Deck 16: Control of Gene Expression
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Deck 16: Control of Gene Expression
1
Regulatory proteins can identify specific sequences on the DNA double helix without unwinding the helix.This is accomplished by inserting
A)RNA promoters into either the major groove or the minor groove of the double helix where the edges of the nitrogen bases protrude.
B)DNA-binding motifs into the minor groove of the double helix where the edges of the nitrogen bases protrude.
C)DNA polymerase into the major groove of the double helix where the edges of the nitrogen bases protrude.
D)RNA polymerase into the major groove of the double helix where the edges of the nitrogen bases protrude.
E)DNA-binding motifs into the major groove of the double helix where the edges of the nitrogen bases protrude.
A)RNA promoters into either the major groove or the minor groove of the double helix where the edges of the nitrogen bases protrude.
B)DNA-binding motifs into the minor groove of the double helix where the edges of the nitrogen bases protrude.
C)DNA polymerase into the major groove of the double helix where the edges of the nitrogen bases protrude.
D)RNA polymerase into the major groove of the double helix where the edges of the nitrogen bases protrude.
E)DNA-binding motifs into the major groove of the double helix where the edges of the nitrogen bases protrude.
E
2
You have been asked to design a synthetic DNA motif,able to bind transcriptional regulatory proteins.The location on this motif that you will design for protein binding is the
A)minor groove of the DNA double helix.
B)major groove of the DNA double helix.
C)phosphate backbone of the DNA double helix.
D)sugar backbone of the DNA double helix.
E)available hydrogen-bonds of bases in single-stranded DNA.
A)minor groove of the DNA double helix.
B)major groove of the DNA double helix.
C)phosphate backbone of the DNA double helix.
D)sugar backbone of the DNA double helix.
E)available hydrogen-bonds of bases in single-stranded DNA.
B
3
Certain proteins can bind to specific DNA regulatory sequences by entering
A)the major groove of the DNA and reading the nucleotide base pairs.
B)the minor groove of the DNA and reading the nucleotide base pairs.
C)the major groove of RNA and reading the nucleotide base pairs.
D)DNA's major groove by using DNA polymerase and reading the nucleotide base pairs.
E)DNA's minor groove by using DNA polymerase and reading the nucleotide base pairs.
A)the major groove of the DNA and reading the nucleotide base pairs.
B)the minor groove of the DNA and reading the nucleotide base pairs.
C)the major groove of RNA and reading the nucleotide base pairs.
D)DNA's major groove by using DNA polymerase and reading the nucleotide base pairs.
E)DNA's minor groove by using DNA polymerase and reading the nucleotide base pairs.
A
4
Through control of gene expression,a bacterial cell responds to changing ________ conditions.
A)internal
B)protein
C)environmental
D)genetic
A)internal
B)protein
C)environmental
D)genetic
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5
Examples of shapes in regulatory proteins that are used to bind to DNA include (check all that apply)
A)zinc finger.
B)TATA box.
C)helix-turn-helix.
D)leucine zipper.
E)Doubledomain.
A)zinc finger.
B)TATA box.
C)helix-turn-helix.
D)leucine zipper.
E)Doubledomain.
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6
In multicellular organisms,the mechanism most directly responsible for directing development and maintaining homeostasis is gene
A)mutation.
B)duplication.
C)deletion.
D)regulation.
A)mutation.
B)duplication.
C)deletion.
D)regulation.
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7
The DNA-binding proteins of almost all regulatory proteins use one of a small set of shapes that enable them to fit into the DNA major groove.These shapes are called
A)structural motifs.
B)DNA prints.
C)fingerprints.
D)repressors.
E)transcriptional domains.
A)structural motifs.
B)DNA prints.
C)fingerprints.
D)repressors.
E)transcriptional domains.
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8
The maintenance of a constant environment in a cell is called ________.
A)active transport
B)homeostasis
C)gene expression
D)translation
A)active transport
B)homeostasis
C)gene expression
D)translation
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9
A site at the 5' end of a gene where RNA polymerase attaches to initiate transcription is called a(n)
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
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10
One of the DNA-binding motifs in many eukaryotic organisms that contains a nearly identical sequence of 60 amino acids is known as the
A)nonhelical zipper.
B)leucine zipper.
C)zinc finger.
D)homeodomain.
E)helix-turn-helix.
A)nonhelical zipper.
B)leucine zipper.
C)zinc finger.
D)homeodomain.
E)helix-turn-helix.
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11
A defining characteristic of eukaryotic organisms is that they
A)have their transcription occurring in the cytoplasm and translation in the nucleus.
B)have their transcription occurring in the nucleus and translation in the cytoplasm.
C)have only operons to assist in gene expression.
D)carry out protein synthesis only in the presence of the cAMP molecule.
E)use the leucine zipper primarily for the production of the amino acid tryptophan.
A)have their transcription occurring in the cytoplasm and translation in the nucleus.
B)have their transcription occurring in the nucleus and translation in the cytoplasm.
C)have only operons to assist in gene expression.
D)carry out protein synthesis only in the presence of the cAMP molecule.
E)use the leucine zipper primarily for the production of the amino acid tryptophan.
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12
Control of gene expression (check all that apply)
A)allows bacteria to adapt to changing environments.
B)is critical for development in multicellular organisms.
C)allows bacteria to replicate without control.
D)allows multicellular organisms to maintain homeostasis.
E)it stops multicellular organisms functioning as a whole.
A)allows bacteria to adapt to changing environments.
B)is critical for development in multicellular organisms.
C)allows bacteria to replicate without control.
D)allows multicellular organisms to maintain homeostasis.
E)it stops multicellular organisms functioning as a whole.
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13
A protein that regulates transcription by binding to the operator is known as the
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
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14
The hallmark of multicellular organisms is their ability to
A)grow and divide rapidly.
B)adjust quickly to outside environment.
C)maintain homeostasis.
D)quickly synthesize amount and type of enzymes according to available nutrients.
E)respond by gene action to oxygen availability.
A)grow and divide rapidly.
B)adjust quickly to outside environment.
C)maintain homeostasis.
D)quickly synthesize amount and type of enzymes according to available nutrients.
E)respond by gene action to oxygen availability.
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15
The most common DNA-binding motif is the
A)nonhelical zipper.
B)leucine zipper.
C)zinc finger.
D)homeodomain.
E)helix-turn-helix.
A)nonhelical zipper.
B)leucine zipper.
C)zinc finger.
D)homeodomain.
E)helix-turn-helix.
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16
As a microbiologist you have been asked to investigate the regulation of a novel gene in a bacterial species.Given what is known about bacteria,the logical place to begin your investigation is
A)transcriptional regulation.
B)translational regulation.
C)alternative splicing regulation.
D)regulation by enhancer elements.
E)regulation by general transcription factors.
A)transcriptional regulation.
B)translational regulation.
C)alternative splicing regulation.
D)regulation by enhancer elements.
E)regulation by general transcription factors.
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17
In order for transcription to be initiated
A)DNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
B)RNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
C)DNA polymerase must have access to the RNA and must also be capable of binding to the gene's promoter.
D)RNA ligase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
E)RNA polymerase must have access to the DNA double helix and also must be capable of binding to the gene's operator.
A)DNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
B)RNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
C)DNA polymerase must have access to the RNA and must also be capable of binding to the gene's promoter.
D)RNA ligase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
E)RNA polymerase must have access to the DNA double helix and also must be capable of binding to the gene's operator.
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18
In some organisms,the primary function of a gene in a cell is to participate in regulating the body as a whole rather than responding to the cell's immediate environment.These organisms would be
A)multicellular.
B)diploid.
C)bacterial.
D)prokaryotic.
A)multicellular.
B)diploid.
C)bacterial.
D)prokaryotic.
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19
You lead a research team challenged with the task of creating a regulatory protein able to shut off transcription of a particular gene.You focus your design around a binding site called an operator that is associated with the promotor.The physical location of the operator most likely to affect transcription would be
A)upstream of the gene promoter.
B)downstream of the gene promoter.
C)internal to the gene itself.
D)internal to the gene promoter.
A)upstream of the gene promoter.
B)downstream of the gene promoter.
C)internal to the gene itself.
D)internal to the gene promoter.
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20
The basic tool of genetic regulation is the ability of certain proteins to bind to specific
A)regulatory RNA sequences.
B)regulatory DNA sequences.
C)repressor parts of the gene.
D)promoter parts of the gene.
E)enzymes of the cell.
A)regulatory RNA sequences.
B)regulatory DNA sequences.
C)repressor parts of the gene.
D)promoter parts of the gene.
E)enzymes of the cell.
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21
A cluster of functionally-related genes that are regulated together and encoded into a single mRNA molecule is called a(n)
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
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22
Enhancers are the binding sites for the
A)promoters of DNA synthesis.
B)suppressor factors.
C)coactivation factors.
D)mediator factors.
E)specific transcription factors.
A)promoters of DNA synthesis.
B)suppressor factors.
C)coactivation factors.
D)mediator factors.
E)specific transcription factors.
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23
In eukaryotes,specific transcription factors have two distinct domains:
A)a DNA-binding domain and a RNA-binding domain.
B)a DNA-binding domain and an activation domain.
C)a DNA-binding domain and a repressor domain.
D)a DNA-binding domain and an enhancer domain.
E)a DNA-binding domain and an operator domain.
A)a DNA-binding domain and a RNA-binding domain.
B)a DNA-binding domain and an activation domain.
C)a DNA-binding domain and a repressor domain.
D)a DNA-binding domain and an enhancer domain.
E)a DNA-binding domain and an operator domain.
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24
The eukaryotic transcription initiation complex is (check all that apply)
A)transcription of virtually all genes transcribed by RNA pol II requiring the same suite of general factors.
B)responsible for highly regulated transcription levels.
C)only associated with RNA polymerase I.
D)interactive with activators through DNA looping.
E)a basal factor associated with RNA pol II after positioning RNA pol II at the start site.
A)transcription of virtually all genes transcribed by RNA pol II requiring the same suite of general factors.
B)responsible for highly regulated transcription levels.
C)only associated with RNA polymerase I.
D)interactive with activators through DNA looping.
E)a basal factor associated with RNA pol II after positioning RNA pol II at the start site.
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25
If a strain of bacteria had a mutation that blocked expression of the lac repressor,what would you expect as a result?
A)The mutant strain would outcompete wildype strains,since it could always utilize lactose.
B)The mutant strain would grow at the same rate as wildtype if lactose was not present.
C)The mutant strain would waste energy producing enzymes in the absence of lactose.
D)The mutant strain would act the same,because it would still require lac activator protein to turn on.
A)The mutant strain would outcompete wildype strains,since it could always utilize lactose.
B)The mutant strain would grow at the same rate as wildtype if lactose was not present.
C)The mutant strain would waste energy producing enzymes in the absence of lactose.
D)The mutant strain would act the same,because it would still require lac activator protein to turn on.
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26
When E.coli cells produce the amino acid tryptophan,a cluster of five genes is transcribed together.This cluster of genes is referred to as the
A)trp transcriptional operator.
B)trp regulator.
C)trp suppressor.
D)trp operon.
E)trp promoter.
A)trp transcriptional operator.
B)trp regulator.
C)trp suppressor.
D)trp operon.
E)trp promoter.
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27
A protein that initiates gene transcription and allows for non-glucose molecules to be used is
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
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28
Transcriptional control-proteins increase the rate of transcription by binding to
A)mRNA sequences within the DNA.
B)tRNA sequences within the DNA.
C)operator sequences within the DNA.
D)promoter sequences within the DNA.
E)enhancer sequences within the DNA.
A)mRNA sequences within the DNA.
B)tRNA sequences within the DNA.
C)operator sequences within the DNA.
D)promoter sequences within the DNA.
E)enhancer sequences within the DNA.
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29
You are studying regulation of a prokaryotic operon.Experiments show that expression of the operon is increased when levels of biosynthetic product from that pathway are low.Based on this information,what is a likely mode of regulation?
A)The product binds a repressor,allowing it to bind the operator.
B)The product binds an activator,helping RNA polymerase to bind the promoter.
C)The product binds a repressor,preventing it from binding the operator.
D)The product binds to the biosynthetic enzymes,blocking them directly.
A)The product binds a repressor,allowing it to bind the operator.
B)The product binds an activator,helping RNA polymerase to bind the promoter.
C)The product binds a repressor,preventing it from binding the operator.
D)The product binds to the biosynthetic enzymes,blocking them directly.
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30
The proteins necessary for the use of lactose in E.coli are collectively called the
A)lac regulator.
B)lac suppressor.
C)lac operon.
D)lac promoter.
E)lac transcriptional operator.
A)lac regulator.
B)lac suppressor.
C)lac operon.
D)lac promoter.
E)lac transcriptional operator.
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31
The TFIID complex is formed by
A)TATA-binding protein TBP and TAFs.
B)TATA-binding protein TBP,TAFs and RNA pol II.
C)TAFs and the core promoter.
D)TATA-binding protein and activators.
A)TATA-binding protein TBP and TAFs.
B)TATA-binding protein TBP,TAFs and RNA pol II.
C)TAFs and the core promoter.
D)TATA-binding protein and activators.
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32
The progesterone receptor (PR)is a steroid hormone receptor and transcription factor.The protein SRC-1,which does not bind DNA by itself,can bind to PR and increase expression of genes regulated by PR.Based on this information,SRC-1 is best described as a(n)
A)coactivator.
B)inducer.
C)general transcription factor.
D)specific transcription factor.
A)coactivator.
B)inducer.
C)general transcription factor.
D)specific transcription factor.
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33
The enzyme beta-galactosidase acts on lactose to form galactose.In turn,the presence of galactose leads to expression of the enzymes responsible for the metabolism of galactose.In this case,lactose is serving as a carbon source and as a(n)
A)inducer.
B)repressor.
C)DNA-binding protein.
D)operon.
A)inducer.
B)repressor.
C)DNA-binding protein.
D)operon.
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34
The lactose analog isopropyl-beta-D-thio-galactoside (IPTG)is often used to regulate gene expression systems in bacteria.IPTG does not act as a substrate for beta-galactosidase,but can bind to,and inactivate,the repressor.In this case,IPTG serves as a(n)
A)inducer.
B)repressor.
C)DNA-binding protein.
D)operon.
A)inducer.
B)repressor.
C)DNA-binding protein.
D)operon.
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35
If you were given a bacterial strain with a mutation in the promoter region of the trp operon the most likely effect would be
A)interference with RNA polymerase binding.
B)to block the repressor from binding.
C)derepression of the trp operon.
D)enabling the trp operon to be expressed in the absence of tryptophan.
E)lack of expression of just the first gene in the operon.
A)interference with RNA polymerase binding.
B)to block the repressor from binding.
C)derepression of the trp operon.
D)enabling the trp operon to be expressed in the absence of tryptophan.
E)lack of expression of just the first gene in the operon.
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36
The main form of glucose repression in the lac operon is
A)induction.
B)repression.
C)inducer exclusion.
D)the CAP/cAMP system.
A)induction.
B)repression.
C)inducer exclusion.
D)the CAP/cAMP system.
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37
A site of negative genetic regulation where binding by repressor blocks transcription is the
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
A)operon.
B)repressor.
C)promoter.
D)operator.
E)CAP.
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38
Your research project involves the characterization of a recently identified transcription factor.As part of your project,you want to determine if this transcription factor binds directly to any of the general transcription factors.Unfortunately,you are having trouble expressing the full-length protein.What approach may help achieve your research goal?
A)You must find a way to express the full-length protein,since transcription factor functions do not lie in separate domains.
B)Identify the RNA that mediates the protein-protein binding.
C)Express and purify only the DNA-binding domain,and test for binding.
D)Express and purify only the activation domain,and test for binding.
A)You must find a way to express the full-length protein,since transcription factor functions do not lie in separate domains.
B)Identify the RNA that mediates the protein-protein binding.
C)Express and purify only the DNA-binding domain,and test for binding.
D)Express and purify only the activation domain,and test for binding.
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39
A bacterial gene regulatory system is likely to have (check all that apply)
A)a coding sequence.
B)an operator.
C)one or more introns.
D)a motor.
E)a ribosome recognition site.
A)a coding sequence.
B)an operator.
C)one or more introns.
D)a motor.
E)a ribosome recognition site.
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40
Control of the lac operon and lactose utilization in bacteria is (check all that apply)
A)not induced in the presence of both glucose and lactose.
B)only induced when there is glucose but not lactose.
C)is a negative control,mediated by a repressor.
D)controlled by the expression of three downstream genes.
E)preferentially utilizing lactose as a carbon source.
A)not induced in the presence of both glucose and lactose.
B)only induced when there is glucose but not lactose.
C)is a negative control,mediated by a repressor.
D)controlled by the expression of three downstream genes.
E)preferentially utilizing lactose as a carbon source.
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41
If you were able to look very closely at a portion of DNA and find methylated histones,you would
A)be mistaken since only DNA can be methylated,not histones.
B)be looking at a region of active chromatin.
C)be looking at a region of inactive chromatin.
D)be looking at a chromatin remodeling complex.
A)be mistaken since only DNA can be methylated,not histones.
B)be looking at a region of active chromatin.
C)be looking at a region of inactive chromatin.
D)be looking at a chromatin remodeling complex.
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42
Although the specific mechanism of RNA interference has not been fully defined,it involves
A)double stranded RNA interference with mRNA.
B)double stranded RNA interference with DNA.
C)double stranded DNA interference with mRNA.
D)double stranded mRNA interference with DNA.
A)double stranded RNA interference with mRNA.
B)double stranded RNA interference with DNA.
C)double stranded DNA interference with mRNA.
D)double stranded mRNA interference with DNA.
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43
What is the difference between a gene that is derepressed and one that is induced?
A)A gene that is derepressed is turned on because a repressor protein is not bound without its cofactor.By comparison,a gene that is induced is turned on because an inducer molecule prevents binding of the repressor.
B)Genes that are derepressed are turned on because an inducer molecule is present.By comparison,a gene that is induced is turned on because a repressor protein is bound to the operator.
C)There is no difference between a gene that is derepressed and one that is induced.
D)A derepressed gene is turned off and an induced gene is activated to be expressed.
A)A gene that is derepressed is turned on because a repressor protein is not bound without its cofactor.By comparison,a gene that is induced is turned on because an inducer molecule prevents binding of the repressor.
B)Genes that are derepressed are turned on because an inducer molecule is present.By comparison,a gene that is induced is turned on because a repressor protein is bound to the operator.
C)There is no difference between a gene that is derepressed and one that is induced.
D)A derepressed gene is turned off and an induced gene is activated to be expressed.
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44
Vertebrate cells possess a protein that binds to clusters of 5-methylcytosine and ensures the gene will stay in the "off" position.This control of gene regulation is a result of
A)translation.
B)enhancer expression.
C)methylation.
D)promoter expression.
E)operator suppression.
A)translation.
B)enhancer expression.
C)methylation.
D)promoter expression.
E)operator suppression.
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45
Elucidation of the histone code might
A)allow us to turn specific genes on or off.
B)allow the determination of nucleosome composition.
C)lead to chromatin remodeling.
D)allow us to control translation.
A)allow us to turn specific genes on or off.
B)allow the determination of nucleosome composition.
C)lead to chromatin remodeling.
D)allow us to control translation.
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46
Production of the iron-storing protein ferritin is regulated by aconitase,which binds to a 30-nucleotide sequence at the beginning of the ferritin mRNA and interferes with ribosome binding.Aconitase must be a
A)transcription repressor protein.
B)translation repressor protein.
C)RNA interference protein.
D)translation initiation protein.
A)transcription repressor protein.
B)translation repressor protein.
C)RNA interference protein.
D)translation initiation protein.
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47
The p300/CBP (CREB-binding protein)proteins are histone acetyltransferases that help regulate the transcription of many genes.Based on this information you can conclude that these proteins
A)acetylate purines.
B)remodel chromatin.
C)recruit helicases.
D)physically connect activator proteins.
A)acetylate purines.
B)remodel chromatin.
C)recruit helicases.
D)physically connect activator proteins.
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48
You are studying the function of a recently identified gene in C.elegans.You perform genetic screens for several months in an attempt to isolate loss-of-function gene mutations,but your efforts are unsuccessful.Your advisor suggests you try another approach to eliminate gene function.The best technique to accomplish this goal would be
A)to design a repressor to bind to the operon of this gene.
B)use a histone deacetylase to induce a transcriptionally inactive state.
C)use a C.elegans strain with a homozygous TFIID mutation to prevent the translation initiation complex from forming.
D)use RNA interference to prevent mRNA translation.
A)to design a repressor to bind to the operon of this gene.
B)use a histone deacetylase to induce a transcriptionally inactive state.
C)use a C.elegans strain with a homozygous TFIID mutation to prevent the translation initiation complex from forming.
D)use RNA interference to prevent mRNA translation.
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49
Within its core a nucleosome contains ________ histones.
A)2
B)4
C)6
D)8
E)64
A)2
B)4
C)6
D)8
E)64
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50
The gene encoding apolipoprotein B exists in two isoforms,APOB100 and APOB48.These two forms are produced as a result of
A)tissue-specific expression.
B)a gene mutation that results in a stop codon.
C)RNA editing.
D)alternative splicing.
A)tissue-specific expression.
B)a gene mutation that results in a stop codon.
C)RNA editing.
D)alternative splicing.
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51
DNA methylation is the only known natural modification of DNA.It affects
A)adenine bases.
B)guanine bases.
C)cytosine bases.
D)thymine bases.
A)adenine bases.
B)guanine bases.
C)cytosine bases.
D)thymine bases.
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52
You work for a pharmaceutical company that designs small RNAs,used to control expression of disease genes.The primary focus area of your research should be
A)transcriptional repression.
B)transcriptional activation.
C)translational repression.
D)translational activation.
A)transcriptional repression.
B)transcriptional activation.
C)translational repression.
D)translational activation.
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53
The enzyme dicer chops dsRNA molecules into small pieces of
A)mRNA and miRNA.
B)miRNA and siRNA.
C)siRNA and rRNA.
D)mRNA and siRNA.
A)mRNA and miRNA.
B)miRNA and siRNA.
C)siRNA and rRNA.
D)mRNA and siRNA.
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54
In vertebrates,DNA methylation - the addition of a methyl group to DNA nucleotides - ensures that
A)the gene functions without interruption.
B)no errors will be made during transcription.
C)the nucleosome will quickly form,which assists in mRNA formation.
D)once that gene is transcribed,the mRNA is saved and used over and over again.
E)once a gene is turned off,it will remain off.
A)the gene functions without interruption.
B)no errors will be made during transcription.
C)the nucleosome will quickly form,which assists in mRNA formation.
D)once that gene is transcribed,the mRNA is saved and used over and over again.
E)once a gene is turned off,it will remain off.
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55
Small RNAs can regulate gene expression.One type,called micro RNA (miRNA),acts by binding directly to
A)mRNA to prevent translation.
B)tRNA to prevent transcription.
C)mRNA to prevent transcription.
D)tRNA to prevent translation.
A)mRNA to prevent translation.
B)tRNA to prevent transcription.
C)mRNA to prevent transcription.
D)tRNA to prevent translation.
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56
You are studying the effects of specific transcription factors on the activation of gene expression.You notice that one particular transcription factor binds far away from the promoter of its target gene.What can explain this?
A)The transcription factor transcribes small RNAs that then bind to the promoter and activate the gene's expression.
B)The transcription factor tags the enhancer with ubiquitin to stimulate transcription.
C)DNA looping brings the transcription factor closer to the promoter and initiates gene transcription.
D)RNA looping brings the transcription factor closer to the promoter and initiates gene transcription.
A)The transcription factor transcribes small RNAs that then bind to the promoter and activate the gene's expression.
B)The transcription factor tags the enhancer with ubiquitin to stimulate transcription.
C)DNA looping brings the transcription factor closer to the promoter and initiates gene transcription.
D)RNA looping brings the transcription factor closer to the promoter and initiates gene transcription.
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57
The primary transcripts in eukaryotes are most accurately described as
A)composed of RNA polymerase and associated histones.
B)having the exons removed and the introns retained for translation.
C)a faithful copy of the entire gene including exons and introns.
D)an exact copy of the gene,but the introns have been removed.
E)an RNA copy,but the noncoding exons and introns have been removed.
A)composed of RNA polymerase and associated histones.
B)having the exons removed and the introns retained for translation.
C)a faithful copy of the entire gene including exons and introns.
D)an exact copy of the gene,but the introns have been removed.
E)an RNA copy,but the noncoding exons and introns have been removed.
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58
Eukaryotic cell mRNA transcripts can remain in the cell for hours because they are
A)stable.
B)long.
C)isolated.
D)analogous.
A)stable.
B)long.
C)isolated.
D)analogous.
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59
Histones that are tightly wound by DNA and are the basic unit of chromatin are called
A)operons.
B)nucleosomes.
C)protein clusters.
D)repressor genes.
E)facilitators sites.
A)operons.
B)nucleosomes.
C)protein clusters.
D)repressor genes.
E)facilitators sites.
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60
You are working to identify enhancer regions of a particular gene.The best place to look is
A)immediately upstream of the promoter.
B)immediately downstream of the promoter.
C)primarily upstream of the promoter,possibly some distance away.
D)primarily downstream of the promoter,including the exons of the coding region.
A)immediately upstream of the promoter.
B)immediately downstream of the promoter.
C)primarily upstream of the promoter,possibly some distance away.
D)primarily downstream of the promoter,including the exons of the coding region.
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61
In your research,you have discovered that Protein X is often ubiquitinated in people with a certain disease.What would you predict about levels of Protein X in these patients?
A)Levels of Protein X will be low due to degradation in the proteasome.
B)Levels of Protein X will be the same,just carrying a ubiquitin tag.
C)Levels of Protein X will be increased due to transcriptional activation.
D)Levels of Protein X will be decreased due to negative feedback on transcription.
E)Levels of Protein X will be increased due to protection from degradation.
A)Levels of Protein X will be low due to degradation in the proteasome.
B)Levels of Protein X will be the same,just carrying a ubiquitin tag.
C)Levels of Protein X will be increased due to transcriptional activation.
D)Levels of Protein X will be decreased due to negative feedback on transcription.
E)Levels of Protein X will be increased due to protection from degradation.
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62
The ubiquitin-proteasome pathway (check all that apply)
A)is used to regulate expression of a number of cell surface receptors.
B)requires only one molecule of ATP to target a protein.
C)targets proteins in a stepwise fashion via ubiquitin ligase adding ubiquitin residues to the protein.
D)is used to digest macromolecules.
E)does not destroy the ubiquitin moiety,but rather cleaves it off for reuse.
A)is used to regulate expression of a number of cell surface receptors.
B)requires only one molecule of ATP to target a protein.
C)targets proteins in a stepwise fashion via ubiquitin ligase adding ubiquitin residues to the protein.
D)is used to digest macromolecules.
E)does not destroy the ubiquitin moiety,but rather cleaves it off for reuse.
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