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
Use the following list of choices for the following question
I.
helicase
II. DNA polymerase III
III. ligase
IV. DNA
polymerase I
V. primase
34) 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
Grains represent radioactive material within the replicating eye.
47) 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, primase
Answer: A