AP Biology Chapter 11/12 Quiz
1. The Hershey–Chase experiment determined that
a. protein and DNA are the hereditary materials of viruses.
c. viruses do not contain hereditary material.
d. DNA, not protein, is the hereditary material of viruses.
e. None of the above
Answer: d
2. The rules formulated by Erwin Chargaff state that
a. A = T and G = C in any molecule of DNA.
b. A = C and G = T in any molecule of DNA.
c. A = G and C = T in any molecule of DNA.
d. A = U and G = C in any molecule of RNA.
e. DNA and RNA are made up of the same four nitrogenous bases.
Answer: a
3. Purines include
a. cytosine, uracil, and thymine.
b. adenine and cytosine.
c. adenine and thymine.
d. cytosine and thymine.
e. adenine and guanine.
Answer: e
4. The structure of DNA is characterized by a
a. right- or left-handed double helix and antiparallel strands.
b. right-handed double helix and antiparallel strands.
c. right-handed single helix.
d. right-handed single helix and parallel strands.
e. All of the above
Answer: b
5. The antiparallel relationship of the two strands of DNA refers to the
a. twisted configuration of the strands.
b. alternative branching pattern of the strands.
c. alignment of the strands, such that one strand starts with a 3′ carbon, the other with a 5′
carbon.
d. view at one end of the molecule–one strand has an A wherever the other has a T, and
one has a G wherever the other has a C.
e. All of the above
Answer: c
6. The nitrogenous bases (and the two strands of the DNA double helix) are held together
by
a. weak van der Waals forces.
b. covalent bonds.
c. hydrogen bonds.
d. Both a and b
e. Both a and c
Answer: c
7. Why are RNA primers used during DNA replication?
a. RNA primase adds bases that act as primers.
b. RNA primase is able to use DNA as a template.
c. RNA primase must be incorporated into the holoenzyme complex.
d. DNA polymerase can only add on to an existing strand.
e. All of the above
Answer: d
8. Semiconservative replication of DNA involves
a. each of the original strands acting as a template for a new strand.
b. only one of the original strands acting as a template for a new strand.
c. the complete separation of the original strands, the synthesis of new strands, and the
reassembly of double-stranded molecules.
d. the use of the original double-stranded molecule as a template.
e. None of the above
Answer: a
9. The molecules that function to replicate DNA in the cell are
a. DNA nucleoside triphosphates.
b. DNA polymerases.
c. nucleoside polymerases.
d. DNAses.
e. ribonucleases.
Answer: b
10. During replication, the new DNA strand is synthesized
a. in the 3′ to 5′ direction.
b. in the 5′ to 3′ direction.
c. in both the 3′ to 5′ and 5′ to 3′ directions from the replication fork.
d. from one end to the other, in the 3′ to 5′ or the 5′ to 3′ direction.
e. None of the above
Answer: b
11. DNA replication in eukaryotes differs from replication in bacteria because
a. synthesis of the new DNA strand is from 3′ to 5′ in eukaryotes and from 5′ to 3′ in
bacteria.
b. synthesis of the new DNA strand is from 5′ to 3′ in eukaryotes and from 3′ to 5′ in
bacteria.
c. there are many replication forks in each eukaryotic chromosome and only one in
bacterial DNA.
d. synthesis of the new DNA strand is from 5′ to 3′ in eukaryotes and is random in
prokaryotes.
e. Okazaki fragments are produced in eukaryotic DNA replication but not in prokaryotic
Answer: c
12. Mutations are
a. heritable changes in the sequence of DNA bases that produce an observable phenotype.
b. heritable changes in the sequence of DNA bases.
c. mistakes in the incorporation of amino acids into proteins.
d. heritable changes in the mRNA of an organism.
e. None of the above
Answer: b
13. In order to show that DNA is the “transforming principle,” Avery, MacLeod, and
McCarty showed that DNA could transform nonvirulent strains of pneumococcus. Their
hypothesis was strengthened by their demonstration that
a. enzymes that destroyed proteins also destroyed transforming activity.
b. enzymes that destroyed nucleic acids also destroyed transforming activity.
c. enzymes that destroyed complex carbohydrates also destroyed transforming activity.
d. the transformation activity was destroyed by boiling.
e. other strains of bacteria also could be transformed successfully.
Answer: b
14. It was shown through experiments that during infection of E. coli cells by
bacteriophage T2,
b. both proteins and nucleic acids enter the cell.
c. only protein from the infecting phage can also be detected in progeny phage.
d. only nucleic acids enter the cell.
e. more than one infecting phage particle is required to produce infection.
Answer: d
15. Bacteriophage nucleic acids were labeled by carrying out an infection of E. coli cells
growing in
a. 14C-labeled CO2.
b. 3H-labeled water.
c. 32P-labeled phosphate.
d. 35S-labeled sulfate.
e. 18O-labeled water.
Answer: c
16. Information sources used by Watson and Crick to determine the structure of DNA
included
a. electron micrographs of individual DNA molecules.
b. light micrographs of bacteriophage particles.
d. nuclear magnetic resonance analysis of DNA.
e. X-ray crystallography of double-stranded DNA.
Answer: e
17. Double-stranded DNA looks a little like a ladder that has been twisted into a helix, or
spiral. The side supports of the ladder are
a. individual nitrogenous bases.
b. alternating bases and sugars.
c. alternating bases and phosphate groups.
d. alternating sugars and phosphates.
e. alternating bases, sugars, and phosphates.
Answer: d
18. The steps of the ladder are
a. individual nitrogenous bases.
b. pairs of bases.
c. alternating bases and phosphate groups.
d. alternating sugars and bases.
e. alternating bases, sugars, and phosphates.
Answer: b
19. If a double-stranded DNA molecule contains 30 percent T, it must contain _______
percent G.
a. 20
b. 30
c. 40
d. 50
e. 60
Answer: a
20. In the Meselson–Stahl experiment, the conservative model of DNA replication is
ruled out by which of the following observations?
a. No completely “heavy” DNA is observed after the first round of replication.
b. No completely “light” DNA ever appears, even after several replications.
c. The product that accumulates after two rounds of replication is completely “heavy.”
d. Completely “heavy” DNA is observed throughout the experiment.
e. Three different DNA densities are observed after a single round of replication.
Answer: a
21. The enzyme DNA ligase is required continuously during DNA replication because
a. fragments of the leading strand must be joined together.
b. fragments of the lagging strand must be joined together.
c. the parental strands must be joined back together.
d. 3′-deoxynucleoside triphosphates must be converted to 5′-deoxynucleoside
triphosphates.
e. the complex of proteins that work together at the replication fork must be kept from
falling apart.
Answer: b
22. In DNA replication, each newly made strand is
a. identical in DNA sequence to the strand from which it was copied.
b. complementary in sequence to the strand from which it was copied.
c. oriented in the same 3′ to 5′ direction as the strand from which it was copied.
d. an incomplete copy of one of the parental strands.
e. a hybrid molecule consisting of both ribo- and deoxyribonucleotides.
Answer: b
23. The Hershey–
a. bacteria can be transformed.
b. DNA is indeed the carrier of hereditary information.
c. DNA replication is semiconservative.
d. the transforming principle requires host factors.
e. All of the above
Answer: b
24. Which of the following features summarizes the molecular architecture of DNA?
a. The two strands run in opposite directions.
b. The molecule twists in the same direction as the threads of most screws.
c. The molecule is a double-stranded helix.
d. It has a uniform diameter.
e. All of the above
Answer: e
25. Chargaff’s rule states that
a. DNA must be replicated before a cell can divide.
b. viruses enter cells without their protein coat.
c. only protein from the infecting phage can also be detected in progeny phage.
d. only nucleic acids enter the cell during infection.
e. the amount of cytosine equals the amount of guanine.
Answer: e
26. In eukaryotic cells, each chromosome has
a. one origin of replication.
b. two origins of replication.
c. many origins of replication.
d. only one origin of replication per nucleus.
e. None of the above
Answer: c
27. When adding the next monomer to a growing DNA strand, the monomer is added to
which carbon of the deoxyribose?
a. 1′
b. 2′
c. 3′
d. 4′
e. 5′
Answer: c
28. The enzyme that restores the phosphodiester linkage between adjacent fragments in
a. DNA ligase.
b. primase.
c. reverse transcriptase.
d. helicase.
e. DNA polymerase I.
Answer: a
29. The building blocks for a new DNA molecule are deoxyribose
a. nucleoside monophosphates.
b. nucleoside diphosphates.
c. nucleoside triphosphates.
d. nucleotide diphosphates.
e. nucleotide triphosphates.
Answer: c
30. The enzyme that unwinds the DNA prior to replication is called
a. DNA polymerase III.
b. DNA ligase.
c. single-stranded DNA binding protein.
d. primase.
e. helicase.
Answer: e
31. The first repair of mistakes during DNA replication is made by
a. the mismatch repair system.
b. DNA polymerase.
c. excision repair.
d. SOS repair.
e. postreplication repair.
Answer: b
32. If Meselson and Stahl had observed one intermediate, slightly smeared band after
growing bacteria for one generation, and then after two generations again had found one
slightly smeared band, what would have been their likely conclusion about the mode of
DNA replication?
a. DNA replicates semiconservatively.
b. DNA replicates conservatively.
c. DNA replicates semidiscontinuously.
d. DNA replicates dispersively.
e. None of the above
Answer: d
33. Which one of the following is not found in DNA?
a. Carbon
b. Oxygen
c. Nitrogen
d. Hydrogen
e. Sulfur
Answer: e
34. The Hershey–Chase experiment
a. proved that DNA replication is semiconservative.
b. used 32P to label protein.
c. used 35S to label DNA.
d. helped prove that DNA is the genetic molecule.
e. Both b and c
Answer: d
35. Ideally, PCR _______ increases the amount of DNA during additional cycles.
a. additively
b. gradually
c. linearly
d. systematically
e. exponentially
Answer: e
36. Why don’t cells last the entire lifetime of an organism?
a. DNA replication leads to a shortening of the chromosome and eventual cell death.
b. The enzyme telomerase is readily destroyed by the environment, resulting in cell death.
c. DNA replication is subject to errors, causing premature cell death.
d. Okazaki fragments disrupt protein synthesis, resulting in premature cell death.
e. The repeating telomeric sequence of TTAGGG interferes with normal DNA replication
and leads to cell death.
Answer: a
37. The strands that make up DNA are antiparallel. This means that
a. one strand is positively charged and the other is negatively charged.
b. the base pairings create unequal spacing between the two DNA strands.
c. the 5′ to 3′ direction of one strand is counter to the 5′ to 3′ direction of the other strand.
d. the twisting of the DNA molecule has shifted the two strands.
e. purines bond with purines and pyrimidines bond with pyrimidines.
Answer: c
38. What accounts for the uniform diameter of the DNA molecule?
a. The two sides of the DNA molecule are held together by hydrogen bonds.
b. A purine always bonds with a pyrimidine.
c. One side of the DNA molecule has an unconnected 5′ phosphate group and the
opposite end has an unconnected 3′ hydroxyl group.
d. The 3′ carbon of one deoxyribose and the 5′ carbon of another deoxyribose bond
together.
e. The alternating sugar and phosphate backbone coils around the outside of the helix.
Answer: b
39. After irradiating Neurospora, Beadle and Tatum collected mutants that require
arginine to grow. These mutants
a. will not grow on minimal media but will grow on minimal media with arginine.
b. will grow on minimal media and on minimal media with arginine.
c. will not grow on minimal media and will not grow on minimal media with arginine.
d. will grow on minimal media but will not grow on minimal media with arginine.
e. None of the above
Answer: a
40. Within a group of mutants with the same growth requirement (i.e., the same overt
phenotype), mapping studies determined that individual mutations were on different
chromosomes. This indicates that
a. the same gene governs all the steps in a particular biological pathway.
b. different genes can govern different individual steps in the same biological pathway.
c. different genes govern the same step in a particular biological pathway.
d. all biological pathways are governed by different genes.
e. genes do not govern steps in biological pathways.
Answer: b
41. The study of Neurospora mutants grown on various supplemented media led to
a. a determination of the steps in biological pathways.
b. the “one-gene, one-enzyme” theory.
c. the idea that genes are “on” chromosomes.
d. Both a and b
e. Both a and c
Answer: d
42. Genes code for
a. enzymes.
b. polypeptides.
c. RNA.
d. All of the above
e. None of the above
Answer: d
43. How does RNA differ from DNA?
a. RNA contains uracil instead of thymine and it is usually single-stranded.
b. RNA contains uracil instead of thymine and it is usually double-stranded.
c. RNA contains thymine instead of uracil and it is usually single-stranded.
d. RNA contains uracil instead of cytosine.
e. None of the above
Answer: a
44. Which of the following statements about the flow of genetic information is true?
a. Proteins encode information that is used to produce other proteins of the same amino
acid sequence.
b. RNA encodes information that is translated into DNA, and DNA encodes information
that is translated into proteins.
c. Proteins encode information that can be translated into RNA, and RNA encodes
information that can be transcribed into DNA.
d. DNA encodes information that is translated into RNA, and RNA encodes information
that is translated into proteins.
e. None of the above
Answer: d
45. Which of the following molecules transfers information from the nucleus to the
cytoplasm?
a. DNA
b. mRNA
c. tRNA
d. Proteins
e. Lipids
Answer: b
46. mRNA is synthesized in the _______ direction, which corresponds to the _______ of
the protein.
a. 5′ to 3′; N terminus to C terminus
b. 3′ to 5′; C terminus to N terminus
c. 5′ to 3′; C terminus to N terminus
d. 3′ to 5′; N terminus to C terminus
e. Examples of all of the above have been found.
Answer: a
47. Which of the following molecules serves as an adapter molecule?
a. DNA
b. mRNA
c. tRNA
d. Proteins
e. Lipids
Answer: c
48. A sequence of three RNA bases can function as a
a. codon.
b. anticodon.
c. gene.
d. Both a and b
e. Both a and c
Answer: d
49. Ribosomes are a collection of _______ that are needed for _______.
a. small proteins; translation.
b. proteins and small RNAs; translation.
c. proteins and tRNAs; transcription.
d. proteins and mRNAs; translation.
e. mRNAs and tRNAs; translation.
Answer: b
50. Retroviruses do not follow the “central dogma” of DNA→RNA→protein because
they
a. contain RNA that is used to make DNA.
b. contain DNA that is used to make more RNA.
c. contain DNA that is used to make tRNA.
d. contain only RNA as the genetic material.
e. do not contain either DNA or RNA as the genetic material.
Answer: a
51. The type of mutation that stops translation of a protein is a(n)
a. missense mutation.
b. nonsense mutation.
c. frame-shift mutation.
d. aberration.
e. None of the above
Answer: b
52. The type of mutation that is an insertion or a deletion of a single base is a(n)
a. missense mutation.
b. nonsense mutation.
c. frame-shift mutation.
d. aberration.
e. None of the above
Answer: c
53. A transcription start signal is called a(n)
a. initiation codon.
b. promoter.
c. origin.
d. operator.
e. nonsense codon.
Answer: b
54. Initiation of transcription requires
a. a temporary stoppage of DNA replication.
c. destruction of one of the strands of the DNA template.
Answer: b
55. The genetic code is best described as
a. redundant
b. Anticodons
c. confusing
d. neither confusing nor redundant.
e. nonsense.
Answer: a
56. The three codons in the genetic code that do not specify amino acids are called
a. missense codons.
b. start codons.
c. stop codons.
d. promoters.
e. initiator codons.
Answer: c
57. Sickle-cell disease is caused by a _______ mutation.
a. nonsense
b. missense
c. frame-shift
d. temperature sensitive
e. silent
Answer: b
58. The direction of synthesis for a new mRNA molecule is _______ from a _______
template strand.
a. 5′ to 3′; 5′ to 3′
b. 5′ to 3′; 3′ to 5′
c. 3′ to 5′; 5′ to 3′
d. 3′ to 5′; 3′ to 5′
e. 5′ to 5′; 3′ to 3′
Answer: b
59. The region of DNA in prokaryotes to which RNA polymerase binds most tightly is
the
a. promoter.
b. poly C center.
c. enhancer.
d. operator site.
e. minor groove.
Answer: a
60. DNA is composed of two strands, only one of which typically is used as a template
for RNA synthesis. By what mechanism is the correct strand chosen?
b. Only one strand has the start codon.
c. The promoter acts to aim the RNA polymerase.
d. A start factor informs the system.
e. It is chosen randomly.
Answer: c
61. The error rate for RNA polymerase is _______ that for most DNA polymerases.
a. less than
b. equal to
c. greater than
d. greater for frame shifts but less for base substitutions than
e. greater for base substitutions but less for frame shifts than
Answer: c
62. _______ is the addition of sugar residues to the protein after translation.
a. Glycation
b. Glycosylation
c. Phosphorylation
d. Proteolysis
e. Exonuclease digestion
Answer: b
63. After translation, some proteins are processed by _______, which is cleavage of the
protein to make a shortened finished protein.
a. glycation
b. glycosylation
c. phosphorylation
d. proteolysis
e. exonuclease digestion
Answer: d
64. The stop codons code for
a. no amino acid.
b. methionine.
c. glycine.
d. halt enzyme.
e. DNA binding protein.
Answer: a
65. Breaking and rejoining of chromosomes can lead to
a. deletions.
b. duplications.
c. inversions.
d. translocations.
e. All of the above
Answer: e
66. The link between mRNA and a protein is
a. tRNA.
b. a promoter.
c. RNA polymerase.
d. DNA polymerase.
e. a start codon.
Answer: a
67. Which of the following do (does) not follow the “central dogma”?
a. Yeast
b. Onion cells
c. Bread mold
d. Skin cells
e. Retroviruses
Answer: e
68. What events must take place to ensure that the protein made is the one specified by
mRNA?
a. tRNA must read mRNA correctly.
b. tRNA must carry the amino acid that is correct for its reading of the mRNA.
c. Covalent bonding between the base pairs must occur.
d. Both a and b
e. All of the above
Answer: d
69. Which of the following statements is true of codons and anticodons?
a. The codon bonds covalently with the anticodon.
b. The base sequences are the same.
c. There are 64 codons and 61 anticodons.
d. Activating enzymes link codons and anticodons.
e. At contact, the codon and the anticodon are antiparallel to each other.
Answer: e
70. The enzyme that charges the tRNA molecules with appropriate amino acids is
a. tRNA chargeatase.
b. amino tRNA chargeatase.
c. transcriptase.
d. aminoacyl-tRNA synthetase.
e. None of the above
Answer: d