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Why does conjugation with an Hfr donor result in a much higher rate of gene transfer than conjugation with an F+ donor?
Why does conjugation with an Hfr donor result in a much higher rate of gene transfer than conjugation with an F+ donor?
The Hfr donor segment must undergo recombination in the recipient.
The F+ donor segment must undergo recombination in the recipient.
An F+ donor makes fewer pili.
The Hfr donor transfers genes from
the chromosome and not the plasmid.
An F+ donor’s T strand is slow to reach the pilus.
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The Hfr donor transfers genes from the chromosome and not the plasmid.
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To integrate an F plasmid’s genes into a host chromosome, there must be DNA recombination at homologous regions on the two circular strands. What is the minimum number of such crossover events needed for integration?
2
4
8
1
3
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1
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A bacterial cell containing an F factor is a threonine prototroph. Conjugation occurs with an F- threonine auxotrophic cell. The exconjugant cell can grow on medium lacking threonine. Where was the F factor in the donor cell?
plasmid
R plasmid
Hfr chromosome
A or B are both possible.
A, B, or C are all possible.
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R plasmid
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Which of the following statements about bacterial genomes is NOT true?
In addition to the main bacterial chromosomes, most bacteria also carry additional mini-chromosomes (plasmids), which carry nonessential genes or provide special functions.
Like eukaryotic chromosomes, bacterial chromosomes are generally linear.
Bacteria tend to have only one main chromosome.
Bacterial genomes tend to be small by comparison to eukaryotic genomes.
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Like eukaryotic chromosomes, bacterial chromosomes are generally linear.
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What is the name of the process in bacteria that involves DNA transfer through a cytoplasmic connection known as a pilus?
conjugation
transformation
generalized transduction
specialized transduction
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conjugation
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How is an Hfr chromosome formed?
by integration of the F plasmid into the bacterial chromosome
by a point mutation in a bacterial gene that is essential for conjugation
by a specific deletion mutation in the bacterial chromosome
by a specific gene duplication in the F plasmid
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by integration of the F plasmid into the bacterial chromosome
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In an experiment similar to the Lederberg and Tatum experiment, researchers grew two strains of E. coli in complete medium.
E. coli strain AX has the genotype AmpSbio+ trp+ leu-.
E. coli strain AY has the genotype AmpRbio- trp- leu+.
The researchers then plated the bacteria on different media types.
Part A
On which type(s) of media will the AX strain grow?
Select all that apply.
leu-
trp-
bio-
leu-trp-bio- + ampicillin
complete medium
complete medium + ampicillin
On which type(s) of media will the AY strain grow?
Select all that apply.
leu-
trp-
bio-
leu-trp-bio- + ampicillin
complete medium
complete medium + ampicillin
Next, the researchers mixed equal amounts of the AX and AY E. coli strains.
On which type(s) of media would growth provide evidence that genetic transfer had occurred between the two strains of bacteria?
Select all that apply.
bio- leu-
bio- trp-
trp- leu-
leu- trp- bio-
none of the above
Next the researchers repeated the experiment in Part C but instead of mixing the AX and AY strains, they put one strain on each side of a glass U-tube. The U-tube contains a glass filter between the two strains of bacteria that prevents the cells from interacting directly.
On which type(s) of media would you expect cells from the AY side of the U-tube to grow?
Select all that apply.
bio- leu-
bio- trp-
trp- leu-
leu- trp- bio-
none of the above
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Bactria that are AmpS cannot grow in the presence of ampicillin. If a strain of bacteria has a mutation in a gene that is necessary to make an essential nutrient (for example, leu-), then the strain requires that nutrient to be provided in the medium. Bactria that are AmpR are resistant to ampicillin and can grow in the presence of ampicillin. If a strain of bacteria is wild-type for a gene that is necessary to make an essential nutrient (for example, leu+), then the strain can grow in media that lacks that nutrient. When gene transfer occurs between bacteria, an auxotropic strain such as AY, that cannot grow on media lacking bio and trp, can become prototropic. Prototropic bacteria do not need nutrients added to minimal medium in order to grow. Bacterial gene transfer by conjugation requires physical contact between cells of the two strains. The glass filter in this experiment prevents cell-cell contact. Without gene transfer, the AY strain remains bio- trp- and cannot grow on any medium that lacks bio and/or trp.
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There are four categories of gene regulation in prokaryotes:
negative inducible control
negative repressible control
positive inducible control
positive repressible control
What is the difference between negative and positive control? If an operon is repressible, how does it respond to signal? If an operon is inducible, how does it respond to signal?
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1. In negative inducible control, the transcription factor is a(n) repressor. Binding of the signal molecule to the transcription factor causes transcription to start. 2. In negative repressible control, the transcription factor is a(n) repressor. Binding of the signal molecule to the transcription factor causes transcription to stop. 3. In positive inducible control, the transcription factor is a(n) activator. Binding of the signal molecule to the transcription factor causes transcription to start. 4. In positive repressible control, the transcription factor is a(n) activator. Binding of the signal molecule to the transcription factor causes transcription to stop.
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What would be the effect of a mutation in the lacI gene that prevented the repressor from binding to lactose?
Hints
The lac Z, Y, and A genes would not be expressed.
The lac Z, Y, and A genes would be induced by lactose.
The lac Z, Y, and A genes would be expressed constitutively.
The lac Z, Y, and A genes would be repressed by lactose.
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The lac Z, Y, and A genes would not be expressed.
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What is the role of glucose in catabolite repression?
Hints
It increases the levels of cAMP in the cell, stimulating transcription from the lac operon.
It decreases the levels of cAMP in the cell, repressing transcription from the lac operon.
It stimulates transcription from the lac operon, causing an increase in cAMP levels in the cell.
It represses transcription from the lac operon, causing a decrease in cAMP levels in the cell.
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It decreases the levels of cAMP in the cell, repressing transcription from the lac operon. Glucose decreases the levels of cAMP in the cell, preventing formation of the CAP-cAMP complexes necessary for the stimulation of transcription from the lac operon.
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What do we call a cluster of genes whose transcription is controlled by a single promoter and associated regulatory sequences?
An operon
A transcription unit
An operator
A polyprotein sequence
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An operon
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Which of the following mutations would result in constitutive expression of the lac operon?
A mutation in lacI that increases the affinity of repressor binding to the operator
A mutation in lacY that prevents transport of lactose into the cell
A mutation in lacI that prevents the repressor from binding lactose
A mutation in lacI that prevents the repressor from binding to the operator
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A mutation in lacI that prevents the repressor from binding to the operator
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Under which of the following conditions will the lac operon be expressed at the highest level?
Glucose absent, lactose present
Glucose present, lactose absent
Glucose and lactose absent
Glucose and lactose present
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Glucose absent, lactose present
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In this problem you will explore how to solve problems involving partial diploid lac operon bacterial strains.
Bacterial strains that are "partially diploid" have two copies of the lac operon because they aquired a plasmid carrying just the lac operon region. One copy of the lac operon region is on the recipient’s bacterial chromosome, and the other copy is on the F’ plasmid that was introduced into the cell by conjugation. Partial diploid genotypes are written with the F’ segment first and the recipient chromosome next.
You create a lac operon partial diploid with this genotype:
F’ I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y−
To determine which genes are transcribed and under what conditions, you need to first consider each genotype separately, and then together.
For each region of the lac operon on the F’ plasmid, I+ P+ Oc Z− Y+ , determine whether the region is wild type (that is, it produces a functional protein or it’s a correct protein binding sequence) or whether the region is mutated.
Select all that apply.
The promoter sequence is correct/functional.
The repressor protein is produced.
The operator sequence is correct/functional.
Beta-galactosidase is produced from the lacZ gene.
Permease is produced from the lacY gene.
Use the information from Part A to determine how the operon is regulated for this genotype (I+ P+ Oc Z− Y+ ).
Operon is noninducible; permease is never produced, even when lactose is present.
Operon is inducible; permease is produced only in the presence of lactose.
Operon is repressible; permease is only produced in the absence of lactose.
Operon is constitutive; permease is produced even when lactose is absent.
Now, for each region of the lac operon on the bacterial chromosome, I− P+ O+ Z+ Y−, determine whether the region is wild type (that is, it produces a functional protein or it’s a correct protein binding sequence) or whether the region is mutated.
Select all that apply.
The promoter sequence is correct/functional.
The repressor protein is produced.
The operator sequence is correct/functional.
Beta-galactosidase is produced from the lacZ gene.
Permease is produced from the lacY gene.
Use the information from Part C to determine how the operon is regulated for this genotype (I- P+ O+ Z+ Y-).
Operon is inducible; lacZ is transcribed only in the presence of lactose.
Operon is noninducible; lacZ is never transcribed, even in the presence of lactose.
Operon is constitutive; lacZ is transcribed even when lactose is absent.
Operon is repressible; lacZ is only expressed in the absence of lactose.
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The promoter sequence is correct/functional. The repressor protein is produced. Permease is produced from the lacY gene. This lac operon I+ P+ Oc Z− Y+ has a normal promoter sequence. Of the two structural genes (lacZ and lacY) only lacY has a wild-type sequence. In terms of regulation in response to lactose, the operon makes functional repressor (I+ ), however the operator sequence is mutated (Oc ). Operon is constitutive; permease is produced even when lactose is absent. For the I+ P+ Oc Z− Y+ genotype, of the two structural genes, only lacY is wild-type sequence, meaning only permease is produced. The gene for beta-galactosidase (lacZ) is mutated, meaning that no functional enzyme is produced. The promoter sequence is normal, so its possible for RNA polymerase to transcribe the operon, and produce permease. In terms of regulation, functional repressor (I+) is made, however the operator region is mutated (Oc). Because the operator sequence is altered, the repressor cannot bind to DNA and prevent transcription. As a result, permease is produced under all conditions; it is "constitutively" expressed. The promoter sequence is correct/functional. The operator sequence is correct/functional. Beta-galactosidase is produced from the lacZ gene. This lac operon, I- P+ O+ Z+ Y-, has a normal promoter sequence. Of the two structural genes (lacZ and lacY), only lacZ has a wild-type sequence. In terms of regulation in response to lactose, the operon does not produce any repressor protein (I-), however the operator sequence is wild type (O+). Operon is constitutive; lacZ is transcribed even when lactose is absent. Of the two structural genes, only lacZ is a wild-type sequence, meaning that only beta-galactosidase can be produced from this operon.The promoter sequence is normal, so it’s possible for RNA polymerase to transcribe the operon, and produce beta-galactosidase. In terms of regulation, the operator region is wild type (O+), but no repressor is made (I- ). Without repressor protein, it’s not possible to prevent transcription. As a result, beta-galactosidase is produced under all conditions; it is "constitutively" expressed.
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In a partial diploid strain, both copies of the lac operon are together in the same cell. To determine the overall regulation of lac operon genes in a partial diploid strain, here are the most important things to understand:
Proteins can diffuse around the cell and potentially interact with either copy of the lac operon (known as trans-acting).
DNA binding sites can only regulate genes on the same copy (known as cis-acting).
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For the partial diploid you’ve created, F’ I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y− , you already know that the promoter sequences for copies are functional, so you can focus on the repressor proteins and operator regions that control expression.
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1. Repressor protein is made from the F’ plasmid. 2. Repressor protein is able to bind to the operator region on the bacterial chromosome. 3. Functional beta-galactosidase protein could be made from the bacterial chromosome . 4. Functional permease protein could be made from the F’ plasmid. Repressor protein is only produced from the F’ plasmid (I+); the bacterial chromosome has a mutated sequence (I-). In terms of operator sequences, only the operator sequence on the bacterial chromosome (O+) can bind the repressor; the F’ plasmid has a mutated operator region (Oc). LacZ gene is wild type (Z+) only on the bacterial chromosome; lacY is wild type (Y+) only on the F’ plasmid.
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Which of the following summarizes expression of the lac genes in this partical diploid: I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y− ?
Beta-galactosidase is made both in the presence of and in the absence of lactose; permease is only made when lactose is present.
Permease is made both in the presence of and in the absence of lactose; beta-galactosidase is only made when lactose is present.
Both permease and beta-galactosidase are made both in the presence of and in the absence of lactose.
Both permease and beta-galactosidase are made only when lactose is present.
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Permease is made both in the presence of and in the absence of lactose; beta-galactosidase is only made when lactose is present. Repressor protein is made from the F’ plasmid and diffuses in the cell. It can bind to the operator region on the bacterial chromosome, but cannot bind to the operator on the region on the F’ plasmid. As a result, the wild-type lacY gene on the F’ plasmid will be expressed constitutively, while the wild-type lacZ gene on the bacterial chromosome will be inducible. If there is no lactose present, beta-galactosidase will be produced. When lactose is present, both beta-galactosidase and permease will be prodcued.
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You are investigating a new bacterial operon that contains five regions (A, B, C, D and E) involved in coordinated regulation of transcription. One is the gene for a regulatory protein, one is the binding site for the regulatory protein, two produce structural enzymes (enzyme 1 and enzyme 2), and one is the promoter for the two enzyme genes.
The table below shows data collected for five different bacterial strains. A "+" indicates that the structural enzyme is produced and a "-" indicates that the enzyme is not produced. The "signal" is a molecule that either represses or induces the operon. Use the information to determine how the operon is regulated, and the identity of each region.
Strain Enzyme 1 Enzyme 1 Enzyme 2 Enzyme 2
-signal +signal -signal +signal
A+B+C+D+E+ + – + –
A-B+C+D+E+ – – – –
A+B-C+D+E+ + + + +
A+B+C-D+E+ + – – –
A+B+C+D-E+ + + + +
A+B+C+D+E- – – + –
In the wild-type operon (A+ B+ C+ D+ E+), how does the presence of the signal affect expression of the structural enzymes ?
The presence of the signal causes the structural enzymes to be produced. Without the signal, no enzymes are made.
The presence of the signal stops production of the structural enzymes. Without the signal, the structural enzymes are produced.
The structural enzymes are never produced. The presence or absence of the signal has no effect on the production of the structural enzymes.
The structural enzymes are produced under all conditions. The presence or absence of the signal has no effect on the production of the structural enzymes.
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The presence of the signal stops production of the structural enzymes. Without the signal, the structural enzymes are produced. If you look at only the wild-type genotype (top row), you can see that enzyme 1 and enzyme 2 are only produced when the signal molecule is absent (-signal). When the signal is present (+signal), the enzymes are not produced. In this operon, the signal stops transcription of the operon; this means that the operon is repressible. In repressible operons, transcription occurs until the level of the final product is high. High levels of the final product suppress further transcription of the operon.
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Parts of an operon that control the expression of both enzyme genes are called "regulatory regions." The genes that produce the enzymes are called "structural genes." Using the data in the table, determine which regions are regulatory regions and which are structural genes.
Sort each region into the correct category.
HelpReset
Regulatory Region(s)
Region ARegion BRegion D
Structural Gene(s)
Region CRegion E
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In the operon, both structural genes (the genes that produce enzyme 1 and enzyme 2) are regulated by the same regulatory protein, the same binding site for the regulatory protein, and the same promoter sequence. Therefore, mutations in these regulatory regions would affect both enzyme 1 and enzyme 2. This is true of mutations in regions A, B, and D, indicating these are regulatory regions.
In contrast, mutations in regions C and E affect only one of the two enzymes, indicating that these are the structural genes that encode the individual enzymes.
Part C
Now that you have determined which are the regulatory regions, identify how this operon is regulated.
Is this operon inducible or repressible? Is it under positive control or negative control?
The operon is inducible (the signal induces/starts expression of the enzymes).
The operon is under negative control (the regulatory protein is a repressor).
The operon is repressible (the signal represses/stops expression of the enzymes).
The operon is under negative control (the regulatory protein is a repressor).
The operon is inducible (the signal induces/starts expression of the enzymes).
The operon is under positive control (the regulatory protein is an activator).
The operon is repressible (the signal represses/stops expression of the enzymes).
The operon is under positive control (the regulatory protein is an activator).
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Incorrect; correct answer displayed
You know from Part A that this operon is repressible; the signal molecule stops transcription.
To figure out whether the operon is under negative or positive control, you must determine whether the regulatory protein is a repressor (negative control) or an activator (positive control). There are three regulatory regions for this operon: the promoter sequence, the gene that produces the regulatory protein, and the binding site for the regulatory protein.
A mutation in region A prevents transcription of both enzyme genes, suggesting this is the promoter region. Both region B and region D mutations give constitutive expression. Eliminating the regulatory protein or its binding site gives expression under all conditions. This suggests the operon is regulated by a repressor protein binding to an operator sequence.
Part D
Given only the data in the table, for which part(s) of the operon is it possible to identify the exact region it corresponds to?
Select all that apply.
the gene for enzyme 1
the gene for enzyme 2
the promoter region
the operator region
the gene for the repressor protein
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Of the five mutant strains, three have distinctive effects.
A mutation in region A eliminates expression of both enzymes. This suggests that region A is the promoter sequence, where RNA polymerase binds to transcribe both enzyme genes.
A mutation in region C eliminates expression of enzyme 2. This suggests that region C is the gene for enzyme 2.
A mutation in region E eliminates the expression of enzyme 1. This suggests that region E is the gene for enzyme 1.
The other two mutations, in region B and in region D, both result in constitutive expression. This suggests that these two regions are involved in negative control — a repressor binding to an operator region. With just this information, however, you cannot determine which region is the gene for the repressor protein and which region is the operator.
Part E
Mutations in region B and region D give the same results – when either of these regions is mutated, the operon is expressed under all conditions (called constitutive expression).
You hypothesize that region B is the operator region and that Region D is the repressor protein.
To test your hypothesis, you create two partial diploid lines by introducing a F’ plasmid with a wild-type lac operon:
Strain 1: F’ A+ B+ C+ D+ E+ / A+ B- C+ D+ E+
Strain 2: F’ A+ B+ C+ D+ E+ / A+ B+ C+ D- E+
What experimental results would be predicted by your hypothesis?
Complete this table with the expected results by dragging the "+" and "-" signs to the appropriate locations.
+
–
+
+
+
+
+
–
+
–
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Correct
The goal of this experiment is to determine which region is the gene for the repressor and which region is the operator. Recall that repressor proteins are trans-acting; proteins are able to diffuse around the cell and interact with any wild-type operator that exists. Introduction of a wild-type repressor gene will restore wild-type repressor protein to the mutated strain, restoring normal regulation.
Unlike the repressor protein, operator regions only regulate genes on that piece of DNA, and cannot regulate genes on other pieces of DNA. Introducing a wild-type operator region will not restore normal regulation to strain with an operator mutation. The wild-type operator sequence on the F’ plasmid cannot regulate the genes on the bacterial chromosome. If there is an operator mutation on the bacterial chromosome, expression will be constitutive even if a normal operon is introduced on F’ plasmid.
If region B is the operator, then the genotype F’ A+B+C+D+E+/A+B-C+D+E+ (Strain 1) should still exhibit constitutive expression of the enzymes.
If region D is the repressor gene, then the genotype F’ A+B+C+D+E+/A+B+C+D-E+ (Strain 2) will exhibit wild-type regulation/expression –enzyme genes will be transcribed when signal is absent, and will stop being transcribed when signal is present.
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The trp operon in E. coli contains five structural genes that share a regulatory region with a promoter (trpP), an operator (trpO), and a leader region (trpL) that contains the attentuator region. The five structural genes are responsible for the synthesis of the amino acid tryptophan. A sixth gene, trpR, encodes the repressor protein, which is not activated until it binds with tryptophan.
Diagram showing the regulatory region and genes of the trp operon
The trp operon is a repressible operon under negative control, and is regulated by two mechanisms.
1) The first mechanism is inhibition mediated by a repressor. Tryptophan, which is the final product of the trp operon genes, acts as a corepressor, activating the repressor (trpR) when tryptophan levels are high. trp operon expression is "repressed" by a high level of tryptophan.
2) The second mechanisms is attenuation, the premature termination of transcription. The trpL region codes for sequences that, when transcribed into mRNA, are able to form stem-loop structures through complementary base-pairing. When tryptophan levels are high, a stem-loop structure forms in the mRNA, which causes transcription to stop before the trp genes can be transcribed.
Part A
trpR+ (wild-type): 8% expression with tryptophan present; 100% with trypthophan absent; trpR- mutant: 33% expression with tryptophan present; 100%with tryptophan absent
Expression of the trp operon is regulated by the level of free tryptophan in the cell both through repressor action and attenuation. This chart shows the percent expression of the trp operon in the presence and absence of tryptophan for wild-type (trpR+ or trp+) and repressor mutants (trpR-).
Use this information to determine the levels of expression for the following genotypes and conditions.
Rank the genotypes and conditions from highest to lowest level of trp operon expression.
HelpReset
LowestHighest
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trp+ / tryptophan absent trpL- / tryptophan present trpR- / tryptophan present trp+ / tryptophan present trpP- / tryptophan absent
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My AnswersGive Up
Correct
The trp operon is a negative repressible system; the presence of tryptophan prevents transcription of the trp operon. Tryptophan acts as a corepressor by binding to and activating the trp repressor protein (trpR). Tryptophan also plays a role in promoting attenuation, which leads to the termination of transcription, through the structure of the leader sequence (trpL).
The highest level of trp operon expression (100%) would be in a wild-type strain (trp+) in the absence of tryptophan. Without tryptophan, the repressor cannot bind to the operator to stop transcription and the folding of the mRNA favors anti-termination, increasing the likelihood of transcribing the trp genes.
To determine the effect of trpR- versus trpL- on expression in the absence of tryptophan, first consider the information in the chart. Full expression of the operon in the absence of tryptophan in a wild-type strain is 100%; when tryptophan is present, expression drops to a basal level of 8%. The chart indicates that when the repressor is mutated (trpR-), expression jumps from 8% to 33%. If the only two repression mechanisms are the repressor protein and attentuation, this suggests that the 67% of expression that is still inhibited must be mediated by attentuation. By this logic, strains mutant in trpL should have higher expression than strains mutant in trpR.
A wild-type genotype would express a very low level of trp genes in the presence of tryptophan. Even though tryptophan would promote binding of the repressor to the operator and increase the likelihood of attenuation, neither process is permanent/irreversible. The repressor protein binds and unbinds, and as a result, a low level of trp genes will be transcribed (8%).
The lowest level of trp expression (0%) would be in the genotype with a promoter mutation (trpP-). Without a wild-type promoter sequence (trpP), RNA polymerase will be unable to bind and transcribe the genes .
Part B
The mRNA that is transcribed from the trpL region can alternately fold to form three possible stem-loop structures. Different stem-loop structures have varying effects on transcription.
Which stem-loop structure is responsible for terminating transcription?
the 1-2 stem loop
the 2-3 stem loop
the 3-4 stem loop
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The formation of stem loops of the trpL region is directly related to the continuation or termination of transcription. In trpL, region 1 is complementary to region 2, region 2 is complementary to region 3, region 3 is complementary to region 4.
The 3-4 stem loop is the termination stem loop, signaling transcription termination. Formation of the 3-4 stem loop halts RNA polymerase in the leader region before it reaches the structural genes. Region 4 is followed by a poly-uracil sequence; this configuration is the same as that in intrinsic termination of transcription in bacteria.
Part C
Under what conditions do the different stem-loop structures occur, and what effect do they have on transcription of the trp genes?
Sort items into the correct bins to associate the specific events with the conditions in which they are seen.
HelpReset
High tryptophan
Region 1 – Region 2 stem loopRegion 3 – Region 4 stem looptrp genes not transcribed
Low tryptophan
Region 2 – Region 3 stem looptrp genes transcribed
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If tryptophan is present, coupled transcription/translation proceeds quickly because there are plenty of tRNAs charged with tryptophan to bind to the UGG codons in the leader sequence. This pulls region 1 and region 2 into the ribosome complex quickly, leaving region 3 to bind to region 4, forming a termination stem-loop. Transcription is terminated and the trp genes are not expressed because tryptophan levels are high, and tryptophan is the final product of the enzymes encoded by the trp operon.
If tryptophan is low/absent, coupled transcription/translation stalls because tRNAs charged with tryptophan are scarce. Slow transcription leaves regions 2, 3 and 4 outside the ribosome complex. This allows region 2 to bind to region 3, preventing the formation of the 3-4 termination stem loop. Transcription continues and the trp genes are expressed.
Part D
Which of the following mutations would result in higher-than-normal expression of the trp genes in presence of tryptophan?
Select all that apply.
mutations in Region 3 that prevent 3-4 stem loop formation
mutations in Region 2 that prevents 2-3 stem loop formation
mutations in Region 1 that prevent 1-2 stem loop formation
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The 3-4 stem loop terminates transcription (termination stem loop). Mutations that reduce the percentage of complementary base pairs binding these two regions destablize the termination stem loop and increase the probability of transcription of the trp operon.
The 2-3 stem loop prevents termination of transcription. Mutations that reduce the percentage of complementary base pairs binding region 2 and region 3 destabilize the 2-3 stem loop. If region 3 cannot bind to region 2, it is more likely to bind to region 4, forming a termination stem loop. As a result, mutations in region 2 that prevent the 2-3 stem loop formation increase the probability of transcription termination.
The 1-2 stem loop causes a minor pause in coupled transcription/translation. Mutations that reduce the percentage of complementary base pairs binding region 1 and region 2 destabilize the 1-2 stem loop. If region 2 cannot bind to region 1, its more likely to bind to region 3, forming an anti-termination loop. This increases the probability that transcription will continue.
Part E
If the UGG codons in Region 1 of trpL were changed to AGG codons, what effect would this have on expression of the trp operon?
The operon would never be expressed.
The operon would be constitutively expressed.
The operon would be regulated by tryptophan, but attentuation would be triggered at higher levels of tryptophan than in wild type.
The operon would be regulated by tryptophan, but attentuation would be triggered at lower levels of tryptophan than in wild type.
The operon would be regulated by arginine levels; high levels of arginine would attenuate expression of trp genes.
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The two tryptophan codons (UGG) are responsible for sensing the level of tryptophan.
If tryptophan is high, transcription/translation continues quickly because tRNAs charged with tryptophan are readily available. Fast transcription/translation pulls region 1 and 2 into the ribosome quickly, leaving only region 3 and region 4 able to complementary base pair and form a stem-loop. A 3-4 stem loop causes termination of transcription (attenuation).
If tryptophan is low, transcription/translation stalls because tRNAs charged with tryptophan are scarce. If transcription/translation is slow, region 2 will be able to bind to region 3, forming an anti-termination loop and transcription continues.
If these codons were changed to arginine codons (AGG), the operon would no longer be sensitive to tryptophan levels, but would instead be sensitive to arginine levels. Quick transcription/translation would be dependent on readily available tRNAs charged with arginine.
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Phage are viruses that must infect bacteria to reproduce. Viral genomes do not contain all the genes needed to produce all proteins necessary for DNA replication, RNA transcription, and protein translation; instead the viruses force bacterial proteins and enzymes to express phage genes instead of bacterial genes.
Phage are very adaptable, and their behavior depends on whether conditions are favorable or unfavorable for making more phage.
When bacteria are plentiful (favorable conditions), viral DNA within the bacteria remains separate from the bacterial chromosome, and the virus forces the bacteria to make more virus. When the amount of virus becomes too much for the cell to hold, the membrane ruptures, and viruses are released to infect surrounding bacteria. This is known as the lytic cycle.
When bacteria are scarce (unfavorable conditions), the viral DNA within the bacteria becomes integrated into the bacterial chromosome. The bacteria survive and are still able to reproduce; viral DNA is transmitted to daughter cells when the bacteria reproduces. This is known as the lysogenic cycle.
Part A
As you just read, phage depend on bacteria to reproduce. Phage can be maintained by infecting a culture of bacteria with a strain of phage, and then plating the bacteria on an agar plate to grow.
Suppose that you isolate two mutant strains of phage – one strain cannot enter the lytic cycle and the other strain cannot enter the lysogenic cycle. You also grow wild-type phage as a control. You see three different phenotypes:
A plate with no missing bacteria- the lawn of bacteria is intact
A plate with spots/circles (called plaques) where there are absolutely no bacteria in the circles
A plate with spots/circles (called plaques) where there are some bacteria in the center of the circles.
Based on what you know about what happens in each cycle, match the plating results with the identity of the strain.
Drag each plate into the correct category.
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Wild-type strain
Strain with no lytic cycle
Strain with no lysogenic cycle
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Wild-type strains of phage are capable of either lysogenic or lytic cycle. When a phage is in the lytic cycle, it uses the bacterial machinery to produce more phage, ultimately rupturing the bacteria and releasing large amount of new virus. When bacteria are less numerous, wild-type phage go into the lysogenic cycle instead, where the phage integrates its DNA into the bacterial chromosome. This is basically a "dormant" phase, but the phage DNA is replicated as a bacteria undergoes cell division.
When you plate wild-type phage on a lawn of bacteria, the phage initially enter the lytic cycle. When bacteria are lysed, plaques appear on the plate where there are no bacteria. Wild-type plaques appear "turbid" with some bacteria in the center, because some phage are in the lysogenic cycle, especially as more and more bacteria get lysed and die.
In strains that cannot enter the lytic cycle, phage are unable to make new viruses and rupture the bacteria. As a result, when phage are plated on a lawn of bacteria, the bacteria are not lysed. The lawn of bacteria remains even and undisturbed.
In strains that cannot enter lysogeny, phage are unable to integrate into the bacterial chromosome, and go "dormant." Because these phage are locked into a lytic cycle, all bacteria infected by these mutant phage are lysed. The plaques formed by this strain are absolutely clear because all bacteria that are infected are lysed.
Part B
The decision to enter the lytic cycle or the lysogenic cycle is determined by environmental conditions, signal transduction, and the regulation of specific sets of genes — some that trigger the lytic cycle and some that trigger the lysogenic cycle.
Genome map of lambda phage
Sort the structural genes on the phage plasmid as to whether they encode proteins that promote the lysogenic cycle or the lytic cycle.
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Lysogenic cycle
integrase (int)DNA recombination proteins (bet, exo)
Lytic cycle
lysis proteins (S,R,R2)enzymes to cut DNA for packaginggenes for head proteinsgenes for tail proteins
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During lysogeny, the phage integrates its DNA into the bacterial chromosome. Genes that are active to promote lysogeny include integrase, the enzyme that integrates phage DNA into the bacterial chromosome, and DNA recombination proteins.
During the lytic cycle, the phage make more viruses and rupture the bacterial host cells to release the new viruses. Genes that promote the lytic cycle encode proteins that build viral particles (heads and tails), enzymes that cut DNA for packaging into viral particles, and enzymes that rupture the bacterial cell.
Part C
Genome map of lambda phage
Notice that the genes for lysogeny are clustered in an operon on the "left" side of the phage plasmid, while the genes for lysis are clustered in an operon on the "right" side of the plasmid.
At the beginning of each infection, transcription begins at two "early" promoters.
"Leftward" genes
Early promoter PL controls leftward transcription of early genes, beginning with the N gene. The N protein plays an anti-termination role in transcription from PL . Leftward transcription is also enhanced by binding of the cII/cIII complex to the PRE promoter, transcribing the cI gene. The cI protein enhances transcription of "leftward" genes and represses transcription of "rightward" genes. Genes under control of PL are involved in the integration of viral DNA into the bacterial chromosome.
"Rightward" genes
Early promoter PR controls rightward transcription of genes, beginning with cro. The cro protein enhances the transcription of the genes that are rightward of the PR , and prevents transcription of the leftward gene, cI. Genes under the control of PR are involved in the synthesis of head and tail proteins as well as proteins that lyse the host cell.
Based on this information, determine the phenotype of each phage mutation.
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No lysogenic cycle
cI-cII-int-N-PL-
No lytic cycle
cro-PR-
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cro promotes transcription of genes for the lytic cycle, producing viral head/tail proteins and enzymes for cutting and packaging viral DNA. cI promotes transcription of genes for the lysogenic cycle, producing integrase and DNA recombination proteins. The lytic and lysogenic cycles are mutually exclusive: cI represses transcription of cro, and cro represses transcription of cI.
Mutations in cro prevent the lytic cycle. Mutations in cI prevent lysogeny.
Mutations in cII also prevent lysogeny because the wild-type cII/cIII complex promotes transcription of cI.
The genes controlled by the PL promoter (leftward) promote the lysogenic cycle; mutations in PL prevent the lysogenic cycle.
The genes controlled by the PR promoter (rightward) promote the lytic cycle. Mutations in PR prevent the lytic cycle.
Mutations in the int gene (integrase) prevent integration of phage DNA into the bacterial chromosome, preventing lysogeny.
The N gene produces an anti-termination factor that facilitates transcription of the "leftward" genes and promotes lysogeny. Mutations in N prevent lysogeny and favor the lytic cycle.
Part D
Phage are able to exit lysogeny and return to the lytic cycle in response to extracellular signals; this process is called "induction." This switch from lysogeny to the lytic cycle is most often triggered by UV damage to DNA, and the resulting activation of the protein RecA.
Use what you know about the regulation of the lysogenic and lytic cycles to predict what would be true of a phage undergoing induction.
Drag the terms on the left to the blanks on the right to complete the sentences about a phage undergoing induction.
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cro
cI
increase
decrease
remain the same
1. The cI protein will be degraded by proteases in response to UV/cellular damage.
2. The transcription level of the cI gene will decrease, leading to the production of the lambda repressor.
3. The transcription level of the cro gene will increase, leading to production of the cro protein.
4. The transcription level of the xis gene will increase, leading to production of the enzyme excisionase.
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When bacterial DNA is damaged by UV radiation, the protein-destroying activity of RecA is activated. The protease targets the amino-terminal segment of the two cI proteins that make up the repressor dimer, cutting the protein into pieces. As a result, transcription of the cI gene decreases.
As cI protein levels drop, cro protein levels increase. This leads to increased expression of the xis gene, which produces the enzyme excisionase; this enzyme excises the phage DNA from its integrated location within the bacterial DNA. This triggers the lytic cycle.
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Bacterial genomes frequently contain groups of genes organized into operons.
Part A
What is the biological advantage of operons to bacteria?
Operons significantly increase the rate of DNA replication, thereby ensuring the required rate of bacteria reproduction.
Operons afford the organism the opportunity to simultaneously regulate transcription of multiple genes, whose products are active in the same process.
Operons provide redundancy in proofreading the most essential parts of genetic code, thereby reducing the susceptibility of the organism to lethal mutations.
Promoters and repressors as parts of operons serve as information integrating units, thereby affording more flexible response to the environmental changes.
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Part B
Identify the cis regulatory components you would expect to find in an operon.
Select the three correct regulatory components.
repressor
regulator
operator
inducer
attenuator
promoter
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Part C
How are the expressed genes of an operon usually arranged?
The expressed genes are located one by one after the regulatory region of the operon.
The expressed genes are located in two groups before and after the regulatory region of the operon.
The expressed genes are located in the regulatory region of the operon.
The expressed genes are located one by one before the regulatory region of the operon.
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Why is it essential that bacterial cells be able to regulate the expression of their genes?
Expression of some genes is required only under certain conditions.
Bacteria have very short reproduction cycles, which require accurate timing of gene expression.
Bacteria are subjected to invasions by bacteriophages, which often incorporate their genes into bacterial DNA.
Bacterial cells do not have isolated cellular compartments for ribosomes.
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Part B
What are the energetic and evolutionary advantages of regulated gene expression?
Select the two correct statements.
Regulated gene expression allows bacteria to adapt their metabolism to match changes in the environment.
Regulated gene expression prevents the expression of genes affected by mutations, which might be detrimental for bacteria.
Regulated gene expression allows the bacteria to select the most appropriate protein product from a particular gene.
Regulated gene expression prevents synthesis of unnecessary proteins, which might be in competition to the detriment of the organism.
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Part C
Is the expression of all bacterial genes subject to regulated expression?
yes
no
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Part D
Compare and contrast the difference between regulated gene expression and constitutive gene expression.
Constitutive gene expressions applies to the genes whose expression is always required, whereas regulated gene expression applies to the genes whose expression is required only under certain conditions.
Constitutive gene expressions applies to the genes whose expression is induced by protein factors, whereas regulated gene expression applies to the genes whose expression is induced by specific environmental substances.
Constitutive gene expression applies to the genes whose expression is required only under certain conditions, whereas regulated gene expression applies to the genes whose expression is always required.
Constitutive gene expressions applies to the genes whose expression is induced by specific environmental substances, whereas regulated gene expression applies to the genes whose expression is induced by protein factors.
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For the following lac operon partial diploids, determine whether the synthesis of lacZ mRNA is "constitutive," "inducible," or "uninducible," and indicate whether the merodiploid is lac+ or lac− (able or not able to utilize lactose).
Drag the appropriate items to their respective bins.
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constitutive
I+P+OCZ+Y+/I+P+O+Z−Y+
inducible
I−P+O+Z+Y+/I+P+O+Z+Y+I+P+O+Z−Y+/I+P−O+Z+Y+
uninducible
ISP+O+Z+Y+/I+P+O+Z+Y+I+P+O+Z+Y−/I+P+O+Z+Y−
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Part B
Drag the appropriate items to their respective bins.
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lac+
I−P+O+Z+Y+/I+P+O+Z+Y+I+P+OCZ+Y+/I+P+O+Z−Y+
lac−
ISP+O+Z+Y+/I+P+O+Z+Y+I+P+O+Z−Y+/I+P−O+Z+Y+I+P+O+Z+Y−/I+P+O+Z+Y−
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Choose which of the following genotypes for lac operon haploids have the following phenotypic characteristics:
Drag the appropriate items to their respective bins.
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I+P+OCZ−Y+I−P+O+Z−Y+I+P+OCZ+Y−
I+P−O+Z+Y+ISP+O+Z+Y+
I+P+O+Z−Y+
I+P+OCZ+Y+I−P+O+Z+Y+
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For the following lac operon haploid genotypes identify the mode of transcription of operon genes and indicate determine whether the strain is lac + or lac- .
Part A
I+P+O+Z+Y−
is lac + and the genes are constitutively transcribed.
is lac- and the genes are inducibly transcribed.
is lac+ and the genes are inducibly transcribed.
is lac+ and the genes are not transcribed.
is lac- and the genes are constitutively transcribed.
is lac- and the genes are not transcribed.
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Part B
I−P−OCZ−Y+
is lac + and the genes are constitutively transcribed.
is lac- and the genes are inducibly transcribed.
is lac + and the genes are inducibly transcribed.
is lac- and the genes are not transcribed.
is lac- and the genes are constitutively transcribed.
is lac + and the genes are not transcribed.
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Part C
I−P+O+Z+Y+
is lac- and the genes are not constitutively transcribed.
is lac + and the genes are inducibly transcribed.
is lac- and the genes are inducibly transcribed.
is lac+ and the genes are constitutively transcribed.
is lac + and the genes are not transcribed.
is lac- and the genes are not transcribed.
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Part D
I−P+OCZ+Y+
is lac + and the genes are inducibly transcribed.
is lac- and the genes are inducibly transcribed.
is lac- and the genes are constitutively transcribed.
is lac+ and the genes are constitutively transcribed.
is lac- and the genes are not transcribed.
is lac + and the genes are not transcribed.
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Part E
I+P+O+Z−Y+
is lac + and the genes are inducibly transcribed.
is lac- and the genes are inducibly transcribed.
is lac + and the genes are not transcribed.
is lac- and the genes are not transcribed.
is lac- and the genes are constitutively transcribed.
is lac + and the genes are constitutively transcribed.
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Part F
I+P−O+Z+Y+
is lac- and the genes are constitutively transcribed.
is lac- and the genes are inducibly transcribed.
is lac + and the genes are not transcribed.
is lac- and the genes are not transcribed.
is lac + and the genes are constitutively transcribed.
is lac + and the genes are inducibly transcribed.
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Part G
I+P−OCZ−Y+
is lac- and the genes are constitutively transcribed.
is lac- and the genes are not transcribed.
is lac+ and the genes are constitutively transcribed.
is lac + and the genes are inducibly transcribed.
is lac- and the genes are inducibly transcribed.
is lac + and the genes are not transcribed.
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In the lac operon, what are the likely effects on operon gene transcription of the mutations identified below?
Part A
Mutation of consensus sequence in the lac promoter.
Transcription is blocked.
Transcription is constitutive.
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Part B
Mutation of the repressor binding site on the operator sequence.
Transcription is blocked.
Transcription is constitutive.
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Part C
Mutation of the lac I gene affecting the allosteric site of the protein.
Transcription is blocked.
Transcription is constitutive.
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Part D
Mutation of the lac I gene affecting the DNA-binding site of the protein.
Transcription is blocked.
Transcription is constitutive.
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Part E
Mutation of the CAP binding site of the lac promoter.
Transcription is blocked.
Transcription is constitutive.
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Attenuation of trp operon transcription is controlled by the formation of stem-loop structures in mRNA. The attenuation function can be disrupted by mutations that alter the sequence of repeat DNA regions 1 to 4 and prevent the formation of mRNA stem loops.
Determine the likely effects on attenuation of each of the following mutations under the conditions specified.
Drag the appropriate items to their respective bins.
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transcription does not occur
bdf
transcription occurs
acegh
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How would a cap− mutation that produces an inactive CAP protein affect transcriptional control of the lac operon?
lac operon transcription in the presence of lactose would be enhanced.
lac operon transcription would be enhanced and become unregulated.
lac operon transcription in the absence of lactose would be enhanced.
lac operon transcription would be minimal.
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The CAP binding site in the lac promoter is the location of positive regulation of gene expression for the operon.
Part A
Identify what binds at this site to produce positive regulation.
polycistronic mRNA
CAP-cAMP complex
CAP-glucose complex
RNA polymerase
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Part B
Under what circumstances does the binding occur?
When lactose is available.
When glucose is available.
When lactose is not available.
When glucose is not available.
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Part C
How does the binding exert a positive effect?
The CAP-cAMP complex prevents the binding of the repressor protein.
The CAP-cAMP complex alters the structure of the promoter and allows RNA polymerase to bind to it.
The CAP-cAMP complex triggers the formation of the DNA loop, which prevents binding of the repressor protein.
The CAP-cAMP complex unbends the DNA loop, thereby allowing binding of RNA polymerase.
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For bacteria that are F+, Hfr, F′, and F−, answer the following:
Part A
Identify the state of the F factor.
F−
F′
F+
Hfr
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F+ F plasmid is present F- not present Hfr – F plasmid is part of chromosome F’ – F plas is present and contain a portion of the chromosome
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Which of these cells are donors?
Select all that apply.
F−
Hfr
F′
F+
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Part C
Which are the recipients?
Select all that apply.
Hfr
F−
F+
F′
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Part D
Which of these cells can convert exconjugants to a donor state?
Select all that apply.
F+
F′
F−
Hfr
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Part E
Which of these donors can transfer a donor gene to exconjugants?
Select all that apply.
F+
F′
Hfr
F−
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Part F
Which of the following is a donor cell that will not change the genotype of the recipient cell other than make it a donor cell?
F′
Hfr
F−
F+
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Part G
Which of the following is a donor cell that will change the genotype of the recipient cell in addition to making it a donor cell?
F′
F+
F−
Hfr
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Part H
Which of the following is a donor cell that will change the genotype of the recipient cell but will not make it a donor cell?
F−
Hfr
F+
F′
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Part I
Choose the description of a "partial diploid" and how it originates.
A partial diploid is a bacterium that has two copies of all genes. It is formed when a recipient cell receives an F′ plasmid from a donor.
A partial diploid is a bacterium that has two copies of some genes but only one copy of most genes. It is formed when a recipient cell receives an Hfr plasmid from a donor.
A partial diploid is a bacterium that has two copies of some genes but only one copy of most genes. It is formed when a recipient cell receives an F′ plasmid from a donor.
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Part A
What is lateral gene transfer (LGT)?
It is the transfer of DNA from parent to offspring by sexual reproduction.
It is the transfer of DNA from parent to offspring by either sexual or asexual reproduction.
It is the transfer of DNA between organisms by a manner independent of asexual or sexual reproduction.
It is the transfer of DNA from parent to offspring by asexual reproduction.
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Part B
How might it take place between two bacterial cells?
Select the three correct answers.
lysogeny
transduction
conjugation
transformation
translocation
transfection
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A 2013 CDC report identified the practice of routinely adding antibiotic compounds to animal feed as a major culprit in the rapid increase in the number of antibiotic-resistant strains. Agricultural practice in recent decades has encouraged the addition of antibiotics to the animal feed to promote growth rather than to treat disease.
Part A
Choose the correct speculations about the process by which feeding antibiotics to animals such as cattle might lead to an increase in the number of antibiotic-resistant strains of bacteria.
Select the two correct answers.
Lateral gene transfer of the antibiotic resistance genes leads to an increase in the number of antibiotic resistant bacterial species.
The presence of antibiotics favors the growth of antibiotic resistant bacteria.
Lateral gene transfer could transmit antibiotic resistance genes into the animal genome, enablling animals to become resistant to bacteria.
The use of antibiotics in animal feed could lead to the accumulation of toxic compounds in agricultural products
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Part B
How might the increase in antibiotic-resistant strains of bacteria in cattle be a threat to human health?
Select the two correct answers.
The favored growth of antibiotic resistant bacteria decreases biological diversity of human and animal symbiotic microorganisms.
Bacteria that are pathogenic to humans could acquire antibiotic resistance from the cattle bacteria by the lateral transfer of antibiotic resistance genes.
Some bacteria in or on cattle are contaminants on food products and can infect consumers of that food. If these bacteria are resistant to antibiotics, it would complicate the treatment of those infections.
Lateral gene transfer could transmit antibiotic resistance genes from bacteria into the human genome, causing mutations and genetic diseases.
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