A) They cannot replicate DNA.
B) They cannot undergo mitosis.
C) They cannot exchange DNA with other cells.
D) They cannot repair thymine dimers.
E) They do not recombine homologous chromosomes during meiosis.
Correct Answer
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Multiple Choice
A) the protein coat from pathogenic cells was able to transform nonpathogenic cells.
B) heat-killed pathogenic cells caused pneumonia.
C) some substance from pathogenic cells was transferred to nonpathogenic cells, making them pathogenic.
D) the polysaccharide coat of bacteria caused pneumonia.
E) bacteriophages injected DNA into bacteria.
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Multiple Choice
A) The twisting nature of DNA creates nonparallel strands.
B) The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.
C) Base pairings create unequal spacing between the two DNA strands.
D) One strand is positively charged and the other is negatively charged.
E) One strand contains only purines and the other contains only pyrimidines.
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Multiple Choice
A) the diameter of the helix
B) the rate of replication
C) the sequence of nucleotides
D) the bond angles of the subunits
E) the frequency of A vs. T nucleotides
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Multiple Choice
A) exonuclease, DNA polymerase III, RNA primase
B) helicase, DNA polymerase I, DNA ligase
C) DNA ligase, nuclease, helicase
D) DNA polymerase I, DNA polymerase III, DNA ligase
E) endonuclease, DNA polymerase I, DNA ligase
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Multiple Choice
A) 5' G C C T A G G 3'
B) 3' G C C T A G G 5'
C) 5' A C G T T A G G 3'
D) 5' A C G U U A G G 3'
E) 5' G C C U A G G 3'
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Multiple Choice
A) One of the daughter cells, but not the other, would have radioactive DNA.
B) Neither of the two daughter cells would be radioactive.
C) All four bases of the DNA would be radioactive.
D) Radioactive thymine would pair with nonradioactive guanine.
E) DNA in both daughter cells would be radioactive.
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Multiple Choice
A) leading strands and Okazaki fragments.
B) lagging strands and Okazaki fragments.
C) Okazaki fragments and RNA primers.
D) leading strands and RNA primers.
E) RNA primers and mitochondrial DNA.
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Multiple Choice
A) double-stranded DNA, four kinds of dNTPs, primers, origins
B) topoisomerases, telomerases, polymerases
C) G-C rich regions, polymerases, chromosome nicks
D) nucleosome loosening, four dNTPs, four rNTPs
E) ligase, primers, nucleases
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Multiple Choice
A) polymerase molecules.
B) ribosomes.
C) histones.
D) a thymine dimer.
E) satellite DNA.
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Multiple Choice
A) DNA polymerase begins adding nucleotides at the 5' end of the template.
B) Okazaki fragments prevent elongation in the 3' to 5' direction.
C) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end.
D) replication must progress toward the replication fork.
E) DNA polymerase can only add nucleotides to the free 3' end.
Correct Answer
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Multiple Choice
A) A = C
B) A = G and C = T
C) A + C = G + T
D) G + C = T + A
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Multiple Choice
A) It synthesizes RNA nucleotides to make a primer.
B) It catalyzes the lengthening of telomeres.
C) It joins Okazaki fragments together.
D) It unwinds the parental double helix.
E) It stabilizes the unwound parental DNA.
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Multiple Choice
A) 8%
B) 16%
C) 31%
D) 42%
E) It cannot be determined from the information provided.
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Multiple Choice
A) replication followed by mitosis
B) replication without separation
C) meiosis followed by mitosis
D) fertilization by multiple sperm
E) special association with histone proteins
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Multiple Choice
A) the evolution of telomerase enzyme
B) DNA polymerase that cannot replicate the leading strand template to its 5' end
C) gaps left at the 5' end of the lagging strand
D) gaps left at the 3' end of the lagging strand because of the need for a primer
E) the "no ends" of a circular chromosome
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Multiple Choice
A) the inactivity of this DNA
B) the low frequency of mutations occurring in this DNA
C) that new evolution of telomeres continues
D) that mutations in telomeres are relatively advantageous
E) that the critical function of telomeres must be maintained
Correct Answer
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