AP Bio Chapter 16-20 Practice test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. In his transformation experiments, what did Griffith observe?
a. Mutant mice were resistant to bacterial infections.
b. Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic
strain can convert some of the living cells into the pathogenic form.
c. Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic
strain makes the pathogenic strain nonpathogenic.
d. Infecting mice with nonpathogenic strains of bacteria makes them resistant to
pathogenic strains.
e. Mice infected with a pathogenic strain of bacteria can spread the infection to other
mice.
____
2. What does transformation involve in bacteria?
a. the creation of a strand of DNA from an RNA molecule
b. the creation of a strand of RNA from a DNA molecule
c. the infection of cells by a phage DNA molecule
d. the type of semiconservative replication shown by DNA
e. assimilation of external DNA into a cell
____
3. The following scientists made significant contributions to our understanding of the structure and
function of DNA. Place the scientists' names in the correct chronological order, starting with the first
scientist(s) to make a contribution.
I. Avery, McCarty, and MacLeod
II. Griffith
III. Hershey and Chase
IV. Meselson and Stahl
V. Watson and Crick
a. V, IV, II, I, III
b. II, I, III, V, IV
c. I, II, III, V, IV
d. I, II, V, IV, III
e. II, III, IV, V, I
____
4. Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. Approximately
what percentage of the nucleotides in this sample will be thymine?
a. 12
b. 24
c. 31
d. 38
e. It cannot be determined from the information provided.
____
5. Chargaff's analysis of the relative base composition of DNA was significant because he was able to
show that
a. the relative proportion of each of the four bases differs within individuals of a
b.
c.
d.
e.
____
species.
the human genome is more complex than that of other species.
the amount of A is always equivalent to T, and C to G.
the amount of ribose is always equivalent to deoxyribose.
transformation causes protein to be brought into the cell.
6. Which of the following can be determined directly from X-ray diffraction photographs of
crystallized DNA?
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
____
7. Why does the DNA double helix have a uniform diameter?
a. Purines pair with pyrimidines.
b. C nucleotides pair with A nucleotides.
c. Deoxyribose sugars bind with ribose sugars.
d. Nucleotides bind with nucleosides.
e. Nucleotides bind with nucleoside triphosphates.
____
8. What kind of chemical bond is found between paired bases of the DNA double helix?
a. hydrogen
b. ionic
c. covalent
d. sulfhydryl
e. phosphate
____
9. In an analysis of the nucleotide composition of DNA, which of the following will be found?
a. A = C
b. A = G and C = T
c. A + C = G + T
d. G + C = T + A
____ 10. Replication in prokaryotes differs from replication in eukaryotes for which of these reasons?
a. The prokaryotic chromosome has histones, whereas eukaryotic chromosomes do
not.
b. Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic
chromosomes have many.
c. The rate of elongation during DNA replication is slower in prokaryotes than in
eukaryotes.
d. Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes
do not.
e. Prokaryotes have telomeres, and eukaryotes do not.
____ 11. What is meant by the description "antiparallel" regarding the strands that make up DNA?
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.
____ 12. An Okazaki fragment has which of the following arrangements?
a. primase, polymerase, ligase
b. 3' RNA nucleotides, DNA nucleotides 5'
c. 5' RNA nucleotides, DNA nucleotides 3'
d. DNA polymerase I, DNA polymerase III
e. 5' DNA to 3'
____ 13. Which enzyme catalyzes the elongation of a DNA strand in the 5'  3' direction?
a. primase
b. DNA ligase
c. DNA polymerase III
d. topoisomerase
e. helicase
____ 14. What determines the nucleotide sequence of the newly synthesized strand during DNA replication?
a. the particular DNA polymerase catalyzing the reaction
b. the relative amounts of the four nucleoside triphosphates in the cell
c. the nucleotide sequence of the template strand
d. the primase used in the reaction
e. the arrangement of histones in the sugar phosphate backbone
____ 15. Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of
which of the following?
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 because of the need for a 3' onto which
nucleotides can attach
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
____ 16. The DNA of telomeres has been found to be highly conserved throughout the evolution of
eukaryotes. What does this most probably reflect?
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
____ 17. At a specific area of a chromosome, the sequence of nucleotides below is present where the chain
opens to form a replication fork:
3' C C T A G G C T G C A A T C C 5'
An RNA primer is formed starting at the underlined T (T) of the template. Which of the following
represents the primer sequence?
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'
____ 18. To repair a thymine dimmer by nucleotide excision repair, in which order do the necessary enzymes
act?
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
____ 19. What is the function of DNA polymerase III?
a. to unwind the DNA helix during replication
b. to seal together the broken ends of DNA strands
c. to add nucleotides to the end of a growing DNA strand
d. to degrade damaged DNA molecules
e. to rejoin the two DNA strands (one new and one old) after replication
____ 20. Which of the following removes the RNA nucleotides from the primer and adds equivalent DNA
nucleotides to the 3' end of Okazaki fragments?
a. helicase
b. DNA polymerase III
c. ligase
d. DNA polymerase I
e. primase
____ 21. Which of the following separates the DNA strands during replication?
a. helicase
b. DNA polymerase III
c. ligase
d. DNA polymerase I
e. primase
____ 22. Which of the following covalently connects segments of DNA?
a. helicase
b. DNA polymerase III
c. ligase
d. DNA polymerase I
e. primase
____ 23. Which of the following synthesizes short segments of RNA?
a. helicase
b. DNA polymerase III
c. ligase
d. DNA polymerase I
e. primase
____ 24. The leading and the lagging strands differ in that
a. the leading strand is synthesized in the same direction as the movement of the
replication fork, and the lagging strand is synthesized in the opposite direction.
b. the leading strand is synthesized by adding nucleotides to the 3' end of the growing
strand, and the lagging strand is synthesized by adding nucleotides to the 5' end.
c. the lagging strand is synthesized continuously, whereas the leading strand is
synthesized in short fragments that are ultimately stitched together.
d. the leading strand is synthesized at twice the rate of the lagging strand.
____ 25. What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
a. synthesize RNA nucleotides to make a primer
b. catalyze the lengthening of telomeres
c. join Okazaki fragments together
d. unwind the parental double helix
e. stabilize the unwound parental DNA
____ 26. Which of the following help to hold the DNA strands apart while they are being replicated?
a. primase
b. ligase
c. DNA polymerase
d. single-strand binding proteins
e. exonuclease
____ 27. Which of the following statements describes histones?
a. Each nucleosome consists of two molecules of histone H1.
b. Histone H1 is not present in the nucleosome bead; instead it is involved in the
formation of higher-level chromatin structures.
c. The carboxyl end of each histone extends outward from the nucleosome and is
called a "histone tail."
d. Histones are found in mammals, but not in other animals or in plants.
e. The mass of histone in chromatin is approximately nine times the mass of DNA.
____ 28. In his work with pneumonia-causing bacteria and mice, Griffith found that
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.
____ 29. In a nucleosome, the DNA is wrapped around
a. polymerase molecules.
b. ribosomes.
c. histones.
d. a thymine dimer.
e. satellite DNA.
____ 30. The nitrogenous base adenine is found in all members of which group?
a. proteins, triglycerides, and testosterone
b. proteins, ATP, and DNA
c. ATP, RNA, and DNA
d. alpha glucose, ATP, and DNA
e. proteins, carbohydrates, and ATP
____ 31. If proteins were composed of only 12 different kinds of amino acids, what would be the smallest
possible codon size in a genetic system with four different nucleotides?
a. 1
b. 2
c. 3
d. 4
e. 12
____ 32. A particular triplet of bases in the template strand of DNA is 5' AGT 3'. The corresponding codon
for the mRNA transcribed is
a. 3' UCA 5'.
b. 3' UGA 5'.
c. 5' TCA 3'.
d. 3'ACU 5'.
e. either UCA or TCA, depending on wobble in the first base.
The following questions refer to Figure 17.2, a table of codons.
Figure 17.2
____ 33. A possible sequence of nucleotides in the template strand of DNA that would code for the
polypeptide sequence phe-leu-ile-val would be
a. 5' TTG-CTA-CAG-TAG 3'.
b. 3' AAC-GAC-GUC-AUA 5'.
c. 5' AUG-CTG-CAG-TAT 3'.
d. 3' AAA-AAT-ATA-ACA 5'.
e. 3' AAA-GAA-TAA-CAA 5'.
____ 34. What amino acid sequence will be generated, based on the following mRNA codon sequence?
5' AUG-UCU-UCG-UUA-UCC-UUG 3'
a. met-arg-glu-arg-glu-arg
b. met-glu-arg-arg-gln-leu
c. met-ser-leu-ser-leu-ser
d. met-ser-ser-leu-ser-leu
e. met-leu-phe-arg-glu-glu
____ 35. A peptide has the sequence NH2-phe-pro-lys-gly-phe-pro-COOH. Which of the following sequences
in the coding strand of the DNA could code for this peptide?
a. 3' UUU-CCC-AAA-GGG-UUU-CCC
b. 3' AUG-AAA-GGG-TTT-CCC-AAA-GGG
c. 5' TTT-CCC-AAA-GGG-TTT-CCC
d. 5' GGG-AAA-TTT-AAA-CCC-ACT-GGG
e. 5' ACT-TAC-CAT-AAA-CAT-TAC-UGA
____ 36. What is the sequence of a peptide based on the following mRNA sequence?
5' . . . UUUUCUUAUUGUCUU 3'
a. leu-cys-tyr-ser-phe
b. cyc-phe-tyr-cys-leu
c. phe-leu-ile-met-val
d. leu-pro-asp-lys-gly
e. phe-ser-tyr-cys-leu
____ 37. RNA polymerase moves in which direction along the DNA?
a. 3'  5' along the template strand
b. 3'  5' along the coding (sense) strand
c. 5'  3' along the template strand
d. 3'  5' along the coding strand
e. 5'  3' along the double-stranded DNA
____ 38. Which of these is the function of a poly (A) signal sequence?
a. It adds the poly (A) tail to the 3' end of the mRNA.
b. It codes for a sequence in eukaryotic transcripts that signals enzymatic cleavage
~10—35 nucleotides away.
c. It allows the 3' end of the mRNA to attach to the ribosome.
d. It is a sequence that codes for the hydrolysis of the RNA polymerase.
e. It adds a 7-methylguanosine cap to the 3' end of the mRNA.
____ 39. Transcription in eukaryotes requires which of the following in addition to RNA polymerase?
a. the protein product of the promoter
b. start and stop codons
c. ribosomes and tRNA
d. several transcription factors (TFs)
e. aminoacyl synthetase
____ 40. A part of the promoter, called the TATA box, is said to be highly conserved in evolution. Which
might this illustrate?
a. The sequence evolves very rapidly.
b. The sequence does not mutate.
c. Any mutation in the sequence is selected against.
d. The sequence is found in many but not all promoters.
e. The sequence is transcribed at the start of every gene.
____ 41. Which of the following help(s) to stabilize mRNA by inhibiting its degradation?
a. TATA box
b. spliceosomes
c. 5' cap and poly (A) tail
d. introns
e. RNA polymerase
____ 42. What is a ribozyme?
a. an enzyme that uses RNA as a substrate
b. an RNA with enzymatic activity
c. an enzyme that catalyzes the association between the large and small ribosomal
subunits
d. an enzyme that synthesizes RNA as part of the transcription process
e. an enzyme that synthesizes RNA primers during DNA replication
____ 43. What are the coding segments of a stretch of eukaryotic DNA called?
a. introns
b. exons
c. codons
d. replicons
e. transposons
____ 44. Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that includes
a. union with ribosomes.
b. fusion into circular forms known as plasmids.
c. linkage to histone molecules.
d. excision of introns.
e. fusion with other newly transcribed mRNA.
____ 45. Introns are significant to biological evolution because
a. their presence allows exons to be shuffled.
b. they protect the mRNA from degeneration.
c. they are translated into essential amino acids.
d. they maintain the genetic code by preventing incorrect DNA base pairings.
e. they correct enzymatic alterations of DNA bases.
____ 46. Which of the following is (are) true of snRNPs?
a. They are made up of both DNA and RNA.
b. They bind to splice sites at each end of the exon.
c. They join together to form a large structure called the spliceosome.
d. They act only in the cytosol.
e. They attach introns to exons in the correct order.
____ 47. A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA
that binds the mRNA codon is
a. TTT.
b. UUA.
c. UUU.
d. AAA.
e. either UAA or TAA, depending on first base wobble.
____ 48. A part of an mRNA molecule with the following sequence is being read by a ribosome: 5'
CCG-ACG 3' (mRNA). The following charged transfer RNA molecules (with their anticodons
shown in the 3' to 5' direction) are available. Two of them can correctly match the mRNA so that a
dipeptide can form.
Figure 17.3
The dipeptide that will form will be
a. cysteine-alanine.
b. proline-threonine.
c. glycine-cysteine.
d. alanine-alanine.
e. threonine-glycine.
____ 49. What is the most abundant type of RNA?
a. mRNA
b. tRNA
c. rRNA
d. pre-mRNA
e. hnRNA
____ 50. Choose the answer that has these events of protein synthesis in the proper sequence.
1. An aminoacyl-tRNA binds to the A site.
2. A peptide bond forms between the new amino acid and a polypeptide chain.
3. tRNA leaves the P site, and the P site remains vacant.
4. A small ribosomal subunit binds with mRNA.
5. tRNA translocates to the P site.
a. 1, 3, 2, 4, 5
b. 4, 1, 2, 5, 3
c. 5, 4, 3, 2, 1
d. 4, 1, 3, 2, 5
e. 2, 4, 5, 1, 3
____ 51. When a tRNA molecule is shown twisted into an L shape, the form represented is
a. its linear sequence.
b. its 2-dimensional shape.
c. its 3-dimensional shape.
d. its microscopic image.
Use the following information to answer the following questions.
A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the
growing polypeptide on the other tRNA (#2) in the ribosome already.
____ 52. Which enzyme causes a covalent bond to attach lysine to the polypeptide?
a. ATPase
b. lysine synthetase
c. RNA polymerase
d. ligase
e. peptidyl transferase
____ 53. Where does tRNA #2 move to after this bonding of lysine to the polypeptide?
a. A site
b. P site
c. E site
d. Exit tunnel
e. Directly to the cytosol
____ 54. Which component of the complex described enters the exit tunnel through the large subunit of the
ribosome?
a. tRNA with attached lysine (#1)
b. tRNA with polypeptide (#2)
c. tRNA that no longer has attached amino acid
d. newly formed polypeptide
e. initiation and elongation factors
____ 55. The process of translation, whether in prokaryotes or eukaryotes, requires tRNAs, amino acids,
ribosomal subunits, and which of the following?
a. polypeptide factors plus ATP
b. polypeptide factors plus GTP
c. polymerases plus GTP
d. SRP plus chaperones
e. signal peptides plus release factor
____ 56. When the ribosome reaches a stop codon on the mRNA, no corresponding tRNA enters the A site. If
the translation reaction were to be experimentally stopped at this point, which of the following
would you be able to isolate?
a. an assembled ribosome with a polypeptide attached to the tRNA in the P site
b. separated ribosomal subunits, a polypeptide, and free tRNA
c. an assembled ribosome with a separated polypeptide
d. separated ribosomal subunits with a polypeptide attached to the tRNA
e. a cell with fewer ribosomes
____ 57. A frameshift mutation could result from
a. a base insertion only.
b. a base deletion only.
c. a base substitution only.
d. deletion of three consecutive bases.
e. either an insertion or a deletion of a base.
____ 58. Which point mutation would be most likely to have a catastrophic effect on the functioning of a
protein?
a. a base substitution
b. a base deletion near the start of a gene
c. a base deletion near the end of the coding sequence, but not in the terminator
codon
d. deletion of three bases near the start of the coding sequence, but not in the initiator
codon
e. a base insertion near the end of the coding sequence, but not in the terminator
codon
____ 59. Which of the following statements are true about protein synthesis in prokaryotes?
a. Extensive RNA processing is required before prokaryotic transcripts can be
translated.
b. Translation can begin while transcription is still in progress.
c. Prokaryotic cells have complicated mechanisms for targeting proteins to the
appropriate cellular organelles.
d. Translation requires antibiotic activity.
e. Unlike eukaryotes, prokaryotes require no initiation or elongation factors.
____ 60. Of the following, which is the most current description of a gene?
a. a unit of heredity that causes formation of a phenotypic characteristic
b. a DNA subunit that codes for a single complete protein
c. a DNA sequence that is expressed to form a functional product: either RNA or
polypeptide
d. a DNA—RNA sequence combination that results in an enzymatic product
e. a discrete unit of hereditary information that consists of a sequence of amino acids
____ 61. What does the operon model attempt to explain?
a. the coordinated control of gene expression in bacteria
b. bacterial resistance to antibiotics
c. how genes move between homologous regions of DNA
d. the mechanism of viral attachment to a host cell
e. horizontal transmission of plant viruses
____ 62. The tryptophan operon is a repressible operon that is
a. permanently turned on.
b. turned on only when tryptophan is present in the growth medium.
c. turned off only when glucose is present in the growth medium.
d. turned on only when glucose is present in the growth medium.
e. turned off whenever tryptophan is added to the growth medium.
____ 63. This protein is produced by a regulatory gene:
a. operon
b. inducer
c. promoter
d. repressor
e. corepressor
____ 64. When this is taken up by the cell, it binds to the repressor so that the repressor no longer binds to the
operator:
a. operon
b. inducer
c. promoter
d. repressor
e. corepressor
____ 65. The lactose operon is likely to be transcribed when
a. there is more glucose in the cell than lactose.
b. the cyclic AMP levels are low.
c. there is glucose but no lactose in the cell.
d. the cyclic AMP and lactose levels are both high within the cell.
e. the cAMP level is high and the lactose level is low.
____ 66. In both eukaryotes and prokaryotes, gene expression is primarily regulated at the level of
a. transcription.
b. translation.
c. mRNA stability.
d. mRNA splicing.
e. protein stability.
____ 67. Eukaryotic cells can control gene expression by which of the following mechanisms?
a. histone acetylation of nucleosomes
b. DNA acetylation
c. RNA induced modification of chromatin structure
d. repression of operons
e. induction of operators in the promoter
____ 68. This binds to a site in the DNA far from the promoter to stimulate transcription:
a. enhancer
b. promoter
c. activator
d. repressor
e. terminator
____ 69. This can inhibit transcription by blocking the binding of positively acting transcription factors to the
DNA:
a. enhancer
b. promoter
c. activator
d. repressor
e. terminator
____ 70. This is the site in the DNA located near the end of the final exon, encoding an RNA sequence that
determines the 3' end of the transcript:
a. enhancer
b. promoter
c. activator
d. repressor
e. terminator
____ 71. In animals, embryonic stem cells differ from adult stem cells in that
a. embryonic stem cells are totipotent, and adult stem cells are pluripotent.
b. embryonic stem cells are pluripotent, and adult stem cells are totipotent.
c. embryonic stem cells have more genes than adult stem cells.
d. embryonic stem cells have fewer genes than adult stem cells.
e. embryonic stem cells are localized to specific sites within the embryo, whereas
adult stem cells are spread throughout the body.
____ 72. Which of the following statements is true about stem cells?
a. Stem cells can continually reproduce and are not subject to mitotic control.
b. Stem cells can differentiate into specialized cells.
c. Stem cells are found only in bone marrow.
d. Stem cells are found only in the adult human brain.
e. Stem cell DNA lacks introns.
____ 73. In most cases, differentiation is controlled at which level?
a. replication of the DNA
b. nucleosome formation
c. transcription
d. translation
e. post-translational activation of the proteins
____ 74. Which of the following can contribute to the development of cancer?
a. random spontaneous mutations
b. mutations caused by X-rays
c. transposition
d. chromosome translocations
e. all of the above
____ 75. Muscle cells differ from nerve cells mainly because they
a. express different genes.
b. contain different genes.
c. use different genetic codes.
d. have unique ribosomes.
e. have different chromosomes.
____ 76. Viral genomes vary greatly in size and may include from four genes to several hundred genes.
Which of the following viral features is most apt to correlate with the size of the genome?
a. size of the viral capsomeres
b. RNA versus DNA genome
c. double versus single strand genomes
d. size and shape of the capsid
e. glycoproteins of the envelope
____ 77. Which of the following molecules make up the viral envelope?
a.
b.
c.
d.
e.
glycoproteins
proteosugars
carbopeptides
peptidocarbs
carboproteins
____ 78. Most human-infecting viruses are maintained in the human population only. However, a zoonosis is
a disease that is transmitted from other vertebrates to humans, at least sporadically, without requiring
viral mutation. Which of the following is the best example of a zoonosis?
a. rabies
b. herpesvirus
c. smallpox
d. HIV
e. hepatitis virus
____ 79. Which of the following is characteristic of the lytic cycle?
a. Many bacterial cells containing viral DNA are produced.
b. Viral DNA is incorporated into the host genome.
c. The viral genome replicates without destroying the host.
d. A large number of phages is released at a time.
e. The virus-host relationship usually lasts for generations.
____ 80. Which of the following terms describes bacteriophage DNA that has become integrated into the host
cell chromosome?
a. intemperate bacteriophages
b. transposons
c. prophages
d. T-even phages
e. plasmids
____ 81. What is the name given to viruses that are single-stranded RNA that acts as a template for DNA
synthesis?
a. retroviruses
b. proviruses
c. viroids
d. bacteriophages
e. lytic phages
____ 82. What is the function of reverse transcriptase in retroviruses?
a. It hydrolyzes the host cell's DNA.
b. It uses viral RNA as a template for DNA synthesis.
c. It converts host cell RNA into viral DNA.
d. It translates viral RNA into proteins.
e. It uses viral RNA as a template for making complementary RNA strands.
____ 83. Which of the following describes plant virus infections?
a. They can be controlled by the use of antibiotics.
b. They are spread throughout a plant by passing through the plasmodesmata.
c. They have little effect on plant growth.
d. They are seldom spread by insects.
e. They can never be inherited from a parent.
____ 84. Which of the following represents a difference between viruses and viroids?
a. Viruses infect many types of cells, whereas viroids infect only prokaryotic cells.
b. Viruses have capsids composed of protein, whereas viroids have no capsids.
c. Viruses contain introns; viroids have only exons.
d. Viruses always have genomes composed of DNA, whereas viroids always have
genomes composed of RNA.
e. Viruses cannot pass through plasmodesmata; viroids can.
____ 85. What are prions?
a. misfolded versions of normal brain protein
b. tiny molecules of RNA that infect plants
c. viral DNA that has had to attach itself to the host genome
d. viruses that invade bacteria
e. a mobile segment of DNA
____ 86. Which of the following is the most probable fate of a newly emerging virus that causes high
mortality in its host?
a. It is able to spread to a large number of new hosts quickly because the new hosts
have no immunological memory of them.
b. The new virus replicates quickly and undergoes rapid adaptation to a series of
divergent hosts.
c. A change in environmental conditions such as weather patterns quickly forces the
new virus to invade new areas.
d. Sporadic outbreaks will be followed almost immediately by a widespread
pandemic.
e. The newly emerging virus will die out rather quickly or will mutate to be far less
lethal.
____ 87. RNA viruses require their own supply of certain enzymes because
a. host cells rapidly destroy the viruses.
b. host cells lack enzymes that can replicate the viral genome.
c. these enzymes translate viral mRNA into proteins.
d. these enzymes penetrate host cell membranes.
e. these enzymes cannot be made in host cells.
____ 88. Which of the following characteristics, structures, or processes is common to both bacteria and
viruses?
a. metabolism
b. ribosomes
c. genetic material composed of nucleic acid
d. cell division
e. independent existence
____ 89.
Figure 20.1
Which enzyme was used to produce the molecule in Figure 20.1?
a. ligase
b. transcriptase
c. a restriction enzyme
d. RNA polymerase
e. DNA polymerase
____ 90. What is the enzymatic function of restriction enzymes?
a. to add new nucleotides to the growing strand of DNA
b. to join nucleotides during replication
c. to join nucleotides during transcription
d. to cleave nucleic acids at specific sites
e. to repair breaks in sugar-phosphate backbones
____ 91. What is the most logical sequence of steps for splicing foreign DNA into a plasmid and inserting the
plasmid into a bacterium?
I. Transform bacteria with recombinant DNA molecule.
II. Cut the plasmid DNA using restriction enzymes.
III. Extract plasmid DNA from bacterial cells.
IV. Hydrogen-bond the plasmid DNA to nonplasmid DNA fragments.
V. Use ligase to seal plasmid DNA to nonplasmid DNA.
a. I, II, IV, III, V
b. II, III, V, IV, I
c. III, II, IV, V, I
d. III, IV, V, I, II
e. IV, V, I, II, III
Use the following information to answer the questions below.
A eukaryotic gene has "sticky ends" produced by the restriction endonuclease EcoRI. The gene is
added to a mixture containing EcoRI and a bacterial plasmid that carries two genes conferring
resistance to ampicillin and tetracycline. The plasmid has one recognition site for EcoRI located in
the tetracycline resistance gene. This mixture is incubated for several hours, exposed to DNA ligase,
and then added to bacteria growing in nutrient broth. The bacteria are allowed to grow overnight and
are streaked on a plate using a technique that produces isolated colonies that are clones of the
original. Samples of these colonies are then grown in four different media: nutrient broth plus
ampicillin, nutrient broth plus tetracycline, nutrient broth plus ampicillin and tetracycline, and
nutrient broth without antibiotics.
____ 92. Bacteria that contain the plasmid, but not the eukaryotic gene, would grow
a. in the nutrient broth plus ampicillin, but not in the broth containing tetracycline.
b.
c.
d.
e.
only in the broth containing both antibiotics.
in the broth containing tetracycline, but not in the broth containing ampicillin.
in all four types of broth.
in the nutrient broth without antibiotics only.
____ 93. Bacteria containing a plasmid into which the eukaryotic gene has integrated would grow in
a. the nutrient broth only.
b. the nutrient broth and the tetracycline broth only.
c. the nutrient broth, the ampicillin broth, and the tetracycline broth.
d. all four types of broth.
e. the ampicillin broth and the nutrient broth.
____ 94. Bacteria that do not take up any plasmids would grow on which media?
a. the nutrient broth only
b. the nutrient broth and the tetracycline broth
c. the nutrient broth and the ampicillin broth
d. the tetracycline broth and the ampicillin broth
e. all four broths
____ 95. A principal problem with inserting an unmodified mammalian gene into a bacterial plasmid, and
then getting that gene expressed in bacteria, is that
a. prokaryotes use a different genetic code from that of eukaryotes.
b. bacteria translate polycistronic messages only.
c. bacteria cannot remove eukaryotic introns.
d. bacterial RNA polymerase cannot make RNA complementary to mammalian
DNA.
e. bacterial DNA is not found in a membrane-bounded nucleus and is therefore
incompatible with mammalian DNA.
____ 96. A gene that contains introns can be made shorter (but remain functional) for genetic engineering
purposes by using
a. RNA polymerase to transcribe the gene.
b. a restriction enzyme to cut the gene into shorter pieces.
c. reverse transcriptase to reconstruct the gene from its mRNA.
d. DNA polymerase to reconstruct the gene from its polypeptide product.
e. DNA ligase to put together fragments of the DNA that codes for a particular
polypeptide.
____ 97. Why are yeast cells frequently used as hosts for cloning?
a. they easily form colonies
b. they can remove exons from mRNA.
c. they do not have plasmids.
d. they are eukaryotic cells
e. only yeast cells allow the gene to be cloned
____ 98. The major advantage of using artificial chromosomes such as YACs and BACs for cloning genes is
that
a. plasmids are unable to replicate in cells.
b. only one copy of a plasmid can be present in any given cell, whereas many copies
of a YAC or BAC can coexist in a single cell.
c. YACs and BACs can carry much larger DNA fragments than ordinary plasmids
can.
d. YACs and BACs can be used to express proteins encoded by inserted genes, but
plasmids cannot.
e. all of the above
____ 99. Which of the following produces multiple identical copies of a gene for basic research or for
large-scale production of a gene product?
a. restriction enzymes
b. gene cloning
c. DNA ligase
d. gel electrophoresis
e. reverse transcriptase
____ 100. Which of the following seals the sticky ends of restriction fragments to make recombinant DNA?
a. restriction enzymes
b. gene cloning
c. DNA ligase
d. gel electrophoresis
e. reverse transcriptase
____ 101. Which of the following is used to make complementary DNA (cDNA) from RNA?
a. restriction enzymes
b. gene cloning
c. DNA ligase
d. gel electrophoresis
e. reverse transcriptase
____ 102. Which of the following cuts DNA molecules at specific locations?
a. restriction enzymes
b. gene cloning
c. DNA ligase
d. gel electrophoresis
e. reverse transcriptase
____ 103. Which of the following separates molecules by movement due to size and electrical charge?
a. restriction enzymes
b. gene cloning
c. DNA ligase
d. gel electrophoresis
e. reverse transcriptase
____ 104. Restriction fragments of DNA are typically separated from one another by which process?
a. filtering
b. centrifugation
c. gel electrophoresis
d. PCR
e. electron microscopy
____ 105. RFLP analysis can be used to distinguish between alleles based on differences in which of the
following?
a. restriction enzyme recognition sites between the alleles
b. the amount of DNA amplified from the alleles during PCR
c. the ability of the alleles to be replicated in bacterial cells
d. the proteins expressed from the alleles
e. the ability of nucleic acid probes to hybridize to the alleles
____ 106. Which was developed by a British researcher and causes DNA sequences to be transferred to a
membrane and identified with a probe?
a. Southern blotting
b. Northern blotting
c. Western blotting
d. Eastern blotting
e. RT-PCR
____ 107. Which describes the transfer of polypeptide sequences to a membrane to analyze gene expression?
a. Southern blotting
b. Northern blotting
c. Western blotting
d. Eastern blotting
e. RT-PCR
____ 108. Which uses reverse transcriptase to make cDNA followed by amplification?
a. Southern blotting
b. Northern blotting
c. Western blotting
d. Eastern blotting
e. RT-PCR
____ 109. In 1997, Dolly the sheep was cloned. Which of the following processes was used?
a. use of mitochondrial DNA from adult female cells of another ewe
b. replication and dedifferentiation of adult stem cells from sheep bone marrow
c. separation of an early stage sheep blastula into separate cells, one of which was
incubated in a surrogate ewe
d. fusion of an adult cell's nucleus with an enucleated sheep egg, followed by
incubation in a surrogate
e. isolation of stem cells from a lamb embryo and production of a zygote equivalent
Use Figure 20.3 to answer the following questions. The DNA profiles below represent four different
individuals.
Figure 20.3
____ 110. Which of the following statements is consistent with the results?
a. B is the child of A and C.
b. C is the child of A and B.
c. D is the child of B and C.
d. A is the child of B and C.
e. A is the child of C and D.
____ 111. Which of the following statements is most likely true?
a. D is the child of A and C.
b. D is the child of A and B.
c. D is the child of B and C.
d. A is the child of C and D.
e. B is the child of A and C.
____ 112. Which of the following are probably siblings?
a. A and B
b. A and C
c. A and D
d. C and D
e. B and D
____ 113. Genetically engineered plants
a. are more difficult to engineer than animals.
b. include a transgenic rice plant that can help prevent vitamin A deficiency.
c. are being rapidly developed, but traditional plant breeding programs are still the
only method used to develop new plants.
d. are able to fix nitrogen themselves.
e. are banned throughout the world.
____ 114. Plants are more readily manipulated by genetic engineering than are animals because
a. plant genes do not contain introns.
b.
c.
d.
e.
more vectors are available for transferring recombinant DNA into plant cells.
a somatic plant cell can often give rise to a complete plant.
genes can be inserted into plant cells by microinjection.
plant cells have larger nuclei.
____ 115. DNA technology has many medical applications. Which of the following is not done routinely at
present?
a. production of hormones for treating diabetes and dwarfism
b. production of viral proteins for vaccines
c. introduction of genetically engineered genes into human gametes
d. prenatal identification of genetic disease genes
e. genetic testing for carriers of harmful alleles
AP Bio Chapter 16-20 Practice test
Answer Section
MULTIPLE CHOICE
1. ANS:
MSC:
2. ANS:
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3. ANS:
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4. ANS:
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5. ANS:
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6. ANS:
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16. ANS:
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17. ANS:
MSC:
18. ANS:
MSC:
19. ANS:
MSC:
20. ANS:
MSC:
21. ANS:
MSC:
22. ANS:
B
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Application/Analysis
C
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Application/Analysis
B
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Synthesis/Evaluation
E
PTS: 1
Synthesis/Evaluation
D
PTS: 1
Synthesis/Evaluation
E
PTS: 1
Application/Analysis
C
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
C
PTS: 1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.1
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
MSC:
23. ANS:
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24. ANS:
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25. ANS:
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26. ANS:
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27. ANS:
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28. ANS:
29. ANS:
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36. ANS:
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37. ANS:
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38. ANS:
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39. ANS:
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40. ANS:
MSC:
41. ANS:
MSC:
42. ANS:
MSC:
43. ANS:
MSC:
44. ANS:
MSC:
45. ANS:
MSC:
46. ANS:
MSC:
47. ANS:
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
C
PTS: 1
C
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Application/Analysis
A
PTS: 1
Application/Analysis
E
PTS: 1
Application/Analysis
D
PTS: 1
Application/Analysis
C
PTS: 1
Application/Analysis
E
PTS: 1
Application/Analysis
A
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Synthesis/Evaluation
C
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
C
PTS: 1
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.2
TOP: Concept 16.3
TOP: Self-Quiz Questions
TOP: Self-Quiz Questions
TOP: Concept 17.1
TOP: Concept 17.1
TOP: Concept 17.1
TOP: Concept 17.1
TOP: Concept 17.1
TOP: Concept 17.1
TOP: Concept 17.1
TOP: Concept 17.2
TOP: Concept 17.2
TOP: Concept 17.2
TOP: Concept 17.2
TOP: Concept 17.3
TOP: Concept 17.3
TOP: Concept 17.3
TOP: Concept 17.3
TOP: Concept 17.3
TOP: Concept 17.3
TOP: Concept 17.4
MSC:
48. ANS:
MSC:
49. ANS:
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50. ANS:
MSC:
51. ANS:
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52. ANS:
MSC:
53. ANS:
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54. ANS:
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66. ANS:
MSC:
67. ANS:
MSC:
68. ANS:
MSC:
69. ANS:
MSC:
70. ANS:
MSC:
71. ANS:
MSC:
Application/Analysis
B
PTS: 1
Application/Analysis
C
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Application/Analysis
D
PTS: 1
Application/Analysis
B
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Synthesis/Evaluation
E
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.4
TOP: Concept 17.5
TOP: Concept 17.5
TOP: Concept 17.6
TOP: Concept 17.6
TOP: Concept 18.1
TOP: Concept 18.1
TOP: Concept 18.1
TOP: Concept 18.1
TOP: Concept 18.1
TOP: Concept 18.2
TOP: Concept 18.2
TOP: Concept 18.2
TOP: Concept 18.2
TOP: Concept 18.2
TOP: Concept 18.4
72. ANS:
MSC:
73. ANS:
MSC:
74. ANS:
MSC:
75. ANS:
76. ANS:
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77. ANS:
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78. ANS:
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79. ANS:
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86. ANS:
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87. ANS:
88. ANS:
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91. ANS:
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92. ANS:
MSC:
93. ANS:
MSC:
94. ANS:
MSC:
95. ANS:
MSC:
96. ANS:
MSC:
97. ANS:
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
A
PTS: 1
D
PTS: 1
Synthesis/Evaluation
A
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Application/Analysis
D
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Synthesis/Evaluation
B
PTS: 1
C
PTS: 1
C
PTS: 1
Application/Analysis
D
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Application/Analysis
E
PTS: 1
Application/Analysis
A
PTS: 1
Application/Analysis
C
PTS: 1
Synthesis/Evaluation
C
PTS: 1
Application/Analysis
D
PTS: 1
TOP: Concept 18.4
TOP: Concept 18.4
TOP: Concept 18.5
TOP: Self-Quiz Questions
TOP: Concept 19.1
TOP: Concept 19.1
TOP: Concept 19.2
TOP: Concept 19.2
TOP: Concept 19.2
TOP: Concept 19.3
TOP: Concept 19.3
TOP: Concept 19.3
TOP: Concept 19.3
TOP: Concept 19.3
TOP: Concept 19.3
TOP: Self-Quiz Questions
TOP: Self-Quiz Questions
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
MSC:
98. ANS:
MSC:
99. ANS:
MSC:
100. ANS:
MSC:
101. ANS:
MSC:
102. ANS:
MSC:
103. ANS:
MSC:
104. ANS:
MSC:
105. ANS:
MSC:
106. ANS:
MSC:
107. ANS:
MSC:
108. ANS:
MSC:
109. ANS:
MSC:
110. ANS:
MSC:
111. ANS:
MSC:
112. ANS:
MSC:
113. ANS:
MSC:
114. ANS:
115. ANS:
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
E
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Knowledge/Comprehension
D
PTS: 1
Knowledge/Comprehension
C
PTS: 1
Knowledge/Comprehension
A
PTS: 1
Application/Analysis
A
PTS: 1
Application/Analysis
C
PTS: 1
Application/Analysis
E
PTS: 1
Application/Analysis
D
PTS: 1
Knowledge/Comprehension
B
PTS: 1
Application/Analysis
B
PTS: 1
Application/Analysis
D
PTS: 1
Application/Analysis
B
PTS: 1
Knowledge/Comprehension
C
PTS: 1
C
PTS: 1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.1
TOP: Concept 20.2
TOP: Concept 20.2
TOP: Concept 20.2
TOP: Concept 20.2
TOP: Concept 20.2
TOP: Concept 20.2
TOP: Concept 20.3
TOP: Concept 20.4
TOP: Concept 20.4
TOP: Concept 20.4
TOP: Concept 20.4
TOP: Self-Quiz Questions
TOP: Self-Quiz Questions