Chapter 8 - Control of Gene Expression

Question 1

Which of the following is considered a housekeeping protein?

• cortisol
• RNA polymerase
• an antibody
• insulin
• hemoglobin

Answer 1

RNA polymerase

(Proteins that are common to all the cells of a multicellular organism are called housekeeping proteins. All cells are involved in gene expression and are therefore undergoing transcription, which requires RNA polymerase.)

Question 2

Which would be the best method for determining which genes are being transcribed in a particular cell type?

• DNA sequencing
• X-ray crystallography
• NMR spectroscopy
• RNA sequencing (RNA seq)
• electron microscopy

Answer 2

RNA sequencing (RNA seq)

(RNA sequencing, also commonly called “RNA seq,” is a new, powerful technique that allows researchers to catalog all of the RNA molecules present in a cell at a given moment in time.)

Question 3

Which form of control directly influences which mRNAs are selected by ribosomes for the synthesis of proteins?

• protein activity control
• transcriptional control
• mRNA degradation control
• mRNA processing and localization control
• translational control

Answer 3

Translational control

If an mRNA fails to associate with the ribosome, then that mRNA will not be translated into protein.

Question 4

Although all of the steps involved in expressing a gene can in principle be regulated, what is the most important stage of control for most genes?

• RNA processing
• mRNA degradation
• RNA transport and localization
• transcription initiation
• mRNA translation

Answer 4

Transcription initiation

(Because transcriptional control is the first step in gene expression, regulation at this level has the most dramatic effects. If an RNA transcript is never synthesized, there is no way for that gene to ever be expressed.)

Question 5

At any given time, a typical differentiated human cell will express how many of its approximately 19,000 protein-coding genes?

• about 200
• all 19,000
• from 5000 to 15,000
• about 20

Answer 5

From 5,000 to 15,000

(Experimental examination of mRNAs in different cell types suggests that differentiated human cells express about this number of genes.)

Question 6

How or where do most transcription regulators bind?

• to a DNA sequence called the homeodomain
• to the minor groove of DNA
• to the major groove of RNA
• as dimers
• to a DNA sequence called a leucine zipper

Answer 6

As dimers

Dimerization roughly doubles the area of contact with the DNA, making the interaction tighter and more specific.

Question 7

Which of the following statements about eukaryotic activator proteins is false?

• They stimulate transcriptional initiation by opening up the double helix.
• They stimulate transcription initiation by aiding in the assembly of general transcription factors and RNA polymerase at the promoter.
• They stimulate transcription initiation by recruiting proteins that modify chromatin structure.
• They stimulate transcription initiation by promoting the assembly of a transcription initiation complex at the promoter.

Answer 7

They stimulate transcriptional initiation by opening up the double helix.

(Eukaryotic activator proteins do not open up the double helix. Rather, eukaryotic transcriptional activators can recruit chromatin-modifying proteins to help initiate gene transcription.)

Question 8

What is an operon?

• a set of genes transcribed as a single mRNA from a single promoter
• a set of genes that is constitutively active
• a set of genes controlled by the binding of two or more transcription regulators
• a sequence of DNA that produces a variety of mRNAs
• a short sequence of DNA to which a transcription regulator binds

Answer 8

A set of genes transcribed as a single mRNA from a single promoter

(Operons are defined by the coordination of expression of their resident genes under the direction of a single promoter. Each operon produces a single mRNA that encodes multiple proteins.)

Question 9

Where do transcription regulators usually bind on a DNA double helix?

• minor groove
• 5’ end
• 3’ end
• single-stranded regions
• major groove

Answer 9

Major groove

(It is the binding of a transcription regulator to the major groove of DNA in a regulatory sequence that acts as the switch to control transcription.)

Question 10

The transcription initiation site of a eukaryotic gene is found at which location?

• where transcription regulators bind
• where RNA polymerase first binds
• where RNA synthesis begins
• where general transcription factors bind

Answer 10

Where RNA synthesis begins

(The transcription initiation site of a eukaryotic promoter is where RNA synthesis first begins, and this location is “downstream” of the core promoter region.)

Question 11

Clinicians and the public are excited about the prospects of replacing damaged and diseased tissues with patient-derived (autologous) cells. Using autologous cells, as opposed to cells from a donor, avoids complications such as immune rejection. What series of steps could lead to the production of smooth muscle cells from the fibroblasts of a patient?

Answer 11

1. Obtain fibroblasts
2. Use transcription factors to convert fibroblasts to iPS cells
3. Grow iPS cells in culture
4. Use transcription factors to convert iPS cells to smooth muscle cells

(Once induced pluripotent stem cells are created, different suites of transcription factors can convert them to specific differentiated cell types.)

Question 12

Consider the following image of a Drosophila leg that has a misplaced eye growing.

Determine whether the following statement is true or false: Master regulators such as Ey in Drosophila are so powerful that they can even activate their regulatory networks outside the normal location.

This statement is

• True
• False

Answer 12

True

(In Drosophila, master regulators such as Ey induce the expression of other regulatory genes, which can result in the development of organs such as the eye outside of their normal location.)

Question 13

To reinforce cell identity, vertebrate cells can methylate which nucleotide?

• cytosine that falls next to guanine in the sequence CG
• any cytosine or guanine
• guanine that falls next to cytosine in the sequence CG
• any guanine
• any cytosine

Answer 13

Cytosine that falls next to guanine in the sequence CG

(In vertebrates, this modification occurs on select cytosine (C) nucleotides that fall next to a guanine (G) in the sequence 5’-CG-3’.)

Question 14

Which is not an example of epigenetic inheritance?

• the inheritance of a single point mutation in a gene
• the inheritance of a regulatory protein that activates its own transcription
• the inheritance of methylation patterns in DNA
• the inheritance of patterns of chromosome condensation

Answer 14

The inheritance of a single point mutation in a gene

(Epigenetic inheritance does not involve changes to the nucleotide sequence of DNA, but a mutation does change the nucleotide sequence.)

Question 15

Which of the following cell types, when fully differentiated, does not divide to form new cells?

• fibroblasts
• liver cells
• smooth muscle cells
• skeletal muscle cells

Answer 15

Skeletal muscle cells

Skeletal muscle is an example of a terminally differentiated cell.

Question 16

The image shows cells from the same organism. Which of the following statements is correct regarding these two human cells?

• The neuron contains more genes as compared to the liver cell.
• Both cells contain the same genes, but they are expressed differently.
• One human cannot contain both of these diverse cell types.
• The neuron contains fewer genes as compared to the liver cell.

Answer 16

Both cells contain the same genes, but they are expressed differently.

(Both cells contain the same genes, but they are expressed differently. If both cells come from the same human organism, then they are both descendants of the initial cell that was created at fertilization and therefore contain the same genome.)

Question 17

What happens to the miRNAs that are bound in a RISC?

• They remain bound to the RISC, where they can target the elimination of multiple mRNAs.
• They are destroyed as soon as they bind to an mRNA.
• If they include a sufficiently extensive region of complementarity with an mRNA, they are diverted to a region of the cytosol where they are eventually degraded.
• Their translation is blocked.

Answer 17

They remain bound to the RISC, where they can target the elimination of multiple mRNAs.

(This feature allows a single miRNA molecule to inactivate many mRNA molecules.)

Question 18

The long noncoding RNA Xist silences genes on the X chromosome by doing what?

• producing siRNAs
• encoding a repressor protein
• acting as an antisense transcript that binds to mRNAs
• producing miRNAs
• promoting the formation of heterochromatin

Answer 18

Promoting the formation of heterochromatin

(Although the molecular mechanisms are not entirely clear, it is thought that Xist recruits enzymes and chromatin-remodeling complexes that promote the formation of a highly condensed form of heterochromatin.)

Question 19

What type of molecule triggers RNA interference (RNAi)?

• double-stranded DNA
• double-stranded miRNAs
• single-stranded RNAs
• foreign, double-stranded RNA
• foreign, single-stranded DNA

Answer 19

Foreign, double-stranded RNA

(Double-stranded RNAs are produced by many viruses and transposable elements, and are thus recognized as “foreign.” This makes them valid targets for elimination by RNAi.)

Question 20

In bacteria, what would be the consequence of a protein binding to and blocking the ribosomal binding site on an mRNA?

• The ribosome will begin translation at a new location on the mRNA.
• The protein will recruit the small ribosomal subunit to the mRNA, increasing translation efficiency.
• The small ribosomal subunit will not be able to bind to the mRNA, and translation will be inhibited.
• The protein will be incorporated into the growing polypeptide chain.

Answer 20

The small ribosomal subunit will not be able to bind to the mRNA, and translation will be inhibited.

(When the ribosomal binding site is covered by a protein, translation is inhibited.)

Question 21

Researchers assayed the activity of enzyme F in three different types of tissue from the same mouse by determining the amount of enzyme product produced per milligram of tissue per unit time. As shown in the graph below, results indicate more product generation in the liver compared to the kidney and muscle samples.

Which of the following factors might explain the different results among the three tissues?

Choose one or more:

• differences in the post-translational modifications of the enzyme among the tissue types
• differences in the DNA content among the tissue types
• differences in the transcription of the gene encoding the enzyme among the tissue types
• differences in the translation of the mRNA encoding the protein among the tissue types

Answer 21

• differences in the post-translational modifications of the enzyme among the tissue types
• differences in the transcription of the gene encoding the enzyme among the tissue types
• differences in the translation of the mRNA encoding the protein among the tissue types

(Control of gene expression can happen at any step along the pathway from DNA to protein, including transcription, mRNA processing and localization, protein production and degradation, and protein activity control.)

Question 22

Using powerful new sequencing technologies, investigators can now catalog every RNA molecule made by a cell and determine at what quantities these RNAs are present. In an experiment, researchers measured the relative quantities of two different mRNAs—one transcribed from gene A, the other from gene B—in two different cell types. Gene B is expressed in both the liver and the brain whereas gene A is expressed in the brain but not in the liver. Which most likely encodes a housekeeping protein?

• gene A only
• both genes
• gene B only
• neither gene

Answer 22

Gene B only

(Because gene B is expressed in both the liver and the brain, it could encode a housekeeping protein, which is the term given to genes that are expressed in all cells of a multicellular organism.)

Question 23

Which of the following statements is not true about the differences between liver cells and kidney cells in the same organism?

• They express different genes.
• They contain different genes.
• They contain the entire set of instructions needed to form the whole organism.
• They contain different proteins.

Answer 23

They contain different genes.

(Specialized cells, such as liver and kidney cells, do contain the same genes. They just express them differently. This is referred to as differential gene expression.)

Question 24

Researchers have created plasmids that only allow expression of inserted genes in response to a metabolite. Researchers can add these plasmids to E. coli cells and increase the expression of the inserted gene by adding the appropriate metabolite to the culture media. Plasmids containing which combination of operator and promoter allow activation of gene expression in response to an added metabolite?

Choose one or more:

• operator recognized by Lac repressor protein
• strong promoter
• weak promoter
• operator recognized by Trp repressor protein

Answer 24

• operator recognized by Lac repressor protein
• strong promoter

(The combination of a strong promoter and the operator sequence from the Lac operon allows transcription in response to lactose.)

Question 25

Drag the labels to their correct targets to identify the correct carbohydrate conditions that must be in place in order to cause the molecular mechanisms that are illustrated in the figure.

Answer 25

+ lactose, +glucose
The operon is simply off

-lactose, +glucose
When lactose is absent, the lac repressor binds to the Lac operator and shuts the expression of the operon. So the operon is OFF

-lactose, -glucose
When glucose is absent, the cell produces cyclic AMP & CAP activator binds to DNA; in this case there is no lactose and thus the lac repressor still binds to the DNA. So the operon is OFF.

+lactose, -glucose
When glucose is absent and there is lactose, the operon is ON. For the operon to be transcribed, glucose must be absent (which allows the CAP activator to bind). The presence of lactose releases the ​repressor. Then, lacZ the first gene of the operon, encodes the enzyme beta-galactosidase, which breaks down lactose to galactose and glucose.

(The Lac operon is controlled by two transcription regulators: the Lac repressor, which responds to lactose presence/absence, and CAP, which responds to glucose presence/absence through cAMP concentration that is inverse to that of glucose. Thus, transcription of the Lac operon takes place when lactose is present but glucose is absent.)

Question 26

In bacterial cells, the tryptophan operon encodes the genes needed to synthesize tryptophan. What happens when the concentration of tryptophan inside a cell is high?

-It activates the tryptophan repressor, allowing transcription of the tryptophan operon.
-It activates the tryptophan repressor, which breaks down excess tryptophan.
-It activates the tryptophan repressor, which shuts down expression of the tryptophan operon.
-It inactivates the tryptophan repressor, which shuts down the tryptophan operon.
It inactivates the tryptophan repressor, allowing transcription of the tryptophan operon.

Answer 26

It activates the tryptophan repressor, which shuts down expression of the tryptophan operon.

(In bacterial cells, when the concentration of tryptophan is high, it activates the tryptophan repressor, which binds to the Trp operator and shuts down expression of the tryptophan operon.)

Question 27

Which of the following statements concerning leucine zipper protein dimerization and DNA binding is correct?

• Leucine zipper proteins function as a dimer with both subunits making contact with the sequence-specific DNA site.
• Leucine zipper protein dimerization is facilitated by polar amino acids in the dimerization domains.
• Leucine zipper proteins contain many leucine amino acids in the DNA-binding region that facilitate sequence-specific DNA binding.
• Leucine zipper proteins use ionic bonds to bind with the sequence-specific DNA site

Answer 27

Leucine zipper proteins function as a dimer with both subunits making contact with the sequence-specific DNA site.

(The leucine and other hydrophobic side chains in the first portion of the helices dimerize along the length of the two helices. The second portion of the helices forms the DNA-binding domain, where amino acid side chains hydrogen-bond with specific bases in the DNA, leading to sequence-specific binding.)

Question 28

Which of the following mutations would be least likely to disrupt the function of the leucine zipper protein in the animation? The structures of relevant amino acids are provided below.

• mutation of a leucine to aspartic acid in the dimerization domain of the protein
• mutation of the DNA-binding arginine to alanine in the DNA-binding domain of the protein
• mutation of a leucine to valine in the dimerization domain of the protein
• mutation of the guanine nucleotide to a thymine nucleotide in the binding site on the DNA

Answer 28

Mutation of a leucine to valine in the dimerization domain of the protein

(This would be least likely of the choices to disrupt the function of the leucine zipper protein. Dimerization of the two subunits occurs through tight packing of the hydrophobic side chains. Since both leucine and valine have hydrophobic side chains, dimerization is likely to still occur.)

Question 29

The maltose operon contains genes that code for proteins that catabolize the disaccharide maltose. Similar to the Lac operon, which is only efficiently transcribed in the presence of lactose, the maltose operon is only efficiently transcribed in the presence of maltose. How might induction of the maltose operon in response to maltose be achieved?

Choose one or more:

• Maltose removes a repressor from an operon with an efficient promoter.
• Maltose causes an activator to bind an operon with an inefficient promoter.
• Maltose causes a repressor to bind an operon with an efficient promoter.
• Maltose causes an activator to bind an operon with an efficient promoter.
• Maltose removes an activator from an operon with an inefficient promoter.

Answer 29

• Maltose removes a repressor from an operon with an efficient promoter.
• Maltose causes an activator to bind an operon with an inefficient promoter.

(There are two ways to accomplish operon induction: (1) the inducer can remove a repressor from an operon with an efficient promoter, or (2) the inducer can cause an activator to bind to an operon with an inefficient promoter.)

Question 30

Which of the following describes the Lac operon in E. coli when lactose, but not glucose, is present in the culture medium?

• The Lac repressor, but not CAP, is bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.
• Neither CAP nor the Lac repressor is bound to the Lac operon’s regulatory DNA, and the Lac operon is expressed.
• CAP, but not the Lac repressor, is bound to the Lac operon’s regulatory DNA, and the Lac operon is expressed.
• CAP and the Lac repressor are both bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.
• Neither CAP nor the Lac repressor is bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.

Answer 30

CAP, but not the Lac repressor, is bound to the Lac operon’s regulatory DNA, and the Lac operon is expressed.

(Under these conditions, only CAP (and not the Lac repressor protein) is bound to the Lac operon’s regulatory DNA. Therefore, the Lac operon is expressed.)

Question 31

Which of the following describes the Lac operon in E. coli when both lactose and glucose are present in the culture medium?

• The Lac repressor, but not CAP, is bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.
• CAP binds to the Lac repressor, preventing it from binding to the Lac operon’s regulatory DNA, and the Lac operon is expressed.
• Neither CAP nor the Lac repressor is bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.
• CAP, but not the Lac repressor, is bound to the Lac operon’s regulatory DNA, and the Lac operon is expressed.
• CAP and the Lac repressor are both bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.

Answer 31

Neither CAP nor the Lac repressor is bound to the Lac operon’s regulatory DNA, and the Lac operon is not expressed.

(Under these conditions, neither CAP nor the Lac repressor is bound to the Lac operon’s regulatory DNA, and the operon is not expressed.)

Question 32

Which of the following statements most accurately describes the expression of the repressor protein of the tryptophan operon?

• The gene for the tryptophan repressor is turned on in response to high levels of tryptophan in the cell.
• The gene for the tryptophan repressor is turned off in response to high levels of tryptophan in the cell.
• The gene for the tryptophan repressor is turned on in response to low levels of tryptophan in the cell.
• The gene for the tryptophan repressor is turned off in response to low levels of tryptophan in the cell.
• The gene for the tryptophan repressor is expressed constitutively.

Answer 32

The gene for the tryptophan repressor is expressed constitutively.

(The gene for the tryptophan repressor is expressed constitutively. The Trp repressor protein must always be present so as to constantly respond to the levels of tryptophan in the cell.)

Question 33

The Drosophila regulatory segment that defines the location of Eve stripe 2 contains binding sites for four different transcription regulators: two repressors (Giant and Krüppel) and two activators (Bicoid and Hunchback). For Eve to be efficiently expressed in stripe 2, both repressors must be absent and both activators present. What would you expect to see in flies that lack the gene that encodes Bicoid? (Assume that Bicoid does not influence the expression of Hunchback, Giant, or Krüppel.)

• Stripe 2 would become narrower.
• Stripe 2 would expand toward the tail of the embryo.
• All 7 stripes would disappear.
• Stripe 2 would expand toward the head of the embryo.
• Stripe 2 would become fainter.

Answer 33

Stripe 2 would become fainter.

Stripe 2 would become faint due to lower level of expression, but would remain at its normal position.

Question 34

The Drosophila regulatory segment that defines the location of Eve stripe 2 contains binding sites for four different transcription regulators: two repressors (Giant and Krüppel) and two activators (Bicoid and Hunchback). For Eve to be efficiently expressed in stripe 2, both repressors must be absent and both activators present. What would you expect to see in flies that lack the gene that encodes Bicoid? (Assume that Bicoid does not influence the expression of Hunchback, Giant, or Krüppel.)

• Stripe 2 would become narrower.
• Stripe 2 would expand toward the tail of the embryo.
• All 7 stripes would disappear.
• Stripe 2 would expand toward the head of the embryo.
• Stripe 2 would become fainter.

Answer 34

Stripe 2 would become fainter.

Stripe 2 would become faint due to lower level of expression, but would remain at its normal position.

Question 35

c-Met is an oncogene that contributes to the development of certain cancers by triggering cell division and tumor growth. In a 2009 article, Yan and colleagues found regions in the 3’ untranslated region of c-Met mRNA complementary to microRNA-1/206. In addition, higher levels of microRNA-1/206 were associated with slower cell proliferation. What is a likely explanation for the inverse correlation between microRNA-1/206 and cell proliferation?

• MicroRNA-1/206 targets c-Met mRNA for destruction via RISC.
• MicroRNA-1/206 codes for a tumor suppressor protein that directly inhibits cell proliferation.
• MicroRNA-1/206 stabilizes c-Met mRNA, leading to enhanced translation.
• MicroRNA-1/206 codes for a protein that directly binds to and inhibits c-Met protein.

Answer 35

MicroRNA-1/206 targets c-Met mRNA for destruction via RISC.

(MicroRNAs are noncoding regulatory RNAs that regulate gene expression by binding complementary mRNAs, leading to the destruction of the targeted mRNAs.)

Question 36

What would happen to the helix-3 interaction with DNA if a mutation occurred that altered this adenine (as shown) to guanine?

• The integrity of the interaction would decrease because one of the two hydrogen bonds would not be able to form.
• The integrity of the interaction would remain the same because the same two hydrogen bonds would still be able to form.
• The integrity of the interaction would increase because the hydrogen bonds would still be able to form and would be stronger.

Answer 36

The integrity of the interaction would decrease because one of the two hydrogen bonds would not be able to form.

(If guanine replaced adenine, the integrity of the protein–DNA interaction would decrease because one of the two hydrogen bonds would not be able to form. This is because adenine contains an amino group that interacts with the asparagine carbonyl group; however, this interaction would not form if guanine were present in place of adenine.)

Question 37

Which of the following statements is/are true of long noncoding RNAs?

• They are involved in X chromosome inactivation.
• They can regulate the translation and stability of mRNAs.
• They can silence genes by promoting the formation of euchromatin.
• They can trigger the activity of histone acetyltransferases.

Answer 37

• They are involved in X chromosome inactivation.
• They can regulate the translation and stability of mRNAs.

(Long noncoding RNAs are involved in X chromosome inactivation and can regulate the translation and stability of mRNAs.)