Deck 18: Control of Gene Expression in Eukaryotes

Question 1

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

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)control of both RNA splicing and chromatin remodeling
E)control of chromatin remodeling,RNA splicing,and transcription

Answer 2

D

Question 3

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.

Answer 3

D

Question 4

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 5

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 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

If the DNA sequence was substantially altered from one of the following,which would prevent the binding of the TATA-binding protein (TBP)?

A)RNA polymerase
B)the point where transcription begins
C)the promoter
D)promoter-proximal elements
E)histone acetyl transferases (HATS)

Question 8

The B-I mutation in anthocyanin (red pigments)produces pigmented corn plants,whereas the B′ mutation in the same gene produces nearly unpigmented corn plants.Normally,when B-I is crossed with recessive colorless alleles of the b gene,the resulting plants are pigmented.However,when B-I and B′ plants are intercrossed,the F₁ plants are essentially unpigmented,like the B′ homozygotes.Thus,B-I is altered by being in the same genome as B′.If this outcome were due simply to the dominance of B′ to B-I,then a self-cross of the F₁ plants should generate B-I-colored homozygotes as approximately 1/4 of the F₂ progeny.Instead,no F₂ are pigmented.Intercrosses of the F₂ and of further generations do not restore the pigmented phenotype.What is the term for this type of inheritance?

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

Question 9

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 10

If you extracted mRNA from the following,which would show the most differences in the sequences of mRNA found?

A)leaf epidermis cells and root xylem cells from the same sugar maple tree
B)leaf epidermis cells from sugar maple trees in Vermont,USA,and Pennsylvania,USA
C)leaf epidermis cells from two very closely related maples,red and sugar,in the same grove of trees
D)leaf epidermis cells from sugar maple trees with varying genotypes in one grove of trees

Question 11

If natural selection over the time period of Eukaryote evolution has favored a DNA architecture in Eukaryotes that keeps genes turned off except when needed,this structure would be _____.

A)nucleosomes
B)terminators
C)exons
D)introns

Question 12

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 promoter regions of DNA.
C)DNase preferentially digests DNA not associated with protein.
D)DNase cuts at specific DNA sequences.

Question 13

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

A)enhancers are transcription factors;promoter-proximal elements are DNA sequences
B)enhancers enhance transcription;promoter-proximal elements inhibit transcription
C)enhancers are part of the promoter;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;promoter-proximal elements are close to the promoter and their position and orientation must be maintained
E)enhancers are DNA sequences;promoter-proximal elements are proteins that bind proximal to the promoter

Question 14

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 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

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 17

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 18

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 19

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 20

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)alternative forms of nucleosomes
D)control of the frequency of translation initiation
E)all 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 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 22

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 23

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 24

If a pharmaceutical company wishes to make a protein-based drug from yeast with a particular carbohydrate tag to ensure correct tissue delivery,which stage of gene expression would be altered to change the tag on the protein?

A)transcription
B)translation
C)post-translation
D)alternative splicing

Question 25

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 26

If a biologist wanted to analyze the evolutionary relationship in a gene type that is found in both eukaryotes and bacteria,which of the following would NOT be useful?

A)histone coding genes
B)transfer RNA coding genes
C)ribosomal RNA coding genes
D)hexokinase coding gene for glycolysis

Question 27

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.

Question 28

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 29

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

Question 30

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 31

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)prevent meiosis
D)prevent mutations in p53

Question 32

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 33

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 from the same gene?

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

Question 34

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 35

Recently,the theobromine synthesis production of coffee plants was knocked down by the insertion of a gene that produced complementary hairpin mRNA.The process that led to these decaffeinated coffee plants is called _____.

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

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

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