequence that is carried within its structure
C)a primosome subunit associated with the telomere structure
D)a unique protein-based primer
E)DNA polymerase III holoenzyme
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35
The replisome contains a protein subunit responsible for unwinding the double helix to enable DNA replication.This subunit/enzyme is named:
A)pol III holoenzyme.
B)the beta clamp.
C)primase.
D)ligase.
E)helicase.
A)pol III holoenzyme.
B)the beta clamp.
C)primase.
D)ligase.
E)helicase.
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36
The replisome contains a protein subunit responsible for attaching free ends of DNA on the newly formed strand.This subunit/enzyme is named:
A)pol III holoenzyme.
B)the beta clamp.
C)primase.
D)ligase.
E)helicase.
A)pol III holoenzyme.
B)the beta clamp.
C)primase.
D)ligase.
E)helicase.
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37
If Escherichia coli,grown for a period of time in 15N,is transferred to 14N for one generation of DNA replication,the resulting DNA should have 14N added to all "new" DNA.If semiconservative replication is occurring,the 14N-containing "new" DNA will compose:
A)one strand of all bacterial chromosomes.
B)both strands of DNA in half of all bacterial chromosomes.
C)regions dispersed throughout all bacterial DNA.
D)none of the new DNA;it will only be found in the old DNA.
E)None of the answer options are correct.
A)one strand of all bacterial chromosomes.
B)both strands of DNA in half of all bacterial chromosomes.
C)regions dispersed throughout all bacterial DNA.
D)none of the new DNA;it will only be found in the old DNA.
E)None of the answer options are correct.
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38
When comparing the three key models of DNA replication,the model that included the synthesis of a brand new double-stranded DNA molecule from an original molecule was named:
A)dispersive replication.
B)conservative replication.
C)semiconservative replication.
D)liberal replication.
E)None of the answer options are correct.
A)dispersive replication.
B)conservative replication.
C)semiconservative replication.
D)liberal replication.
E)None of the answer options are correct.
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39
The Meselson-Stahl experiment made clear predictions regarding experimental outcomes if dispersive,conservative,or semiconservative DNA replication was occurring in their Escherichia coli cells.To begin the experiment,cells were grown for a period of time in media containing __________ to label the chromosome.
A)heavy nitrogen,15N
B)radioactive phosphate,32P
C)chemically modified nitrogenous bases
D)only sulfur and no nitrogen
E)None of the answer options are correct.
A)heavy nitrogen,15N
B)radioactive phosphate,32P
C)chemically modified nitrogenous bases
D)only sulfur and no nitrogen
E)None of the answer options are correct.
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40
The Meselson-Stahl experiment made clear predictions regarding experimental outcomes if dispersive,conservative,or semiconservative DNA replication was occurring in their Escherichia coli cells.This experiment enables the detection of "new" and "old" DNA by assessing the _______________ of DNA molecules in the cells.
A)intensity of radioactive labeling
B)density
C)charge distribution along the double helix
D)number of mutations
E)size (length)
A)intensity of radioactive labeling
B)density
C)charge distribution along the double helix
D)number of mutations
E)size (length)
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41
Telomeres have been a recent focus in experiments regarding both aging and cloning.For example,it has been shown that in the absence of telomerase,the chromosome becomes shorter and shorter after each cell division.When chromosomes reach a certain length,the cell may cease to divide and die.Hence,maintaining telomerase action could be a "fountain of youth" for a cell.
Excess telomerase activity is associated with cancer cells,providing unlimited numbers of cell divisions (immortality)to these cells and presenting a dangerous threat to the organism.
Find a recent news or research article on telomeres,and discuss what the experiment tells you about the role of telomerase in the cell.Could activation of this enzyme in humans be beneficial with regard to age-related disease?
Hint: "http://www.wired.com/wiredscience/2010/11/mouse-aging-reversal/ (and similar articles)
Excess telomerase activity is associated with cancer cells,providing unlimited numbers of cell divisions (immortality)to these cells and presenting a dangerous threat to the organism.
Find a recent news or research article on telomeres,and discuss what the experiment tells you about the role of telomerase in the cell.Could activation of this enzyme in humans be beneficial with regard to age-related disease?
Hint: "http://www.wired.com/wiredscience/2010/11/mouse-aging-reversal/ (and similar articles)
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42
Primase and telomerase enzymes are both considered types of:
A)DNA polymerases.
B)retroviruses.
C)transcription factors.
D)reverse transcriptases.
A)DNA polymerases.
B)retroviruses.
C)transcription factors.
D)reverse transcriptases.
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43
Erwin Chargaff provided experimental data (Chargaff's rules)that helped Watson and Crick to gain important insight into the structure of DNA.What did Chargaff determine,and what did it mean to Watson?
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44
A-form DNA has very little external exposure of its nucleotide bases as compared to the B-form.B-DNA is the biologically significant form.Why do you think this is so?
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45
The number of proteins in the prokaryotic replisome is 13,while the number in a eukaryotic cell is 27.What are important protein components of the replisome in general? Why might a eukaryotic cell have more components?
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46
Which of the following conditions is/are associated with abnormal telomerase activity resulting in premature aging?
A)Down syndrome
B)Patau syndrome
C)Cri-du chat syndrome
D)Werner syndrome
A)Down syndrome
B)Patau syndrome
C)Cri-du chat syndrome
D)Werner syndrome
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47
Imagine it is the early 1900s and the nature of genetic material is not yet known.You believe that proteins (polypeptides)are the most likely candidates for storing heritable information,rather than DNA (which you regard as some kind of macromolecule used to store hydrocarbons).Make an argument supporting this position.
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48
Imagine it is the year 2050.Because the Martian landers discovered that liquid water was present on Mars,another probe was sent to look for signs of life.Cells containing double-stranded DNA were found,and using an automated variant of the Meselson-Stahl experiment,the original DNA was labeled with 14N and then transferred to 15N (note that this is the reverse of the original M-S experiment).In the low temperatures of Mars,the replication machinery works slower,so each cell division takes 60 hours.
Assume you do not know if replication machinery uses a dispersive,conservative,or semiconservative approach.Predict what the centrifuge tube patterns should be at 60,120,180,and 240 hours.
Assume you do not know if replication machinery uses a dispersive,conservative,or semiconservative approach.Predict what the centrifuge tube patterns should be at 60,120,180,and 240 hours.
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49
Telomeres regulate the replication of the ends of chromosomes in eukaryotes.Why is this structure implicated in human aging?
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50
An epidemic disease affecting sheep grazing near the hot springs of Thermopolis,Wyoming,was found to be due to a new virus.This virus could infect sheep kidney cells cultured in vitro.The virus appeared to contain four chemically defined biomolecules,which we will call W,X,Y,and Z.Investigators wished to determine which of these components carried the genetic information of the virus.
An experiment was conducted in which one of the four components was radioactively labeled in different batches of kidney cells cultured and infected with the virus.Thus,four batches of labeled virus were generated (labeled W,X,Y,or Z).
Radioactive virus from each of these batches was allowed to attach to nonradioactive kidney cells.The cells were centrifuged down to remove the unattached viruses.The cells were then briefly exposed to a vigorous agitation to release viral particles on the cell surface,and again centrifuged.The supernatant (containing shaved-off viral parts)and the pelleted infected cells were examined for radioactivity,with the following results:
% Radioactivity
a)On the basis of these results,which of these components does NOT carry genetic information? Why?
Other experiments revealed that one of these purified biomolecules was contaminating sheep DNA,and the two remaining components did appear to be derived from a virus.The nitrogenous base compositions were as follows:
b)What can you say about the type of nucleic acid found in each of these three components?
c)Can you determine which of these purified biomolecules is most likely sheep DNA and which are more likely to belong to the virus? Explain your response.
An experiment was conducted in which one of the four components was radioactively labeled in different batches of kidney cells cultured and infected with the virus.Thus,four batches of labeled virus were generated (labeled W,X,Y,or Z).
Radioactive virus from each of these batches was allowed to attach to nonradioactive kidney cells.The cells were centrifuged down to remove the unattached viruses.The cells were then briefly exposed to a vigorous agitation to release viral particles on the cell surface,and again centrifuged.The supernatant (containing shaved-off viral parts)and the pelleted infected cells were examined for radioactivity,with the following results:
% Radioactivity
a)On the basis of these results,which of these components does NOT carry genetic information? Why?
Other experiments revealed that one of these purified biomolecules was contaminating sheep DNA,and the two remaining components did appear to be derived from a virus.The nitrogenous base compositions were as follows:
b)What can you say about the type of nucleic acid found in each of these three components?
c)Can you determine which of these purified biomolecules is most likely sheep DNA and which are more likely to belong to the virus? Explain your response.
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