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.
Answer: B
2) How do we describe transformation 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
Answer: E
3) After mixing a heat-killed, phosphorescent strain of
bacteria with a living nonphosphorescent strain, you discover that some of the
living cells are now phosphorescent. Which observations would provide the best
evidence that the ability to fluoresce is a heritable trait?
A) DNA passed from the heat-killed strain to the living
strain.
B) Protein passed from the heat-killed strain to the
living strain.
C) The phosphorescence in the living strain is especially
bright.
D) Descendants of the living cells are also
phosphorescent.
E) Both DNA and protein passed from the heat-killed
strain to the living strain.
Answer: D
4) In trying to determine whether DNA or protein is the
genetic material, Hershey and Chase made use of which of the following facts?
A) DNA contains sulfur, whereas protein does not.
B) DNA contains phosphorus, whereas protein does not.
C) DNA contains nitrogen, whereas protein does not.
D) DNA contains purines, whereas protein includes
pyrimidines.
E) RNA includes ribose, whereas DNA includes deoxyribose
sugars.
Answer: B
5) Which of the following investigators was/were
responsible for the following discovery?
In DNA from any species, the amount of adenine equals the
amount of thymine, and the amount of guanine equals the amount of cytosine.
A) Frederick Griffith
B) Alfred Hershey and Martha Chase
C) Oswald Avery, Maclyn McCarty, and Colin MacLeod
D) Erwin Chargaff
E) Matthew Meselson and Franklin Stahl
Answer: D
6) Cytosine makes up 42% of the nucleotides in a sample
of DNA from an organism. Approximately what percentage of the nucleotides in
this sample will be thymine?
A) 8%
B) 16%
C) 31%
D) 42%
E) It cannot be determined from the information provided.
Answer: A
7) 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
Answer: A
8) It became apparent to Watson and Crick after
completion of their model that the DNA molecule could carry a vast amount of
hereditary information in which of the following?
A) sequence of bases
B) phosphate-sugar backbones
C) complementary pairing of bases
D) side groups of nitrogenous bases
E) different five-carbon sugars
Answer: A
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
Answer: C
10) Replication in prokaryotes differs from replication
in eukaryotes for which of the following reasons?
A) Prokaryotic chromosomes have 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.
Answer: B
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.
Answer: B
12) Suppose you are provided with an actively dividing
culture of E. coli bacteria to which radioactive thymine has been added. What
would happen if a cell replicates once in the presence of this radioactive
base?
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.
Answer: E
13) 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'
Answer: C
14) In E. coli, there is a mutation in a gene called dnaB
that alters the helicase that normally acts at the origin. Which of the
following would you expect as a result of this mutation?
A) No proofreading will occur.
B) No replication fork will be formed.
C) The DNA will supercoil.
D) Replication will occur via RNA polymerase alone.
E) Replication will require a DNA template from another
source.
Answer: B
15) 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
Answer: C
16) 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
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
Answer: C
17) The enzyme telomerase solves the problem of
replication at the ends of linear chromosomes by which method?
A) adding a single 5' cap structure that resists
degradation by nucleases
B) causing specific double-strand DNA breaks that result
in blunt ends on both strands
C) causing linear ends of the newly replicated DNA to
circularize
D) adding numerous short DNA sequences such as TTAGGG,
which form a hairpin turn
E) adding numerous GC pairs which resist hydrolysis and
maintain chromosome integrity
Answer: D
18) 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
Answer: E
19) 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'
Answer: D
20) Polytene chromosomes of Drosophila salivary glands
each consist of multiple identical DNA strands that are aligned in parallel
arrays. How could these arise?
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
Answer: B
21) To repair a thymine dimer 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
Answer: E
22) 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 3' 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
Answer: C
23) The difference between ATP and the nucleoside
triphosphates used during DNA synthesis is that
A) the nucleoside triphosphates have the sugar
deoxyribose; ATP has the sugar ribose.
B) the nucleoside triphosphates have two phosphate
groups; ATP has three phosphate groups.
C) ATP contains three high-energy bonds; the nucleoside
triphosphates have two.
D) ATP is found only in human cells; the nucleoside
triphosphates are found in all animal and plant cells.
E) triphosphate monomers are active in the nucleoside
triphosphates, but not in ATP.
Answer: A
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.
Answer: A
25) A new DNA strand elongates only in the 5' to 3'
direction because
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.
Answer: E
26) What is the function of topoisomerase?
A) relieving strain in the DNA ahead of the replication
fork
B) elongating new DNA at a replication fork by adding
nucleotides to the existing chain
C) adding methyl groups to bases of DNA
D) unwinding of the double helix
E) stabilizing single-stranded DNA at the replication
fork
Answer: A
27) What is the role of DNA ligase in the elongation of
the lagging strand during DNA replication?
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.
Answer: C
28) Which of the following help(s) 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
Answer: D
29) Individuals with the disorder xeroderma pigmentosum
are hypersensitive to sunlight. This occurs because their cells are impaired in
what way?
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.
Answer: D
30) Which of the following would you expect of a
eukaryote lacking telomerase?
A) a high probability of somatic cells becoming cancerous
B) production of Okazaki fragments
C) inability to repair thymine dimers
D) a reduction in chromosome length in gametes
E) high sensitivity to sunlight
Answer: D
Use the following list of choices for the following
question
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
31) Which of the enzymes removes the RNA nucleotides from
the primer and adds equivalent DNA nucleotides to the 3' end of Okazaki
fragments?
A) I
B) II
C) III
D) IV
E) V
Answer: D
Use the following list of choices for the following
question
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
32) Which of the enzymes separates the DNA strands during
replication?
A) I
B) II
C) III
D) IV
E) V
Answer: A
Use the following list of choices for the following
question
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
33) Which of the enzymes covalently connects segments of
DNA?
A) I
B) II
C) III
D) IV
E) V
Answer: C
34. Use the following list of choices for the following
question
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
Which of the enzymes synthesizes short segments of RNA?
A) I
B) II
C) III
D) IV
E) V
Answer: E
35) Which of the following sets of materials are required
by both eukaryotes and prokaryotes for replication?
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
Answer: A
36) Studies of nucleosomes have shown that histones
(except H1) exist in each nucleosome as two kinds of tetramers: one of 2 H2A
molecules and 2 H2B molecules, and the other as 2 H3 and 2 H4 molecules. Which
of the following is supported by this data?
A) DNA can wind itself around either of the two kinds of
tetramers.
B) The two types of tetramers associate to form an
octamer.
C) DNA has to associate with individual histones before
they form tetramers.
D) Only H2A can form associations with DNA molecules.
E) The structure of H3 and H4 molecules is not basic like
that of the other histones.
Answer: B
37) In a linear eukaryotic chromatin sample, which of the
following strands is looped into domains by scaffolding?
A) DNA without attached histones
B) DNA with H1 only
C) the 10-nm chromatin fiber
D) the 30-nm chromatin fiber
E) the metaphase chromosome
Answer: D
38) Which of the following statements describes the
eukaryotic chromosome?
A) It is composed of DNA alone.
B) The nucleosome is its most basic functional subunit.
C) The number of genes on each chromosome is different in
different cell types of an organism.
D) It consists of a single linear molecule of
double-stranded DNA plus proteins.
E) Active transcription occurs on heterochromatin but not
euchromatin.
Answer: D
39) If a cell were unable to produce histone proteins,
which of the following would be a likely effect?
A) There would be an increase in the amount of
"satellite" DNA produced during centrifugation.
B) The cell's DNA couldn't be packed into its nucleus.
C) Spindle fibers would not form during prophase.
D) Amplification of other genes would compensate for the
lack of histones.
E) Pseudogenes would be transcribed to compensate for the
decreased protein in the cell.
Answer: B
40) Which of the following statements is true of
histones?
A) Each nucleosome consists of two molecules of histone
H1.
B) Histone H1 is not present in the nucleosome bead;
instead, it draws the nucleosomes together.
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 or fungi.
E) The mass of histone in chromatin is approximately nine
times the mass of DNA.
Answer: B
41) Why do histones bind tightly to DNA?
A) Histones are positively charged, and DNA is negatively
charged.
B) Histones are negatively charged, and DNA is positively
charged.
C) Both histones and DNA are strongly hydrophobic.
D) Histones are covalently linked to the DNA.
E) Histones are highly hydrophobic, and DNA is
hydrophilic.
Answer: A
42) Which of the following represents the order of
increasingly higher levels of organization of chromatin?
A) nucleosome, 30-nm chromatin fiber, looped domain
B) looped domain, 30-nm chromatin fiber, nucleosome
C) looped domain, nucleosome, 30-nm chromatin fiber
D) nucleosome, looped domain, 30-nm chromatin fiber
E) 30-nm chromatin fiber, nucleosome, looped domain
Answer: A
43) Which of the following statements describes
chromatin?
A) Heterochromatin is composed of DNA, whereas
euchromatin is made of DNA and RNA.
B) Both heterochromatin and euchromatin are found in the
cytoplasm.
C) Heterochromatin is highly condensed, whereas
euchromatin is less compact.
D) Euchromatin is not transcribed, whereas
heterochromatin is transcribed.
E) Only euchromatin is visible under the light
microscope.
Answer: C
44) In the late 1950s, Meselson and Stahl grew bacteria
in a medium containing "heavy" nitrogen (¹⁵N) and then transferred
them to a medium containing ¹⁴N. Which of the results in the figure above would
be expected after one round of DNA replication in the presence of ¹⁴N?
A) A
B) B
C) C
D) D
E) E
Answer: D
45) A space probe returns with a culture of a
microorganism found on a distant planet. Analysis shows that it is a
carbon-based life-form that has DNA. You grow the cells in ¹⁵N medium for
several generations and then transfer them to ¹⁴N medium. Which pattern in the
figure above would you expect if the DNA was replicated in a conservative
manner?
A) A
B) B
C) C
D) D
E) E
Answer: B
46) Once the pattern found after one round of replication
was observed, Meselson and Stahl could be confident of which of the following
conclusions?
A) Replication is semi-conservative.
B) Replication is not dispersive.
C) Replication is not semi-conservative.
D) Replication is not conservative.
E) Replication is neither dispersive nor conservative.
Answer: D
47. Grains represent radioactive material within the
replicating eye.
In an experiment, DNA is allowed to replicate in an
environment with all necessary enzymes, dATP, dCTP, dGTP, and radioactively
labeled dTTP (³H thymidine) for several minutes and then switched to
nonradioactive medium. It is then viewed by electron microscopy and
autoradiography. The figure above represents the results.
Which of the following is the most likely interpretation?
A) There are two replication forks going in opposite
directions.
B) Thymidine is only being added where the DNA strands
are furthest apart.
C) Thymidine is only added at the very beginning of
replication.
D) Replication proceeds in one direction only.
Answer: A
48) For a science fair project, two students decided to
repeat the Hershey and Chase experiment, with modifications. They decided to
label the nitrogen of the DNA, rather than the phosphate. They reasoned that
each nucleotide has only one phosphate and two to five nitrogens. Thus,
labeling the nitrogens would provide a stronger signal than labeling the
phosphates. Why won't this experiment work?
A) There is no radioactive isotope of nitrogen.
B) Radioactive nitrogen has a half-life of 100,000 years,
and the material would be too dangerous for too long.
C) Avery et al. have already concluded that this
experiment showed inconclusive results.
D) Although there are more nitrogens in a nucleotide,
labeled phosphates actually have 16 extra neutrons; therefore, they are more
radioactive.
E) Amino acids (and thus proteins) also have nitrogen
atoms; thus, the radioactivity would not distinguish between DNA and proteins.
Answer: E
49) You briefly expose bacteria undergoing DNA replication
to radioactively labeled nucleotides. When you centrifuge the DNA isolated from
the bacteria, the DNA separates into two classes. One class of labeled DNA
includes very large molecules (thousands or even millions of nucleotides long),
and the other includes short stretches of DNA (several hundred to a few
thousand nucleotides in length). These two classes of DNA probably represent
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.
Answer: A
50) 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.
Answer: C
51) What is the basis for the difference in how the
leading and lagging strands of DNA molecules are synthesized?
A) The origins of replication occur only at the 5' end.
B) Helicases and single-strand binding proteins work at
the 5' end.
C) DNA polymerase can join new nucleotides only to the 3'
end of a growing strand.
D) DNA ligase works only in the 3' → 5' direction.
E) Polymerase can work on only one strand at a time.
Answer: C
52) In analyzing the number of different bases in a DNA
sample, which result would be consistent with the base-pairing rules?
A) A = G
B) A + G = C + T
C) A + T = G + T
D) A = C
E) G = T
Answer: B
53) The elongation of the leading strand during DNA
synthesis
A) progresses away from the replication fork.
B) occurs in the 3' → 5' direction.
C) produces Okazaki fragments.
D) depends on the action of DNA polymerase.
E) does not require a template strand.
Answer: D
54) In a nucleosome, the DNA is wrapped around
A) polymerase molecules.
B) ribosomes.
C) histones.
D) a thymine dimer.
E) satellite DNA.
Answer: C
55) E. coli cells grown on ¹⁵N medium are transferred to
¹⁴N medium and allowed to grow for two more generations (two rounds of DNA
replication). DNA extracted from these cells is centrifuged. What density
distribution of DNA would you expect in this experiment?
A) one high-density and one low-density band
B) one intermediate-density band
C) one high-density and one intermediate-density band
D) one low-density and one intermediate-density band
E) one low-density band
Answer: D
56) A biochemist isolates, purifies, and combines in a
test tube a variety of molecules needed for DNA replication. When she adds some
DNA to the mixture, replication occurs, but each DNA molecule consists of a
normal strand paired with numerous segments of DNA a few hundred nucleotides
long. What has she probably left out of the mixture?
A) DNA polymerase
B) DNA ligase
C) nucleotides
D) Okazaki fragments
E) primase
Answer: B
57) The spontaneous loss of amino groups from adenine in
DNA results in hypoxanthine, an uncommon base, opposite thymine. What
combination of proteins could repair such damage?
A) nuclease, DNA polymerase, DNA ligase
B) telomerase, primase, DNA polymerase
C) telomerase, helicase, single-strand binding protein
D) DNA ligase, replication fork proteins, adenylyl
cyclase
E) nuclease, telomerase, primaseAnswer : A
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