Download GENE EXPRESSION: CONTROL IN BACTERIA AND PHAGES

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Hyde Chapter 16—Solutions
16
GENE EXPRESSION:
CONTROL IN BACTERIA
AND PHAGES
CHAPTER SUMMARY QUESTIONS
2. The enzyme adenylyl cyclase catalyzes the conversion of ATP to cAMP. Adenylyl
cyclase is inhibited glucose. Therefore, when glucose levels are high, adenylyl
cyclase is inhibited, and cAMP levels in the cell are low. When glucose levels are
low, adenylyl cyclase is active, and cAMP levels in the cell are high. Thus, cAMP
levels are inversely correlated to glucose levels. The catabolite activator protein
(CAP) binds cAMP; this binding is required for CAP to activate the transcription of
operons encoding enzymes that metabolize other sugars. When glucose levels are
high, cAMP is low, CAP is inactive, and operons that encode enzymes for the
metabolism of sugars other than glucose are not activated. This ensures that the cell
will metabolize glucose when it is available. When glucose levels are low, cAMP is
high, CAP is active, and operons encoding enzymes that metabolize other sugars may
be activated.
4. In the trp operator, the TrpR repressor protein cannot bind to the operator sequence in
the absence of tryptophan; in this situation, the lack of repression permits transcription
of the genes of the trp operon. When tryptophan is present, it acts as a corepressor; in
other words the TrpR protein binds to tryptophan and then binds to the operator to
repress transcription.
The attenuator system is also responsive to tryptophan concentrations. When
tryptophan is scarce, the levels of charged tRNATrp are low. In the leader peptide, 2
out of 14 amino acids are tryptophan. When tRNATrp levels are low, the ribosome
pauses at the two tryptophan codons, allowing the mRNA to form the antiterminator
stem. Formation of this mRNA secondary structure prevents formation of a terminator
structure in the mRNA, and the remaining genes of the trp operon will be transcribed.
When tryptophan is abundant, tRNATrp levels are not limiting, so the ribosome will
not pause at the tryptophan codons. In this situation, the terminator stem-loop structure
forms. This structure is a rho-independent transcription terminator, and transcription
of the operon will be terminated before the structural genes of the trp operon are
transcribed. (See figures 16.21, 16.22, 16.23, and 16.24.)
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In the trp operon the operator and attenuator systems work together. When
tryptophan levels are low, the operon is derepressed because the TrpR repressor
cannot bind to the operator, and transcription is not attenuated. Both mechanisms
ensure that the enzymes for tryptophan synthesis will be transcribed when tryptophan
levels are low. When tryptophan levels are high, the operon is repressed because
tryptophan acts as a corepressor for TrpR, and transcription is attenuated. Again, these
systems work together to repress transcription of the structural genes when tryptophan
concentrations are high.
When other amino acids are scarce, transcription of the trp operon is attenuated, as
shown in figure 16.24.
6. Antisense RNA is RNA complementary to the 5' end of an mRNA. Antisense RNA
works by binding to the mRNA through complementary base pairing and preventing
translation of the mRNA. The most obvious source of the regulatory RNA is
transcription of the nontemplate strand of the gene. Antisense RNA could be used to
treat disease by preventing transcription of disease alleles.
8. The repressor favors lysogeny and represses lysis; therefore, a phage that enters
an E. coli cell containing high concentrations of the repressor will likely undergo
lysogeny. The cro gene product favors lysis and represses lysogeny, so if the same
phage enters an E. coli cell containing high concentrations of Cro, it will probably
enter the lytic cycle.
10. OR1: The OR1 site is the operator site with the highest affinity for the repressor;
binding at this site enhances binding at the OR2 site. If OR1 is mutated, the repressor
will not bind efficiently, and the right operon will fail to be repressed, making lysis
more likely.
OR3: This site has the lowest affinity for the repressor; binding of the repressor to
OR3 results in repression of the cI gene itself. If OR3 is mutated, cI will fail to be
repressed and levels of the repressor will rise, so lysogeny will be favored.
PL: Mutation of the left promoter will prevent transcription of all genes transcribed
from the left promoter. The genes transcribed from the left promoter are required
for recombination and integration; in the absence of these gene products, lysis will
occur.
PRE: This promoter site is required for transcription of the repressor at high levels.
If this promoter is mutated, the repressor will not be transcribed, and lysis will be
favored.
PRM: This promoter is required for repressor maintenance. If it is mutated, levels of
the repressor will fail to be maintained, and lysogeny will occur.
tL1: This sequence acts as a terminator, so transcription will continue, and cIII
will be transcribed. The phage will be more likely to undergo lysogeny.
Hyde Chapter 16—Solutions
121
nutL: As in the preceding, transcription will continue through tL1, which will favor
lysogeny.
nutR: Transcription will continue through t1 to the cII gene, which will favor
lysogeny.
12.
UV light induces the SOS repair response in bacterial cells. One of the components
of this response is conversion of the RecA protein (which is normally involved in
recombination) into a protease (RecA*). RecA* cleaves the repressor, which
makes operator sites available for the Cro protein. This response eventually leads to
transcription of the genes for the lytic cycle.
14.
One mechanism that controls the rate of protein degradation is the N-end rule.
Under this mechanism, identity of the N-terminal amino acid determines the halflife of the protein. Regulation of the removal of the N-terminal methionine or other
N-terminal amino acids can regulate the half-life of the protein. A second model for
control of the rate of protein degradation is the PEST hypothesis. Under this
hypothesis, protein degradation is regulated by regions of a protein that are rich in
proline, glutamic acid, serine, and threonine. Proteins rich in these regions tend to
be degraded more rapidly than proteins that do not have these regions.
EXERCISES AND PROBLEMS
16.
Consider one DNA molecule at a time. If a molecule cannot make the functional
lacZ gene product because it is lacZ–, it can be ignored. The Oc mutant acts in cis,
so lacZ+ alleles that are on the same DNA molecule as the Oc allele will be
expressed constitutively. The lacI gene encodes a diffusible protein, so it will act in
trans and affect both alleles. With these concepts in mind, the merozygotes will
have the following phenotypes:
a.
b.
c.
d.
e.
f.
18.
Inducible
Inducible
Inducible
Inducible
Constitutive
Constitutive
The lacY gene encodes the permease that is responsible for transporting lactose into
the bacterial cell. If the permease were nonfunctional, no lactose would enter the
cell, and the inducer would never be synthesized. Therefore, the genes of the lac
operon would be constitutively repressed, and the cell would fail to respond to the
addition of lactose.
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20.
a. They could be defective in either the repressor or the operator. Both the I–
mutant and the Oc mutant are constitutively active.
b. If the mutation is I–, the merodiploid should be inducible because the wild-type
repressor will bind to the operator on either DNA molecule. If the mutation is
Oc, the merodiploid will still be constitutive because the operator mutation is
cis-dominant.
22.
The mutations could affect adenylyl cyclase or the catabolite activator protein
(CAP). If adenylyl cylcase is mutated, then no cAMP will be made, and all four
operons will be uninducible. If the gene encoding CAP is mutated, the cells will be
unresponsive to cAMP.
24.
This mutant will be constitutive for lacZ because the constitutive operator sequence
cannot be bound by any form of the repressor.
26.
The attenuator system requires only one binding event (charged tRNA) rather than
11; therefore, it can respond more sensitively to changes in tryptophan
concentration.
28.
Repression and attenuation allow the cell to monitor both amino acid levels and
charged tRNA levels.
30.
The mutant strain could have a mutation in the CI repressor. This would result in
the phage always causing lysis of the host cells.
32.
When the bacteria are shifted to 42°C, the CI repressor will denature and no longer
function, which will make the operator sites available for the Cro protein. The
phage DNA will be excised from the bacterial genome, and lysis will occur.
34.
Transcriptional control is more efficient than translational control because it
ensures that a cell does not expend energy transcribing a gene that is not needed.
Translational control can be used to directly respond to levels of a molecule such as
a metabolite without the need for other protein factors.
36.
The data are consistent with the presence of two promoters: an early promoter that
produces a polycistronic message including both rIIA and rIIB and a late promoter
that transcribes only rIIB. A nonsense mutation in rIIA could cause ribosomes to
dissociate from the polycistronic transcript, meaning that early expression of the
rIIB gene would be reduced. The late promoter presumably overlaps with the 3' end
of the rIIA gene, so that a small deletion near the end of the rIIA gene would delete
the late promoter and prevent transcription of rIIB from the late promoter.
Transcription of rIIB from the early promoter would not be affected. The following
figure diagrams the structure of these genes.
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PE
PL
AUG
Stop
codon AUG
Stop
codon
Transcript
from PE
AUG
Stop
codon
Transcript
from PL
38.
First, it is more efficient to utilize one sugar preferentially and to express the genes
for the metabolism of other sugars only when the first sugar is absent. Second,
many of the alternative sugars (such as lactose) are converted to glucose or its
derivatives before metabolism. For example, lactose is cleaved by -galactosidase
to produce glucose and galactose. The galactose is then converted to glucose 6phosphate, which is the first compound formed when metabolizing glucose. Thus,
metabolizing lactose still requires all the enzymes that are used in glucose
metabolism. Because these sugars pass through the glucose metabolic pathway, the
cell prefers to metabolize any glucose (if it is present) before expressing additional
enzymes that will convert other sugars to glucose or one of its derivatives.
40.
The mutation is probably in one of the tryptophan codons in the leader peptide. The
tryptophan codon is TGG; TGT and TGC are both cysteine codons. If one of the
tryptophan codons is mutated to a cysteine codon, the bacteria would be less
sensitive than usual to tryptophan starvation; however, when starved for both of
these amino acids, the response would be normal.
42.
The terms positive and negative regulation refer to the effect that a regulatory
protein has on transcription. The terms inducible and repressible describe whether
a system is regulated by a substrate or a product. An inducible system is regulated
by a substrate or a molecule related to a substrate. If the activator protein cannot
bind DNA in the absence of the inducer but can bind DNA to activate transcription
in the presence of the inducer molecule, the inducible system will exhibit positive
regulation. If a repressor protein blocks transcription in the absence of the inducer
molecule and fails to block transcription when the inducer molecule is present, the
inducible system will show negative regulation (see figure 16.3).
A repressible system will exhibit negative regulation if a co-repressor molecule
is required to block transcription, as is the case in the trp operon. However, a
repressible system could also exhibit positive regulation if an activator protein
promotes transcription in the absence of the co-repressor molecule but fails to
activate transcription when the co-repressor molecule is present.
44.
The LacI protein binds to two of the three operator sites as a tetramer. Maximum
repression is achieved only when all three sites are bound by a repressor. When the
repressor is bound to two of these sites, the intervening DNA, which includes the
promoter, is looped out and is inaccessible for RNA polymerase. Use of sites that
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are separated from one another allows DNA looping to occur. The use of multiple
sites also allows the system to be more sensitive to change in levels of the inducer.
46.
a. This is the coding strand of DNA and therefore the nontemplate strand.
b. There are 20 amino acids, so the probability of finding a phenylalanine at any
position would be 1/20. The probability of the tripeptide phe-phe-phe will be
(1/20)(1/20)(1/20) = 1/8000. Therefore, a polypeptide of 8000 amino acids will be
expected to have one phe-phe-phe sequence.
c. This discrepancy could be related to the levels of charged (aminoacylated)
tRNAs in the cell. Amino acids, such as tryptophan, that normally have low
cellular levels of charged tRNAs are likely to have few codons in the leader
mRNA. If instead they had many codons, ribosome stalling would take place
even at the normal low levels of charged tRNAs. In contrast, amino acids, such
as phenylalanine, that normally have high cellular levels of charged tRNAs are
likely to have many codons in the leader mRNA. If they had few codons,
ribosome stalling would not take place as often as needed.
d. i. Attenuation will not occur. When charged phe-tRNAs are scarce, the
ribosome would stall on the phe codons in region 1, which is prevented from
base-pairing with region 2. Region 2 is then free to base-pair with region 3,
thereby precluding the formation of the terminator stem loop (between
regions 3 and 4).
ii. Attenuation will occur. No translation means that stem loops are formed
between regions 1 and 2, and between regions 3 and 4. Stem loop 3–4 will
cause rho-independent termination of transcription.
iii. Attenuation will not occur. Region 1 cannot base-pair with region 2, and so
stem loop 2–3 (antiterminator) forms.
iv. Attenuation will occur. The ribosome would stall on the phe codons.
However, region 2 cannot base-pair with region 3, thereby causing stem loop
3–4 to form.
v. Attenuation will not occur, simply because the stem loop 3–4 transcription
terminator cannot form.
e. The mutant leader peptide sequence and encoded amino acids are given here:
5'-ATG AAA CAC ATA CCG TTT TCT TCG CAT TCT TTT TTA CCT TCC CCA TGA-3'
Met Lys His Ile Pro Phe Ser Ser His Ser Phe Leu Pro Ser Pro
Two phe codons remain in the leader peptide sequence, so low levels of
phenylalanine may still prevent attenuation by causing the ribosome to stall in
region 1. However, the mutant sequence now contains four ser and three pro
codons. Therefore, low levels of serine and proline would also prevent
attenuation in this mutant.
Hyde Chapter 16—Solutions
125
CHAPTER INTEGRATION PROBLEM
a. There may be a prophage inserted between the met+ and trp+ genes. As the singlestranded DNA is preparing to enter the recipient cell, the CI protein would be
removed (it only binds to the double-stranded OR and OL sequences). The transferred
DNA is converted to double-stranded DNA in the recipient cell. If the entire
+
+
genome transferred into the recipient cell (met transferred, but not trp ), the
absence of the CI protein would allow the double-stranded prophage to excise from
the bacterial genome and enter the lytic cycle in the recipient cell.
b. At 42°C, the CI repressor would denature and the prophage would excise from the
bacterial genome. With the -DNA being absent from the bacterial genome, it could
not be transferred by conjugation and the recipient cell could not enter the lytic cycle.
The absence of the -DNA would also result in a normal conjugation distance
between the met+ and trp+ genes.
c. The kanamycin-resistant strain already contains a prophage. This would result in
CI repressor protein already being present in the recipient cell. Thus, when the donor
prophage enters the recipient cell and becomes double-stranded, the endogenous
CI protein would bind to the newly entered OR and OL sites and prevent the excision
of the entering prophage.
d. The leu+, met+, bio+, trp– cells would grow on plates that lacked leucine, would grow
on plates that lacked methionine, but could not grow on plates that lacked tryptophan.
e. The trp+ gene was deleted in strain F.
f. An inversion may have switched the order of the trp+ and met+ genes.
g. Strain F appears to contain a deletion of much of the 10-kb region. Several bands are
missing entirely from the Southern blot. Strain M has two bands that are smaller than
those in the original strain, again consistent with a deletion. The results are also
consistent with an inversion with breakpoints located in the two bands of altered size.