CHAPTER 5
DNA AND CHROMOSOMES
 2009 Garland Science Publishing
The Structure and Function of DNA
5-1
Using terms from the list below, fill in the blanks in the following brief description of the
experiment with Streptococcus pneumoniae that identified which biological molecule
carries heritable genetic information. Some terms may be used more than once.
Cell-free extracts from S-strain cells of S. pneumoniae were fractionated
to __________________ DNA, RNA, protein, and other cell components.
Each fraction was then mixed with __________________ cells of S.
pneumoniae. Its ability to change these into cells with
__________________ properties resembling the __________________
cells was tested by injecting the mixture into mice. Only the fraction
containing __________________ was able to __________________ the
__________________ cells to __________________ (or
__________________ ) cells that could kill mice.
carbohydrate
DNA
identify
label
lipid
nonpathogenic
pathogenic
purify
R-strain
RNA
S-strain
transform
5-2
Many of the breakthroughs in modern biology came after Watson and Crick
published their model of DNA in 1953. In what decade did scientists first identify
chromosomes?
(a)
1880s
(b)
1920s
(c)
1940s
(d)
1780s
5-3
Mitotic chromosomes were first visualized in the 1880s with the use of very simple tools:
a basic light microscope and some dyes. Which of the following characteristics of mitotic
chromosomes reflects how they were named?
(a)
motion
(b)
color
(c)
shape
(d)
location
5-4
In a DNA double helix, _____________________.
(a)
the two DNA strands are identical
(b)
(c)
(d)
purines pair with purines
thymine pairs with cytosine
the two DNA strands run antiparallel
5-5
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A.
DNA molecules, like proteins, consist of a single, long polymeric chain that is
assembled from small monomeric subunits.
B.
The polarity of a DNA strand results from the polarity of the nucleotide subunits.
C.
There are five different nucleotides that become incorporated into a DNA strand.
D.
Hydrogen bonds between each nucleotide hold individual DNA strands together.
5-6
Several experiments were required to demonstrate how traits are inherited. Which
scientist or team of scientists first demonstrated that cells contain some component that
can be transferred to a new population of cells and permanently cause changes in the new
cells?
(a)
Griffith
(b)
Watson and Crick
(c)
Avery, MacLeod, and McCarty
(d)
Hershey and Chase
5-7
Several experiments were required to demonstrate how traits are inherited. Which
scientist or team of scientists obtained definitive results demonstrating that DNA is the
genetic molecule?
(a)
Griffith
(b)
Watson
(c)
Crick
(d)
Hershey and Chase
5-8
Which of the following chemical groups is not used to construct a DNA molecule?
(a)
five-carbon sugar
(b)
phosphate
(c)
nitrogen-containing base
(d)
six-carbon sugar
5-9
Which of the following sequences can fully base-pair with itself?
(a)
5′-AAGCCGAA-3′
(b)
5′-AAGCCGTT-3′
(c)
5′-AAGCGCAA-3′
(d)
5′-AAGCGCTT-3′
5-10
The DNA from two different species can often be distinguished by a difference in the
______________________.
(a)
ratio of A + T to G + C
(b)
ratio of A + G to C + T
(c)
ratio of sugar to phosphate
(d)
5-11
presence of bases other than A, G, C, and T
For a better understanding of DNA structure, it helps to be able to compare physical
characteristics evident from a side view of double-stranded DNA with those of individual
base pairs.
A.
Use brackets to designate the major and minor grooves on Figure Q5-11A and
shade in the surface that will be exposed in the major grove in Figure Q5-11B.
B.
If base pairs were aligned and stacked directly on top of each other, the major and
minor grooves would be linear depressions all along the DNA. Explain why this is
not the actual conformation of a DNA molecule.
Figure Q5-11
5-12
Which DNA base pair is represented in Figure Q5-12?
(a)
A-T
(b)
T-A
(c)
G-C
(d)
C-G
Figure Q5-12
5-13
Use the terms listed to fill in the blanks in Figure Q5-13.
A.
A-T base pair
B.
G-C base pair
C.
deoxyribose
D.
phosphodiester bonds
E.
purine base
F.
pyrimidine base
Figure Q5-13
5-14
The structures of the four bases in DNA are given in Figure Q5-14.
Figure Q5-14
A.
B.
Which are purines and which are pyrimidines?
Which bases pair with each other in double-stranded DNA?
5-15
Using the structures in Figure Q5-15 as a guide, sketch the hydrogen bonds
between the base pairs in DNA. Hint: The bases in the figure are all drawn with
the –NH– that attaches to the sugar at the bottom of the structure.
5-16
Because hydrogen bonds hold the two strands of a DNA molecule together, the
strands can be separated without breaking any covalent bonds. Every unique DNA
molecule “melts” at a different temperature. In this context, Tm, melting
temperature, is the point at which two strands separate, or become denatured.
Order the DNA sequences listed below according to relative melting temperatures
(from lowest Tm to highest Tm). Assume that they all begin as stable doublestranded DNA molecules. Explain your answer.
A.
GGCGCACC
B.
TATTGTCT
C.
GACTCCTG
D.
CTAACTGG
5-17
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A.
Each strand of DNA contains all the information needed to create a new doublestranded DNA molecule with the same sequence information.
B.
All functional DNA sequences inside a cell code for protein products.
C.
Gene expression is the process of duplicating genes during DNA replication.
D.
Gene sequences correspond exactly to the respective protein sequences produced
from them.
5-18
The complete set of information found in a given organism’s DNA is called its
____________.
(a)
genetic code
(b)
coding sequence
(c)
gene
(d)
genome
5-19
The manner in which a gene sequence is related to its respective protein sequence
is referred to as the _________ code.
(a)
protein
(b)
genetic
(c)
translational
(d)
expression
5-20
Given the sequence of one strand of a DNA helix as
5′-GCATTCGTGGGTAG-3′,
give the sequence of the complementary strand and label the 5′ and 3′ ends.
5-21
When double-stranded DNA is heated, the two strands separate into single strands in a
process called melting or denaturation. The temperature at which half of the duplex DNA
molecules are intact and half have melted is defined as the Tm.
A.
Do you think Tm is a constant, or can it depend on other small molecules in the
solution? Do you think high salt concentrations increase, decrease, or have no
effect on Tm?
B.
Under standard conditions, the expected melting temperature in degrees Celsius
can be calculated from the equation Tm = 59.9 + 0.41 [%(G + C)] – [675/length of
duplex]. Does the Tm increase or decrease if there are more G + C (and thus fewer
A + T) base pairs? Does the Tm increase or decrease as the length of DNA
increases? Why?
C.
Calculate the predicted Tm for a stretch of double helix that is 100 nucleotides
long and contains 50% G + C content.
5-22
Consider the structure of the DNA double helix.
A.
You and a friend want to split a double-stranded DNA molecule so you each have
half. Is it better to cut the length of DNA in half so each person has a shorter
length, or to separate the strands and each take one strand? Explain.
B.
In the original 1953 publication describing the discovery of the structure of DNA,
Watson and Crick wrote, “It has not escaped our notice that the specific pairings
we have postulated immediately suggest a possible copying mechanism for the
genetic material.” What did they mean?
5-23
A.
B.
In principle, what would be the minimum number of consecutive nucleotides
necessary to correspond to a single amino acid to produce a workable genetic
code? Assume that each amino acid is encoded by the same number of
nucleotides. Explain your reasoning.
On average, how often would the nucleotide sequence CGATTG be expected to
occur in a DNA strand 4000 bases long? Explain your reasoning.
The Structure of Eucaryotic Chromosomes
5-24
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or phrase
should be used only once.
In eucaryotic __________________, DNA is complexed with proteins to
form __________________. The paternal and maternal copies of human
Chromosome 1 are __________________, whereas the paternal copy of
Chromosome 1 and the maternal copy of Chromosome 3 are
__________________. Cytogeneticists can determine large-scale
chromosomal abnormalities by looking at a patient’s
__________________. Fluorescent molecules can be used to paint a
chromosome by a technique that employs DNA __________________,
and thereby to identify each chromosome by microscopy.
bands
chromatin
chromosomes
condensation
5-25
A.
B.
C.
extended
homologous
hybridization
karyotype
kinetochore
nonhomologous
Define a gene.
Consider two different species of yeast that have similar genome sizes. Is it likely
that they contain the same number of genes? A similar number of chromosomes?
Figure 5-15 in the textbook shows the G + C content and genes found along a
single chromosome. Is there any relationship between the G + C content and the
locations of genes?
5-26
The human genome has enough DNA to stretch more than 2 m. However, this DNA is
not contained in a single molecule; it is divided into linear segments and packaged into
structures called chromosomes. What is the total number of chromosomes found in each
of the somatic cells in your body?
(a)
22
(b)
23
(c)
44
(d)
46
5-27
The number of cells in an average-sized adult human is on the order of 1014. Use
this information, and the estimate that the length of DNA contained in each cell is
2 m, to do the following calculations (look up the necessary distances and show
your working):
A.
Over how many miles would the total DNA from the average human stretch?
B.
How many times would the total DNA from the average human wrap around the
planet Earth at the Equator?
C.
How many times would the total DNA from the average human stretch from
Earth to the Sun and back?
D.
How many times would the total DNA from the average human stretch from the
Earth to Pluto and back?
5-28
The process of sorting human chromosomes pairs by size and morphology is called
karyotyping. A modern method employed for karyotyping is called chromosome
painting. How are individual chromosomes “painted”?
(a)
with a laser
(b)
using fluorescent antibodies
(c)
using fluorescent DNA molecules
(d)
using green fluorescent protein
5-29
The human genome comprises 23 pairs of chromosomes found in nearly every cell in the
body. Answer the quantitative questions below by choosing one of the numbers in the
following list:
23
46
A.
B.
C.
69
92
>200
>109
How many centromeres are in each cell? What is the main function of the
centromere?
How many telomeres are in each cell? What is their main function?
How many replication origins are in each cell? What is their main function?
5-30
Explain the differences between chromosome painting and the older, more traditional
method of staining chromosomes being prepared for karyotyping. Highlight the way in
which each method identifies chromosomes by the unique sequences they contain.
5-31
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A.
Comparing the relative number of chromosome pairs is a good way to determine
whether two species are closely related.
B.
Chromosomes exist at different levels of condensation, depending on the stage of
the cell cycle.
C.
Eucaryotic chromosomes contain many different sites where DNA replication can
be initiated.
D.
The telomere is a specialized DNA sequence where microtubules from the mitotic
spindle attach to the chromosome so that duplicate copies move to opposite ends
of the dividing cell.
5-32
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or phrase
should be used only once.
Each chromosome is a single molecule of __________________ whose
extraordinarily long length can be compacted by as much as
__________________-fold during __________________ and tenfold
more during __________________. This is accomplished by binding to
__________________ that help package the DNA in an orderly manner so
it can fit in the small space delimited by the __________________. The
structure of the DNA–protein complex, called __________________, is
highly __________________ over time.
10,000
100
1000
cell cycle
cell wall
chromatin
chromosome
different
DNA
dynamic
interphase
lipids
mitosis
nuclear envelope
nucleolus
proteins
similar
static
5-33
The images of chromosomes we typically see are isolated from mitotic cells. These
mitotic chromosomes are in the most highly condensed form. Interphase cells contain
chromosomes that are less densely packed and __________________________.
(a)
occupy discrete territories in the nucleus
(b)
share the same nuclear territory as their homolog
(c)
are restricted to the nucleolus
(d)
are completely tangled with other chromosomes
5-34
Figure Q5-34 clearly depicts the nucleolus, a nuclear structure that looks like large dark
region when stained. The other dark speckled regions in this image are the locations of
particularly compact chromosomal segments called ____________.
(a)
euchromatin
(b)
heterochromatin
(c)
nuclear pores
(d)
nucleosomes
Figure Q5-34
5-35
Mitotic chromosomes are _____ times more compact than a DNA molecule in its
extended form.
(a)
10,000
(b)
100,000
(c)
(d)
1000
100
5-36
Interphase chromosomes are about______ times less compact than mitotic chromosomes,
but still are about______ times more compact than a DNA molecule in its extended form.
(a)
10; 1000
(b)
20; 500
(c)
5; 2000
(d)
50; 200
5-37
For each of the following sentences, choose one of the options enclosed in square
brackets to make a correct statement about nucleosomes.
A.
Nucleosomes are present in [procaryotic/eucaryotic] chromosomes, but not in
[procaryotic/eucaryotic] chromosomes.
B.
A nucleosome contains two molecules each of histones [H1 and H2A/H2A and
H2B] as well as of histones H3 and H4.
C.
A nucleosome core particle contains a core of histone with DNA wrapped around
it approximately [twice/three times/four times].
D.
Nucleosomes are aided in their formation by the high proportion of
[acidic/basic/polar] amino acids in histone proteins.
E.
Nucleosome formation compacts the DNA into approximately [one-third/onehundredth/one-thousandth] of its original length.
5-38
The classic “beads-on-a-string” structure is the most decondensed chromatin structure
possible and is produced experimentally. Which chromatin components are not retained
when this structure is generated?
(a)
linker histones
(b)
linker DNA
(c)
nucleosome core particles
(d)
core histones
5-39
Nucleosomes are formed when DNA wraps _____ times around the histone octamer in a
______ coil.
(a)
2.0; right-handed
(b)
2.5; left-handed
(c)
1.7; left-handed
(d)
1.3; right-handed
5-40
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or phrase
should be used only once.
Interphase chromosomes contain both darkly staining
__________________ and more lightly staining __________________.
Genes that are being transcribed are thought to be packaged in a
__________________ condensed type of euchromatin. Nucleosome core
particles are separated from each other by stretches of
__________________ DNA. A string of nucleosomes coils up with the
help of __________________ to form the more compact structure of the
__________________. A __________________ model describes the
structure of the 30 nm fiber. The 30 nm chromatin fiber is further
compacted by the formation of __________________ that emanate from a
central __________________.
30 nm fiber
active chromatin
axis
beads-on-a-string
euchromatin
heterochromatin
histone H1
histone H3
histone H4
less
linker
loops
more
synaptic complex
zigzag
5-41
The octameric histone core is composed of four different histone proteins, assembled in a
stepwise manner. Once the core octamer has been formed, DNA wraps around it to form
a nucleosome core particle. Which of the following histone proteins does not form part of
the octameric core?
(a)
H4
(b)
H2A
(c)
H3
(d)
H1
5-42
The core histones are small, basic proteins that have a globular domain at the C-terminus
and a long extended conformation at the N-terminus. Which of the following is not true
of the N terminal “tail” of these histones?
(a)
It is subject to covalent modifications,
(b)
It extends out of the nucleosome core.
(c)
It binds to DNA in a sequence-specific manner.
(d)
It helps DNA pack tightly.
5-43
Stepwise condensation of linear DNA happens in five different packing processes. Which
of the following four processes has a direct requirement for histone H1?
(a)
formation of “beads-on-a-string”
(b)
formation of the 30 nm fiber
(c)
looping of the 30 nm fiber
(d)
packing of loops to form interphase chromosomes
5-44
Evidence suggests that the replication of DNA packaged into heterochromatin occurs
later than the replication of other chromosomal DNA. What is the simplest possible
explanation for this phenomenon?
The Regulation of Chromosome Structure
5-45
Although the chromatin structure of interphase and mitotic chromosomes is very
compact, DNA-binding proteins and protein complexes must be able to gain access to the
DNA molecule. Chromatin-remodeling complexes provide this access by
__________________.
(a)
recruiting other enzymes
(b)
modifying the N-terminal tails of core histones
(c)
using the energy of ATP hydrolysis to move nucleosomes
(d)
denaturing the DNA by interfering with hydrogen-bonding between base pairs
5-46
You are studying a newly identified chromatin-remodeling complex, which you call
NICRC. You decide to run an in vitro experiment to characterize the activity of the
purified complex. Your molecular toolbox includes: (1) a 400-base-pair DNA molecule
that has a single recognition site for the restriction endonuclease EcoRI, an enzyme that
cleaves internal sites on double-stranded DNA (dsDNA); (2) purified EcoRI enzyme; (3)
purified DNase I, a DNA endonuclease that will cleave dsDNA at nonspecific sites if
they are exposed; and (4) core octamer histones. You are able to assemble core
nucleosomes on this DNA template and test for NICRC activity. Figure Q5-46A
illustrates the DNA template used and indicates both the location of the EcoRI cleavage
site and the size of the DNA fragments that are produced when it cuts. Figure Q5-46B
illustrates how the DNA molecules in your experiment looked after separation according
to size by using gel electrophoresis. Your experiment had a total of six samples, each of
which was treated according to the legend below the gel. The sizes of the DNA fragments
observed are indicated on the left side of the gel.
Figure Q5-46
A
Explain the results in lanes 1–4 and why it is important to have this information
before you begin to test your remodeling complex.
B.
What can you conclude about your purified remodeling complex from the results
in lanes 5 and 6?
5-47
The N-terminal tail of histone H3 can be extensively modified, and depending on the
number, location, and combination of these modifications, these changes may promote
the formation of heterochromatin. What is the result of heterochromatin formation?
(a)
increase in gene expression
(b)
gene silencing
(c)
recruitment of remodeling complexes
(d)
displacement of histone H1
5-48
Methylation and acetylation are common changes made to histone H3, and the specific
combination of these changes is sometimes referred to as the “histone code.” Which of
the following patterns will probably lead to gene silencing?
(a)
lysine 9 methylation
(b)
lysine 4 methylation and lysine 9 acetylation
(c)
lysine 14 acetylation
(d)
lysine 9 acetylation and lysine 14 acetylation
5-49
When there is a well-established segment of heterochromatin on an interphase
chromosome, there is usually a special barrier sequence that prevents the heterochromatin
from expanding along the entire chromosome. Gene A, which is normally expressed, has
been moved by DNA recombination near an area of heterochromatin. None of the
daughter cells produced after this recombination event express gene A, even though its
DNA sequence is unchanged. What is this the best way to describe what has happened to
the function of gene A in these cells?
(a)
barrier destruction
(b)
heterochromatization
(c)
epigenetic inheritance
(d)
euchromatin depletion
5-50
Your friend is working in a lab to study how yeast cells adapt to growth on different
carbon sources. He grew half of his cells in the presence of glucose and the other half in
the presence of galactose. Then he harvested the cells and isolated their DNA with a
gentle procedure that leaves nucleosomes and some higher-order chromatin structures
intact. He treated the DNA briefly with a low concentration of M-nuclease, a special
enzyme that easily degrades protein-free stretches of DNA. After removing all the
proteins, he separated the resulting DNA on the basis of length. Finally, he used a
procedure to visualize only those DNA fragments from a region near a particular gene
called Sweetie or another gene called Salty. The separated DNA fragments are shown in
Figure Q5-50. Each vertical column, called a lane, is from a different sample. DNA spots
near the top of the figure represent DNA molecules that are longer than those near the
bottom. Darker spots contain more DNA than fainter spots. The lanes are as follows:
1.
2.
“marker” containing known DNA fragments of indicated lengths
cells grown in glucose, DNA visualized near Sweetie gene
3.
4.
5.
cells grown in galactose, DNA visualized near Sweetie gene
cells grown in glucose, DNA visualized near Salty gene
cells grown in galactose, DNA visualized near Salty gene
Figure Q5-50
A.
B.
C.
D.
E.
The lowest spot (as observed in lanes 2, 4, and 5) has a length of about 150
nucleotides. Can you propose what it is and how it arose?
What are the spots with longer lengths? Why is there a ladder of spots?
Notice the faint spots and extensive smearing in lane 3, suggesting the DNA could
be cut almost anywhere near the Sweetie gene after growth of the cells in
galactose. This was not observed in the other lanes. What probably happened to
the DNA to change the pattern between lanes 2 and 3?
What kinds of enzyme might have been involved in changing the chromatin
structure between lanes 2 lane 3?
Do you think that gene expression of Sweetie is higher, lower, or the same in
galactose compared to glucose? What about Salty?
CHAPTER 6
DNA REPLICATION, REPAIR, AND RECOMBINATION
 2009 Garland Science Publishing
DNA Replication
6-1
The process of DNA replication requires that each of the parental DNA strands be used
as a ___________________ to produce a duplicate of the opposing strand.
(a)
catalyst
(b)
competitor
(c)
template
(d)
copy
6-2
DNA replication is considered semiconservative because
____________________________.
(a)
after many rounds of DNA replication, the original DNA double helix is still
intact
(b)
each daughter DNA molecule consists of two new strands copied from the parent
DNA molecule
(c)
each daughter DNA molecule consists of one strand from the parent DNA
molecule and one new strand
(d)
new DNA strands must be copied from a DNA template
6-3
The classic experiments conducted by Meselson and Stahl demonstrated that DNA
replication is accomplished by employing a ________________ mechanism.
(a)
continuous
(b)
semiconservative
(c)
dispersive
(d)
conservative
6-4
If the genome of the bacterium E. coli requires about 20 minutes to replicate itself, how
can the genome of the fruit fly Drosophila be replicated in only 3 minutes?
(a)
The Drosophila genome is smaller than the E. coli genome.
(b)
Eucaryotic DNA polymerase synthesizes DNA at a much faster rate than
procaryotic DNA polymerase.
(c)
The nuclear membrane keeps the Drosophila DNA concentrated in one place in
the cell, which increases the rate of polymerization.
(d)
Drosophila DNA contains more origins of replication than E. coli DNA.
6-5
Meselson and Stahl grew cells in media that contained different isotopes of nitrogen (15N
and 14N) so that the DNA molecules produced from these different isotopes could be
distinguished by mass.
A.
Explain how “light” DNA was separated from “heavy” DNA in the Meselson and
Stahl experiments.
B.
C.
Describe the three existing models for DNA replication when these studies were
begun, and explain how one of them was ruled out definitively by the experiment
you described for part A.
What experimental result eliminated the dispersive model of DNA replication?
6-6
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A.
When DNA is being replicated inside a cell, local heating occurs, allowing the
two strands to separate.
B.
DNA replication orgins are typically rich in G-C base pairs.
C.
Meselson and Stahl ruled out the dispersive model for DNA replication.
D.
DNA replication is a bidirectional process that is initiated at multiple locations
along chromosomes in eucaryotic cells.
6-7
Answer the following questions about DNA replication.
A.
On a DNA strand that is being synthesized, which end is growing—the 3′ end, the
5′ end, or both ends? Explain your answer.
B.
On a DNA strand that is being used as a template, where is the copying occurring
relative to the replication origin—3′ of the origin, 5′, or both?
6-8
How does the total number of replication origins in bacterial cells compare with the
number of origins in human cells?
(a)
1 versus 100
(b)
5 versus 500
(c)
10 versus 1000
(d)
1 versus 10,000
6-9
The chromatin structure in eucaryotic cells is much more complicated than that observed
in procaryotic cells. This is thought to be the reason that DNA replication occurs much
faster in procaryotes. How much faster is it?
(a)
2×
(b)
5×
(c)
10×
(d)
100×
6-10
DNA polymerase catalyzes the joining of a nucleotide to a growing DNA strand. What
prevents this enzyme from catalyzing the reverse reaction?
(a)
hydrolysis of PPi to Pi + Pi
(b)
release of PPi from the nucleotide
(c)
hybridization of the new strand to the template
(d)
loss of ATP as an energy source
6-11
Figure Q6-11 shows a replication bubble.
Figure Q6-11
A.
B.
C.
D.
E.
F.
On the figure, indicate where the origin of replication was located (use O).
Label the leading-strand template and the lagging-strand template of the righthand fork [R] as X and Y, respectively.
Indicate by arrows the direction in which the newly made DNA strands (indicated
by dark lines) were synthesized.
Number the Okazaki fragments on each strand 1, 2, and 3 in the order in which
they were synthesized.
Indicate where the most recent DNA synthesis has occurred (use S).
Indicate the direction of movement of the replication forks with arrows.
6-12
Which of the following statements about the newly synthesized strand of a human
chromosome is true?
(a)
It was synthesized from a single origin solely by continuous DNA synthesis.
(b)
It was synthesized from a single origin by a mixture of continuous and
discontinuous DNA synthesis.
(c)
It was synthesized from multiple origins solely by discontinuous DNA synthesis.
(d)
It was synthesized from multiple origins by a mixture of continuous and
discontinuous DNA synthesis.
6-13
You have discovered an “Exo–” mutant form of DNA polymerase in which the 3′-to-5′
exonuclease function has been destroyed but the ability to join nucleotides together is
unchanged. Which of the following properties do you expect the mutant polymerase to
have?
(a)
It will polymerize in both the 5′-to-3′ direction and the 3′-to-5′ direction.
(b)
It will polymerize more slowly than the normal Exo+ polymerase.
(c)
It will fall off the template more frequently than the normal Exo+ polymerase.
(d)
It will be more likely to generate mismatched base pairs.
6-14
A molecule of bacterial DNA introduced into a yeast cell is imported into the nucleus but
fails to replicate the yeast DNA. Where do you think the block to replication arises?
Choose the protein or protein complex below that is most probably responsible for the
failure to replicate bacterial DNA. Give an explanation for your answer.
(a)
primase
(b)
helicase
(c)
DNA polymerase
(d)
initiator proteins
6-15
Most cells in the body of an adult human lack the telomerase enzyme because its gene is
turned off and is therefore not expressed. An important step in the conversion of a normal
cell into a cancer cell, which circumvents normal growth control, is the resumption of
telomerase expression. Explain why telomerase might be necessary for the ability of
cancer cells to divide over and over again.
6-16
Which diagram accurately represents the directionality of DNA strands at one side of a
replication fork?
6-17
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A.
Primase is needed to initiate DNA replication on both the leading strand and the
lagging strand.
B.
The sliding clamp is loaded once on each DNA strand, where it remains
associated until replication is complete.
C.
Telomerase is a DNA polymerase that carries its own RNA molecule to use as a
primer at the end of the lagging strand.
D.
Primase requires a proofreading function that ensures there are no errors in the
RNA primers used for DNA replication.
6-18
Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the
parental strands are antiparallel, DNA replication is accomplished with the use of
two mechanisms: continuous and discontinuous replication. Indicate whether the
following items relate to (1) continuous replication, (2) discontinuous replication,
or (3) both modes of replication.
______ primase
______ single-strand binding protein
______ sliding clamp
______ RNA primers
______ leading strand
______ lagging strand
______ Okazaki fragments
______ DNA helicase
______ DNA ligase
6-19
The synthesis of DNA in living systems occurs in the 5′-to-3′ direction. However,
scientists synthesize short DNA sequences needed for their experiments on an instrument
dedicated to this task.
A.
The chemical synthesis of DNA by this instrument proceeds from the 3′-to-5′
direction. Draw a diagram to show how this is possible and explain the process.
B.
Although 3′-to-5′ synthesis of DNA is chemically possible, it does not occur in
living systems. Why not?
6-20
DNA polymerases are processive, which means that they remain tightly associated with
the template strand while moving rapidly and adding nucleotides to the growing daughter
strand. Which piece of the replication machinery accounts for this characteristic?
(a)
helicase
(b)
sliding clamp
(c)
single-strand binding protein
(d)
primase
6-21
Use the components in the list below to label the diagram of a replication fork in Figure
Q6-21.
A.
DNA polymerase
B.
single-strand binding protein
C.
Okazaki fragment
D.
primase
E.
sliding clamp
F.
RNA primer
G.
DNA helicase
Figure Q6-21
6-22
Researchers have isolated a mutant strain of E. coli that carries a temperature-sensitive
variant of the enzyme DNA ligase. At the permissive temperature, the mutant cells grow
just as well as the wild-type cells. At the nonpermissive temperature, all of the cells in the
culture tube die within 2 hours. DNA from mutant cells grown at the nonpermissive
temperature for 30 minutes is compared with the DNA isolated from cells grown at the
permissive temperature. The results are shown in Figure Q6-22, where DNA molecules
have been separated by size by means of electrophoresis (P, permissive; NP,
nonpermissive). Explain the appearance of a distinct band with a size of 200 base pairs
(bp) in the sample collected at the nonpermissive temperature.
Figure Q6-22
6-23
Telomeres serve as caps at the ends of linear chromosomes. Which of the following is not
true regarding the replication of telomeric sequences?
(a)
The lagging strand telomeres are not completely replicated by DNA polymerase.
(b)
Telomeres are made of repeating sequences.
(c)
Additional repeated sequences are added to the template strand.
(d)
The leading strand doubles back on itself to form a primer for the lagging strand.
DNA Repair
6-24
Sickle-cell anemia is an example of an inherited disease. Individuals with this disorder
have misshapen (sickle-shaped) red blood cells caused by a change in the sequence of the
β-globin gene. What is the nature of the change?
(a)
chromosome loss
(b)
base-pair change
(c)
gene duplication
(d)
base-pair insertion
6-25
Even though DNA polymerase has a proofreading function, it still introduces errors in the
newly synthesized strand at a rate of 1 per 107 nucleotides. To what degree does the
mismatch repair system decrease the error rate arising from DNA replication?
(a)
2-fold
(b)
5-fold
(c)
10-fold
(d)
100-fold
6-26
Which of the choices below represents the correct way to repair the mismatch shown in
Figure Q6-26?
Figure Q6-26
6-27
A mismatched base pair causes a distortion in the DNA backbone. If this were the only
indication of an error in replication, the overall rate of mutation would be much higher.
Explain why.
6-28
Beside the distortion in the DNA backbone caused by a mismatched base pair, what
additional mark is there on eucaryotic DNA to indicate which strand needs to be
repaired?
(a)
a nick in the template strand
(b)
a chemical modification of the new strand
(c)
a nick in the new strand
(d)
a sequence gap in the new strand
6-29
A pregnant mouse is exposed to high levels of a chemical. Many of the mice in her litter
are deformed, but when they are interbred with each other, all their offspring are normal.
Which two of the following statements could explain these results?
(a)
In the deformed mice, somatic cells but not germ cells were mutated.
(b)
The original mouse’s germ cells were mutated.
(c)
In the deformed mice, germ cells but not somatic cells were mutated.
(d)
The toxic chemical affects development but is not mutagenic.
6-30
The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a
multistep process. Which choice below describes the known sequence of events in this
process?
(a)
DNA damage is recognized, the newly synthesized strand is identified by an
existing nick in the backbone, a segment of the new strand is removed by repair
proteins, the gap is filled by DNA polymerase, and the strand is sealed by DNA
ligase.
(b)
DNA repair polymerase simultaneously removes bases ahead of it and
polymerizes the correct sequence behind it as it moves along the template. DNA
ligase seals the nicks in the repaired strand.
(c)
DNA damage is recognized, the newly synthesized strand is identified by an
existing nick in the backbone, a segment of the new strand is removed by an
exonuclease, and the gap is repaired by DNA ligase.
(d)
A nick in the DNA is recognized, DNA repair proteins switch out the wrong base
and insert the correct base, and DNA ligase seals the nick.
6-31
You are examining the DNA sequences that code for the enzyme phosphofructokinase in
skinks and Komodo dragons. You notice that the coding sequence that actually directs the
sequence of amino acids in the enzyme is very similar in the two organisms but that the
surrounding sequences vary quite a bit. What is the most likely explanation for this?
(a)
Coding sequences are repaired more efficiently.
(b)
Coding sequences are replicated more accurately.
(c)
Coding sequences are packaged more tightly in the chromosomes to protect them
from DNA damage.
(d)
Mutations in coding sequences are more likely to be deleterious to the organism
than mutations in noncoding sequences.
6-32
Sometimes chemical damage to DNA can occur just before DNA replication begins, not
giving the repair system enough time to correct the error before the DNA is duplicated.
This gives rise to mutation. If the cytosine in the sequence TCAT is deaminated and not
repaired, which of the following is the point mutation you would observe after this
segment has undergone two rounds of DNA replication?
(a)
TTAT
(b)
TUAT
(c)
TGAT
(d)
TAAT
6-33
During DNA replication in a bacterium, a C is accidentally incorporated instead of an A
into one newly synthesized DNA strand. Imagine that this error was not corrected and
that it has no effect on the ability of the progeny to grow and reproduce.
A.
After this original bacterium has divided once, what proportion of its progeny
would you expect to contain the mutation?
B.
What proportion of its progeny would you expect to contain the mutation after
three more rounds of DNA replication and cell division?
6-34
Sometimes chemical damage to DNA can occur just before DNA replication begins, not
giving the repair system enough time to correct the error before the DNA is duplicated.
This gives rise to mutation. If the adenosine in the sequence TCAT is depurinated and not
repaired, which of the following is the point mutation you would observe after this
segment has undergone two rounds of DNA replication?
(a)
TCGT
(b)
TAT
(c)
TCT
(d)
TGTT
6-35
Which of the following statements is not an accurate statement about thymidine dimers?
(a)
Thymidine dimers can cause the DNA replication machinery to stall.
(b)
Thymidine dimers are covalent links between thymidines on opposite DNA
strands.
(c)
Prolonged exposure to sunlight causes thymidine dimers to form.
(d)
Repair proteins recognize thymidine dimers as a distortion in the DNA backbone.
6-36
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A.
Ionizing radiation and oxidative damage can cause DNA double-strand breaks.
B.
After damaged DNA has been repaired, nicks in the phosphate backbone are
maintained as a way to identify the strand that was repaired.
C.
Depurination of DNA is a rare event that is caused by ultraviolet irradiation.
D.
Nonhomologous end joining is a mechanism that ensures that DNA double-strand
breaks are repaired with a high degree of fidelity to the original DNA sequence.
6-37
Several members of the same family were diagnosed with the same kind of cancer when
they were unusually young. Which one of the following is the most likely explanation for
this phenomenon? It is possible that the individuals with the cancer have
_______________________.
(a)
inherited a cancer-causing gene that suffered a mutstion in an ancestor’s somatic
cells
(b)
inherited a mutation in a gene required for DNA synthesis
(c)
inherited a mutation in a gene required for mismatch repair
(d)
inherited a mutation in a gene required for the synthesis of purine nucleotides
6-38
You have made a collection of mutant fruit flies that are defective in various aspects of
DNA repair. You test each mutant for its hypersensitivity to three DNA-damaging
agents: sunlight, nitrous acid (which causes deamination of cytosine), and formic acid
(which causes depurination). The results are summarized in Table Q6-38, where a “yes”
indicates that the mutant is more sensitive than a normal fly, and blanks indicate normal
sensitivity.
Table Q6-38
A.
B.
Which mutant is most likely to be defective in the DNA repair polymerase?
What aspect of repair is most likely to be affected in the other mutants?
Homologous Recombination and Mobile Genetic Elements and Viruses
6-39
Homologous recombination is an important mechanism in which organisms use a “backup” copy of the DNA as a template to fix double-strand breaks without loss of genetic
information. Which of the following is not necessary for homologous recombination to
occur?
(a)
3′ DNA strand overhangs
(b)
5′ DNA strand overhangs
(c)
a long stretch of sequence similarity
(d)
nucleases
6-40
In addition to the repair of DNA double-strand breaks, homologous recombination is a
mechanism for generating genetic diversity by swapping segments of parental
chromosomes. During which process does swapping occur?
(a)
DNA replication
(b)
DNA repair
(c)
meiosis
(d)
transposition
6-41
Recombination has occurred between the chromosome segments shown in Figure Q6-41.
The genes A and B, and the recessive alleles a and b, are used as markers on the maternal
and paternal chromosomes, respectively. After alignment and homologous
recombination, the specific arrangements of A, B, a, and b have changed.
Figure Q6-41
Which of the choices below correctly indicates the gene combination from the replication
products of the maternal chromosome?
(a)
(b)
(c)
(d)
6-42
AB and aB
ab and Ab
AB and Ab
aB and Ab
The events listed below are all necessary for homologous recombination to occur
properly:
A.
Holliday junction cut and ligated
B.
strand invasion
C.
DNA synthesis
D.
DNA ligation
E.
double-strand break
F.
nucleases create uneven strands
Which of the following is the correct order of events during homologous recombination?
(a)
E, B, F, D, C, A
(b)
B, E, F, D, C, A
(c)
C, E, F, B, D, A
(d)
E, F, B, C, D, A
6-43
Homologous recombination is initiated by double-strand breaks (DSBs) in a
chromosome. DSBs arise from DNA damage caused by harmful chemicals or by
radiation (for example x-rays). During meiosis, the specialized cell division that produces
gametes (sperm and eggs) for sexual reproduction, the cells intentionally cause DSBs so
as to stimulate crossover homologous recombination. If there is not at least one
occurrence of crossing-over within each pair of homologous chromosomes during
meiosis, those noncrossover chromosomes will not segregate properly.
Figure Q6-43
A.
B.
C.
Consider the copy of chromosome 3 that you received from your mother. Is it
identical to the chromosome 3 that she received from her mother (her maternal
chromosome) or identical to the chromosome 3 she received from her father (her
paternal chromosome), or neither? Explain.
Starting with the representation in Figure Q6-43 of the double-stranded maternal
and paternal chromosomes found in your mother, draw two possible
chromosomes you may have received from your mother.
What does this indicate about your resemblance to your grandfather and
grandmother?
6-44
Mobile genetic elements are sometimes called “jumping genes,” because they move from
place to place throughout the genome. The exact mechanism by which they achieve this
mobility depends on the genes contained within the mobile element. Which of the
following mobile genetic elements carry both a transposase gene and a reverse
transcriptase gene?
(a)
L1
(b)
B1
(c)
Alu
(d)
Tn3
6-45
Which of the following is true of a retrovirus but not of the Alu retrotransposon?
(a)
It requires cellular enzymes to make copies.
(b)
It can be inserted into the genome.
(c)
It can be excised and moved to a new location in the genome.
(d)
It encodes its own reverse transcriptase.
6-46
Which of the following DNA sequences is not commonly carried on mobile genetic
elements? You may choose more than one option.
(a)
transposase gene
(b)
Holliday junction
(c)
recognition site for transposase
(d)
antibiotic resistance gene
6-47
HIV is a human retrovirus that integrates into the host cell’s genome and will eventually
replicate, produce viral proteins, and ultimately escape the host cell. Which of the
following proteins is not encoded in the HIV genome?
(a)
reverse transcriptase
(b)
envelope protein
(c)
RNA polymerase
(d)
capsid protein
6-48
Some retrotransposons and retroviruses integrate preferentially into regions of the
chromosome that are packaged in euchromatin and are also located outside the coding
regions of genes that contain information for making a protein. Why might these mobile
genetic elements have evolved this strategy?
CHAPTER 7
FROM DNA TO PROTEIN: HOW CELLS READ THE GENOME
 2009 Garland Science Publishing
From DNA to RNA
7-1
RNA in cells differs from DNA in that ___________________.
(a)
it contains the base uracil, which pairs with cytosine
(b)
it is single-stranded and cannot form base pairs
(c)
it is single-stranded and can fold up into a variety of structures
(d)
the sugar ribose contains fewer oxygen atoms than does deoxyribose
7-2
Transcription is similar to DNA replication in that ___________________.
(a)
an RNA transcript is synthesized discontinuously and the pieces are then
joined together
(b)
it uses the same enzyme as that used to synthesize RNA primers during
DNA replication
(c)
the newly synthesized RNA remains paired to the template DNA
(d)
nucleotide polymerization occurs only in the 5′-to-3′ direction
7-3
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; use each word
or phrase only once.
For a cell’s genetic material to be used, the information is first
copied from the DNA into the nucleotide sequence of RNA in a
process called __________________. Various kinds of RNA are
produced, each with different functions. __________________
molecules code for proteins, __________________ molecules act
as adaptors for protein synthesis, __________________ molecules
are integral components of the ribosome, and
__________________ molecules are important in the splicing of
RNA transcripts.
incorporation
mRNA
pRNA
translation
7-4
rRNA
snRNA
transcription
Match the following structures with their names.
transmembrane
tRNA
proteins
Figure Q7-4
7-5
Which of the following statements is false?
(a)
A new RNA molecule can begin to be synthesized from a gene before the
previous RNA molecule’s synthesis is completed.
(b)
If two genes are to be expressed in a cell, these two genes can be
transcribed with different efficiencies.
(c)
RNA polymerase is responsible for both unwinding the DNA helix and
catalyzing the formation of the phosphodiester bonds between nucleotides.
(d)
Unlike DNA, RNA uses a uracil base and a deoxyribose sugar.
7-6
Which one of the following is the main reason that a typical eucaryotic gene is
able to respond to a far greater variety of regulatory signals than a typical
procaryotic gene or operon?
(a)
Eucaryotes have three types of RNA polymerase.
(b)
Eucaryotic RNA polymerases require general transcription factors.
(c)
The transcription of a eucaryotic gene can be influenced by proteins that
bind far from the promoter.
(d)
Procaryotic genes are packaged into nucleosomes.
7-7
Match the following types of RNA with the main polymerase that transcribes
them.
7-8
List three ways in which the process of eucaryotic transcription differs from the
process of bacterial transcription.
7-9
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or
phrase should be used only once.
In eucaryotic cells, general transcription factors are required for the
activity of all promoters transcribed by RNA polymerase II. The
assembly of the general transcription factors begins with the
binding of the factor __________________ to DNA, causing a
marked local distortion in the DNA. This factor binds at the DNA
sequence called the __________________ box, which is typically
located 25 nucleotides upstream from the transcription start site.
Once RNA polymerase II has been brought to the promoter DNA,
it must be released to begin making transcripts. This release
process is facilitated by the addition of phosphate groups to the tail
of RNA polymerase by the factor __________________. It must
be remembered that the general transcription factors and RNA
polymerase are not sufficient to initiate transcription in the cell and
are affected by proteins bound thousands of nucleotides away from
the promoter. Proteins that link the distantly bound transcription
regulators to RNA polymerase and the general transcription factors
include the large complex of proteins called
the__________________. The packing of DNA into chromatin
also affects transcriptional initiation, and histone
__________________ is an enzyme that can render the DNA less
accessible to the general transcription factors.
activator
CAP
deacetylase
enhancer
7-10
lac
ligase
mediator
TATA
TFIIA
TFIID
TFIIH
You have a piece of DNA that includes the following sequence:
5′-ATAGGCATTCGATCCGGATAGCAT-3′
3′-TATCCGTAAGCTAGGCCTATCGTA-5′
Which of the following RNA molecules could be transcribed from this piece of
DNA?
(a)
(b)
(c)
(d)
7-11
5′-UAUCCGUAAGCUAGGCCUAUGCUA-3′
5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′
5′-UACGAUAGGCCUAGCUUACGGAUA-3′
none of the above
You have a segment of DNA that contains the following sequence:
5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′
3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′
If you know that the RNA transcribed from this segment contains the following
sequence:
5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA–3′
Which of the following choices best describes how transcription occurs?
(a)
The top strand is the template strand; RNA polymerase moves along this
strand from 5′ to 3′.
(b)
The top strand is the template strand; RNA polymerase moves along this
strand from 3′ to 5′.
(c)
The bottom strand is the template strand; RNA polymerase moves along
this strand from 5′ to 3′.
(d)
The bottom strand is the template strand; RNA polymerase moves along
this strand from 3′ to 5′.
7-12
Which of the following molecules of RNA would you predict to be the most
likely to fold into a specific structure as a result of intramolecular base pairing?
(a)
(b)
(c)
(d)
7-13
5′CCCUAAAAAAAAAAAAAAAAUUUUUUUUUUUUUUUUAGGG-3′
5′UGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUG-3′
5′AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA-3′
5′GGAAAAGGAGAUGGGCAAGGGGAAAAGGAGAUGGGCAAGG-3′
Imagine that an RNA polymerase is transcribing a segment of DNA that contains
the following sequence:
5′-AGTCTAGGCACTGA-3′
3′-TCAGATCCGTGACT 5′
A.
B.
7-14
If the polymerase is transcribing from this segment of DNA from left to
right, which strand (top or bottom) is the template?
What will be the sequence of that RNA (be sure to label the 5′ and 3′ ends
of your RNA molecule)?
The sigma subunit of bacterial RNA polymerase ___________________.
(a)
contains the catalytic activity of the polymerase
(b)
remains part of the polymerase throughout transcription
(c)
recognizes promoter sites in the DNA
(d)
recognizes transcription termination sites in the DNA
7-15
Which of the following might decrease the transcription of only one specific gene
in a bacterial cell?
(a)
a decrease in the amount of sigma factor
(b)
a decrease in the amount of RNA polymerase
(c)
a mutation that introduced a stop codon into the DNA that precedes the
gene’s coding sequence
(d)
a mutation that introduced extensive sequence changes into the DNA that
precedes the gene’s transcription start site
7-16
There are several reasons why the primase used to make the RNA primer for
DNA replication is not suitable for gene transcription. Which of the statements
below is not one of those reasons?
(a)
Primase initiates RNA synthesis on a single-stranded DNA template.
(b)
Primase can initiate RNA synthesis without the need for a base-paired
primer.
(c)
Primase synthesizes only RNAs of about 5–20 nucleotides in length.
(d)
The RNA synthesized by primase remains base-paired to the DNA
template.
7-17
You have a bacterial strain with a mutation that removes the transcription
termination signal from the Abd operon. Which of the following statements
describes the most likely effect of this mutation on Abd transcription?
(a)
The Abd RNA will not be produced in the mutant strain.
(b)
The Abd RNA from the mutant strain will be longer than normal.
(c)
Sigma factor will not dissociate from RNA polymerase when the Abd
operon is being transcribed in the mutant strain.
(d)
RNA polymerase will move in a backwards fashion at the Abd operon in
the mutant strain.
7-18
Transcription in bacteria differs from transcription in a eucaryotic cell because
__________________________.
(a)
RNA polymerase (along with its sigma subunit) can initiate transcription
on its own
(b)
RNA polymerase (along with its sigma subunit) requires the general
transcription factors to assemble at the promoter before polymerase can
begin transcription
(c)
The sigma subunit must associate with the appropriate type of RNA
polymerase to produce mRNAs
(d)
RNA polymerase must be phosphorylated at its C-terminal tail for
transcription to proceed
7-19
Which of the following does not occur before a eucaryotic mRNA is exported
from the nucleus?
(a)
The ribosome binds to the mRNA.
(b)
The mRNA is polyadenylated at its 3′ end.
(c)
(d)
7-20
7-methyl-G is added in a 5′ to 5′ linkage to the mRNA.
RNA polymerase dissociates.
Use the numbers in the choices below to indicate where in the schematic diagram
of a eucaryotic cell (Figure Q7-20) those processes take place.
Figure Q7-20
1.
2.
3.
4.
5.
transcription
translation
RNA splicing
polyadenylation
RNA capping
7-21
Total nucleic acids are extracted from a culture of yeast cells and are then mixed
with resin beads to which the polynucleotide 5′TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached. After a
short incubation, the beads are then extracted from the mixture. When you
analyze the cellular nucleic acids that have stuck to the beads, which of the
following is most abundant?
(a)
DNA
(b)
tRNA
(c)
rRNA
(d)
mRNA
7-22
Name three covalent modifications that can be made to an RNA molecule in
eucaryotic cells before the RNA molecule becomes a mature mRNA.
7-23
Which of the following statements about RNA splicing is false?
(a)
Conventional introns are not found in bacterial genes.
(b)
(c)
(d)
For a gene to function properly, every exon must be removed from the
primary transcript in the same fashion on every mRNA molecule produced
from the same gene.
Small RNA molecules in the nucleus perform the splicing reactions
necessary for the removal of introns.
Splicing occurs after the 5′ cap has been added to the end of the primary
transcript.
7-24
The length of a particular gene in human DNA, measured from the start site for
transcription to the end of the protein-coding region, is 10,000 nucleotides,
whereas the length of the mRNA produced from this gene is 4000 nucleotides.
What is the most likely reason for this difference?
7-25
Why is the old dogma “one gene—one protein” not always true for eucaryotic
genes?
7-26
Is this statement true or false? Explain your answer.
“Since introns do not contain protein coding information, they do not have
to be removed precisely (meaning, a nucleotide here and there should not
matter) from the primary transcript during RNA splicing.”
7-27
You have discovered a gene (Figure Q7-27A) that is alternatively spliced to
produce several forms of mRNA in various cell types, three of which are shown
in Figure Q7-27B. The lines connecting the exons that are included in the mRNA
indicate the splicing. From your experiments, you know that protein translation
begins in exon 1. For all forms of the mRNA, the encoded protein sequence is the
same in the regions of the mRNA that correspond to exons 1 and 10. Exons 2 and
3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of
the following statements about exons 2 and 3 is the most accurate? Explain your
answer.
Figure Q7-27
(a)
(b)
(c)
(d)
Exons 2 and 3 must have the same number of nucleotides.
Exons 2 and 3 must contain an integral number of codons (that is, the
number of nucleotides divided by 3 must be an integer).
Exons 2 and 3 must contain a number of nucleotides that when divided by
3, leaves the same remainder (that is, 0, 1, or 2).
Exons 2 and 3 must have different numbers of nucleotides.
From RNA to Protein
7-28
Which of the following statements about the genetic code is correct?
(a)
All codons specify more than one amino acid.
(b)
The genetic code is redundant.
(c)
All amino acids are specified by more than one codon.
(d)
All codons specify an amino acid.
NOTE: The following codon table is to be used for Problems Q7-29 to Q7-36, Q7-43,
and Q7-47.
7-29
The piece of RNA below includes the region that codes the binding site for the
initiator tRNA needed in translation.
5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′
Which amino acid will be on the tRNA that is the first to bind to the A-site of the
ribosome?
(a)
methionine
(b)
arginine
(c)
cystine
(d)
valine
7-30
The following DNA sequence includes the beginning of a sequence coding for a
protein. What would be the result of a mutation that changed the C marked by an
asterisk to an A?
5′-AGGCTATGAATGGACACTGCGAGCCC….
*
7-31
Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to
carry?
(a)
lysine
(b)
glutamic acid
(d)
leucine
(d)
phenylalanine
7-32
Which of the following pairs of codons might you expect to be read by the same
tRNA as a result of wobble?
(a)
CUU and UUU
(b)
GAU and GAA
(c)
CAC and CAU
(d)
AAU and AGU
7-33
Below is a segment of RNA from the middle of an mRNA.
5′-UAGUCUAGGCACUGA-3′
If you were told that this segment of RNA was part of the coding region of an
mRNA for a large protein, give the amino acid sequence for the protein that is
encoded by this segment of mRNA. Write your answer using the one-letter amino
acid code.
7-34
Below is the sequence from the 3′ end of an mRNA.
5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′
If you were told that this sequence contains the stop codon for the protein encoded
by this mRNA, what is the anticodon on the tRNA in the P-site of the ribosome
when release factor binds to the A-site?
(a)
5′-CCA-3′
(b)
5′-CCG-3′
(c)
5′-UGG-3′
(d)
5′-UUA-3′
7-35
One strand of a section of DNA isolated from the bacterium E. coli reads:
5′-GTAGCCTACCCATAGG-3′
A.
B.
C.
7-36
Suppose that an mRNA is transcribed from this DNA using the
complementary strand as a template. What will be the sequence of the
mRNA in this region (make sure you label the 5′ and 3′ ends of the
mRNA)?
How many different peptides could potentially be made from this
sequence of RNA, assuming that translation initiates upstream of this
sequence?
What are these peptides? (Give your answer using the one-letter amino
acid code.)
A strain of yeast translates mRNA into protein inaccurately. Individual molecules
of a particular protein isolated from this yeast have variations in the first 11 amino
acids compared with the sequence of the same protein isolated from normal yeast
cells, as listed in Figure Q7-36. What is the most likely cause of this variation in
protein sequence?
Figure Q7-36
(a)
(b)
(c)
(d)
a mutation in the DNA coding for the protein
a mutation in the anticodon of the isoleucine tRNA (tRNAIle)
a mutation in the isoleucyl-tRNA synthetase that decreases its ability to
distinguish between different amino acids
a mutation in the isoleucyl-tRNA synthetase that decreases its ability to
distinguish between different tRNA molecules
7-37
Which of the following statements is true?
(a)
Ribosomes are large RNA structures composed solely of rRNA.
(b)
Ribosomes are synthesized entirely in the cytoplasm.
(c)
rRNA contains the catalytic activity that joins amino acids together.
(d)
A ribosome binds one tRNA at a time.
7-38
Figure Q7-38A shows the stage in translation when an incoming aminoacyl-tRNA
has bound to the A-site on the ribosome. Using the components shown in Figure
Q7-38A as a guide, show on Figure Q7-38B and Q7-38C what happens in the
next two stages to complete the addition of the new amino acid to the growing
polypeptide chain.
Figure Q7-38
7-39
A poison added to an in vitro translation mixture containing mRNA molecules
with the sequence 5′-AUGAAAAAAAAAAAAUAA-3′ has the following effect:
the only product made is a Met-Lys dipeptide that remains attached to the
ribosome. What is the most likely way in which the poison acts to inhibit protein
synthesis?
(a)
It inhibits peptidyl transferase activity.
(b)
It inhibits movement of the small subunit relative to the large subunit.
(c)
It inhibits release factor.
(d)
It mimics release factor.
7-40
In eucaryotes, but not in procaryotes, ribosomes find the start site of translation by
____________________________.
(a)
binding directly to a ribosome-binding site preceding the initiation codon
(b)
scanning along the mRNA from the 5′ end
(c)
recognizing an AUG codon as the start of translation
(d)
binding an initiator tRNA
7-41
Which of the following statements about procaryotic mRNA molecules is false?
(a)
A single procaryotic mRNA molecule can be translated into several
proteins.
(b)
Ribosomes must bind to the 5′ cap before initiating translation.
(c)
mRNAs are not polyadenylated.
(d)
Ribosomes can start translating an mRNA molecule before transcription is
complete.
7-42
Figure Q7-42 shows an mRNA molecule.
Figure Q7-42
A.
B.
Match the labels given in the list below with the label lines in Figure Q742.
(a)
ribosome-binding site
(b)
initiator codon
(c)
stop codon
(d)
untranslated 3′ region
(e)
untranslated 5′ region
(f)
protein-coding region
Is the mRNA shown procaryotic or eucaryotic? Explain your answer.
7-43
A mutation in the tRNA for the amino acid lysine results in the anticodon
sequence 5′-UAU-3′ (instead of 5′-UUU-3′). Which of the following aberrations
in protein synthesis might this tRNA cause? (Refer to the codon table provided
above Q7-29.)
(a)
read-through of stop codons
(b)
substitution of lysine for isoleucine
(c)
substitution of lysine for tyrosine
(d)
substitution of lysine for phenylalanine
7-44
You have discovered a protein that inhibits translation. When you add this
inhibitor to a mixture capable of translating human mRNA and centrifuge the
mixture to separate polyribosomes and single ribosomes, you obtain the results
shown in Figure Q7-44. Which of the following interpretations is consistent with
these observations?
Figure Q7-44
(a)
(b)
(c)
(d)
The protein binds to the small ribosomal subunit and increases the rate of
initiation of translation.
The protein binds to sequences in the 5′ region of the mRNA and inhibits
the rate of initiation of translation.
The protein binds to the large ribosomal subunit and slows down
elongation of the polypeptide chain.
The protein binds to sequences in the 3′ region of the mRNA and prevents
termination of translation.
7-45
The concentration of a particular protein X in a normal human cell rises gradually
from a low point, immediately after cell division, to a high point, just before cell
division, and then drops sharply. The level of its mRNA in the cell remains fairly
constant throughout this time. Protein X is required for cell growth and survival,
but the drop in its level just before cell division is essential for division to
proceed. You have isolated a line of human cells that grow in size in culture but
cannot divide, and on analyzing these mutants, you find that levels of X mRNA in
the mutant cells are normal. Which of the following mutations in the gene for X
could explain these results?
(a)
the introduction of a stop codon that truncates protein X at the fourth
amino acid
(b)
a change of the first ATG codon to CCA
(c)
the deletion of a sequence that encodes sites at which ubiquitin can be
attached to the protein
(d)
a change at a splice site that prevents splicing of the RNA
7-46
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; use each word
or phrase only once.
Once an mRNA is produced, its message can be decoded on
ribosomes. The ribosome is composed of two subunits: the
__________________ subunit, which catalyzes the formation of
the peptide bonds that link the amino acids together into a
polypeptide chain, and the __________________ subunit, which
matches the tRNAs to the codons of the mRNA. During the chain
elongation process of translating an mRNA into protein, the
growing polypeptide chain attached to a tRNA is bound to the
__________________-site of the ribosome. An incoming
aminoacyl-tRNA carrying the next amino acid in the chain will
bind to the __________________-site by forming base pairs with
the exposed codon in the mRNA. The __________________
enzyme catalyzes the formation of a new peptide bond between the
growing polypeptide chain and the newly arriving amino acid. The
end of a protein-coding message is signaled by the presence of a
stop codon, which binds the __________________ called release
factor. Eventually, most proteins will be degraded by a large
complex of proteolytic enzymes called the __________________.
A
central
DNA
E
large
7-47
medium
P
peptidyl transferase
polymerase
protein
proteasome
RNA
small
T
ubiquitin
After treating cells with a mutagen, you isolate two mutants. One carries alanine
and the other carries methionine at a site in the protein that normally contains
valine. After treating these two mutants again with mutagen, you isolate mutants
from each that now carry threonine at the site of the original valine (see Figure
Q7-47). Assuming that all mutations caused by the mutagen are due to single
nucleotide changes, deduce the codons that are used for valine, alanine,
methionine, and threonine at the affected site. (Refer to the codon table
provided above Q7-29.)
Figure Q7-47
7-48
Which of the following methods is not used by cells to regulate the amount of a
protein in the cell?
(a)
Genes can be transcribed into mRNA with different efficiencies.
(b)
Many ribosomes can bind to a single mRNA molecule.
(c)
(d)
7-49
Proteins can be tagged with ubiquitin, marking them for degradation.
Nuclear pore complexes can regulate the speed at which newly
synthesized proteins are exported from the nucleus into the cytoplasm.
Which of the following statements about the proteasome is false?
(a)
Ubiquitin is a small protein that is covalently attached to proteins to mark
them for delivery to the proteasome.
(b)
Proteases reside in the central cylinder of a proteasome.
(c)
Misfolded proteins are delivered to the proteasome, where they are
sequestered from the cytoplasm and can attempt to refold.
(d)
The protein stoppers that surround the central cylinder of the proteasome
use the energy from ATP hydrolysis to move proteins into the proteasome
inner chamber.
RNA and the Origins of Life
7-50
Which of the following molecules is thought to have arisen first during evolution?
(a)
protein
(b)
DNA
(c)
RNA
(d)
All came to be at the same time.
7-51
According to current thinking, the minimum requirement for life to have
originated on Earth was the formation of a _______________.
(a)
molecule that could provide a template for the production of a
complementary molecule
(b)
double-stranded DNA helix
(c)
molecule that could direct protein synthesis
(d)
molecule that could catalyze its own replication
7-52
Ribozymes catalyze which of the following reactions?
(a)
DNA synthesis
(b)
transcription
(c)
RNA splicing
(d)
protein hydrolysis
7-53
You are studying a disease that is caused by a virus, but when you purify the virus
particles and analyze them you find they contain no trace of DNA. Which of the
following molecules are likely to contain the genetic information of the virus?
(a)
high-energy phosphate groups
(b)
RNA
(c)
lipids
(d)
carbohydrates
7-54
Give a reason why DNA makes a better material than RNA for the storage of
genetic information, and explain your answer.
How We Know: Cracking the Genetic Code
7-55
You have discovered an alien life form that surprisingly uses DNA as its genetic
material, makes RNA from DNA, and reads the information from RNA to make
protein using ribosomes and tRNAs, which read triplet codons. Because it is your
job to decipher the genetic code for this alien, you synthesize some artificial RNA
molecules and examine the protein products produced from these RNA molecules
in a cell-free translation system using purified alien tRNAs and ribosomes. You
obtain the results shown in Table Q7-55.
Table Q7-55
From this information, which of the following peptides can be produced from
poly UAUC?
(a)
Ile-Phe-Val-Tyr
(b)
Tyr-Ser-Phe-Ala
(c)
Ile-Lys-His-Tyr
(d)
Cys-Pro-Lys-Ala
7-56
An extraterrestrial organism (ET) is discovered whose basic cell biology seems
pretty much the same as that of terrestrial organisms except that it uses a different
genetic code to translate RNA into protein. You set out to break the code by
translation experiments using RNAs of known sequence and cell-free extracts of
ET cells to supply the necessary protein-synthesizing machinery. In experiments
using the RNAs below, the following results were obtained when the 20 possible
amino acids were added either singly or in different combinations of two or three:
RNA 1: 5′-GCGCGCGCGCGCGCGCGCGCGCGCGCGC-3′
RNA 2: 5′-GCCGCCGCCGCCGCCGCCGCCGCCGCCGCC-3′
Using RNA 1, a polypeptide was produced only if alanine and valine were added
to the reaction mixture. Using RNA 2, a polypeptide was produced only if leucine
and serine and cysteine were added to the reaction mixture. Assuming that protein
synthesis can start anywhere on the template, that the ET genetic code is
nonoverlapping and linear, and that each codon is the same length (like the
terrestrial triplet code), how many nucleotides does an ET codon contain?
(a)
2
(b)
3
(c)
4
(d)
5
(e)
6
7-57
NASA has discovered an alien life form. You are called in to help NASA
scientists to deduce the genetic code for this alien. Surprisingly, this alien life
form shares many similarities with life on Earth: this alien uses DNA as its
genetic material, makes RNA from DNA, and reads the information from RNA to
make protein using ribosomes and tRNAs. Even more amazing, this alien uses the
same 20 amino acids, like the organisms found on Earth, and also codes for each
amino acid by a triplet codon. However, the scientists at NASA have found that
the genetic code used by the alien life form differs from that used by life on Earth.
NASA scientists drew this conclusion after creating a cell-free protein synthesis
system from alien cells and adding an mRNA made entirely of uracil (poly U).
They found that poly U directs the synthesis of a peptide containing only glycine.
NASA scientists have synthesized a poly AU mRNA and observe that it codes for
a polypeptide of alternating serine and proline amino acids. From these
experiments, can you determine which codons code for serine and proline?
Explain.
Bonus question. Can you propose a mechanism for how the alien’s physiology is
altered so that it uses a different genetic code from life on Earth, despite all the
similarities?
CHAPTER 8
CONTROL OF GENE EXPRESSION
 2009 Garland Science Publishing
An Overview of Gene Expression
8-1
The distinct characteristics of different cell types in a multicellular organism
result mainly from the differential regulation of the _________________.
(a)
replication of specific genes
(b)
transcription of genes transcribed by RNA polymerase II
(c)
transcription of housekeeping genes
(d)
packing of DNA into nucleosomes in some cells and not others
8-2
The human genome encodes about 24,000 genes. Approximately how many genes
does the typical differentiated human cell express at any one time?
(a)
24,000—all of them
(b)
between 21,500 and 24,000—at least 90% of the genes
(c)
between 5000 and 15,000
(d)
less than 2500
8-3
Which of the following statements about differentiated cells is true?
(a)
Cells of distinct types express nonoverlapping sets of transcription factors.
(b)
Once a cell has differentiated, it can no longer change its gene expression.
(c)
Once a cell has differentiated, it will no longer need to transcribe RNA.
(d)
Some of the proteins found in differentiated cells are found in all cells of a
multicellular organism.
8-4
Investigators performed nuclear transplant experiments to determine whether
DNA is altered irreversibly during development. Which of the following
statements about these experiments is true?
(a)
Because the donor nucleus is taken from an adult animal, the
chromosomes from the nucleus must undergo recombination with the
DNA in the egg for successful development to occur.
(b)
The embryo that develops from the nuclear transplant experiment is
genetically identical to the donor of the nucleus.
(c)
The meiotic spindle of the egg must interact with the chromosomes of the
injected nuclei for successful nuclear transplantation to occur.
(d)
Although nuclear transplantation has been successful in producing
embryos in some mammals with the use of foster mothers, evidence of
DNA alterations during differentiation has not been obtained for plants.
8-5
In principle, a eucaryotic cell can regulate gene expression at any step in the
pathway from DNA to the active protein. Place the types of control listed below at
the appropriate places on the diagram in Figure Q8-5.
Figure Q8-5
A.
B.
C.
D.
translation control
transcriptional control
RNA processing control
protein activity control
How Transcriptional Switches Work
8-6
Fill in the blanks with the best word or phrase selected from the list below. Not all
words or phrases will be used; each word or phrase should be used only once.
The genes of a bacterial __________________ are transcribed into
a single mRNA. Many bacterial promoters contain a region known
as a(n) __________________, to which a specific transcription
regulator binds. Genes in which transcription is prevented are said
to be __________________. The interaction of small molecules,
such as tryptophan, with __________________ DNA-binding
proteins, such as the tryptophan repressor, regulates bacterial
genes. Genes that are being __________________ expressed are
being transcribed all the time.
allosteric
constitutively
induced
8-7
negatively
operator
operon
positively
promoter
repressed
Which of the following statements about transcriptional regulators is false?
(a)
Transcriptional regulators usually interact with the sugar-phosphate
backbone on the outside of the double helix to determine where to bind on
the DNA helix.
(b)
Transcriptional regulators will form hydrogen bonds, ionic bonds, and
hydrophobic interactions with DNA.
(c)
The DNA-binding motifs of transcriptional regulators usually bind in the
major groove of the DNA helix.
(d)
The binding of transcriptional regulators generally does not disrupt the
hydrogen bonds that holds the double helix together.
8-8
Operons ___________________________.
(a)
are commonly found in eucaryotic cells
(b)
are transcribed by RNA polymerase II
(c)
contain a cluster of genes transcribed as a single mRNA
(d)
can only be regulated by gene activator proteins
8-9
The tryptophan operator ___________________________.
(a)
is an allosteric protein
(b)
binds to the tryptophan repressor when the repressor is bound to
tryptophan
(c)
is required for production of the mRNA encoded by the tryptophan operon
(d)
is important for the production of the tryptophan repressor
8-10
Which of the following statements about the Lac operon is false?
(a)
The Lac repressor binds when lactose is present in the cell.
(b)
Even when the CAP activator is bound to DNA, if lactose is not present,
the Lac operon will not be transcribed.
(c)
The CAP activator can only bind DNA when it is bound to cAMP.
(d)
The Lac operon only produces RNA when lactose is present and glucose is
absent.
8-11
You are interested in examining the regulation of the gene that encodes an
enzyme, Tre-ase, important in metabolizing trehalose into glucose in bacteria.
Trehalose is a disaccharide formed of two glucose units. It is known that two
DNA binding proteins, TreA and TreB, are important for binding to the promoter
of the Tre-ase gene and are involved in regulating the transcription of the Tre-ase
gene: TreA binds to the “A” site in the promoter region, and TreB binds to the
“B” site. You make mutations in the TreA and TreB genes to create cells lacking
these genes, observe what happens to transcription of the Tre-ase gene, and obtain
the results in Table Q8-11.
Table Q8-11
A.
B.
What is the role for TreA in controlling Tre-ase expression? Explain.
What is the role for TreB in controlling Tre-ase expression? Explain.
C.
From these data, what do you predict will happen to Tre-ase transcription
(compared with that in normal cells) in the presence of trehalose if you
were to create a version of the TreA protein that will constitutively bind to
the “A” site in the Tre-ase promoter?
Note: Questions 8-12 to 8-15 use the following information and the data in Table
Q8-12. These questions may be used independently, or as a group.
You are interested in examining the Psf gene. It is known that Psf is normally produced
when cells are exposed to high levels of both calcium (Ca2+) and magnesium (Mg2+).
MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are
involved in regulating its transcription. MetA binds to the “A” site in the promoter
region, MetB to the “B” site, and MetC to the “C” site. You create binding site mutations
in the A, B, and C sites and observe what happens to transcription of the Psf gene. Your
results are summarized in Table Q8-12.
Table Q8-12
For this table: -, no transcription of Psf; +, low level of transcription of Psf; ++, high
levels of transcription of Psf.
8-12
Which of the following proteins are likely to act as gene activators?
(a)
MetA only
(b)
MetB only
(c)
MetC only
(d)
Both MetA and MetC
8-13
Which of the following proteins are likely to act as gene repressors?
(a)
MetA only
(b)
MetB only
(c)
MetC only
(d)
Both MetA and MetC
8-14
Which transcription factors are normally bound to the Psf promoter in the
presence of Mg2+ only?
(a)
none
(b)
MetA only
(c)
MetA and Met B
(d)
MetA, MetB, and MetC
8-15
Which transcription factors are normally bound to the Psf promoter in the
presence of both Mg2+ and Ca2+?
(a)
MetA and MetB
(b)
MetB and MetC
(c)
MetA and MetC
(d)
MetA, MetB, and MetC
Note: Questions 8-16 to 8-19 use the following information and the data in Figure
Q8-16. These questions may be used independently, or as a group.
You are interested in understanding the gene regulation of Lkp1, a protein that is
normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1
gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene,
place it upstream of the gene for green fluorescent protein (GFP), and insert this entire
piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells
but not in heart cells, an expression pattern similar to the normal expression of the LKP1
gene. Further experiments demonstrate that there are three regions in the promoter,
labeled A, B, and C in Figure Q8-16, that contribute to this expression pattern. Assume
that a single and unique transcription factor binds each site such that protein X binds site
A, protein Y binds site B, and protein Z binds site C. You want to determine which
region is responsible for tissue-specific expression, and create mutations in the promoter
to determine the function of each of these regions. In Figure Q8-16, if the site is missing,
it is mutated such that it cannot bind its corresponding transcription factor.
Figure Q8-16
8-16
Which of the following proteins are likely to act as gene repressors?
(a)
factor X
(b)
factor Y
(c)
factor Z
(d)
none of the above
8-17
Which of the following proteins are likely to act as gene activators?
(a)
factors X and Y
(b)
factors X and Z
(c)
factors Y and Z
(d)
factor X only
8-18
Experiment 1 in Figure Q8-16 is the positive control, demonstrating that the
region of DNA upstream of the gene for GFP results in a pattern of expression
that we normally find for the LKP1 gene. Experiment 2 shows what happens
when the sites for binding factors X, Y, and Z are removed. Which experiment
above demonstrates that factor X alone is sufficient for expression of LPK1 in the
kidney?
(a)
experiment 3
(b)
experiment 5
(c)
experiment 6
(d)
experiment 7
8-19
In what tissue is factor Z normally present and bound to the DNA?
(a)
kidney
(b)
liver
(c)
heart
(d)
none of the above
8-20
An allosteric transcription regulator called HisP regulates the enzymes for
histidine biosynthesis in the bacterium E. coli. Histidine modulates HisP activity.
On binding histidine, HisP alters its conformation, markedly changing its affinity
for the regulatory sequences in the promoters of the genes for the histidine
biosynthetic enzymes.
A.
If HisP functions as a gene repressor, would you expect that HisP would
bind more tightly or less tightly to the regulatory sequences when histidine
is abundant? Explain your answer.
B.
If HisP functions as a gene activator, would you expect that HisP would
bind more tightly or less tightly to the regulatory sequences when histidine
levels are low? Explain your answer.
8-21
Bacterial cells can take up the amino acid tryptophan from their surroundings, or,
if the external supply is insufficient, they can synthesize trytophan by using
enzymes in the cell. In some bacteria, the control of glutamine synthesis is similar
to that of tryptophan synthesis, such that the glutamine repressor inhibits the
transcription of the glutamine operon, which contains the genes that code for the
enzymes required for glutamine synthesis. On binding to cellular glutamine, the
glutamine repressor binds to a site in the promoter of the operon.
A.
Why is glutamine-dependent binding to the operon a useful property for
the glutamine repressor?
B.
What would you expect to happen to the regulation of the enzymes that
synthesize glutamine in cells expressing a mutant form of the glutamine
repressor that cannot bind to DNA?
C.
What would you expect to happen to the regulation of the enzymes that
synthesize glutamine in cells expressing a mutant form of the glutamine
repressor that binds to DNA even when no glutamine is bound to it?
8-22
In the absence of glucose, E. coli can proliferate by using the pentose sugar
arabinose. As shown in Figure Q8-22, the arabinose operon regulates the ability
of E. coli to use arabinose. The araA, araB, and araD genes encode enzymes for
the metabolism of arabinose. The araC gene encodes a transcription regulator that
binds adjacent to the promoter of the arabinose operon. To understand the
regulatory properties of the AraC protein, you engineer a mutant bacterium in
which the araC gene has been deleted and look at the effect of the presence or
absence of the AraC protein on the AraA enzyme.
Figure Q8-22
A.
B.
If the AraC protein works as a gene repressor, would you expect araA
RNA levels to be high or low in the presence of arabinose in the araCmutant cells? What about in the araC- mutant cells in the absence of
arabinose? Explain your answer.
Your findings from the experiment are summarized in Table Q8-22.
Table Q8-22
Do the results in Table Q8-22 indicate that the AraC protein regulates
arabinose metabolism by acting as a gene repressor or a gene activator?
Explain your answer.
8-23
The CAP activator protein and the Lac repressor both control the Lac operon (see
Figure Q8-23). You create cells that are mutant in the gene coding for the Lac
repressor so that these cells lack the Lac repressor under all conditions. For these
mutant cells, state whether the Lac operon will be switched on or off in the
following situations and explain why.
Figure Q8-23
A.
B.
C.
D.
8-24
in the presence of glucose and lactose
in the presence of glucose and the absence of lactose
in the absence of glucose and the absence of lactose
in the absence of glucose and the presence of lactose
You have discovered an operon in a bacterium that is turned on only when
sucrose is present and glucose is absent. You have also isolated three mutants that
have changes in the upstream regulatory sequences of the operon and whose
behavior is summarized in the Table Q8-24. You hypothesize that there are two
gene regulatory sites in the upstream regulatory sequence, A and B, which are
affected by the mutations. For this question, a plus (+) indicates a normal site and
a minus (-) indicates a mutant site that no longer binds its transcription regulator.
Table Q8-24
A.
B.
C.
8-25
If mutant 1 has sites A- B+, which of these sites is regulated by sucrose and
which by glucose?
Give the state (+ or -) of the A and B sites in mutants 2 and 3.
Which site is bound by a repressor and which by an activator?
Label the following structures in Figure Q8-25.
Figure Q8-25
A.
B.
C.
D.
activator protein
RNA polymerase
general transcription factors
mediator
8-26
How are most eucaryotic transcription regulators able to affect transcription when
their binding sites are far from the promoter?
(a)
by binding to their binding site and sliding to the site of RNA polymerase
assembly
(b)
by looping out the intervening DNA between their binding site and the
promoter
(c)
by unwinding the DNA between their binding site and the promoter
(d)
by attracting RNA polymerase and modifying it before it can bind to the
promoter
8-27
The expression of the BRF1 gene in mice is normally quite low, but mutations in
a gene called BRF2 lead to increased expression of BRF1. You have a hunch that
nucleosomes are involved in the regulation of BRF1 expression and so you
investigate the position of nucleosomes over the TATA box of BRF1 in normal
mice and in mice that lack either the BRF2 protein (BRF2-) or part of histone H4
(HHF-) (histone H4 is encoded by the HHF gene). Table Q8-27 summarizes your
results. A normal functional gene is indicated by a plus sign (+).
Table Q8-27
Which of the following conclusions cannot be drawn from your data? Explain
your answer.
(a)
BRF2 is required for the repression of BRF1.
(b)
BRF2 is required for the specific pattern of nucleosome positions over the
BRF1 upstream region.
(c)
The specific pattern of nucleosome positioning over the BRF1 upstream
region is required for BRF1 repression.
(d)
The part of histone H4 missing in HHF- mice is not required for the
formation of nucleosomes.
8-28
The yeast GAL4 gene encodes a transcriptional regulator that can bind DNA
upstream of genes required for the metabolism of the sugar galactose and turns
them on. Gal4 has a DNA-binding domain and an activation domain. The DNAbinding domain allows it to bind to the appropriate sites in the promoters of the
galactose metabolism genes. The activation domain attracts histone-modifying
enzymes and also binds to a component of the RNA polymerase II enzyme
complex, attracting it to the promoter so that the regulated genes can be turned on
when Gal4 is also bound to the DNA. When Gal4 is expressed normally, the
genes can be maximally activated. You decide to try to produce more of the
galactose metabolism genes by overexpressing the Gal4 protein at levels fiftyfold
greater than normal. You conduct experiments to show that you are
overexpressing the Gal4 protein and that it is properly localized in the nucleus of
the yeast cells. To your surprise, you find that too much Gal4 causes the galactose
genes to be transcribed only at a low level. What is the most likely explanation for
your findings?
8-29
For each of the following sentences, fill in the blanks with the best word or phrase
in the list below. Not all words or phrases will be used; use each word or phrase
only once.
During transcription in __________________ cells, transcriptional
regulators that bind to DNA thousands of nucleotides away from a
gene’s promoter can affect a gene’s transcription. The
__________________ is a complex of proteins that links distantly
bound transcription regulators with the proteins bound closer to the
transcriptional start site. Transcriptional activators can also interact
with histone __________________s, which alter chromatin by
modifying lysines in the tail of histone proteins to allow greater
accessibility to the underlying DNA. Gene repressor proteins can
reduce the efficiency of transcription initiation by attracting
histone __________________s. Sometimes, many contiguous
genes can become transcriptionally inactive as a result of
chromatin remodeling, like the __________________ found in
interphase chromosomes.
viral
acetylase
centrosome
helicase
procaryotic
mediator
telomere
eucaryotic
peroxidase
deoxidase
enhancer
operator
deacetylase
heterochromatin
leucine zipper
The Molecular Mechanisms that Create Specialized Cell Types
8-30
In principle, how many different cell types can an organism having four different
types of transcription regulator and thousands of genes create?
(a)
up to 4
(b)
up to 8
(c)
up to 16
(d)
thousands
8-31
From the sequencing of the human genome, we believe that there are
approximately 24,000 protein-coding genes in the genome, for which there are an
estimated 1500–3000 transcription factors. If every gene has a tissue-specific and
signal-dependent transcription pattern, how can such a small number of
transcriptional regulatory proteins generate a much larger set of transcriptional
patterns?
8-32
Combinatorial control of gene expression __________________________.
(a)
involves every gene using a different combination of transcriptional
regulators for its proper expression
(b)
involves groups of transcriptional regulators working together to
determine the expression of a gene
(c)
involves only the use of gene activators used together to regulate genes
appropriately
(d)
is seen only when genes are arranged in operons
8-33
You are studying a set of mouse genes whose expression increases when cells are
exposed to the hormone cortisol, and you believe that the same cortisolresponsive transcriptional activator regulates all of these genes. Which of the
following statements below should be true if your hypothesis is correct?
(a)
The cortisol-responsive genes share a DNA sequence in their regulatory
regions that binds the cortisol-responsive transcriptional activator.
(b)
The cortisol-responsive genes must all be in an operon.
(c)
The transcriptional regulators that bind to the regulatory regions of the
cortisol-responsive genes must all be the same.
(d)
The cortisol-responsive genes must not be transcribed in response to other
hormones.
8-34
The MyoD transcriptional regulator is normally found in differentiating muscle
cells and participates in the transcription of genes that produce muscle-specific
proteins, such as those needed in contractile tissue. Amazingly, expression of
MyoD in fibroblasts causes these cells derived from skin connective tissue to
produce proteins normally only seen in muscles. However, some other cell types
do not transcribe muscle-specific genes when MyoD is expressed in them. Which
of the following statements below is the best explanation of why MyoD can cause
fibroblasts to express muscle-specific genes?
(a)
Unlike some other cell types, fibroblasts have not lost the muscle-specific
genes from their genome.
(b)
The muscle-specific genes must be in heterochromatin in fibroblasts.
(c)
During their developmental history, fibroblasts have accumulated some
transcriptional regulators in common with differentiating muscle cells.
(d)
The presence of MyoD is sufficient to activate the transcription of musclespecific genes in all cell types.
8-35
A virus produces a protein X that activates only a few of the virus’s own genes
(V1, V2, and V3) when it infects cells. The cellular proteins A (a zinc finger
protein) and the cellular protein B (a homeodomain protein) are known to be
repressors of the viral genes V1, V2, and V3. You examine the complete upstream
gene regulatory sequences of these three viral genes and find the following:
1.
2.
3.
V1 and V2 contain binding sites for the zinc finger protein A only.
V3 contains a binding site for the homeodomain protein B only.
The only sequence that all three genes have in common is the TATA box.
Label each of the choices below as likely or unlikely as an explanation for your
findings. For each choice you label as unlikely, explain why.
A.
B.
C.
D.
E.
8-36
Protein X binds nonspecifically to the DNA upstream of V1, V2, and V3
and activates transcription.
Protein X binds to a repressor and prevents the repressor from binding
upstream of V1, V2, and V3.
Protein X activates transcription by binding to the TATA box.
Protein X activates transcription by binding to and sequestering proteins A
and B.
Protein X represses transcription of the genes for proteins A and B.
In mammals, individuals with two X chromosomes are female, and individuals
with an X and a Y chromosome are male. It had long been known that a gene
located on the Y chromosome was sufficient to induce the gonads to form testes,
which is the main male-determining factor in development, and researchers
sought the product of this gene, the so-called testes-determining factor (TDF). For
several years, the TDF was incorrectly thought to be a zinc finger protein encoded
by a gene called BoY. Which of the following observations would most strongly
suggest that BoY might not be the TDF? Explain your answer.
(a)
Some XY individuals that develop into females have mutations in a
different gene, SRY, but are normal at BoY.
(b)
BoY is not expressed in the adult male testes.
(c)
Expression of BoY in adult females does not masculinize them.
(d)
A few of the genes that are known to be expressed only in the testes have
binding sites for the BoY protein in their upstream regulatory sequences,
but most do not.
8-37
Which of the following is not a general mechanism that cells use to maintain
stable patterns of gene expression as cells divide?
(a)
a positive feedback loop, mediated by a transcriptional regulator that
activates transcription of its own gene in addition to other cell-type
specific genes
(b)
faithful propagation of condensed chromatin structures as cells divide
(c)
inheritance of DNA methylation patterns when cells divide
(d)
proper segregation of housekeeping proteins when cells divide
8-38
Which of the following statements about DNA methylation in eucaryotes is false?
(a)
Appropriate inheritance of DNA methylation patterns involves
maintenance methyltransferase.
(b)
DNA methylation involves a covalent modification of cytosine bases.
(c)
Methylation of DNA attracts proteins that block gene expression.
(d)
Immediately after DNA replication, each daughter helix contains one
methylated DNA strand, which corresponds to the newly synthesized
strand.
8-39
Which of the following statements about the Ey transcriptional regulator is false?
(a)
Expression of Ey in cells that normally form legs in the fly will lead to the
formation of an eye in the middle of the legs.
(b)
The Ey transcription factor must bind to the promoter of every eyespecific gene in the fly.
(c)
Positive feedback loops ensure that Ey expression remains switched on in
the developing eye.
(d)
A homolog of Ey is found in vertebrates; this homolog is also used during
eye development.
Post-Transcriptional Controls
8-40
Which of the following statements about riboswitches is false?
(a)
Riboswitches can block the production of mRNAs.
(b)
Riboswitches can control the translation of mRNAs.
(c)
Riboswitches are made from rRNAs.
(d)
Riboswitches can bind metabolites.
8-41
Which of the following is not involved in post-transcriptional control?
(a)
the spliceosome
(b)
dicer
(c)
mediator
(d)
RISC
8-42
MicroRNAs ____________________.
(a)
are produced from a precursor miRNA transcript
(b)
are found only in humans
(c)
control gene expression by base-pairing with DNA sequences
(d)
can degrade RNAs by using their intrinsic catalytic activity
8-43
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; use each word
or phrase only once.
MicroRNAs are noncoding RNAs that are incorporated into a
protein complex called __________________, which searches the
__________________s in the cytoplasm for sequence
complementary to that of the miRNA. When such a molecule is
found, it is then targeted for __________________. RNAi is
triggered by the presence of foreign __________________
molecules, which are digested by the __________________
enzyme into shorter fragments approximately 23 nucleotide pairs
in length.
tRNA
RISC
procaryotic
double-stranded RNA
single-stranded RNA
phosphorylation
destruction
methylation
mRNA
rRNA
riboswitch
acetylation
DNA
dicer
mitochondria
8-44
The extent of complementarity of a miRNA with its target mRNA determines
___________________________.
(a)
whether the mRNA will be degraded or transported elsewhere in the cell
(b)
whether the mRNA will be transported to the nucleus
(c)
whether RISC is degraded
(d)
whether the miRNA synthesizes a complementary strand
8-45
Which of the following statements about miRNAs is false?
(a)
One miRNA can regulate the expression of many genes.
(b)
miRNAs are transcribed in the nucleus from genomic DNA.
(c)
miRNAs are produced from rRNAs.
(d)
miRNAs are made by RNA polymerase.
8-46
Which of the following statements about RNAi is true?
(a)
The RNAi mechanism is found only in plants and animals.
(b)
RNAi is induced when double-stranded RNA is present in the cell.
(c)
RISC uses the siRNA duplex to locate complementary foreign RNA
molecules.
(d)
siRNAs bind to miRNAs to induce RNAi.
How We Know: Gene Regulation – The Story of Eve
8-47
The gene for a hormone necessary for insect development contains binding sites
for three transcription regulators called A, B, and C. Because the binding sites for
A and B overlap, A and B cannot bind simultaneously. You make mutations in the
binding sites for each of the proteins and measure hormone production in cells
that contain equal amounts of the A, B, and C proteins. Figure Q8-47 summarizes
your results. In each of the following sentences, choose one of the phrases within
square brackets to make the statement consistent with the results.
Figure Q8-47
A.
B.
C.
8-48
Protein A binds to its DNA-binding site [more tightly/less tightly] than
protein B binds to its DNA-binding site.
Protein A is a [stronger/weaker] activator of transcription than protein B.
Protein C is able to prevent activation by [protein A only/protein B
only/both protein A and protein B].
The Drosophila Eve gene has a complex promoter containing multiple binding
sites for four transcription regulators: Bicoid, Hunchback, Giant, and Krüppel.
Bicoid and Hunchback are activators of Eve transcription, whereas Giant and
Krüppel repress Eve transcription. Figure Q8-48A shows the patterns of
expression of these regulators.
Figure Q8-48
The eve promoter contains modules that control expression in various stripes. You
construct a reporter gene that contains the DNA 5 kb upstream of the eve gene, so
that this reporter contains the stripe 3 module, the stripe 2 module, the stripe 7
module, and the TATA box, all fused to the LacZ reporter gene (which encodes
the β-galactosidase enzyme), as shown in Figure Q8-50B. This construct results in
expression of the β-galactosidase enzyme in three stripes, which correspond to the
normal positions of stripes 3, 2, and 7.
A.
B.
By examining the overlap of sites on the stripe 2 module, as depicted in
Figure Q8-48B, what is the biological effect of having some of the
transcription regulator binding sites overlap?
You make two mutant versions in which several of the binding sites in the
Eve stripe 2 module have been deleted, as detailed in items (i) and (ii)
below. Refer to Figure Q8-48B for the positions of the binding sites.
(Note, however, that because many of the binding sites overlap, it is not
possible to delete all of one kind of site without affecting some of the
other sites.) Match the appropriate mutant condition with the most likely
pattern of Eve expression shown in Figure Q8-48C. Explain your choices.
(i)
(ii)
deletion of the Krüppel-binding sites in stripe 2
deletion of the two Bicoid-binding sites in the stripe 2 module that
are marked with an asterisk (*) in Figure Q8-48B