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Chapter 28 Regulation of Gene Expression
Multiple Choice Questions
1. Principles of gene regulation
Page: 1082
Difficulty: 2 Ans: B
“Housekeeping genes” in bacteria are commonly expressed constitutively, but not all of these genes
are expressed at the same level (the same number of molecules per cell). The primary mechanism
responsible for variations in the level of constitutive enzymes from different genes is that:
A)
B)
C)
D)
E)

all constitutive enzymes are synthesized at the same rate, but are not degraded equally.
their promoters have different affinities for RNA polymerase holoenzyme.
some constitutively expressed genes are more inducible than others.
some constitutively expressed genes are more repressible than others.
the same number of mRNA copies are made from each gene, but are translated at different rates.

2. Principles of gene regulation
Pages: 1082-1083 Difficulty: 2 Ans: D
Which of the following statements correctly describes promoters in E. coli?
A)
B)
C)
D)

A promoter may be present on either side of a gene or in the middle of it.
All promoters have the same sequence that is recognized by RNA polymerase holoenzyme.
Every promoter has a different sequence, with little or no resemblance to other promoters.

Many promoters are similar and resemble a consensus sequence, which has the highest affinity
for RNA polymerase holoenzyme.
E) Promoters are not essential for gene transcription, but can increase its rate by two- to three-fold.
3. Principles of gene regulation
Pages: 1083-1084 Difficulty: 2 Ans: D
The operator region normally can be bound by:
A)
B)
C)
D)
E)

attenuator.
inducer.
mRNA.
repressor.
suppressor tRNA.

4. Principles of gene regulation
Pages: 1083-1084 Difficulty: 2 Ans: C
Small signal molecules that regulate transcription are not known to:
A) cause activator proteins to bind DNA sites.
B) cause repressor proteins to bind DNA sites.
C) directly bind to DNA sites.
D) prevent activator proteins from binding to DNA sites.
E) release repressor proteins from DNA sites.


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327

5. Principles of gene regulation
Pages: 1083-1085 Difficulty: 2 Ans: E
The diagram below represents a hypothetical operon in the bacterium E. coli. The operon consists of
two structural genes (A and B), which code for the enzymes A-ase and B-ase, respectively, and also
includes P (promoter) and O (operator) regions as shown.

When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes A-ase
and B-ase are both synthesized at a 50-fold higher rate than in the absence of X. (X has a molecular
weight of about 200.) Which of the following statements is true of the operon decribed above?
A) All four genes (A, B, O, and P) will be transcribed into an mRNA that will then be translated into
four different proteins.
B) The 3' end of the mRNA from the operon will correspond to the left end of the operon as drawn.
C) The 5' end of the messenger from this operon will correspond to the right end of the operon as
drawn.
D) The repressor for this operon binds just to the right of A.
E) When RNA polymerase makes mRNA from this operon, it begins RNA synthesis just to the left
of gene A.
6. Principles of gene regulation
Pages: 1083-1085 Difficulty: 2 Ans: A
The diagram below represents a hypothetical operon in the bacterium E. coli. The operon consists of
two structural genes (A and B) that code for the enzymes A-ase and B-ase, respectively, and also
includes P (promoter) and O (operator) regions as shown.

When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes A-ase
and B-ase are both synthesized at a 50-fold higher rate than in the absence of X (which has a
molecular weight of about 200). Which one of the following statements is true of such an operon?
A)
B)

C)
D)

Adding X to the growth medium causes a repressor protein to be released from the O region.
Adding X to the growth medium causes a repressor protein to bind tightly to the O region.
Synthesis of the mRNA from this operon is not changed by the addition of compound X.
The mRNA copied from this operon will be covalently linked to a short piece of DNA at the 5'
end.
E) Two mRNA molecules are made from this operon, one from gene A the other from gene B.
7. Principles of gene regulation
Pages: 1085-1087 Difficulty: 2 Ans: B
Transcription of the lactose operon in E. coli is stimulated by:
A)
B)
C)
D)
E)

a mutation in the repressor gene that strengthens the affinity of the repressor for the operator.
a mutation in the repressor gene that weakens the affinity of the repressor for the operator.
a mutation in the repressor gene that weakens the affinity of the repressor for the inducer.
binding of the repressor to the operator.
the presence of glucose in the growth medium.


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Chapter 28 Regulation of Gene Expression

8. Principles of gene regulation

Page: 1088
Difficulty: 3 Ans: D
Protein amino acid side chains can hydrogen bond in the major groove of DNA, and discriminate
between each of the four possible base pairs. In which one of the following groups of amino acids
can all three members potentially be used in such DNA-protein recognition?
A)
B)
C)
D)
E)

Ala, Asn, Glu
Arg, Gln, Leu
Asn, Gln, Trp
Asn, Glu, Lys
Glu, Lys, Pro

9. Principles of gene regulation
Pages: 1088-1089 Difficulty: 2 Ans: A
The DNA binding motif for many prokaryotic regulatory proteins, such as the lac repressor, is:
A)
B)
C)
D)
E)

helix-turn-helix.
homeobox.
homeodomain.
leucine zipper.

zinc finger.

10. Principles of gene regulation
Pages: 1088-1091 Difficulty: 2
Ans: D
Protein structural motifs often have general functions in common. Which one of the following motifs
is known to be involved in protein dimer formation, but not in direct protein-DNA interactions?
A)
B)
C)
D)
E)

β-barrel
helix-turn-helix
homeodomain
leucine zipper
zinc finger

11. Regulation of gene expression in prokaryotes
Pages: 1093-1094 Difficulty: 2 Ans: A
Which of the following statements about regulation of the lac operon is true?
A)
B)
C)
D)
E)

Glucose in the growth medium decreases the inducibility by lactose.
Glucose in the growth medium does not affect the inducibility by lactose.

Glucose in the growth medium increases the inducibility by lactose.
Its expression is regulated mainly at the level of translation.
The lac operon is fully induced whenever lactose is present.


Chapter 28 Regulation of Gene Expression

329

12. Regulation of gene expression in prokaryotes
Page: 1093
Difficulty: 2 Ans: A
The binding of CRP (cAMP receptor protein of E. coli) to DNA in the lac operon:
A)
B)
C)
D)
E)

assists RNA polymerase binding to the lac promoter.
is inhibited by a high level of cAMP.
occurs in the lac repressor region.
occurs only when glucose is present in the growth medium.
prevents repressor from binding to the lac operator.

13. Regulation of gene expression in prokaryotes
Page: 1093
Difficulty: 2 Ans: C
Consider the lac operon of E. coli. When there is neither glucose nor lactose in the growth medium:
A)

B)
C)
D)
E)

CRP protein binds to the lac operator.
CRP protein displaces the Lac repressor from the lac promoter.
repressor is bound to the lac operator.
RNA polymerase binds lac promoter and transcribes the lac operon.
the operon is fully induced.

14. Regulation of gene expression in prokaryotes
Page: 1094
Difficulty: 1 Ans: B
A regulon is a(n):
A)
B)
C)
D)
E)

group of related triplet codons.
network of operons with a common regulator.
operon that is subject to regulation.
protein that regulates gene expression.
ribosomal protein that regulates translation.

15. Regulation of gene expression in prokaryotes
Pages: 1094-1095 Difficulty: 2 Ans: C
The tryptophan operon of E. coli is repressed by tryptophan added to the growth medium. The

tryptophan repressor probably:
A)
B)
C)
D)
E)

binds to RNA polymerase when tryptophan is present.
binds to the trp operator in the absence of tryptophan.
binds to the trp operator in the presence of tryptophan.
is a DNA sequence.
is an attenuator.

16. Regulation of gene expression in prokaryotes
Pages: 1094-1097 Difficulty: 2 Ans: A
Which one of the following statements about the transcription attenuation mechanism is true?
A)
B)
C)
D)
E)

In some operons (e.g., the his operon), attenuation may be the only regulatory mechanism.
Sequences of the trp operon leader RNA resemble an operator.
The leader peptide acts by a mechanism that is similar to that of a repressor protein.
The leader peptide gene of the trp operon includes no Trp codons.
The leader peptide is an enzyme that catalyzes transcription attenuation.


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Chapter 28 Regulation of Gene Expression

17. Regulation of gene expression in prokaryotes
Pages: 1094-1097 Difficulty: 2 Ans: D
Which of the following statements is true of the attenuation mechanism used to regulate the
tryptophan biosynthetic operon in E. coli?
A) Attenuation is the only mechanism used to regulate the trp operon.
B) One of the enzymes in the Trp biosynthetic pathway binds to the mRNA and blocks translation
when tryptophan levels are high.
C) The leader peptide plays a direct role in causing RNA polymerase to attenuate transcription.
D) Trp codons in the leader peptide gene allow the system to respond to tryptophan levels in the cell.
E) When tryptophan levels are low, the trp operon transcripts are attenuated (halted) before the
operon’s structural genes are transcribed.
18. Regulation of gene expression in prokaryotes
Pages: 1094-1097 Difficulty: 3 Ans: B
Attenuation in the trp operon of E. coli:
A)
B)
C)
D)
E)

can adjust transcription of the structural genes upwards when tryptophan is present.
can fine-tune the transcription of the operon in response to small changes in Trp availability.
is a mechanism for inhibiting translation of existing (complete) trp mRNAs.
results from the binding of the Trp repressor to the operator.
results from the presence of short leader peptides at the 5' end of each structural gene.

19. Regulation of gene expression in prokaryotes

Pages: 1097-1098 Difficulty: 2
Ans: D
RecA protein provides the functional link between DNA damage and the SOS response by displacing
the LexA protein from its operator sites on the SOS genes. RecA does so by:
A)
B)
C)
D)
E)

associating with polymerase holoenzyme to help it remove LexA from operator.
bending LexA operator DNA to force dissociation of LexA repressor.
binding to LexA protein to weaken directly its affinity for operator sites.
causing self-cleavage of LexA, thus inactivating its binding to operator.
competitively binding to LexA operators and serving as an activator.

20. Regulation of gene expression in prokaryotes
Pages: 1098-1099 Difficulty: 2
Ans: D
An example of coordinate control is the down-regulation of ribosomal RNA synthesis in response to
amino acid starvation, which will cause synthesis of ribosomal proteins to be limited. What is the
correct order of the following events that participate in the signaling process?
1. Binding of stringent factor to the ribosome.
2. Formation of the unusual nucleotide ppGpp.
3. Formation of the unusual nucleotide pppGpp.
4. Uncharged tRNA binds in the ribosomal A-site.
A)
B)
C)
D)

E)

1, 4, 2, 3
1, 4, 3, 2
4, 1, 2, 3
4, 1, 3, 2
4, 2, 1, 3


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21. Regulation of gene expression in eukaryotes
Page: 1103-1104 Difficulty: 2 Ans: B
Which one of the following statements about eukaryotic gene regulation is correct?
A)
B)
C)
D)
E)

Large polycistronic transcripts are common.
Most regulation is positive, involving activators rather than repressors.
Transcription and translation are mechanistically coupled.
Transcription does not involve promoters.
Transcription occurs without major changes in chromosomal organization.

22. Regulation of gene expression in eukaryotes
Page: 1104

Difficulty: 1 Ans: C
Which of the following is a DNA sequence?
A)
B)
C)
D)
E)

Coactivator
Corepressor
Enhancer
Inducer
Transactivator

23. Regulation of gene expression in eukaryotes
Page: 1104
Difficulty: 2 Ans: E
Which one of the following types of eukaryotic regulatory proteins interact with enhancers?
A)
B)
C)
D)
E)

Basal transcription factors
Coactivators
Repressors
TATA-binding proteins
Transactivators

24. Regulation of gene expression in eukaryotes
Page: 1108
Difficulty: 2 Ans: A
Which one of the following is not involved in steroid hormone action?
A)
B)
C)
D)
E)

Cell surface receptors
Hormone-receptor complexes
Specific DNA sequences
Transcription activation and repression
Zinc fingers

25. Regulation of gene expression in eukaryotes
Page: 1112
Difficulty: 2 Ans: C
Which one of the following classes is expressed in the unfertilized egg and is involved in directing the
spatial organization of the Drosophila embryo early in development?
A)
B)
C)
D)
E)

Gap genes
Homeotic genes
Maternal genes

Segment polarity genes
Segmentation genes


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Chapter 28 Regulation of Gene Expression

26. Regulation of gene expression in eukaryotes
Pages: 1112, 1116
Difficulty: 2 Ans: B
Which one of the following classes of genes is involved in specifying the localization of organs in the
Drosophila embryo?
A)
B)
C)
D)
E)

Gap genes
Homeotic genes
Maternal genes
Segment polarity genes
Segmentation genes

27. Regulation of gene expression in eukaryotes
Pages: 1112, 1116 Difficulty: 2 Ans: C
In the development of the fly Drosophila, homeotic genes:
A)
B)

C)
D)
E)

are transcribed during egg production; their mRNAs lie dormant in the egg until it is fertilized.
determine the number of body segments that will form.
are expressed late and determine the detailed structure of each body segment.
generally have no introns.
are not translated into proteins.

Short Answer Questions
28. Principles of gene regulation
Page: 1083
Difficulty: 2
Usually, a mutation in the promoter region of an operon causes reduced levels of synthesis of the
proteins encoded by that operon. Occasionally, a mutation in the promoter region actually causes
increased levels of synthesis. Can you suggest (briefly) a plausible explanation?
Ans: It is possible for a mutation to result in a promoter with higher-than-normal affinity for RNA
polymerase (i.e., a promoter that more closely matches the promoter consensus sequence), resulting in
increased expression in the mutant of the genes in this operon.
29. Principles of gene regulation
Page: 1084
Difficulty: 2
Describe and contrast positive regulation and negative regulation of gene expression.
Ans: Positive regulation involves an activator protein that, when bound in the vicinity of the
regulated gene, facilitates the binding of RNA polymerase to its promoter. The affinity of the
activator for the DNA may be either increased or decreased by a signal molecule. Negative
regulation involves a repressor protein that, when bound near the regulated gene, hinders access of
RNA polymerase to its promoter or inhibits its activity. Repressor affinity for its binding site (the
operator) may be increased or decreased by a signal molecule. (See Fig. 28-4, p. 1084.)

30. Principles of gene regulation
Pages: 1085
Difficulty: 2
Define operon and polycistronic mRNA.
Ans: An operon is several contiguous structural genes that are coordinately regulated by promoter


Chapter 28 Regulation of Gene Expression

333

and operator regions, as depicted in Fig. 28-5, p. 1085. All of the genes in an operon are transcribed
in a polycistronic mRNA—a single mRNA molecule encoding more than one structural gene. Note:
operons are not found in eukaryotic organisms.
31. Principles of gene regulation
Pages: 1088-1091 Difficulty: 2
Match the protein or structural feature on the left with one appropriate description on the right.
____ activator
____ helix-turn-helix
____ leucine zipper
____ repressor
____ zinc finger

(a) a positive regulator
(b) a negative regulator
(c) facilitates transcription only when bound to a signal
molecule
(d) a DNA-binding structural motif found in many
prokaryotic regulatory proteins
(e) a structural feature involved in protein-protein interactions

between some regulatory protein monomers
(f) a protein that dissociates from DNA when bound to a
signal molecule
(g) a DNA-binding structural motif found in many eukaryotic
regulatory proteins

Ans: a; d; e; b; g
32. Regulation of gene expression in prokaryotes
Pages: 1085-1086, 1093-1094
Difficulty: 2
E. coli cells are placed in a growth medium containing lactose. Indicate how the following
circumstances would affect the expression of the lactose operon (increase/decrease/no change).
(a) Addition of high levels of glucose
(b) A Lac repressor mutation that prevents dissociation of Lac repressor from the operator
(c) A mutation that inactivates β-galactosidase
(d) A mutation that inactivates galactoside permease
(e) A mutation that prevents binding of CRP to its binding site near the lac promoter
Ans: (a) decrease; (b) decrease; (c) decrease (this enzyme converts lactose into allolactose, the
inducer); (d) decrease (because external lactose would not enter the cell, and allolactose would not be
present to induce the operon); (e) decrease
33. Regulation of gene expression in prokaryotes
Pages: 1086, 1093-1094
Difficulty: 2
Draw a simple map of the lactose operon indicating the relative positions of promoter, operator, CRPbinding site, repressor gene (I), and the structural genes of the operon (A, Y, Z). Indicate where the
CRP protein binds within this operon. When it is bound to this site, does the CRP protein have a
positive or negative effect on gene expression in this system?
Ans: Bound CRP stimulates (has a positive effect on) transcription of this operon. (See Fig. 28-7, p.
1086, and Fig. 28-17, p. 1093.)
34. Regulation of gene expression in prokaryotes
Pages: 1093-1094 Difficulty: 2

Briefly explain (a) why there is a lag in cell growth when bacteria are switched from a medium
containing glucose to one containing lactose. (b) When the growth medium contains both lactose and
glucose, what proteins will be bound to the lac operon regulatory region? (c) If only lactose is in the


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Chapter 28 Regulation of Gene Expression

growth medium, what proteins will be bound to the lac operon regulatory region?
Ans: (a) The lag represents the time required for lactose to enter the cell, be converted into the
inducer allolactose, and turn on the synthesis of the genes of the lac operon that are essential for
growth on lactose. (b) When both glucose and lactose are present, neither Lac repressor nor CRP will
be bound. (c) If only lactose is present, CRP will be bound. (See Fig. 28-18, p.1094.)
35. Regulation of gene expression in prokaryotes
Pages: 1094-1097 Difficutly: 3
In prokaryotes such as E. coli, many operons that encode enzymes involved in amino acid
biosynthesis begin with a sequence coding for a leader peptide. This peptide has no known enzymatic
function and is rich in the amino acid that is synthesized by the enzymes coded for in the operon.
What is the function of this leader peptide?
Ans: The leader peptide is integral to the regulatory mechanism called transcription attenuation. In
prokaryotes, transcription and translation are closely coupled. The ability to translate the leader
peptide signals that the relevant amino acid is readily available in the cell and that transcription of the
operon is undesirable at this time. Translation through the leader peptide (when the amino acid is
abundant) allows the formation of a hairpin attenuator structure downstream in the nascent RNA
chain. This attenuator acts as a transcription terminator, causing dissociation of the RNA polymerase.
When the relevant amino acid is in short supply, the ribosome stalls trying to synthesize the leader
peptide. A different secondary structure forms in the nascent RNA, preventing formation of the
terminator and allowing continued transcription of the operon. (See Fig. 28-21, p.1096.)
36. Regulation of gene expression in prokaryotes

Page: 1097-1098 Difficulty: 3
The SOS response in E. coli is triggered by extensive damage to the cell’s DNA and increases the
capacity for repairing such DNA. What molecular events bring about expression of the SOS genes?
Ans: The SOS genes are located in several operons, all under the regulation of the LexA repressor,
which normally represses these operons. When DNA is damaged extensively, RecA protein binds to
single-stranded DNA at the damaged sites, and bound RecA activates specific proteolysis of the LexA
protein, relieving the repression of the SOS operons. (See Fig. 28-22, p. 1097.)
37. Regulation of gene expression in prokaryotes
Pages: 1098-1099 Difficulty: 3
Explain how synthesis of ribosomal proteins in E. coli is regulated at the level of translation.
Ans: Ribosomal proteins are encoded in several operons. One protein encoded in each operon, when
not part of an assembled ribosome, binds tightly to that operon’s mRNA, preventing its translation.
Thus, when an excess of that protein is present, it slows its own translation. (See Fig. 28-23, p. 1099.)
38. Regulation of gene expression in eukaryotes
Pages: 1102-1103 Difficulty: 2
Describe briefly the relationship between chromatin structure and transcription in eukaryotes.
Ans: Heterochromatin is highly condensed and transcriptionally inert because the chromatin proteins
make promoters inaccessible. The less condensed euchromatin has undergone alteration of its
structure (“remodeling”), allowing some regions to be transcribed. These alterations include covalent
modification (such as acetylation) of histones and displacement of nucleosomes, creating exposed
regions of DNA (hypersensitive sites) that are probably binding sites for regulatory proteins.


Chapter 28 Regulation of Gene Expression

335

39. Regulation of gene expression in eukaryotes
Pages: 1102-1104 Difficulty: 2
Define each in about two sentences: (a) hypersensitive sites in eukaryotic chromosomes; (b)

enhancers (upstream activator sequences); (c) chromatin remodeling.
Ans: (a) A hypersensitive site is a region of eukaryotic DNA that is being associated with actively
transcribed DNA and is unusually sensitive to degradation by nucleases. (b) Enhancers are DNA
sequences that regulate the transcription of certain genes by allowing binding of specific
transactivator proteins. These sequences may occur thousands of base pairs away from the gene
whose transcription they regulate. (c) Chromatin remodeling reflects structural changes resulting
from enzymes that covalently modify the core histone complex in nucleosomes (such as histone
acetyltransferases) and those that move or displace nucleosomes using the energy of ATP hdyrolysis.
(See Table 28-2, p. 1103.)
40. Regulation of gene expression in eukaryotes
Page: 1104
Difficulty: 1
Describe in one or two sentences the role of each of the following types of proteins in the regulation
of gene expression in eukaryotes: (a) basal transcription factors; (b) transactivators; (c) coactivators.
Ans: (a) Basal transcription factors interact with specific DNA sequences and/or each other and/or
with RNA polymerase to form a complex at promoters. (b) Transactivators interact with both target
sites on DNA and with other regulatory proteins such as coactivators to activate transcription. (c)
Coactivators interact with transactivators and proteins at the promoters to activate transcription.
41. Regulation of gene expression in eukaryotes
Pages: 1106-1108 Difficulty: 2
DNA-binding transactivating proteins often possess a domain separate from their DNA-binding
domains that serves as a docking site for interactions with the transcription complex, coactivators,
corepressors, or even chromatin remodeling proteins, to regulate gene transcription. Describe three
known kinds of such domains, and provide an example of each.
Ans: (1) Gal4p (in yeast) has an acidic activation domain rich in aspartic and glutamic acid, (2) Sp1
in higher eukaryotes has two activation domains comprised of about 25% glutamine, and (3) CTF1
has a proline-rich (>20%) activation domain.
42. Regulation of gene expression in eukaryotes
Pages: 1108-1109 Difficulty: 2
Describe briefly the process by which steroid hormones affect gene expression.

Ans: Steroid hormones usually enter a cell by diffusion and then bind to a nuclear receptor protein.
The binding changes the conformation of the receptor protein such that the hormone-receptor
complex can bind to specific DNA sequences (hormone response elements). This binding can either
activate or repress the expression of adjacent genes.
43. Regulation of gene expression in eukaryotes
Page: 1110
Difficulty: 1
What are three mechanisms of translational repression that are known to exist in eukaryotes?
Ans: (a) inactivation of initiation factors usually by phosphorylation; (b) binding of repressor proteins


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Chapter 28 Regulation of Gene Expression

to the mRNA, thereby interfering with initiation factors or the ribosome; (c) interference with
elongation by the binding of proteins to elongation factors.
44. Regulation of gene expression in eukaryotes
Pages: 1110-1111 Difficulty: 2
Large numbers of micro-RNAs (miRNAs), also known as small temporal RNAs (stRNAs) have now
been discovered in higher eukaryotes. Describe their characteristics and general function.
Ans: Their primary transcripts are about 70 nucleotides, with self-complementary internal sequences
that can form hairpin structures. These are cleaved into 20-25 nt partial duplexes, one strand of which
can bind to complementary stretches in cellular mRNAs. This can then inhibit gene expression by
blocking translation of the target mRNA or facilitating its degradation.
45. Regulation of gene expression in eukaryotes
Pages: 1112-1116 Difficulty: 1
Describe briefly the general role of the protein products of each of the following types of genes in the
embryonic development of the Drosophila: (a) maternal genes; (b) segmentation genes; (c) homeotic
genes.

Ans: (a) A maternal gene in Drosophila is a gene that is expressed in the unfertilized egg (see Fig.
28-36, p. 1114). These genes are involved in directing the spatial organization of the Drosophila
embryo early in development. (b) Segmentation genes direct the formation of the correct number of
body segments, and include the gap genes, pair-rule genes, and segment polarity genes as subclasses.
(c) Homeotic genes help specific localization of organs and appendages in the segments of the
embryo. They regulate their target genes by highly conserved DNA-binding homeodomains, depicted
in Fig. 28-13, p. 1090.