Chapter 13 Flashcards

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.

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

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.

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.

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

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.

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

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

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

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.

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.

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.

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'

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.

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

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'

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

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

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

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.

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.

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.

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

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.

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

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.

Which of the enzymes removes the RNA nucleotides from the primer and adds equivalent
DNA nucleotides to the 3' end of Okazaki fragments?

I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase

Which of the enzymes separates the DNA strands during replication?

I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase

Which of the enzymes covalently connects segments of DNA?

I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase

Which of the enzymes synthesizes short segments of RNA?

I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase

Given the damage caused by UV radiation, the kind of gene affected in those with XP 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 single-strand damage and replace it.
D) the removal of double-strand damaged areas.
E) causing affected skin cells to undergo apoptosis.

Which of the following sets of materials is required by both eukaryotes and prokaryotes for
replication?
A) double-stranded DNA, four kinds of dNTPs, primers, origins of replication
B) topoisomerases, telomerases, polymerases
C) G-C rich regions, polymerases, chromosome nicks
D) nucleosome loosening, four dNTPs, four rNTPs
E) ligase, primers, nucleases

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.

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

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.

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.

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.

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.

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.

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

Which of the following modifications is least 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) acetylating the cytosine bases within the DNA fragment
C) increasing the length of the DNA fragment
D) decreasing the length of the DNA fragment
E) neutralizing the negative charges within the DNA fragment

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.

How does a bacterial cell protect its own DNA from restriction enzymes?
A) by adding methyl groups to adenines and cytosines
B) by using DNA ligase to seal the bacterial DNA into a closed circle
C) by adding histones to protect the double-stranded DNA
D) by forming "sticky ends" of bacterial DNA to prevent the enzyme from attaching
E) by reinforcing the bacterial DNA structure with covalent phosphodiester bonds

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 a 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

Why is it so important to be able to amplify DNA fragments when studying genes?
A) DNA fragments are too small to use individually.
B) A gene may represent only a millionth of the cell's DNA.
C) Restriction enzymes cut DNA into fragments that are too small.
D) A clone requires multiple copies of each gene per clone.
E) It is important to have multiple copies of DNA in the case of laboratory error.A gene may represent only a millionth of the cell's DNA.

The reason for using Taq polymerase for PCR is that
A) it is heat stable and can withstand the heating step of PCR.
B) only minute amounts are needed for each cycle of PCR.
C) it binds more readily than other polymerases to the primers.
D) it has regions that are complementary to the primers.
E) it is heat stable, and it binds more readily than other polymerases to the primers.

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.

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 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.

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) only in the broth containing both antibiotics.
C) in the broth containing tetracycline, but not in the broth containing ampicillin.
D) in all four types of broth.
E) in the nutrient broth without antibiotics only.

Bacteria containing a plasmid into which the eukaryotic gene has integrated would grow
A) in the nutrient broth only.
B) in the nutrient broth and the tetracycline broth only.
C) in the nutrient broth, the ampicillin broth, and the tetracycline broth.
D) in all four types of broth.
E) in the ampicillin broth and the nutrient broth.

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 three broths

A group of six students has taken samples of their own cheek cells, purified the DNA, and used a
restriction enzyme known to cut at zero, one, or two sites in a particular gene of interest.
6) Why might they be conducting such an experiment?
A) to find the location of this gene in the human genome
B) to prepare to isolate the chromosome on which the gene of interest is found
C) to find which of the students has which alleles
D) to collect population data that can be used to assess natural selection
E) to collect population data that can be used to study genetic drift

Analysis of the data obtained shows that two students each have two fragments, two students
each have three fragments, and two students each have one only. What does this demonstrate?
A) Each pair of students has a different gene for this function.
B) The two students who have two fragments have one restriction site in this region.
C) The two students who have two fragments have two restriction sites within this gene.
D) The students with three fragments are said to have "fragile sites."
E) Each of these students is heterozygous for this gene.

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

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.

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

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.

In a nucleosome, the DNA is wrapped around
A) polymerase molecules.
B) ribosomes.
C) histones.
D) a thymine dimer.
E) satellite DNA.

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

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

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) topoisomerase, primase, DNA polymerase
C) topoisomerase, helicase, single-strand binding protein
D) DNA ligase, replication fork proteins, adenylyl cyclase
E) nuclease, topoisomerase, primase