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CHAPTER 17
Concept check questions (in figure legends)
FIGURE 17.1
Concept check: Which of these levels is the most energy-efficient way to regulate gene
expression?
Answer: Transcriptional regulation is the most energy-efficient, because a cell avoids
wasting energy making RNA or protein.
FIGURE 17.3
Concept check: Explain how an alpha helix is able to function as a recognition helix in a
transcription factor protein.
Answer: An alpha helix can bind into the major groove of DNA and recognize a specific
sequence of bases.
FIGURE 17.4
Concept check: If a repressor prevents TFIID from binding to the TATA box, why does
this inhibit transcription?
Answer: If TFIID cannot bind to the TATA box, RNA polymerase will not be recruited
to the core promoter, and therefore transcription will not begin.
FIGURE 17.5
Concept check: When an activator protein interacts with mediator, how does this affect
the function of RNA polymerase?
Answer: When an activator interacts with mediator, it causes mediator to phosphorylate
CTD, which causes RNA polymerase to proceed to the elongation phase of transcription.
FIGURE 17.7
Concept check: Explain why the glucocorticoid receptor binds specifically next to the
core promoter of certain genes, but not next to the core promoter of most genes.
Answer: The glucocorticoid receptor binds only next to genes that have a GRE by their
core promoters.
FIGURE 17.8
Concept check: How might nucleosome eviction affect transcription?
Answer: Nucleosome eviction may allow certain proteins access to binding to particular
sites in the DNA.
FIGURE 17.9
Concept check: Describe two different ways that histone modifications may alter
chromatin structure.
Answer: Histone modifications may directly affect the interaction between histones and
the DNA, or they may affect the binding of other proteins to the chromatin.
FIGURE 17.10
Concept check: Why is an NFR needed at the core promoter for transcription to occur?
Answer: An NFR is needed at the core promoter so that transcription factors can
recognize enhancers and so the preinitiation complex can form.
FIGURE 17.11
Concept check: Explain why histone eviction or displacement is needed for the
elongation phase of transcription.
Answer: Histone eviction or displacement is needed for elongation because RNA
polymerase cannot transcribe though nucleosomes. It needs to unwind the DNA for
transcription to take place.
FIGURE 17.13
Concept check: Explain why the events shown in part (a) inhibit transcription.
Answer: In part (a), DNA methylation is preventing an activator protein from binding to
the DNA. This prevents transcriptional activation.
FIGURE 17.14
Concept check: What is the difference between de novo methylation versus maintenance
methylation?
Answer: De novo methylation occurs on unmethylated DNA, whereas maintenance
methylation occurs on hemimethylated DNA.
FIGURE 17.15
Concept check: Why are insulators important for gene regulation in eukaryotes?
Answer: Insulators prevent one gene from regulating a neighboring gene. This allows
each gene to control its own regulation.
FIGURE 17.18
Concept check: A pre-mRNA is recognized by just one splicing repressor that binds to
the 3’-end of the third intron. The third intron is located between exon 3 and exon 4.
After splicing is complete, would you expect the mRNA to contain exon 3 and/or exon 4
in the presence of the splicing repressor?
Answer: Exon 3 would be spliced out. Exon 4 would still be within the mRNA.
FIGURE 17.20
Concept check: Why is the production of double-stranded RNA more likely if multiple
copies of a gene are inserted into a plant cell’s genome?
Answer: If multiple copies of a gene are inserted into a genome, it becomes more likely
that an endogenous promoter will transcribe the antisense strand of that gene.
FIGURE 17.22
Concept check: Explain why RISC binds to a specific mRNA. What type of bonding
occurs?
Answer: A RISC binds to a specific mRNA because the small RNA within RISC is
complementary to that mRNA. The bonding is hydrogen bonding between
complementary bases.
FIGURE 17.24
Concept check: If a mutation prevented IRP from binding to the IRE in the ferritin
mRNA, how would the mutation affect the regulation of ferritin synthesis. Do you think
there would be too much or too little ferritin?
Answer: The mutation would cause the over production of ferritin, because ferritin
synthesis would occur even if iron levels were low.