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CHAPTER 8: CONTROL OF GENE EXPRESSION
© 2014 GARLAND SCIENCE PUBLISHING
8-1 A neuron and a white blood cell have very different functions. For example, a neuron can receive and respond to electrical signals while a white blood cell defends the body against infection. This is because ______.
(a) the proteins found in a neuron are completely different from the proteins found in a white blood cell.
(b) the neuron and the white blood cell within an individual have the same genome.
(c) the neuron expresses some mRNAs that the white blood cell does not
(d) neurons and white blood cells are differentiated cells and thus no longer need to transcribe and translate genes
8-2 The distinct characteristics of different cell types in a multicellular organism result mainly from the differential regulation of the _________________.
(a) replication of specific genes.
(b) transcription of genes transcribed by RNA polymerase II.
(c) transcription of housekeeping genes.
(d) proteins that directly bind the TATA box of eukaryotic genes.
8-3 The human genome encodes about 21,000 protein-coding genes. Approximately how many such genes does the typical differentiated human cell express at any one time?
(a) 21,000 all of them
(b) between 18,900 and 21,000 at least 90% of the genes
(c) between 5000 and 15,000
(d) less than 2100
8-4 Which of the following is not a good example of a housekeeping protein?
(a) DNA repair enzymes
(b) histones
(c) ATP synthase
(d) hemoglobin
8-5 Which of the following statements about differentiated cells is true?
(a) Cells of distinct types express nonoverlapping sets of transcription factors.
(b) Once a cell has differentiated, it can no longer change its gene expression.
(c) Once a cell has differentiated, it will no longer need to transcribe RNA.
(d) Some of the proteins found in differentiated cells are found in all cells of a multicellular organism.
8-6 Investigators performed nuclear transplant experiments to determine whether DNA is altered irreversibly during development. Which of the following statements about these experiments is true?
(a) Because the donor nucleus is taken from an adult animal, the chromosomes from the nucleus must undergo recombination with the DNA in the egg for successful development to occur.
(b) The embryo that develops from the nuclear transplant experiment is genetically identical to the donor of the nucleus.
(c) The meiotic spindle of the egg must interact with the chromosomes of the injected nuclei for successful nuclear transplantation to occur.
(d) Although nuclear transplantation has been successful in producing embryos in some mammals with the use of foster mothers, evidence of DNA alterations during differentiation has not been obtained for plants.
8-7 In principle, a eukaryotic cell can regulate gene expression at any step in the pathway from DNA to the active protein. Place the types of control listed below at the appropriate places on the diagram in Figure Q8-7.
8-8
A. translation control
B. transcriptional control
C. RNA processing control
D. protein activity control
Which of these method(s) of controlling eukaryotic gene expression is NOT employed in prokaryotic cells?
A. controlling how often a gene is transcribed
B. controlling how an RNA transcript is spliced
C. controlling which mRNAs are exported from the nucleus to the cytosol
D. controlling which mRNAs are translated into protein by the ribosomes
E. controlling how rapidly proteins are destroyed once they are made
8-9 Fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.
The genes of a bacterial __________________ are transcribed into a single mRNA. Many bacterial promoters contain a region known as a(n) __________________, to which a specific transcription regulator binds. Genes in which transcription is prevented are said to be __________________. The interaction of small molecules, such as tryptophan, with __________________ DNA-binding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being __________________ expressed are being transcribed all the time.
allosteric negatively positively constitutively operator promoter induced operon repressed
8-10 Which of the following statements about transcriptional regulators is false?
(a) Transcriptional regulators usually interact with the sugar–phosphate backbone on the outside of the double helix to determine where to bind on the DNA helix.
(b) Transcriptional regulators will form hydrogen bonds, ionic bonds, and hydrophobic interactions with DNA.
(c) The DNA-binding motifs of transcriptional regulators usually bind in the major groove of the DNA helix.
(d) The binding of transcriptional regulators generally does not disrupt the hydrogen bonds that hold the double helix together.
8-11 Operons ___________________________.
(a) are commonly found in eukaryotic cells.
(b) are transcribed by RNA polymerase II
(c) contain a cluster of genes transcribed as a single mRNA.
(d) can only be regulated by gene activator proteins.
8-12 The tryptophan operator ___________________________.
(a) is an allosteric protein
(b) binds to the tryptophan repressor when the repressor is bound to tryptophan.
(c) is required for production of the mRNA encoded by the tryptophan operon
(d) is important for the production of the tryptophan repressor.
8-13 Which of the following statements about the Lac operon is false?
(a) The Lac repressor binds when lactose is present in the cell.
(b) Even when the CAP activator is bound to DNA, if lactose is not present, the Lac operon will not be transcribed.
(c) The CAP activator can only bind DNA when it is bound to cAMP.
(d) The Lac operon only produces RNA when lactose is present and glucose is absent.
8-14 The CAP activator protein and the Lac repressor both control the Lac operon (see Figure Q8-14). You create cells that are mutant in the gene coding for the Lac repressor so that
these cells lack the Lac repressor under all conditions. For these mutant cells, state whether the Lac operon will be switched on or off in the following situations, and explain why.
A. in the presence of glucose and lactose
B. in the presence of glucose and the absence of lactose
C. in the absence of glucose and the absence of lactose
D. in the absence of glucose and the presence of lactose
8-15 What do you predict would happen if you replace the Lac operator DNA from the Lac operon with the DNA from the operator region from the tryptophan operon?
(a) The presence of lactose will not cause allosteric changes to the Lac repressor.
(b) The Lac operon will not be transcribed when tryptophan levels are high.
(c) The lack of glucose will no longer allow CAP binding to the DNA.
(d) RNA polymerase will only bind to the Lac promoter when lactose is present.
8-16 You are interested in examining the regulation of the gene that encodes an enzyme, Trease, important in metabolizing trehalose into glucose in bacteria. Trehalose is a disaccharide formed of two glucose units. It is known that two DNA-binding proteins, TreA and TreB, are important for binding to the promoter of the Tre-ase gene and are involved in regulating the transcription of the Tre-ase gene: TreA binds to the “A” site in the promoter region, and TreB binds to the “B” site. You make mutations in the TreA and TreB genes to create cells lacking these genes, observe what happens to transcription of the Tre-ase gene, and obtain the results in Table Q8-16.
A. What is the role for TreA in controlling Tre-ase expression? Explain.
B. What is the role for TreB in controlling Tre-ase expression? Explain.
C. From these data, what do you predict will happen to Tre-ase transcription (compared with that in normal cells) in the presence of trehalose if you were to
create a version of the TreA protein that will constitutively bind to the “A” site in the Tre-ase promoter?
Note: Questions 8-17 to 8-20 use the following information and the data in Table Q8-17. These questions may be used independently or as a group.
You are interested in examining the Psf gene. It is known that Psf is normally produced when cells are exposed to high levels of both calcium (Ca2+) and magnesium (Mg2+). MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are involved in regulating its transcription. MetA binds to the “A” site in the promoter region, MetB to the “B” site, and MetC to the “C” site. You create binding-site mutations in the A, B, and C sites and observe what happens to transcription of the Psf gene. Your results are summarized in Table Q8-17.
8-17 Which of the following proteins are likely to act as gene activators?
(a) MetA only
(b) MetB only
(c) MetC only
(d) Both MetA and MetC
8-18 Which of the following proteins are likely to act as gene repressors?
(a) MetA only
(b) MetB only
(c) MetC only
(d) Both MetA and MetC
8-19 Which transcription factors are normally bound to the Psf promoter in the presence of Mg2+ only?
(a) none
(b) MetA only
(c) MetA and Met B
(d) MetA, MetB, and MetC
8-20 Which transcription factors are normally bound to the Psf promoter in the presence of both Mg2+ and Ca2+?
(a) MetA and MetB
(b) MetB and MetC
(c) MetA and MetC
(d) MetA, MetB, and MetC
Note: Questions 8-21 to 8-24 use the following information and the data in Figure Q8-21. These questions may be used independently or as a group.
You are interested in understanding the gene regulation of Lkp1, a protein that is normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1 gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene, place it upstream of the gene for green fluorescent protein (GFP), and insert this entire piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells but not in heart cells, an expression pattern similar to the normal expression of the LKP1 gene. Further experiments demonstrate that there are three regions in the promoter, labeled A, B, and C in Figure Q8-21, that contribute to this expression pattern. Assume that a single and unique transcription factor binds each site such that protein X binds site A, protein Y binds site B, and protein Z binds site C. You want to determine which region is responsible for tissue-specific expression, and create mutations in the promoter to determine the function of each of these regions. In Figure Q8-21, if the site is missing, it is mutated such that it cannot bind its corresponding transcription factor.
8-21 Which of the following proteins is likely to act as a gene repressor?
(a) factor X
(b) factor Y
(c) factor Z
(d) none of the above
8-22 Which of the following proteins are likely to act as gene activators?
(a) factors X and Y
(b) factors X and Z
(c) factors Y and Z
(d) factor X only
8-23 Experiment 1 in Figure Q8-21 is the positive control, demonstrating that the region of DNA upstream of the gene for GFP results in a pattern of expression that we normally find for the LKP1 gene. Experiment 2 shows what happens when the sites for binding factors X, Y, and Z are removed. Which experiment above demonstrates that factor X alone is sufficient for expression of LPK1 in the kidney?
(a) experiment 3
(b) experiment 5
(c) experiment 6
(d) experiment 7
8-24 In what tissue is factor Z normally present and bound to the DNA?
(a) kidney
(b) liver
(c) heart
(d) none of the above
8-25 An allosteric transcription regulator called HisP regulates the enzymes for histidine biosynthesis in the bacterium E. coli. Histidine modulates HisP activity. On binding histidine, HisP alters its conformation, markedly changing its affinity for the regulatory sequences in the promoters of the genes for the histidine biosynthetic enzymes.
A. If HisP functions as a gene repressor, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine is abundant? Explain your answer.
B. If HisP functions as a gene activator, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine levels are low? Explain your answer.
8-26 Bacterial cells can take up the amino acid tryptophan from their surroundings, or, if the external supply is insufficient, they can synthesize tryptophan by using enzymes in the cell. In some bacteria, the control of glutamine synthesis is similar to that of tryptophan synthesis, such that the glutamine repressor inhibits the transcription of the glutamine operon, which contains the genes that code for the enzymes required for glutamine synthesis. On binding to cellular glutamine, the glutamine repressor binds to a site in the promoter of the operon.
A. Why is glutamine-dependent binding to the operon a useful property for the glutamine repressor?
B. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells expressing a mutant form of the glutamine repressor that cannot bind to DNA?
C. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells expressing a mutant form of the glutamine repressor that binds to DNA even when no glutamine is bound to it?
8-27 In the absence of glucose, E. coli can proliferate by using the pentose sugar arabinose. As shown in Figure Q8-27, the arabinose operon regulates the ability of E. coli to use arabinose. The araA, araB, and araD genes encode enzymes for the metabolism of arabinose. The araC gene encodes a transcription regulator that binds adjacent to the promoter of the arabinose operon. To understand the regulatory properties of the AraC protein, you engineer a mutant bacterium in which the araC gene has been deleted and look at the effect of the presence or absence of the AraC protein on the AraA enzyme.
A. If the AraC protein works as a gene repressor, would you expect araA RNA levels to be high or low in the presence of arabinose in the araC– mutant cells? What about in the araC– mutant cells in the absence of arabinose? Explain your answer.
B. Your findings from the experiment are summarized in Table Q8-27.
Do the results in Table Q8-27 indicate that the AraC protein regulates arabinose metabolism by acting as a gene repressor or a gene activator? Explain your answer.
8-28 You have discovered an operon in a bacterium that is turned on only when sucrose is present and glucose is absent. You have also isolated three mutants that have changes in the upstream regulatory sequences of the operon and whose behavior is summarized in the Table Q8-28. You hypothesize that there are two gene regulatory sites, A and B, in the upstream regulatory sequencethat are affected by the mutations. For this question, a plus (+) indicates a normal site and a minus (–) indicates a mutant site that no longer binds its transcription regulator.
A. If mutant 1 has sites A– B+, which of these sites is regulated by sucrose and which by glucose?
B. Give the state (+ or –) of the A and B sites in mutants 2 and 3.
C. Which site is bound by a repressor and which by an activator?
Note: Questions 8-29 to 8-32 use the following information, Figure Q8-29, and the data in Table Q8-29. These questions may be used independently or as a group.
You are interested in the regulation of gene Q. Proteins G, H, and J are proteins that are important for regulating gene Q, and bind to its promoter region in a sequence-specific fashion. Proteins G and H both bind to site “A” but cannot bind to site “A” at the same time. Protein J binds to site “B” on the promoter. The promoter region is diagrammed in Figure Q8-29.
You develop a cell-free transcriptional system to study the effects of proteins G, H, and J on the transcription of gene Q. Using this system, you can examine the effects of adding these proteins to the transcriptional system in equal amounts and measuring how much gene Q is produced. When you add these proteins to the system, you get the results shown in Table Q8-29.
8-29 Which proteins are likely to act as gene activators?
(a) G
(b) H
(c) J
(d) both H and J
8-30 Which proteins are likely to act as gene repressors?
(a) G
(b) H
(c) J
(d) both H and J
8-31 Your colleague looks at your data above and predicts that protein G will bind more strongly to the DNA at site A, compared to protein H. Which experiment above is critical for this prediction?
(a) #2
(b) #3
(c) #5
(d) #6
8-32 Which proteins do you predict are bound to the promoter in experiment #8?
(a) only H and J
(b) only G and H
(c) only G and J
(d) only J
Note: Questions 8-33 to 8-34 use the following information, Figure Q8-33, and the data in Table Q8-33. These questions may be used independently or as a group.
You are interested in studying the transcriptional regulation of the Gip1 promoter. The Gip1 promoter contains a binding site for the Jk8 protein that overlaps with the binding site for the Pa5 protein. Jk8 and Pa5 cannot bind DNA at the same time, but both proteins are present at high levels in adult liver cells. The binding sites for Jk8 and Pa5 are shown in Figure Q8-33.
Jk8 binds to site A while Pa5 binds to site B. You create mutations that remove the nonoverlapping sequences of either binding site A or B, and examine Gip1 mRNA production in adult liver cells that contain these mutations. The data you obtain from these experiments are shown in Table Q8-33.
8-33 Which gene regulatory protein is bound in experiment 1 (the normal situation)? Explain.
8-34 You know that Gip1 is only expressed in adult liver cells and not in the liver of embryos. You also know that Jk8 and Pa5 behave similarly on other promoters in the embryo or in the adult, in terms of whether they act as repressors or gene activators. Given the data, use of which of the following mechanisms would make the most sense for regulating the Jk8 and Pa5 proteins:
(a) Jk8 is ubiquitylated and targeted for destruction in adult cells
(b) Jk8, but not Pa5, is transcribed in embryonic liver cells.
(c) Jk8 binds to the promoter of the gene that encodes Jk8 in embryonic liver cells.
(d) Pa5 binds to the promoter of the gene that encodes Jk8 in embryonic liver cells.
8-35 Label the following structures in Figure Q8-35.
A. activator protein
B. RNA polymerase
C. general transcription factors
D. Mediator
8-36 How are most eukaryotic transcription regulators able to affect transcription when their binding sites are far from the promoter?
(a) by binding to their binding site and sliding to the site of RNA polymerase assembly
(b) by looping out the intervening DNA between their binding site and the promoter
(c) by unwinding the DNA between their binding site and the promoter
(d) by attracting RNA polymerase and modifying it before it can bind to the promoter
8-37 The expression of the BRF1 gene in mice is normally quite low, but mutations in a gene called BRF2 lead to increased expression of BRF1. You have a hunch that nucleosomes are involved in the regulation of BRF1 expression and so you investigate the position of nucleosomes over the TATA box of BRF1 in normal mice and in mice that lack either the BRF2 protein (BRF2–) or part of histone H4 (HHF–) (histone H4 is encoded by the HHF gene). Table Q8-37 summarizes your results. A normal functional gene is indicated by a plus sign (+).
Which of the following conclusions cannot be drawn from your data? Explain your answer.
(a) BRF2 is required for the repression of BRF1.
(b) BRF2 is required for the specific pattern of nucleosome positions over the BRF1 upstream region.
(c) The specific pattern of nucleosome positioning over the BRF1 upstream region is required for BRF1 repression.
(d) The part of histone H4 missing in HHF– mice is not required for the formation of nucleosomes.
8-38 The yeast GAL4 gene encodes a transcriptional regulator that can bind DNA upstream of genes required for the metabolism of the sugar galactose and turn them on. Gal4 has a DNA-binding domain and an activation domain. The DNA-binding domain allows it to bind to the appropriate sites in the promoters of the galactose metabolism genes. The activation domain attracts histone-modifying enzymes and also binds to a component of the RNA polymerase II enzyme complex, attracting it to the promoter so that the regulated genes can be turned on when Gal4 is also bound to the DNA. When Gal4 is expressed normally, the genes can be maximally activated. You decide to try to produce more of the galactose metabolism genes by overexpressing the Gal4 protein at levels fiftyfold greater than normal. You conduct experiments to show that you are overexpressing the Gal4 protein and that it is properly localized in the nucleus of the yeast cells. To your surprise, you find that too much Gal4 causes the galactose genes to be transcribed only at a low level. What is the most likely explanation for your findings?
8-39 For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or phrases will be used; use each word or phrase only once.
During transcription in __________________ cells, transcriptional regulators that bind to DNA thousands of nucleotides away from a gene’s promoter can affect a gene’s transcription. The __________________ is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcriptional start site. Transcriptional activators can also interact with histone __________________s, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone __________________s. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the __________________ found in interphase chromosomes.
acetylase eukaryotic operator centrosome helicase peroxidase deacetylase heterochromatin prokaryotic deoxidase leucine zipper telomere enhancer Mediator viral
8-40 Which of the following statements about nucleosomes is true?
(a) Nucleosomes activate transcription when bound to the promoter.
(b) Although RNA polymerase can access DNA packed within nucleosomes, the general transcription factors and transcriptional regulators cannot.
(c) Histone acetyltransferases affect transcription by both altering chromatin structure to allow accessibility to the DNA and by adding acetyl groups to histones that can bind proteins that promote transcription.
(d) Histone deacetylases remove lysines from histone tails.
8-41 In principle, how many different cell types can an organism having four different types of transcription regulator and thousands of genes create?
(a) up to 4
(b) up to 8
(c) up to 16
(d) thousands
8-42 From the sequencing of the human genome, we believe that there are approximately 21,000 protein-coding genes in the genome, of which 1500–3000 are transcription factors. If every gene has a tissue-specific and signal-dependent transcription pattern, how can such a small number of transcriptional regulatory proteins generate a much larger set of transcriptional patterns?
8-43 Combinatorial control of gene expression __________________________.
(a) involves every gene using a different combination of transcriptional regulators for its proper expression
(b) involves groups of transcriptional regulators working together to determine the expression of a gene
(c) involves only the use of gene activators used together to regulate genes appropriately.
(d) is seen only when genes are arranged in operons
8-44 You are studying a set of mouse genes whose expression increases when cells are exposed to the hormone cortisol, and you believe that the same cortisol-responsive transcriptional activator regulates all of these genes. Which of the following statements below should be true if your hypothesis is correct?
(a) The cortisol-responsive genes share a DNA sequence in their regulatory regions that binds the cortisol-responsive transcriptional activator.
(b) The cortisol-responsive genes must all be in an operon.
(c) The transcriptional regulators that bind to the regulatory regions of the cortisolresponsive genes must all be the same.
(d) The cortisol-responsive genes must not be transcribed in response to other hormones.
8-45 Which of the following statements about iPS cells is false?
(a) iPS cells are created by adding a combination of transcription regulators to a fibroblast.
(b) iPS cells created from mouse cells can differentiate into almost any human cell type.
(c) Stimulation by extracellular signal molecules causes iPS cells to differentiate.
(d) During the de-differentiation process to become an iPS, the fibroblast will undergo changes to its gene expression profile.
8-46 Which of the following statements about how fruit flies can develop an eye in the middle of a leg is true?
(a) When the Ey gene is expressed in adult leg cells, these cells de-differentiate and become eye cells.
(b) The Ey gene encodes a transcription regulator that is the only transcription regulator used to produce a fruit-fly eye.
(c) When the Ey gene is introduced into cells that would normally give rise to a leg, the transcription regulators used to control its expression in the leg are different from those that are normally used to control Ey expression in the eye.
(d) All the eye cells found in the adult leg are a single cell type and have identical characteristics.
8-47 Which of the following statements about the Ey transcriptional regulator is false?
(a) Expression of Ey in cells that normally form legs in the fly will lead to the formation of an eye in the middle of the legs.
(b) The Ey transcription factor must bind to the promoter of every eye-specific gene in the fly.
(c) Positive feedback loops ensure that Ey expression remains switched on in the developing eye.
(d) A homolog of Ey is found in vertebrates; this homolog is also used during eye development.
8-48 The MyoD transcriptional regulator is normally found in differentiating muscle cells and participates in the transcription of genes that produce muscle-specific proteins, such as those needed in contractile tissue. Amazingly, expression of MyoD in fibroblasts causes these cells derived from skin connective tissue to produce proteins normally only seen in muscles. However, some other cell types do not transcribe muscle-specific genes when MyoD is expressed in them. Which of the following statements below is the best explanation of why MyoD can cause fibroblasts to express muscle-specific genes?
(a) Unlike some other cell types, fibroblasts have not lost the muscle-specific genes from their genome.
(b) The muscle-specific genes must be in heterochromatin in fibroblasts.
(c) During their developmental history, fibroblasts have accumulated some transcriptional regulators in common with differentiating muscle cells.
(d) The presence of MyoD is sufficient to activate the transcription of muscle-specific genes in all cell types.
8-49 A virus produces a protein X that activates only a few of the virus’s own genes (V1, V2, and V3) when it infects cells. The cellular proteins A (a zinc finger protein) and B (a homeodomain protein) are known to be repressors of the viral genes V1, V2, and V3. You examine the complete upstream gene regulatory sequences of these three viral genes and find the following:
1. V1 and V2 contain binding sites for the zinc finger protein, A, only.
2. V3 contains a binding site for the homeodomain protein, B, only.
3. The only sequence that all three genes have in common is the TATA box.
Label each of the choices below as likely or unlikely as an explanation for your findings. For each choice you label as unlikely, explain why.
A. Protein X binds nonspecifically to the DNA upstream of V1, V2, and V3 and activates transcription.
B. Protein X binds to a repressor and prevents the repressor from binding upstream of V1, V2, and V3.
C. Protein X activates transcription by binding to the TATA box.
D. Protein X activates transcription by binding to and sequestering proteins A and B.
E. Protein X represses transcription of the genes for proteins A and B.
8-50 In mammals, individuals with two X chromosomes are female, and individuals with an X and a Y chromosome are male. It had long been known that a gene located on the Y chromosome was sufficient to induce the gonads to form testes, which is the main maledetermining factor in development, and researchers sought the product of this gene, the so-called testes-determining factor (TDF). For several years, the TDF was incorrectly thought to be a zinc finger protein encoded by a gene called BoY. Which of the following
observations would most strongly suggest that BoY might not be the TDF? Explain your answer.
(a) Some XY individuals that develop into females have mutations in a different gene, SRY, but are normal at BoY.
(b) BoY is not expressed in the adult male testes.
(c) Expression of BoY in adult females does not masculinize them.
(d) A few of the genes that are known to be expressed only in the testes have binding sites for the BoY protein in their upstream regulatory sequences, but most do not.
8-51 Which of the following is not a general mechanism that cells use to maintain stable patterns of gene expression as cells divide?
(a) a positive feedback loop, mediated by a transcriptional regulator that activates transcription of its own gene in addition to other cell-type-specific genes
(b) faithful propagation of condensed chromatin structures as cells divide
(c) inheritance of DNA methylation patterns when cells divide
(d) proper segregation of housekeeping proteins when cells divide
8-52 Which of the following statements about DNA methylation in eukaryotes is false?
(a) Appropriate inheritance of DNA methylation patterns involves maintenance methyltransferase.
(b) DNA methylation involves a covalent modification of cytosine bases.
(c) Methylation of DNA attracts proteins that block gene expression.
(d) Immediately after DNA replication, each daughter helix contains one methylated DNA strand, which corresponds to the newly synthesized strand.
8-53 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
The transmission of information important for gene regulation from parent to daughter cell, without altering the actual nucleotide sequence, is called _________________ inheritance. This type of inheritance is seen with the inheritance of the covalent modifications on ____________ proteins bound to DNA; these modifications are important for reestablishing the pattern of chromatin structure found on the parent chromosome. Another way to inherit chromatin structure involves DNA __________, a covalent modification that occurs on cytosine bases that typically turns off the transcription of a gene. Gene transcription patterns can also be transmitted across generations through positive _____________ loops that can involve a transcription regulator activating its own transcription in addition to other genes. These mechanisms all allow for cell ________________, a property involving the maintenance of gene expression patterns important for cell identity.
combinatorial feedback phosphorylation deacetylation histone pluripotency differential leucine zipper proliferation epigenetic memory receptor
8-54 Which of the following statements about mRNA half-life is false?
(a) The half-life of mRNAs produced from different genes will vary more than the half-life of mRNAs produced from the same gene.
(b) The half-life of most eukaryotic-cell mRNAs is >24 hours.
(c) The half-life of most bacterial mRNAs is shorter than the half-life of a typical eukaryotic mRNA.
(d) The 5′ and 3′ untranslated regions of an mRNA often contain specific sequences that determine the lifetime of the mRNA molecule.
8-55 Using genetic engineering techniques, you remove the sequences that code for the ribosome-binding sequences of the bacterial LacZ gene. The removal of these sequences will lead to ___________.
(a) more LacZ protein produced due to faster ribosome movement across the LacZ mRNA.
(b) transcriptional repression, resulting in fewer mRNA molecules produced from this gene.
(c) a longer half-life for the LacZ mRNA.
(d) translational inhibition of the LacZ mRNA.
8-56 miRNAs, tRNAs, and rRNAs all _____________.
(a) do not code for proteins
(b) act in the nucleus.
(c) are packaged with other proteins to form RISC.
(d) form base pairs with mRNA molecules.
8-57 Which of the following is not involved in post-transcriptional control?
(a) the spliceosome
(b) Dicer
(c) Mediator
(d) RISC
8-58 MicroRNAs ____________________.
(a) are produced from a precursor miRNA transcript
(b) are found only in humans.
(c) control gene expression by base-pairing with DNA sequences
(d) can degrade RNAs by using their intrinsic catalytic activity
8-59 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.
MicroRNAs are noncoding RNAs that are incorporated into a protein complex called __________________, which searches the __________________s in the cytoplasm for sequence complementary to that of the miRNA. When such a molecule is found, it is then targeted for __________________. RNAi is triggered by the presence of foreign __________________ molecules, which are digested by the __________________ enzyme into shorter fragments approximately 23 nucleotide pairs in length.
acetylation methylation riboswitch destruction mitochondria RISC
Dicer mRNA rRNA
DNA phosphorylation single-stranded RNA
double-stranded RNA prokaryotic tRNA
8-60 The extent of complementarity of a miRNA with its target mRNA determines ___________________________.
(a) whether the mRNA will be immediately degraded or whether the mRNA will first be transported elsewhere in the cell before degradation.
(b) whether the mRNA will be transported to the nucleus.
(c) whether RISC is degraded
(d) whether the miRNA synthesizes a complementary strand.
8-61 Which of the following statements about miRNAs is false?
(a) One miRNA can regulate the expression of many genes.
(b) miRNAs are transcribed in the nucleus from genomic DNA.
(c) miRNAs are produced from rRNAs.
(d) miRNAs are made by RNA polymerase.
8-62 Which of the following statements about RNAi is true?
(a) The RNAi mechanism is found only in plants and animals.
(b) RNAi is induced when double-stranded, foreign RNA is present in the cell.
(c) RISC uses the siRNA duplex to locate complementary foreign RNA molecules.
(d) siRNAs bind to miRNAs to induce RNAi.
8-63 The owners of a local bakery ask for your help in improving a special yeast strain they use to make bread. They would like you to help them design experiments using RNA interference to turn off genes, to allow them to test their hypothesis that certain genes are important for the good flavors found in their bread. Of the components in the following list, which is the most important to check for in this yeast strain if you’d like this project to succeed?
(a) the presence of foreign double-stranded RNA
(b) the presence of genes in the genome that code for RISC proteins
(c) the presence of miRNA genes in the genome
(d) the presence of single-stranded siRNAs within the cell
8-64 The modular nature of the Eve gene’s regulatory region means that ______.
(a) there are seven regulatory elements and each element is sufficient for driving expression in a single stripe.
(b) all the regulatory elements for each stripe use the same transcriptional activators.
(c) the E. coli LacZ gene is normally only expressed in a single stripe unlike Eve, which is expressed in seven stripes.
(d) transcription regulators only bind to the stripe 2 regulatory DNA segment in stripe 2.
8-65 The gene for a hormone necessary for insect development contains binding sites for three transcription regulators called A, B, and C. Because the binding sites for A and B overlap, A and B cannot bind simultaneously. You make mutations in the binding sites for each of the proteins and measure hormone production in cells that contain equal amounts of the A, B, and C proteins. Figure Q8-65 summarizes your results. In each of the following sentences, choose one of the phrases within square brackets to make the statement consistent with the results.
A. Protein A binds to its DNA binding site [more tightly/less tightly] than protein B binds to its DNA binding site.
B. Protein A is a [stronger/weaker] activator of transcription than protein B.
C. Protein C is able to prevent activation by [protein A only/protein B only/both protein A and protein B].
8-66 The Drosophila Eve gene has a complex promoter containing multiple binding sites for four transcription regulators: Bicoid, Hunchback, Giant, and Krüppel. Bicoid and Hunchback are activators of Eve transcription, whereas Giant and Krüppel repress Eve transcription. Figure Q8-66A shows the patterns of expression of these regulators.
The Eve promoter contains modules that control expression in various stripes. You construct a reporter gene that contains the DNA 5 kb upstream of the Eve gene, so that this reporter contains the stripe 3 module, the stripe 2 module, the stripe 7 module, and the TATA box, all fused to the LacZ reporter gene (which encodes the β-galactosidase enzyme), as shown in Figure Q8-66B. This construct results in expression of the βgalactosidase enzyme in three stripes, which correspond to the normal positions of stripes 3, 2, and 7.
A. By examining the overlap of sites on the stripe 2 module, as depicted in Figure Q8-66B, what is the biological effect of having some of the transcription regulator binding sites overlap?
B. You make two mutant versions in which several of the binding sites in the Eve stripe 2 module have been deleted, as detailed in items (i) and (ii) below. Refer to Figure Q8-66B for the positions of the binding sites. (Note, however, that because many of the binding sites overlap, it is not possible to delete all of one kind of site
without affecting some of the other sites.) Match the appropriate mutant condition with the most likely pattern of Eve expression shown in Figure Q8-66C. Explain your choices.
(i) deletion of the Krüppel-binding sites in stripe 2
(ii) deletion of the two Bicoid-binding sites in the stripe 2 module that are marked with an asterisk (*) in Figure Q8-66B
8-1 Choice (c) is correct. There are proteins common to all cells in multicellular organisms [choice (a)]. Although it is true that the neuron and white blood cell within an individual have the same genome, this does not explain why these two cells have different functions [choice (b)]. Differentiated cells still need to transcribe and translate genes [choice (d)].
8-2 (b) The major cause of differences between different cell types is in the differential expression of protein-coding genes transcribed by RNA polymerase II, because these genes encode not only the specific proteins characteristic of different cell types but also the transcription regulators required to maintain and control this pattern of expression. All genes are replicated equally when cells divide [choice (a)]. Expression of housekeeping genes do not differ much from cell to cell, because they mainly encode the proteins that are necessary for all cells to live [choice (c)]. The general transcription factors bind to the TATA box and do not mediate differential gene expression [choice (d)]
8-3 (c)
8-4 (d)
8-5 (d) The housekeeping proteins are proteins common to all cells and are used in processes important to the basic function of cells. Some transcription factors, particularly those used to regulate housekeeping genes, are found in all cell types [choice (a)]. Even though a cell is differentiated, it may be able to respond to changes in its environment by changing its gene expression pattern [choices (b) and (c)]. A differentiated cell usually continues to express genes and make new proteins [choice (c)].
8-6 Choice (b) is correct. The meiotic spindle and the nucleus in the egg are usually removed in these experiments [choices (a) and (c)]. Some differentiated plant cells have the ability to de-differentiate and regenerate an entire adult plant, providing evidence that DNA is not irreversibly altered during development [choice (d)].
8-7 See Figure A8-7.
8-8 B, C
8-9 The genes of a bacterial operon are transcribed into a single mRNA. Many bacterial promoters contain a region known as an operator, to which a specific transcription regulator binds. Genes in which transcription is prevented are said to be repressed. The interaction of small molecules, such as tryptophan, with allosteric DNA-binding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being constitutively expressed are being transcribed all the time.
8-10 (a) Transcriptional regulators generally recognize specific sequences in the DNA by binding to the edges of the bases.
8-11 (c)
8-12 (b)
8-13 (a) The Lac repressor binds when lactose is not present in the cell.
8-14 A. Operon off. CAP will not bind in the presence of glucose.
B. Operon off. Although normally the Lac repressor would bind in the absence of lactose, the lack of the Lac repressor in this case does not matter because the presence of glucose means that the CAP protein will not bind and activate transcription.
C. Operon on. Normally in the absence of both glucose and lactose, the operon would be off. However, because the cells lack the Lac repressor, the cells cannot sense the absence of lactose. Because the CAP protein will bind and activate transcription, the operon will be on.
D. Operon on. The CAP protein will bind and activate transcription because of the presence of glucose. It does not matter whether the Lac repressor gene is mutant, because there is lactose available.
8-15 (b) Replacing the Lac operator with the tryptophan operon operator will now allow binding of the tryptophan repressor instead of the Lac repressor. Since the tryptophan repressor binds to DNA when tryptophan levels are high, the tryptophan repressor is predicted to block mRNA production from the Lac operon when tryptophan is abundant in the cell. Changing the operator sequences at the Lac operon should not affect how the presence of lactose changes the Lac repressor [choice (a)], nor should it affect the ability of CAP to bind to the DNA at low glucose levels, since the CAP-binding site is on the other side of the RNA polymerase binding site [choice (c)]. Removing the Lac operator will make it so that the Lac promoter is no longer responsive to lactose levels in the cell [choice (d)].
8-16 A. TreA is a gene repressor involved in turning off the Tre-ase gene when glucose is present. When cells lack TreB, the Tre-ase gene is not expressed when glucose and trehalose are present in the growth medium. Normally, Tre-ase is not
produced when glucose is present, suggesting that TreA is required to keep the Tre-ase gene switched off when glucose is present.
B. TreB is a gene activator involved in turning on the Tre-ase gene when trehalose is present. When cells lack TreB, the Tre-ase gene is not transcribed.
C. If TreA is constitutively bound to the “A” site, no transcription of the Tre-ase gene will occur. TreA is a gene repressor involved in turning off the Tre-ase gene when glucose is present. Transcription of the Tre-ase gene is seen inappropriately in cells lacking TreA when both glucose and trehalose are present in the growth medium, suggesting that the TreB gene activator can activate transcription of the Tre-ase gene in those circumstances and normally does not do so because of the repression mediated by TreA. Thus, if TreA were always bound, it would overcome the activation normally due to TreB, and the Tre-ase gene would not be transcribed.
8-17 (d)
8-18 (b)
8-19 (c) Both MetA and MetB must be bound, because Psf is not normally transcribed in the presence of Mg2+ only. When the binding site for MetA is present without any other binding sites, low levels of Psf transcription occur in the presence of Mg2+ only, suggesting that the gene activator MetA binds in the presence of Mg2+. Because no transcription occurs normally, the gene repressor MetB must also bind to DNA in the presence of Mg2+ .
8-20 (c) MetA and MetC are both gene activators. MetA binds in the presence of Ca2+ , whereas MetC binds in the presence of Mg2+. Because MetB is a gene repressor, it is unlikely to be bound when both Mg2+ and Ca2+ are present.
8-21 (c)
8-22 (a)
8-23 Choice (a) is correct. Experiment 5 [choice (b)] does not say anything about factor X, because factor X binds to site A. Experiment 6 [choice (c)] and experiment 7 [choice (d)] both examine the effect of factor X in combination with another factor, and thus would not demonstrate that factor X alone is sufficient for expression of the gene encoding Lkp1 in the kidney.
8-24 (c) Experiments 1 and 2 demonstrate that without factors X, Y, or Z, the gene encoding Lkp1 will be switched on in the heart, where it should not be on. Experiment 5 shows that when factor Z can bind, there is no longer expression in the heart. We confirm this hypothesis by comparing the results of experiment 6 with those of experiments 7 and 8. Taken together, these results suggest that factor Z expression is normally needed in the heart to repress the expression of LKP1
8-25 A. If HisP functions as a gene repressor, it would bind more tightly to the regulatory sequences when histidine is abundant, because the histidine biosynthetic genes should be turned off when the cell has enough histidine.
B. If HisP functions as a gene activator, it should bind more tightly to the regulatory sequences when histidine levels are low. When histidine levels are low, the cell needs to synthesize more histidine.
8-26 A. If sufficient glutamine is present in cells, the glutamine repressor will block the synthesis of enzymes that would make more glutamine. Similarly, if cells are starved for glutamine, the unoccupied repressor would not bind to the DNA, and the enzymes that synthesize glutamine would be induced. These conditions permit a direct connection between the levels of glutamine and the expression of glutamine-synthesizing enzymes.
B. The glutamine synthesis enzymes would be permanently switched on, regardless of the level of glutamine in the cells.
C. The glutamine synthesis enzymes would be always switched off, regardless of the level of glutamine in the cells, because the repressor is always bound to the DNA. These cells will not be able to grow unless glutamine is added to the medium.
8-27 A. If the AraC protein acts as a gene repressor for the arabinose operon, araA RNA levels should be high in the presence or absence of arabinose when there is no AraC protein around. In fact, the araA RNA levels should be high all the time, regardless of the presence or absence of arabinose, because the AraA gene should be transcribed under all conditions in the absence of AraC.
B. The results are consistent with AraC acting as a gene activator for the arabinose operon. A gene activator must bind to the promoter regions of the arabinose genes so as to stimulate their transcription. Thus, if the gene for the regulatory protein is deleted, the arabinose genes cannot be turned on.
8-28 A. Site A is regulated by sucrose, and site B by glucose.
B. Mutant 2 (A+ B–); mutant 3 (A– B–) or (A– B+).
C. Site A is bound by an activator, and site B by a repressor.
8-29 (d)
8-30 (a)
8-31 (c) Experiment #5 examines binding in the presence of both G and H. Because no Q mRNA is produced, the repressor, G, is likely bound.
8-32 (c) From experiment #5, we predict that G is bound to site A. Because we see Q mRNA being produced, J must be bound at site B as well.
8-33 Protein Jk8 is bound to the Gip1 promoter because Gip1 mRNA is produced. Jk8 is a gene activator, as seen in experiment #2 when the A binding site is present (and thus Jk8
is bound to the promoter) and Gip1 mRNA is produced. Pa5 acts as a gene repressor, since experiment #3 shows that when Pa5 is bound, there is no Gip1 mRNA produced.
8-34 Choice (d) is correct. Jk8 must be present in adult cells to act as a gene activator [choice (a)], and should not be produced in embryonic liver cells [choices (b) and (c)]
8-35 A 1; B 4; C 3; D 2 Also see Figure A8-35.
8-36 (b) Most eukaryotic transcription regulators that act at a distance are thought to do so by looping out the intervening DNA while at the same time binding, via the Mediator, to proteins that form the initiation complex at the promoter.
8-37 Choice (c) is correct. All the other conclusions can be drawn from the data. Because the BRF2+ HHF– mutant does not show the specific pattern of nucleosome positioning yet still has a low level of BRF1 expression, and because the BRF2– HHF– mutant has high levels of BRF1 expression (indicating that HHF is not required for BRF1 expression), it seems that repression of BRF1 could take place in the absence of nucleosome positioning. Because nucleosomes are formed in all cases, the missing portion of histone H4 is not required for their formation.
8-38 For Gal4 to work properly, the DNA-bound Gal4 must attract histone-modifying enzymes and recruit RNA polymerase to the promoter. If there is too much Gal4 in the cell, the non-DNA-bound Gal4 (or free Gal4) will compete with the DNA-bound Gal4 for binding to histone-modifying enzymes and RNA polymerase. The excess amount of Gal4 forms nonproductive complexes with histone-modifying enzymes and RNA polymerase, preventing their recruitment to the promoter and lowering the level of transcription.
8-39 During transcription in eukaryotic cells, transcriptional regulators that bind to DNA thousands of nucleotides away from a gene’s promoter can affect a gene’s transcription. The Mediator is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcriptional start site. Transcriptional activators can also interact with histone acetylases, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone deacetylases. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the heterochromatin found in interphase chromosomes.
8-40 Choice (c) is correct. The binding of nucleosomes to the promoter inhibits transcription [choice (a)] by blocking RNA polymerase, the general transcription factors, and transcriptional regulators from accessing the DNA [choice (b)]. Histone deacetylases remove acetyl groups that are attached to lysines in the tail of histone proteins [choice (d)].
8-41 (c) The type of cell is determined by the particular combination of transcription regulators active within it. With four different proteins available, there is one possibility with no proteins at all and one with all four proteins. There are four possibilities with one protein each, six possible combinations of two different proteins, and four possible combinations of three different proteins.
8-42 Transcription regulators are generally used in combinations, thereby increasing the possible regulatory repertoire of gene expression with a limited number of proteins.
8-43 (b)
8-44 (a)
8-45 (b) When iPS cells differentiate, the cells follow a program dictated by their genome. Thus, mouse cells can only differentiate into mouse cells and cannot become human cells.
8-46 (c) In order to express Ey in cells that would normally become a leg, scientists must use a different promoter (and thus, involve different transcription regulators) to force Ey expression in an inappropriate cell. In this experiment, Ey is expressed in the embryo in leg-cell precursors, and so the transformation into eye tissue does not involve dedifferentiation and re-differentiation [choice (a)]. When Ey is expressed in the leg-cell precursors, the Ey transcriptional regulator controls the expression of many different genes, including other transcription regulators that assist in regulating genes important for differentiation into an eye [choice (b)]. When Ey is expressed in the leg-cell precursors, an entire organ of several different cell types will develop [choice (d)]
8-47 (b) Ey turns on the transcription of other transcriptional regulators; it is the combined action of the genes directly regulated by Ey and the genes regulated by the regulators that Ey turns on that forms the eye.
8-48 Choice (c) is correct. In general, genes are not lost from the genome during differentiation [choice (a)]. Heterochromatin is generally associated with transcriptionally silent regions of the chromosome [choice (b)]. The presence of MyoD is not sufficient for the expression of muscle-specific genes, because some cell types do not transcribe muscle-specific genes even in the presence of MyoD.
8-49 A. Unlikely. If protein X were to bind nonspecifically to DNA, it would not specifically regulate a particular subset of genes.
B. Unlikely. If a single repressor were to bind upstream of V1, V2, and V3, we would expect to have found a binding site common to all three genes, but there is none.
C. Unlikely. If a protein X that activated gene transcription were to bind to the TATA box, it would be likely to activate most genes transcribed by RNA polymerase II, because most genes contain TATA boxes.
D. Likely.
E. Likely.
8-50 (a) XY individuals that develop as females presumably lack the testes-determining factor (TDF). If BoY is normal in these individuals, it would strongly suggest that BoY is not the TDF. Although expression of TDF is necessary for testes development, this does not mean that it must be expressed in adult males once the gonads have already formed. Similarly, even though TDF expression is sufficient to induce testes formation, once the structures have been formed, TDF may not be able to exert any additional effect; thus, choices (b) and (c) are not considered strong evidence against BoY being TDF. Choice (d) is not compelling evidence against BoY being the TDF, because the TDF will not necessarily bind upstream of all of the genes whose expression it influences; some of the genes it regulates directly probably encode other transcription regulators that bind to regulatory sites different from the TDF site.
8-51 (d)
8-52 (d)
8-53 The transmission of information important for gene regulation from parent to daughter cell, without altering the actual nucleotide sequence, is called epigenetic inheritance. This type of inheritance is seen with the inheritance of the covalent modifications on histone proteins bound to DNA; these modifications are important for reestablishing the pattern of chromatin structure found on the parent chromosome. Another way to inherit chromatin structure involves DNA methylation, a covalent modification that occurs on cytosine bases that typically turns off the transcription of a gene. Gene transcription patterns can also be transmitted across generations through positive feedback loops that can involve a transcription regulator activating its own transcription in addition to other genes. These mechanisms all allow for cell memory, a property involving the maintenance of gene expression patterns important for cell identity.
8-54 (b) Although there are mRNAs with long half-lives, the typical eukaryotic mRNA has a half-life of less than 30 minutes.
8-55 (d)
8-56 (a)
8-57 (c)
8-58 (a)
8-59 MicroRNAs are noncoding RNAs that are incorporated into a protein complex called RISC, which searches the mRNAs in the cytoplasm for sequence complementary to that of the miRNA. When such a molecule is found, it is then targeted for destruction. RNAi is triggered by the presence of foreign double-stranded RNA molecules, which are digested by the Dicer enzyme into shorter fragments approximately 23 nucleotide pairs in length.
8-60 (a)
8-61 (c)
8-62 (b)
8-63 (b) Without RISC proteins, the RNA degradation necessary to turn off genes would not occur. Foreign double-stranded RNA induces RNAi and cells with an intact RNAi response would not be expected to have foreign double-stranded RNA stably present [choice (a)]. miRNA genes could be present in the genome of cells without an intact RNAi response if the ancestral cell did have RNAi capabilities that were lost though evolutionary time [choice (c)]. Although the presence of single-stranded siRNAs within the cell may indicate the presence of Dicer, Dicer alone cannot carry out RNAi without RISC. In fact, if Dicer were absent but RISC present, it may still be possible to carry out RNAi by providing the cell with short, 23-nucleotide-pair siRNAs [choice (d)]
8-64 Choice (a) is correct. Different transcriptional activators are located in different regions of the Drosophila embryo and are differentially responsible for expression in different stripes [choice (b)]. The E. coli LacZ gene is a reporter gene and is not normally expressed in flies; its expression depends on the specific regulatory element engineered into its promoter [choice (c)]. In other stripes, transcriptional repressors may bind to the stripe 2 regulatory DNA segment while transcriptional activators will bind to this segment in stripe 2 [choice (d)].
8-65 A. Protein A binds to its DNA binding site more tightly than protein B binds to its DNA binding site.
B. Protein A is a weaker activator of transcription than protein B.
C. Protein C is able to prevent activation by both protein A and protein B.
8-66 A. Binding sites for the repressor proteins Krüppel and Giant do not seem to overlap. Binding sites for the activator proteins Bicoid and Hunchback also do not seem to overlap. Instead, the binding sites for repressor proteins seem to overlap with the binding sites for activator proteins. These overlapping binding sites cause repressor and activator proteins to compete for binding to the DNA. It is thought that the binding of a repressor and an activator is mutually exclusive. The overlap between the repressor- and activator-binding sites allows Eve expression to be exquisitely sensitive to the levels of repressors and activators in the cell, and suggests that the repressors function by preventing activator binding. In fact, the repressors and activators can antagonize each other, allowing the creation of sharp stripes of transcription from smooth gradients of protein regulatory factors.
B. (i) Mutant embyo (b). When the Krüppel-binding sites are removed, the effects of the Krüppel repressor are eliminated. Stripe 2 expression now expands slightly in the posterior direction, which is to be expected because Hunchback and Bicoid expression extend slightly beyond the posterior end of stripe 2.
(ii) Mutant embryo (c). When two of the Bicoid-binding sites are removed, expression from the promoter is less sensitive to the effects of the Bicoid activator. Thus, stripe 2 appears at its normal position, but the expression of β-galactosidase is decreased.