Deck 19: Control of Gene Expression in Eukaryotes

Question 1

A pattern of inheritance in which heritable differences in phenotype are due to something other than differences in DNA sequence is

A) organelle genome inheritance.
B) Mendelian inheritance.
C) allelic inheritance.
D) epigenetic inheritance.

Answer 1

D

Question 2

Which method is utilized by eukaryotes to control their gene expression that is not used in bacteria?

A) control of chromatin remodeling
B) control of RNA splicing
C) transcriptional control
D) A and B
E) all of the above

Answer 2

D

Question 3

Which of the following is most critical for the association between histones and DNA?

A) Histones are synthesized in the cytoplasm.
B) Histones are positively charged.
C) Histones are highly conserved i.e., histones are very similar in every eukaryote).
D) There are at least five different histone proteins in every eukaryote.
E) Histones are small proteins.

Answer 3

Histones are positively charged.

Question 4

If the DNA sequences of a particular gene in a skin cell and a liver cell were compared, there would be

A) a small number of differences.
B) a very large number of differences.
C) so many differences that it would be impossible to identify corresponding genes in skin and liver cells.
D) no differences.

Question 5

The primary difference between an enhancer and a promoter- proximal element is that

A) enhancers are part of the promoter, while promoter- proximal elements are regulatory sequences distinct from the promoter.
B) enhancers are transcription factors, while promoter- proximal elements are DNA sequences.
C) enhancers are at considerable distances from the promoter and can be moved or inverted and still function, while promoter- proximal elements are close to the promoter and their position and orientation must be maintained.
D) enhancers enhance transcription, while promoter- proximal elements inhibit transcription.
E) enhancers are DNA sequences, while promoter- proximal elements are proteins that bind proximal to the promoter.

Question 6

If a cell were unable to produce histone proteins, which of the following would be a likely effect?

A) The cell would immediately divide.
B) Spindle fibres would not form during prophase.
C) The cell's DNA couldn't be packed into its nucleus.
D) Amplification of other genes would compensate for the lack of histones.
E) There would be an increase in the amount of "satellite" DNA produced during centrifugation.

Question 7

Which of the following allows more than one type of protein to be produced from one gene?

A) control of mRNA stability
B) alternative forms of RNA splicing
C) control of the frequency of translation initiation
D) alternative forms of chromatin remodeling
E) all of the above

Question 8

What is a key property of DNase that makes it useful for assessing whether chromatin is in a closed tightly condensed) or open loosely packed) configuration?

A) DNase cuts at specific DNA sequences.
B) DNase digests only mammalian DNA.
C) DNase preferentially digests DNA not associated with protein.
D) DNase is a protein.

Question 9

How do chromatin- remodeling complexes recognize the genes they should act on?

A) Chromatin- remodeling complexes recognize specific transcription factors bound to regulatory sequences of DNA.
B) Chromatin- remodeling complexes bind to the basal transcription complex.
C) Chromatin- remodeling complexes are activated by specific extracellular signals that direct them to particular genes.
D) Chromatin- remodeling complexes recognize specific promoters when they are phosphorylated, methylated, or acetylated.

Question 10

Imagine you've isolated a yeast mutant that contains a constitutively constantly) active histone deacetylase. What phenotype do you predict for this mutant?

A) The mutant will require galactose for growth.
B) The mutant will show low levels of gene expression.
C) The mutant will show high levels of gene expression.
D) The mutant will grow rapidly.

Question 11

Imagine you've isolated a yeast mutant that contains histones resistant to acetylation. What phenotype do you predict for this mutant?

A) The mutant will require galactose for growth.
B) The mutant will grow rapidly.
C) The mutant will show low levels of gene expression.
D) The mutant will show high levels of gene expression.

Question 12

Figure 19.1

-Predict what would occur in the experiment shown in Figure 19.1 if Tonegawa and colleagues had mistakenly inserted the antibody gene enhancer in reverse orientation backward) into the fi- globin gene.

A) There would be transcription of the fi- globin gene, but on the opposite strand of DNA from the one normally transcribed.
B) There would be minute levels of fi- globin gene expression in the antibody- producing cells.
C) There would be little difference in the results of this experiment and the one shown in the figure.
D) There would be no fi- globin gene expression in the antibody- producing cells.

Question 13

The TATA- binding protein TBP) binds to

A) promoter- proximal elements.
B) RNA polymerase.
C) the promoter.
D) the point where transcription begins.
E) enhancers.

Question 14

Twenty- five years ago, when Oshima and colleagues discovered that a mutation in the GAL4 gene led to the inability to synthesize all five enzymes required for galactose catabolism breakdown), they couldn't be blamed for wanting to apply a bacterial model to explain this finding. What they expected, but did not find, was

A) for chromatin decondensation to play an important role in regulating the genes of galactose catabolism.
B) for all five genes to constitute an operon.
C) five widely separated genes, each containing a GAL4 binding site in its regulatory region.
D) for transcription to be important in regulating the genes of galactose catabolism.

Question 15

The reason for differences in the sets of proteins expressed in a nerve and a pancreatic cell of the same individual is that nerve and pancreatic cells contain different

A) promoter- proximal elements.
B) sets of regulatory proteins.
C) genes.
D) promoters.
E) regulatory sequences.

Question 16

Histone acetyl transferases exert their effect on gene activity by

A) introducing negative charges on the glutamic acids of histones.
B) neutralizing positive charges on the lysines of histones.
C) increasing the affinity of transcriptional activators for DNA.
D) increasing the affinity of transcriptional inhibitors for DNA.
E) modifying the DNA sequence of the promoter.

Question 17

The association of DNA with nucleosomes means that the default state for eukaryotic genes is to be

A) repressed by silencers.
B) turned on.
C) activated by enhancers.
D) turned off.

Question 18

Figure 19.1
In the experiment shown in the figure above, Tonegawa and his colleagues were able to express fi- globin in an antibody- producing cell that normally does not express fi- globin. They achieved this result by splicing an enhancer from an antibody- producing gene into the protein- coding portion of the fi- globin gene. They then introduced this recombinant gene into cultured antibody- producing cells. Why was the choice of
Antibody- producing cells-rather than, say, muscle or skin cells-critical for the success of this experiment?

A) Only the antibody- producing cells have the correct regulatory transcription factors to bind to the enhancer.
B) Only the antibody- producing cells are capable of expressing recombinant genes.
C) Only the antibody- producing cells have the correct set of enhancers, promoter- proximal elements, and promoters.
D) Only the antibody- producing cells have the correct set of enhancers.
E) Only the antibody- producing cells have the correct set of promoter- proximal elements.

Question 19

Ovalbumin, the major protein of egg white, is secreted by cells that line the oviduct as the bird egg moves down the oviduct. Imagine you're repeating the classic Weintraub- Groudine experiment, but with a twist: You're assaying the DNase sensitivity of the promoter regions of the fi- globin and ovalbumin genes in oviduct cells of laying hens. In this case you expect to find that

A) the fi- globin and ovalbumin promoters are equally resistant to DNase treatment.
B) the ovalbumin promoter is much more sensitive to DNase treatment.
C) the fi- globin promoter is much more sensitive to DNase treatment.
D) the fi- globin and ovalbumin promoters are equally sensitive to DNase treatment.

Question 20

Imagine that you are studying the control of fi- globin gene expression in immature red blood cells mature red blood cells contain fi- globin protein, but lack a nucleus and, therefore, the fi- globin gene). If you deleted a sequence of DNA outside the protein- coding region of the fi- globin gene and found that this increased the rate of transcription, the deleted sequence likely functions as an)

A) enhancer.
B) promoter.
C) silencer.
D) promoter- proximal element.
E) any of the above.

Question 21

In the roundworm C. elegans, the lin- 4 gene produces an RNA that forms a hairpin structure. One of the strands in the double- stranded region of lin- 4 hairpin RNA is complementary to the mRNA of a protein- coding gene, lin- 14. Predict the effect of expressing lin- 4 RNA during development.

A) The lin- 14 expression will rise when lin- 4 expression begins.
B) The lin- 4 RNA will bind the promoter of the lin- 14 gene, destroying the lin- 14 gene.
C) The lin- 4 RNA will bind to the enhancer of lin- 14, increasing lin- 14 transcription.
D) The lin- 14 expression will fall when lin- 4 expression begins.

Question 22

Alternative splicing takes place in the

A) nucleus.
B) rough endoplasmic reticulum.
C) cytoplasm.
D) polysome.
E) ribosome.

Question 23

Not long ago, it was believed that a count of the number of protein- coding genes would provide a count of the number of proteins produced in any given eukaryotic species. This is incorrect, largely due to the discovery of widespread

A) alternative splicing.
B) translational control.
C) transcriptional control.
D) chromatin condensation control.

Question 24

One way to detect alternative splicing of transcripts from a given gene is to

A) compare the sequences of different mRNAs made from this gene.
B) compare the sequences of different primary transcripts made from this gene.
C) compare the DNA sequence of this gene to that of a gene known to be constitutively spliced.
D) measure the relative rates of transcription of this gene compared to that of a gene known to be constitutively spliced.

Question 25

In colorectal cancer, several genes must be mutated for a cell to develop into a cancer cell. Which of the following kinds of genes would you expect to be mutated?

A) genes that have multiple copies
B) genes of the bacteria, which are abundant in the colon
C) genes coding for enzymes that act in the colon
D) genes that are especially susceptible to mutation
E) genes involved in control of the cell cycle

Question 26

Gene expression is often assayed by measuring the level of mRNA produced from a gene. If one is interested in knowing the amount of a final active gene product, a potential problem of this method is that it ignores the possibility of

A) transcriptional control.
B) chromatin condensation control.
C) alternative splicing.
D) translational control.
E) all of the above.

Question 27

If DNA were inflexible, which of the following would not function?

A) enhancers
B) promoters
C) the TATA- binding protein
D) basal transcription factors
E) promoter- proximal elements

Question 28

Why can gene expression be altered more easily at the level of post- transcriptional processing in eukaryotes than in prokaryotes? Because

A) eukaryotic exons may be spliced in alternative patterns.
B) prokaryotes use ribosomes of different structure and size.
C) eukaryotic mRNAs get 5' caps and 3' tails.
D) eukaryotic mRNAs are more stable.
E) prokaryotic genes are expressed as mRNA, which is more stable in the cell.

Question 29

Use the following information when answering the corresponding questions).
A group of enzymes known as cytosine- specific DNA methylases recognize CpG dinucleotide sequences-that is, a cytosine followed by a guanosine G) in one DNA strand-and add methyl groups to the cytosine. Many proteins bind to methylated CpG, including the methyl- CpG binding protein 2 MeCP2). MeCP2 binding leads to the formation of a closed state of chromatin, thus silencing gene expression. A recent paper by Chen et al. reported an interesting mechanism of regulating transcription of a gene via MeCP2 W. Chen, Q. Chang, Y. Lin, A. Meissner, A. E. West, E. C. Griffith, R. Jaenish, and M. Greenberg. 2004. Derepression of BDNF transcription involves calcium- dependent phosphorylation of MeCP2, Science 302:885-89). This gene, called BDNF, encodes brain- derived neurotrophic factor BDNF)-a protein that plays an important role in nerve cell and central nervous system function, including memory and learning. Remarkably, most cases of Rett syndrome, an important cause of mental retardation in females, are due to loss- of- function mutations of MeCP2.
A scientist has a vial containing DNA and tries to digest it with DNA digesting enzymes. After analyzing it, she finds that no digestion occurred. What would be the most likely reason for this?

A) The DNA was packed into regions called euchromatin.
B) The DNA was embedded in the plasma membrane and was not accessible for the enzyme.
C) The DNA was tightly packed into 30nm chromatin fibres.
D) The DNA was tightly packed into 10nm chromatin fibres.

Question 30

The rate of translation can be slowed drastically when a ribosomal protein is phosphorylated in response to a sudden temperature increase or viral infection. In this case, the phosphorylated ribosomal protein is under
Control, and the phosphorylation of this ribosomal protein leads to widespread control.

A) translational; post- translational
B) transcriptional; post- translational
C) post- translational; transcriptional
D) transcriptional; translational
E) post- translational; translational

Question 31

Use the following information when answering the corresponding questions).
A group of enzymes known as cytosine- specific DNA methylases recognize CpG dinucleotide sequences-that is, a cytosine followed by a guanosine G) in one DNA strand-and add methyl groups to the cytosine. Many proteins bind to methylated CpG, including the methyl- CpG binding protein 2 MeCP2). MeCP2 binding leads to the formation of a closed state of chromatin, thus silencing gene expression. A recent paper by Chen et al. reported an interesting mechanism of regulating transcription of a gene via MeCP2 W. Chen, Q. Chang, Y. Lin, A. Meissner, A. E. West, E. C. Griffith, R. Jaenish, and M. Greenberg. 2004. Derepression of BDNF transcription involves calcium- dependent phosphorylation of MeCP2, Science 302:885-89). This gene, called BDNF, encodes brain- derived neurotrophic factor BDNF)-a protein that plays an important role in nerve cell and central nervous system function, including memory and learning. Remarkably, most cases of Rett syndrome, an important cause of mental retardation in females, are due to loss- of- function mutations of MeCP2.
A patient is undergoing genetic screening for cancer. The results show the patient has many proto- oncogenes. Based on this information only, which of the following statements is correct regarding the probability that this patient will develop cancer in the very near future?

A) The chances are low because the patient would have to show an increased number of oncogenes.
B) The chances are high because the patient most likely also shows an increase in the number of tumor suppressors.
C) Due to the increased number of proto- oncogenes, the chances are high.

Question 32

Which of the following best explains the function of proto- oncogenes in eukaryotic cells?

A) Proto- oncogenes first arose from viral infections.
B) Proto- oncogenes are genetic "junk."
C) Cells produce proto- oncogenes as they age.
D) Proto- oncogenes are mutant versions of normal genes.
E) Proto- oncogenes normally help regulate cell division.

Question 33

Use the following information when answering the corresponding questions).
A group of enzymes known as cytosine- specific DNA methylases recognize CpG dinucleotide sequences-that is, a cytosine followed by a guanosine G) in one DNA strand-and add methyl groups to the cytosine. Many proteins bind to methylated CpG, including the methyl- CpG binding protein 2 MeCP2). MeCP2 binding leads to the formation of a closed state of chromatin, thus silencing gene expression. A recent paper by Chen et al. reported an interesting mechanism of regulating transcription of a gene via MeCP2 W. Chen, Q. Chang, Y. Lin, A. Meissner, A. E. West, E. C. Griffith, R. Jaenish, and M. Greenberg. 2004. Derepression of BDNF transcription involves calcium- dependent phosphorylation of MeCP2, Science 302:885-89). This gene, called BDNF, encodes brain- derived neurotrophic factor BDNF)-a protein that plays an important role in nerve cell and central nervous system function, including memory and learning. Remarkably, most cases of Rett syndrome, an important cause of mental retardation in females, are due to loss- of- function mutations of MeCP2.
Elsewhere in the paper, Chen et al. state that "BDNF is encoded by a complex gene with four well- characterized promoters that give rise to at least eight different mRNAs." What mechanism could account for the production of these different BDNF mRNAs?

A) alternative splicing
B) translational control
C) post- translational control
D) chromatin condensation control

Question 34

Regulatory transcription factors

A) influence the assembly of the basal transcription complex.
B) open the two strands of DNA, so RNA polymerase can begin transcription.
C) influence the binding of sigma factor to DNA.
D) influence the degree of unwinding of DNA at the promoter.

Question 35

p53 activates genes that

A) arrest the cell cycle.
B) promote metastasis.
C) prevent DNA damage.
D) increase the rate of endocytosis.
E) increase mutation rate.

Question 36

Which of the following types of mutation would convert a proto- oncogene into an oncogene?

A) a mutation that creates an unstable proto- oncogene mRNA
B) a mutation that blocks transcription of the proto- oncogene
C) a mutation that greatly increases the amount of the proto- oncogene protein
D) a deletion of most of the proto- oncogene coding sequence

Question 37

An example of a basal transcription factor is

A) an enhancer- binding transcription factor.
B) RNA polymerase.
C) a silencer- binding transcription factor.
D) a promoter- proximal- binding transcription factor.
E) the TATA- binding protein.

Question 38

Use the following information when answering the corresponding questions).
A group of enzymes known as cytosine- specific DNA methylases recognize CpG dinucleotide sequences-that is, a cytosine followed by a guanosine G) in one DNA strand-and add methyl groups to the cytosine. Many proteins bind to methylated CpG, including the methyl- CpG binding protein 2 MeCP2). MeCP2 binding leads to the formation of a closed state of chromatin, thus silencing gene expression. A recent paper by Chen et al. reported an interesting mechanism of regulating transcription of a gene via MeCP2 W. Chen, Q. Chang, Y. Lin, A. Meissner, A. E. West, E. C. Griffith, R. Jaenish, and M. Greenberg. 2004. Derepression of BDNF transcription involves calcium- dependent phosphorylation of MeCP2, Science 302:885-89). This gene, called BDNF, encodes brain- derived neurotrophic factor BDNF)-a protein that plays an important role in nerve cell and central nervous system function, including memory and learning. Remarkably, most cases of Rett syndrome, an important cause of mental retardation in females, are due to loss- of- function mutations of MeCP2.
The authors state, "In this study, we report that, in the absence of neuronal activity, MeCP2 binds specifically to BDNF promoter III and functions as a negative regulator of BDNF expression. In response to neuronal
Activity- dependent calcium influx into neurons, MeCP2 becomes phosphorylated and is released from the BDNF promoter, thereby permitting BDNF promoter III- dependent transcription." Reading this statement in isolation, it would be easy to have the impression that MeCP2 works as a negatively acting transcription factor. However, based on the summary of the paper provided above, MeCP2 acts as a

A) translational regulator.
B) gene- specific regulator of chromatin condensation.
C) positively acting transcription factor.
D) splicing regulator.

Question 39

Which of the following is FALSE?

A) All individuals possess many proto- oncogenes.
B) Mutations that inactivate tumor suppressor genes are important in cancer.
C) Cancer is a single disease with one underlying molecular cause.
D) Most agents that cause cancer also cause mutations.
E) Uncontrolled cell growth alone is insufficient for the development of most cancers.

Question 40

The normal function of a tumor suppressor gene is to

A) prevent progression of the cell cycle unless conditions are right for moving forward.
B) suppress the growth of tumors already present in all multicellular eukaryotes.
C) promote progress through the cell cycle.