CHAPTER 16—REGULATION OF GENE EXPRESSION
MULTIPLE CHOICE
1. The primary difference between different nucleated cells in a developing zygote is
a.
the genetic code of each cell.
b.
the expression of housekeeping genes.
c.
the expression of a gene or group of genes.
d.
the number of chromosomes present in the nucleus.
e.
the presence or absence of gene regulators.
2. In prokaryotes, the genes for metabolic pathways are
a.
always on for early steps in the pathway, but genes for later steps are generally off.
b.
always on so the bacteria can respond rapidly to changing conditions.
c.
turned on and off as conditions change.
d.
independently regulated.
e.
expressed in low levels all of the time, but will be expressed in higher levels when
conditions warrant.
3. The genes in an operon are
a.
each transcribed into their own mRNA molecule.
b.
all transcribed into a single mRNA molecule.
c.
transcribed in subsets.
d.
transcribed as a single unit in prokaryotes, but as individual genes in eukaryotes.
e.
located at different locations on the chromosome and so can never be transcribed to the
same mRNA molecule.
4. What is the relationship between operons and transcription units?
a.
They are two terms for the same thing.
b.
An operon is made up of a transcription unit and associated regulatory DNA sequences.
c.
A transcription unit is made up of an operon and associated regulatory DNA sequences.
d.
An operon is comprised of multiple transcription units.
e.
A transcription unit is comprised of multiple operons.
5. What is the relationship between operators, transcription units, and operons?
a.
Operators are comprised of transcription units and operons.
b.
Operons are comprised of transcription units and operators.
c.
Operators are comprised of multiple transcription units associated with a single operon.
d.
Operons are comprised of multiple operators associated with a single transcription unit.
e.
Operons are comprised of multiple transcription units associated with a single operator.
6. Where does RNA polymerase first bind to on the E. coli lac operon ?
a.
the lac I repressor
b.
the promoter
c.
the operator
d.
the transcription initiation site
e.
the first codon of the lacZ gene
7. Which statement makes a correct comparison between the regulation of an operon and controlling the
motion of a car?
a.
The removal of a repressor is like stepping on the gas pedal of a car.
b.
The removal of a repressor is like removing your foot from the brake pedal of a car.
c.
The removal of a repressor is like moving your foot from the brake pedal to the gas pedal
of a car.
d.
The removal of a repressor is like moving your foot from the gas pedal to the brake pedal
of a car.
e.
The removal of a repressor is like removing your left foot from the brake pedal of a car
while your right foot remains on the gas pedal.
8. The E. coli lac operon encodes three genes. These genes
a.
cleave lactose into glucose and galactose, produce transacetylase, and actively transport
lactose into the cell.
b.
encode for an enzyme that cleaves glucose, a transport protein for active transport, and a
transacetylase enzyme for solubilizing lactose in the cytosol.
c.
encode for an enzyme that cleaves lactose, a transport protein for active transport of
lactose, and a transacetylase enzyme whose function is still unknown.
d.
catalyze the synthesis of lactose from glucose and galactose, produce transacetylase, and
actively export lactose from the cell.
e.
directly interact with the lac repressor and initiate their own transcription.
9. Suppose the lacI repressor gene were permanently silenced by an alteration to the DNA sequence.
What would be the impact on the expression of the lac operon?
a.
The lac operon would be transcribed but at a low level.
b.
The lac operon would be transcribed at a high level.
c.
There would be no real impact on lac operon expression.
d.
The lac operon would only be expressed when lactose was present.
e.
The lac operon would not be expressed even when lactose was present.
10. How is lac operon transcription regulated by the presence or absence of lactose?
a.
When lactose is not available, the repressor will bind to the promoter and block RNA
polymerase from transcribing the genes of the lac operon.
b.
When lactose is available, the repressor will bind to the lactose instead of blocking RNA
polymerase.
c.
When lactose is not available, the repressor will bind to allolactose.
d.
When lactose if available, allolactose is produced and binds to the operator, blocking RNA
polymerase.
e.
In the absence of lactose, transcription occurs at a low level resulting in the production of
allolactose which in turn stimulates the cell to take up lactose.
11. Put the following steps in order:
1.
allolactose binds to the lac repressor protein
2.
RNA polymerase binds to the promoter
3.
the lac repressor is expressed
4.
beta-galactosidase cleaves lactose and generates allolactose
5.
mRNA is transcribed from the lac operon at low levels
6.
beta-galactosidase, permease, and transacetylase are produced at high levels
a.
1, 3, 2, 5, 4, 6
b.
4, 1, 2, 3, 5, 6
c.
3, 5, 4, 1, 2, 6
d.
3, 4, 1, 6, 2, 5
e.
5, 3, 1, 2, 4, 6
12. Which of the following is NOT involved in turning off the lac operon?
a.
the short lifespan of mRNA molecules
b.
the short lifespan of the lac operon gene products
c.
the depletion of lactose from the cytosol
d.
the lack of allolactose to bind to the repressor
e.
binding of repressor to the operator
13. Where is the trp repressor gene located?
a.
just upstream of the trp operon
b.
on a region of the chromosome quite distant from the trp operon
c.
adjacent to the lac repressor gene
d.
just downstream of the trp operon
e.
directly adjacent to the promoter region
14. Why does E. coli shut down the trp operon if tryptophan is available in the environment?
a.
Synthesizing an amino acid takes energy so it is a waste of energy to make something that
is already available.
b.
Environmental tryptophan is of higher quality than what the E. coli can make itself.
c.
The trp operon encodes genes that export tryptophan from the cell; if there is already
tryptophan in the environment, further export isn’t necessary.
d.
It doesn’t shut down the trp operon but only lowers the level of trp operon activity.
e.
E. coli actually turns the trp operon on when tryptophan is present; tryptophan is an energy
source for the cell, and there is no reason to make enzymes to degrade something that is
lacking in the environment.
15. In prokaryotes, the concentration of cAMP
a.
increases with increased glucose concentration.
b.
decreases with increased glucose concentration.
c.
negatively regulates operons such as the lac operon.
d.
makes CAP.
e.
degrades CAP.
16. At which level(s) is eukaryotic gene expression regulated?
a.
transcription
b.
translation
c.
posttranscription
d.
posttranslation
e.
all of these
17. If the histones are rearranged to control access to a gene, this is called
a.
transcription.
b.
chromatin remodeling.
c.
DNA condensation.
d.
remodeling complex formation.
e.
acetylation.
18. Acetylation
a.
adds an acetyl group (CH3CO−) to the cytosine nucleotides of DNA.
b.
adds an acetyl group (CH3CH2−) to the DNA of a promoter sequence.
c.
adds an acetyl group (CH3CO−) to the histone protecting the transcription unit of a gene.
d.
adds an acetyl group (CH3CO−) to the histone protecting the promoter region of a gene.
e.
adds an acetyl group (CH3CH2−) to the RNA polymerase that will initiate transcription at
the promoter region.
19. Which of the following is the single most important stage when regulating gene expression?
a.
translational regulation
b.
degradation of mRNA
c.
initiation of transcription
d.
RNA interference
e.
removal of masking segments
20. How does RNA polymerase II get involved in transcription?
a.
It binds directly to the eukaryotic promoter sequence.
b.
It binds directly to the TATA box.
c.
It binds to transcription factors in the nucleus, and the complex then binds to the promoter.
d.
Transcription factors bind to the promoter, and they in turn bind to RNA polymerase II.
e.
Some transcription factors bind to the promoter, others bind to RNA polymerase II, and
then the two groups of proteins bind to each other.
Use the figure above for the following question(s).
21. Which number identifies a sequence that is transcribed but not translated?
a.
2
b.
3
c.
4
d.
5
e.
6
22. Which number identifies the promoter?
a.
1
b.
2
c.
3
d.
4
e.
5
23. Which number identifies the proximal promoter?
a.
1
b.
2
c.
3
d.
4
e.
5
24. Which number identifies the site where the transcriptional complex begins to form?
a.
1
b.
2
c.
3
d.
4
e.
5
25. Which number identifies a region near the promoter that increases the rate of transcription?
a.
2
b.
3
c.
4
d.
5
e.
6
26. Which number identifies the site where activators bind?
a.
2
b.
3
c.
4
d.
5
e.
6
27. Which number identifies a regulatory sequence that can increase the rate of transcription?
a.
1
b.
3
c.
4
d.
5
e.
6
28. Which number identifies the sequence most directly impacted by chromatin remodeling?
a.
1
b.
2
c.
3
d.
4
e.
5
29. Other than the proteins that they encode, what is a major difference between cell-specific genes and
housekeeping genes? (PPE = promoter proximal element)
a.
Cell-specific genes are expressed only in a single cell type; housekeeping genes are
expressed in all cell types.
b.
Cell-specific genes have PPEs and enhancers; housekeeping genes have only PPEs.
c.
Cell-specific genes have PPEs that use specific activators; housekeeping genes have
“universal” activators for their PPEs.
d.
Cell-specific genes are only found in certain cell types; housekeeping genes are found in
all cell types.
e.
Cell-specific genes have PPEs that use “universal” activators; housekeeping genes have
PPEs that respond only to specific activators.
30. What is/are the function(s) of a coactivator?
a.
form a bridge between the enhancer and the promoter proximal region
b.
cause the DNA to form a loop
c.
bind to RNA polymerase
d.
increase the rate of transcription
e.
all of these
31. In eukaryotic cells,
a.
the bulk of the control of gene expression is at the level of transcription just like in
prokaryotes
b.
mRNAs tend to remain stable for about the same length of time
c.
mRNAs can have their activities altered through modifications such as phosphorylation
d.
promoters of genes that are transcribed remain part of heterochromatin
e.
genes that are involved in similar processes are clustered together in operons
32. How do transcription repressors work?
a.
They bind to the same sequence where activators normally bind.
b.
They bind to and disable the activator.
c.
They recruit histone acetylation enzymes and thus interfere with chromatin remodeling.
d.
They bind to RNA polymerase II and prevent it from binding to the transcription factor
complex
e.
They methylate DNA.
33. The way a eukaryotic cell efficiently regulates the expression of multiple genes is to
a.
have individual activators for individual genes.
b.
have a few activators that can interact with small subsets of genes.
c.
have activators that serve as catalysts for further activator protein synthesis.
d.
have a few activators that work in different combinations to activate different genes.
e.
have a single activator that undergoes modifications to enable interaction with different
promoters.
34. Steroid hormones can trigger gene expression in a select number of cells because
a.
only target cells allow the steroid hormone to cross the plasma membrane.
b.
nontarget cells have a steroid hormone response element encoded in their DNA.
c.
only target cells have the correct receptor in their cytosol to bind to the hormone.
d.
nontarget cells lack the genes found in target cells.
e.
only target cells have the genes that steroid hormones activate.
35. Methylation regulates transcription
a.
via the addition of a methyl group to cytosine bases of DNA.
b.
via the addition of a methyl group to cysteine residues on RNA polymerase II.
c.
via the addition of a methyl group to cysteine bases of DNA.
d.
via the addition of a methyl group to cytosine residues on RNA polymerase II.
e.
by interfering with the chromatin remodeling process.
36. When a DNA promoter sequence is methylated,
a.
it is permanently silenced because the methyl group can never be removed.
b.
it is temporarily silenced because the methyl group can be removed.
c.
it is perpetually transcribed because the methyl group can never be removed.
d.
it is temporarily transcribed because the methyl group can be removed.
e.
the impact on expression is dependent on and specific to a particular gene.
37. When a DNA sequence is permanently rendered inaccessible to transcription via methylation and
chromatin modifications, we call this
a.
silencing.
b.
imprinting.
c.
a Barr body.
d.
inactivation.
e.
quiescence.
38. One function of the 5′ UTR (untranslated region) of mRNA is to
a.
control the half-life of mRNA.
b.
extend the half-life of mRNA.
c.
decrease the half-life of mRNA.
d.
stabilize the mRNA structure.
e.
bind ribosomes for initiating translation.
39. Micro RNAs (miRNAs)
a.
are short sequences only 22 or so bases long.
b.
are noncoding RNA sequences that bind to a protein complex.
c.
can regulate mRNA translation by cleaving the mRNA.
d.
can prevent translation of its target mRNA by blocking the ribosome.
e.
all of these
40. Which statement accurately describes the relationship between RNA interference, small interfering
RNA, and micro RNAs?
a.
They all describe different methods of post-translational gene regulation.
b.
Micro RNAs can be broken down into two types: RNA interference molecules and small
interfering RNAs.
c.
RNA interference occurs in two ways: by micro RNA or by small interfering RNAs.
d.
Small interfering RNAs bind to dicer protein, microRNAs do not; both are types of RNA
interference.
e.
MicroRNAs are transcribed from viruses, small interfering RNAs are transcribed from
nuclear DNA; both are types of RNA interference.
41. A small molecule of RNA is transcribed in the nucleus. It is folded, cleaved by dicer protein, and then
binds to a target molecule of mRNA. This molecule of RNA must be
a.
miRNA.
b.
siRNA.
c.
an mRNA that was not properly capped.
d.
an mRNA that never received its poly-A tail.
e.
a transcript for a masking protein.
42. Posttranslational modification includes
a.
cleavage of poly-A tails from mRNA.
b.
the binding of miRNAs to mRNA.
c.
the chemical modification, processing, and degradation of proteins.
d.
chromatin remodeling.
e.
splicing of mRNAs.
43. Pepsin, a digestive enzyme that degrades proteins in the stomach, is synthesized as pepsinogen and
converted to active pepsin in the stomach by the removal of several amino acids. The activation of
pepsin is an example of
a.
transcriptional regulation.
b.
translational regulation.
c.
posttranscriptional regulation.
d.
posttranslational regulation.
e.
RNA interference.
44. Why is ubiquitin nicknamed the “doom tag”?
a.
When ubiquitin binds to a target protein, the ubiquitin enzyme degrades the protein and
destroys it.
b.
When ubiquitin binds to a target protein, the complex is sent to the proteasome and the
complex is hydrolyzed to amino acids.
c.
If a cell is deficient in ubiquitin, the cell will die.
d.
High levels of ubiquitin in a cell cause the cell to destroy vital proteins, leading to
premature cell death.
e.
Ubiquitin catalyzes the conversion of pepsinogen to active pepsin, an enzyme vital to
protein digestion in the stomach of mammals. A lack of ubiquitin in the stomach causes an
organism to starve.
45. The vast majority of proteins expressed early in an animal’s development
a.
are translated from preexisting mRNAs present in the unfertilized egg.
b.
are translated from preexisting mRNAs delivered by the sperm that fertilized an egg.
c.
are translated from mRNAs that were transcribed from the zygote’s DNA in the first three
rounds of cell division.
d.
are transcribed and translated from the unfertilized egg‘s DNA postfertilization.
e.
were present as inactive proteins in the unfertilized egg.
46. The anterior-posterior direction of a mammal is initially controlled by
a.
maternal effect genes
b.
gap genes
c.
pair rule genes
d.
segment polarity genes
e.
homeotic genes
47. When cells differentiate they
a.
lose DNA
b.
remain totipotent
c.
express a specific subset of genes
d.
stop growing
e.
gain DNA
48. These genes encode proteins that generally regulate the cell cycle in normal cells.
a.
tumor suppressor genes
b.
proto-oncogenes
c.
oncogenes
d.
tumor suppressor genes and proto-oncogenes
e.
tumor suppressor genes, proto-oncogenes, and oncogenes
49. Many proto-oncogenes encode
a.
extracellular receptors.
b.
intracellular receptors.
c.
tumor suppressor proteins.
d.
protein kinases.
e.
extracellular receptors and protein kinases.
50. The role of a tumor suppressor protein in a cell is to
a.
promote cell division of healthy cells.
b.
halt cell division in healthy cells experiencing difficulties in DNA replication.
c.
promote cell division of abnormal cells.
d.
trigger DNA replication in preparation for cell division.
e.
activate cyclin-dependent protein kinases.