Deck 18: Control of Gene Expression in Eukaryotes

Question 1

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

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

Answer 1

A

Question 2

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

Answer 2

C

Question 3

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

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

Answer 3

D

Question 4

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

Question 5

The TATA-binding protein (TBP)binds to

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

Question 6

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

Question 7

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

A)enhancers are transcription factors,while promoter-proximal elements are DNA sequences.
B)enhancers enhance transcription,while promoter-proximal elements inhibit transcription.
C)enhancers are part of the promoter,while promoter-proximal elements are regulatory sequences distinct from the promoter.
D)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.
E)enhancers are DNA sequences,while promoter-proximal elements are proteins that bind proximal to the promoter.

Question 8

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

Question 9

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

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

Question 10

Figure 18.1
Predict what would occur in the experiment shown in Figure 18.1 if Tonegawa and colleagues had mistakenly inserted the antibody gene enhancer in reverse orientation (backward)into the β-globin gene.

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

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

Question 12

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

Question 13

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

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

Question 14

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 β-globin and ovalbumin genes in oviduct cells of laying hens.In this case you expect to find that

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

Question 15

Histone acetyl transferases exert their effect on gene activity by

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

Question 16

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

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

Question 17

If cells of an individual contain the same set of genes,how do these cells become different from each other during development?

A)Different cell type-specific regulatory elements in DNA are created during development.
B)Different cell type-specific regulatory elements in DNA are selectively lost during development.
C)Differences in extracellular signals received by each cell lead to differences in the types of regulatory proteins present in each cell.
D)Differences develop in promoter sequences that lead to different signals being produced by each type of cell.

Question 18

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

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

Question 19

Figure 18.1
In the experiment shown in the figure above,Tonegawa and his colleagues were able to express β-globin in an antibody-producing cell that normally does not express β-globin.They achieved this result by splicing an enhancer from an antibody-producing gene into the protein-coding portion of the β-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 have the correct set of enhancers.
C)Only the antibody-producing cells have the correct set of promoter-proximal elements.
D)Only the antibody-producing cells have the correct set of enhancers,promoter-proximal elements,and promoters.
E)Only the antibody-producing cells are capable of expressing recombinant genes.

Question 20

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

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

Question 21

The normal function of a tumor suppressor gene is to

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

Question 22

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

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

Question 23

Alternative splicing takes place in the

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

Question 24

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

Question 25

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

A)a mutation that blocks transcription of the proto-oncogene
B)a mutation that creates an unstable proto-oncogene mRNA
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 26

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

Question 27

An example of a basal transcription factor is

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

Question 28

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

Question 29

An important way to regulate the stability of mRNA is through

A)controlling the rate of transcription.
B)activating chromatin remodeling complexes.
C)controlling the rate of translation.
D)RNA interference.

Question 30

p53 activates genes that

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

Question 31

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

Question 32

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

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

Question 33

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

Question 34

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

Question 35

Use the following information when answering the corresponding question(s).
A group of enzymes known as cytosine-specific DNA methylases recognize CpG dinucleotide sequences-that is, a cytosine (C)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)gene-specific regulator of chromatin condensation.
B)positively acting transcription factor.
C)splicing regulator.
D)translational regulator.

Question 36

Regulatory transcription factors

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

Question 37

A gene is a

A)protein-coding sequence of DNA.
B)sequence of DNA capable of directing the synthesis of a polypeptide.
C)sequence of DNA capable of directing the synthesis of one or more related polypeptides or RNAs.
D)sequence of DNA capable of directing the synthesis of one or more related biological molecules of any type.

Question 38

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

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)Uncontrolled cell growth alone is insufficient for the development of most cancers.
E)Most agents that cause cancer also cause mutations.