Ch 8 Flashcards
Bacterial cells can take up the amino acid tryptophan (Trp) from their surroundings, or if there is an insufficient external supply they can synthesize tryptophan from other small molecules. The Trp repressor is a transcription regulator that shuts off the transcription of genes that code for the enzymes required for the synthesis of tryptophan (see Figure 8–7). A. W hat would happen to the regulation of the tryptophan operon in cells that express a mutant form of the tryptophan repressor that (1) cannot bind to DNA, (2) cannot bind tryptophan, or (3) binds to DNA even in the absence of tryptophan? B. W hat would happen in scenarios (1), (2), and (3) if the cells, in addition, produced normal tryptophan repressor protein from a second, normal gene?
A. Transcription of the tryptophan operon would no longer
be regulated by the absence or presence of tryptophan;
the enzymes would be permanently turned on in
scenarios (1) and (2) and permanently shut off in
scenario (3).
B. In scenarios (1) and (2), the normal tryptophan repressor
molecules would completely restore the regulation
of the tryptophan biosynthesis enzymes. In contrast,
expression of the normal protein would have no effect
in scenario (3), because the tryptophan operator would
remain occupied by the mutant protein, even in the
presence of tryptophan.
Explain how DNA-binding proteins can make sequence-specific contacts to a double-stranded DNA molecule without breaking the hydrogen bonds that hold the bases together. Indicate how, through such contacts, a protein can distinguish a T-A from a C-G pair. Indicate the parts of the nucleotide base pairs that could form noncovalent interactions— hydrogen bonds, electrostatic attractions, or hydrophobic interactions (see Panel 2–7, pp. 78–79)—with a DNA-binding protein. The structures of all the base pairs in DNA are given in Figure 5–6.
Contacts can form between the protein
and the edges of the base pairs that are exposed in the
major groove of the DNA (Figure A8–2). These sequencespecific
contacts can include hydrogen bonds with the
highlighted oxygen, nitrogen, and hydrogen atoms, as
well as hydrophobic interactions with the methyl group on
thymine (yellow). Note that the arrangement of hydrogenbond
donors (blue) and hydrogen-bond acceptors (red) of
a T-A pair is different from that of a C-G pair. Similarly, the
arrangement of hydrogen-bond donors and hydrogen-bond
acceptors of A-T and G-C pairs would be different from one
another and from the two pairs shown in the figure. These
differences allow recognition of specific DNA sequences
via the major groove. In addition to the contacts shown in
the figure, electrostatic attractions between the positively
charged amino acid side chains of the protein and the
negatively charged phosphate groups in the DNA backbone
usually stabilize DNA–protein interactions.
Some transcription regulators bind to DNA and cause the double helix to bend at a sharp angle. Such “bending proteins” can stimulate the initiation of transcription without contacting either the RNA polymerase, any of the general transcription factors, or any other transcription regulators. Can you devise a plausible explanation for how these proteins might work to modulate transcription? Draw a diagram that illustrates your explanation
Bending proteins can help to bring distant DNA regions together that normally would contact each other only inefficiently (Figure A8–3). Such proteins are found in both prokaryotes and eukaryotes and are involved in many examples of transcriptional regulation
A virus that grows in bacteria (bacterial viruses are called
bacteriophages) can replicate in one of two ways. In the
prophage state, the viral DNA is inserted into the bacterial
chromosome and is copied along with the bacterial genome
each time the cell divides. In the lytic state, the viral DNA
is released from the bacterial chromosome and replicates
many times in the cell. This viral DNA then produces
viral coat proteins that together with the replicated viral
DNA form many new virus particles that burst out of the
bacterial cell. These two forms of growth are controlled by
two transcription regulators, called c1 (“c one”) and Cro,
that are encoded by the virus. In the prophage state, cI is
expressed; in the lytic state, Cro is expressed. In addition to
regulating the expression of other genes, c1 represses the
Cro gene, and Cro represses the c1 gene (Figure Q8–4).
When bacteria containing a phage in the prophage state are
briefly irradiated with UV light, c1 protein is degraded.
A. W hat will happen next?
B. W ill the change in (A) be reversed when the UV light is
switched off?
C. W hy might this response to UV light have evolved?
A. UV light throws the switch from the prophage to the
lytic state: when cI protein is destroyed, Cro is made and
turns off the further production of cI. The virus produces
coat proteins, and new virus particles are made.
B. When the UV light is switched off, the virus remains in
the lytic state. Thus, cI and Cro form a gene regulatory
switch that “memorizes” its previous setting.
C. This switch makes sense in the viral life cycle: UV light
tends to damage the bacterial DNA (see Figure 6–24),
thereby rendering the bacterium an unreliable host for
the virus. A prophage will therefore switch to the lytic
state and leave the “sinking ship” in search for new host
cells to infect.
Which of the following statements are correct? Explain your
answers.
A. I n bacteria, but not in eukaryotes, many mRNAs contain
the coding region for more than one gene.
B. M ost DNA-binding proteins bind to the major groove of
the DNA double helix.
C. O f the major control points in gene expression
(transcription, RNA processing, RNA transport, translation,
and control of a protein’s activity), transcription initiation is
one of the most common.
A. True. Prokaryotic mRNAs are often transcripts of entire
operons. Ribosomes can initiate translation at internal
AUG start sites of these “polycistronic” mRNAs (see
Figures 7–36 and 8–6).
B. True. The major groove of double-stranded DNA is
sufficiently wide to allow a protein surface, such as
one face of an α helix, access to the base pairs. The
sequence of H-bond donors and acceptors in the major
groove can then be “read” by the protein to determine
the sequence and orientation of the DNA.
C. True. It is advantageous to exert control at the earliest
possible point in a pathway. This conserves metabolic
energy because unnecessary products are not made in
the first place.
Your task in the laboratory of Professor Quasimodo is
to determine how far an enhancer (a binding site for an
activator protein) could be moved from the promoter
of the straightspine gene and still activate transcription.
You systematically vary the number of nucleotide pairs
between these two sites and then determine the amount of
transcription by measuring the production of Straightspine
mRNA. At first glance, your data look confusing (Figure
Q8–6). What would you have expected for the results of
this experiment? Can you save your reputation and explain
these results to Professor Quasimodo?
From our knowledge of enhancers, one
would expect their function to be relatively independent of
their distance from the promoter—possibly weakening as
this distance increases. The surprising feature of the data
(which have been adapted from an actual experiment) is
the periodicity: the enhancer is maximally active at certain
distances from the promoter (50, 60, or 70 nucleotides),
but almost inactive at intermediate distances (55 or 65
nucleotides). The periodicity of 10 suggests that the mystery
can be explained by considering the structure of doublehelical
DNA, which has 10 base pairs per turn. Thus, placing
an enhancer on the side of the DNA opposite to that of the
promoter (Figure A8–6) would make it more difficult for the
activator that binds to it to interact with the proteins bound
at the promoter. At longer distances, there is more DNA to
absorb the twist, and the effect is diminished.
The λ repressor binds as a dimer to critical sites on the λ
genome to repress the virus’s lytic genes. This is necessary
to maintain the prophage (integrated) state. Each molecule
of the repressor consists of an N-terminal DNA-binding
domain and a C-terminal dimerization domain (Figure
Q8–7). Upon induction (for example, by irradiation with
UV light), the genes for lytic growth are expressed, λ
progeny are produced, and the bacterial cell is lysed (see
Question 8–4). Induction is initiated by cleavage of the
λ repressor at a site between the DNA-binding domain
and the dimerization domain, which causes the repressor
to dissociate from the DNA. In the absence of bound
repressor, RNA polymerase binds and initiates lytic growth.
Given that the number (concentration) of DNA-binding
domains is unchanged by cleavage of the repressor, why do
you suppose its cleavage results in its dissociation from the
DNA?
The affinity of the dimeric λ repressor
for its binding site is the sum of the interactions made by
each of the two DNA-binding domains. A single DNAbinding
domain can make only half the contacts and provide
just half the binding energy compared with the dimer.
Thus, although the concentration of binding domains is
unchanged, they are no longer coupled, and their individual
affinities for DNA are sufficiently weak that they cannot
remain bound. As a result, the genes for lytic growth are
turned on.
The genes that encode the enzymes for arginine
biosynthesis are located at several positions around the
genome of E. coli, and they are regulated coordinately
by a transcription regulator encoded by the ArgR gene.
The activity of the ArgR protein is modulated by arginine.
Upon binding arginine, ArgR alters its conformation,
dramatically changing its affinity for the DNA sequences in
the promoters of the genes for the arginine biosynthetic
enzymes. Given that ArgR is a repressor protein, would you
expect that ArgR would bind more tightly or less tightly
to the DNA sequences when arginine is abundant? If ArgR
functioned instead as an activator protein, would you expect
the binding of arginine to increase or to decrease its affinity
for its regulatory DNA sequences? Explain your answers.
The function of these Arg genes is to
synthesize arginine. When arginine is abundant, expression
of the biosynthetic genes should be turned off. If ArgR acts
as a gene repressor (which it does in reality), then binding
of arginine should increase its affinity for its regulatory
sites, allowing it to bind and shut off gene expression. If
ArgR acted as a gene activator instead, then the binding
of arginine would be predicted to reduce its affinity for its
regulatory DNA, preventing its binding and thereby shutting
off expression of the Arg genes
When enhancers were initially found to influence
transcription many thousands of nucleotide pairs from
the promoters they control, two principal models were
invoked to explain this action at a distance. In the “DNA
looping” model, direct interactions between proteins bound
at enhancers and promoters were proposed to stimulate
transcription initiation. In the “scanning” or “entry-site”
model, RNA polymerase (or another component of the
transcription machinery) was proposed to bind at the
enhancer and then scan along the DNA until it reached the
promoter. These two models were tested using an enhancer
on one piece of DNA and a β-globin gene and promoter on
a separate piece of DNA (Figure Q8–9). The β-globin gene
was not expressed from the mixture of pieces. However,
when the two segments of DNA were joined via a linker
(made of a protein that binds to a small molecule called
biotin), the β-globin gene was expressed.
Does this experiment distinguish between the DNA
looping model and the scanning model? Explain your
answer.
The results of this experiment favor DNA
looping, which should not be affected by the protein bridge
(so long as it allowed the DNA to bend, which it does). The
scanning or entry-site model, however, is predicted to be
affected by the nature of the linkage between the enhancer
and the promoter. If the proteins enter at the enhancer
and scan to the promoter, they would have to traverse the
protein linkage. If such proteins are geared to scan on DNA,
they would likely have difficulty scanning across such a
barrier.
Differentiated cells of an organism contain the same genes.
(Among the few exceptions to this rule are the cells of
the mammalian immune system, in which the formation of
specialized cells is based on limited rearrangements of the
genome.) Describe an experiment that substantiates the
first sentence of this question, and explain why it does.
The most definitive result is one showing
that a single differentiated cell taken from a specialized
tissue can re-create a whole organism. This proves that the
cell must contain all the information required to produce a
whole organism, including all of its specialized cell types.
Experiments of this type are summarized in Figure 8–2.
Figure 8–17 shows a simple scheme by which three
transcription regulators are used during development
to create eight different cell types. How many cell types
could you create, using the same rules, with four different
transcription regulators? As described in the text, MyoD is
a transcription regulator that by itself is sufficient to induce
muscle-specific gene expression in fibroblasts. How does
this observation fit the scheme in Figure 8–17
In principle, you could create 16 different
cell types with 4 different transcription regulators (all the
8 cell types shown in Figure 8–17, plus another 8 created
by adding an additional transcription regulator). MyoD by
itself is sufficient to induce muscle-specific gene expression
only in certain cell types, such as some kinds of fibroblasts.
The action of MyoD is therefore consistent with the model
shown in Figure 8–17: if muscle cells were specified, for
example, by the combination of transcription regulators
1, 3, and MyoD, then the addition of MyoD would convert
only two of the cell types of Figure 8–17 (cells F and H) to
muscle
Imagine the two situations shown in Figure Q8–12. In
cell I, a transient signal induces the synthesis of protein
A, which is a transcriptional activator that turns on many
genes including its own. In cell II, a transient signal induces
the synthesis of protein R, which is a transcriptional
repressor that turns off many genes including its own. In
which, if either, of these situations will the descendants of
the original cell “remember” that the progenitor cell had
experienced the transient signal? Explain your reasoning.
The induction of a transcriptional activator
protein that stimulates its own synthesis can create a
positive feedback loop that can produce cell memory.
The continued self-stimulated synthesis of activator A can
in principle last for many cell generations, serving as a
constant reminder of an event that took place in the past.
By contrast, the induction of a transcriptional repressor that
inhibits its own synthesis creates a negative feedback loop
which ensures that the response to the transient stimulus
Figure A8–6 will be similarly transient. Because repressor R shuts off its
own synthesis, the cell will quickly return to the state that
existed before the signal.
Discuss the following argument: “If the expression of every
gene depends on a set of transcription regulators, then the
expression of these regulators must also depend on the
expression of other regulators, and their expression must
depend on the expression of still other regulators, and so
on. Cells would therefore need an infinite number of genes,
most of which would code for transcription regulators.”
How does the cell get by without having to achieve the
impossible?
in the cell; that is, their expression is
constitutive and the activity of the protein is controlled by
signals from inside or outside the cell (e.g., the availability
of nutrients, as for the tryptophan repressor, or by
hormones, as for the glucocorticoid receptor), thereby
adjusting the transcriptional program to the physiological
needs of the cell. Moreover, a given transcription
regulator usually controls the expression of many different
genes. Transcription regulators are often used in various
combinations and can affect each other’s activity, thereby
further increasing the possibilities for regulation with a
limited set of proteins. Nevertheless, most cells devote
a large fraction of their genomes to the control of
transcription: about 10% of genes in eukaryotic cells code
for transcription regulators.
