Campbell Biology in Focus, 2e (Urry)
Chapter 13 The Molecular Basis of Inheritance
13.1 Multiple-Choice Questions
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.
2) How is transformation in bacteria most accurately described?
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) After mixing a heat-killed, phosphorescent (light-emitting) strain of bacteria with a living,
nonphosphorescent strain, you discover that some of the living cells are now phosphorescent.
Which observation(s) would provide the best evidence that the ability to phosphoresce 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.
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.
5) In DNA from any species, the amount of adenine equals the amount of thymine, and the
amount of guanine equals the amount of cytosine. Which of the following investigators was
(were) responsible for this discovery?
A) Frederick Griffith
B) Alfred Hershey and Martha Chase
C) Oswald Avery
D) Erwin Chargaff
E) Matthew Meselson and Franklin Stahl
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%
7) Which of the following characteristics 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 frequency of A versus T nucleotides
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 based on which of the following
characteristics?
A) the sequence of bases
B) phosphate-sugar backbones
C) complementary pairing of bases
D) different five-carbon sugars
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) What is meant by the description “antiparallel” regarding the two strands of nucleic acids 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 contains only purines and the other contains only pyrimidines.
11) Which of the following statements describes one difference between DNA replication in
prokaryotes and DNA replication in eukaryotes?
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.
12) Suppose you are provided with an actively dividing culture of Escherichia 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.
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′
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) The DNA strands at the origin of replication will separate, but replication will not progress.
C) The DNA will supercoil.
D) Replication will require a DNA template from another source.
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
16) At a specific area of a chromosome, the following sequence of nucleotides 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′
17) Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA
strands that are aligned in parallel structures. 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
18) In which order do the necessary enzymes act to repair a thymine dimer by nucleotide
excision repair?
A) nuclease, DNA polymerase, RNA primase
B) helicase, DNA polymerase, DNA ligase
C) DNA ligase, nuclease, helicase
D) DNA polymerase, RNA polymerase, DNA ligase
E) nuclease, DNA polymerase, 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 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
20) 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.
21) The leading and the lagging strands of DNA formed during DNA replication 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.
22) 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) replication must progress toward the replication fork.
D) DNA polymerase can only add nucleotides to the free 3′ end.
23) 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
24) 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.
25) 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) nuclease
26) 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.
27) During DNA replication, which of the following enzymes 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
28) During DNA replication, which of the following enzymes separates the DNA strands during
replication?
A) helicase
B) DNA polymerase III
C) ligase
D) DNA polymerase I
E) primase
29) During DNA replication, which of the following enzymes covalently connects segments of
DNA?
A) helicase
B) DNA polymerase III
C) ligase
D) DNA polymerase I
E) primase
30) Which of the following enzymes synthesizes short segments of RNA?
A) helicase
B) DNA polymerase III
C) ligase
D) DNA polymerase I
E) primase
31) Given the damage caused by UV radiation, the kind of gene affected in those with xeroderma
pigmentosum is one whose product is involved with
A) mending of double-strand breaks in the DNA backbone.
B) breakage of cross-strand covalent bonds.
C) the ability to excise damage to single-stranded DNA and replace it.
D) the removal of damaged areas of double-stranded DNA.
E) causing affected skin cells to undergo apoptosis.
32) Which of the following sets of materials is required by both eukaryotes and prokaryotes for
DNA replication?
A) double-stranded DNA, four kinds of DNA nucleotides, primers, origins of replication
B) topoisomerases, telomerases, polymerases
C) G-C rich regions, polymerases, chromosome nicks
D) nucleosome loosening, four kinds of DNA nucleotides, four kinds of RNA nucleotides
E) ligase, primers, nucleases
33) 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
34) Which of the following statements best 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.
35) If a cell were unable to produce histone proteins, which of the following would be a likely
effect?
A) The cell’s DNA couldn’t be packed into its nucleus.
B) Spindle fibers would not form during prophase.
C) Expression of other genes would compensate for the lack of histones.
D) DNA polymerase I would not function properly.
36) Which of the following statements correctly describes one characteristic 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.
37) 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.
38) Which of the following lists of structures correctly 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
39) Which of the following statements correctly 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.
40) Which of the following modifications is most likely to alter the rate at which a DNA
fragment moves through a gel during electrophoresis?
A) altering the nucleotide sequence of the DNA fragment without adding or removing
nucleotides
B) radioactively labeling the cytosine bases within the DNA fragment
C) increasing the length of the DNA fragment
D) leaving the length of the DNA fragment the same
41) Assume that you are trying to insert a gene into a plasmid. Someone gives you a preparation
of genomic DNA that has been cut with restriction enzyme X. The gene you wish to insert has
sites on both ends for cutting by restriction enzyme Y. You have a plasmid with a single site for
Y, but not for X. Your strategy should be to
A) insert the fragments cut with restriction enzyme X directly into the plasmid without cutting
the plasmid.
B) cut the plasmid with restriction enzyme X and insert the fragments cut with restriction
enzyme Y into the plasmid.
C) cut the DNA again with restriction enzyme Y and insert these fragments into the plasmid cut
with the same enzyme.
D) cut the plasmid twice with restriction enzyme Y and ligate the two fragments onto the ends of
the DNA fragments cut with restriction enzyme X.
E) cut the plasmid with restriction enzyme X and then insert the gene into the plasmid.