7.28 Spring ’01
Problem Set #2 Answers
1a) You are studying the function of RecA in vitro along with a lab partner. You run several reactions to examine the requirements for RecA activity. For each reaction below give the expected products after RecA has been removed by phenol extraction.
Regions of homology are shown as thick lines.
1b) After you carry out these reactions, your partner runs a non-denaturing acryl amide gel to analyze the products. For each reaction, she runs 3 lanes. Lane (a) is a control reaction run without RecA. Lane (b) is the complete reaction loaded directly into the gel.
Lane (c) is the complete reaction that has been phenol-extracted to remove protein prior to loading on the gel. She exposes the resulting 15-lane gel to X-ray film. She has to

7.28 Spring ’01
Problem Set #2 Answers leave at this point, and after developing the film, you sit down to write up the lab report.
Unfortunately, you realize that you do not know what order she loaded the reactions in.
The film is as follows:
Which reaction numbers (1-5) correspond to the reactions on the film (U-Z)? Explain the differences between lanes a, b, and c for each reaction.
Reaction U = 3, W = 5, X = 2, and Y or Z = 1 or 4 (same bands on gel)
1 & 4 a: These have the same size circular ssDNA, apparently runs at 3 kb b: recA binds both DNA molecules, get a band for 2 DNA's + protein shift (8kb) c: Components missing indicated in (a), so no new stable DNA product is formed (3 kb)
2 a: 4 kb linear hot dsDNA b and c: No ssDNA, recA can't bind, same size observed (4 kb)
3 a: Same hot DNA molecule as 1 and 4 (3 kb) b: recA binds and forms new DNA product combining the 2 DNA's, remains bound (8 kb) c: Final product stable without recA combines two substrate DNA's (6 kb)
5 a: Initial hot DNA is linear ssDNA, apparently runs faster than circular(2kb) b: recA binds and forms new DNA product combining the 2 DNA's, remains bound (8 kb) c: Final hot product is just 4 kb dsDNA.

7.28 Spring ’01
Problem Set #2 Answers
2a) Some transposable elements in E. coli, for example Tn10, transpose at a high frequency immediately after a replication fork has passed. This induction of transposition by replication is a large effect, such that most transposition events occur in a short period of time right after replication. What mechanism could Tn10 use to couple transposition to passage of the replication fork? Propose a simple in vivo experiment to test if this mechanism is true.
Transposition could be coupled to replication by the same system of dam - dependent
DNA methylation which couples mismatch repair to replication. In this case, transposition would be induced in transposons containing hemi-methylated sequences.
[Actually, the Tn10 transposase does not bind fully methylated DNA but does bind hemi-methylated or unmethylated DNA, thus activating transposition after passage of the replication fork.]
An in vivo experiment might involve comparing the frequency of transposition in wt, dam - and dam methylase overexpressing strains. Tn10 should transpose more frequently than normal in dam - cells (where the sites should always be unmethylated) and less frequently than normal in dam overexpressors (where the sites are methylated more rapidly after replication) . Transposition frequency could be measured by
Southern blots with a Tn10 probe as described in part b).
2b) You have a strain of E. coli (strain 708A) that has a single copy of Tn10 in the genome, integrated into the hisB gene. You isolate a derivative of this strain (which you name 708A+) in which the Tn10 has transposed and caused a mutation in the gene required for growth in the absence of leucine. Below is a Southern blot of DNA from a strain 708, which is related to 708A but lacks Tn10, 708A and 708A+. The membrane has been probed with 3 different radioactive probes: (1) to the hisB gene, (2) to the leu gene and (3) to Tn10 sequences. Given the pattern shown for the other strains, draw the bands you would expect to see for the 708A+ strain on each of the films (below):

7.28 Spring ’01
Problem Set #2 Answers
2c) You declare, in group meeting, that Tn10 is a replicative transposon. However, your professor informs you that based only on the experiment performed in part b), you cannot tell if Tn10 is a replicative or non-replicative transposon. Why not? Briefly explain you r professor’s reasoning.
After the replication fork passes through the Tn10 sequence, there are temporarily two copies of Tn10 in the cell, on the parent and on the daughter chromosome, in the same location (the His gene).
(1) If one copy transposes by a replicative mechanism, the resulting progeny cell will carry two Tn10 transposons at two different sites.
(2) If one copy of the transposon excises by a
“ cut and paste
”
, non-replicative process, it will leave a double-stranded break behind. This break will be repaired using the homologous region on the opposite chromosome, which contains the newly replicated copy of the Tn10 sequence. Thus the transposon will also be found both at the old location and at the new target site in the progeny cell.
3a) You are studying the problem of antibiotic resistance spreading throughout a hospital bacterial population. You have reason to suspect that some previously uncharacterized phage are causing the problem. These phage carry antibiotic resistance genes, thereby causing the infected cells to become resistant when the phage integrates into the cell’s chromosomal DNA.
As a first step in trying to prevent further spread of these phage, you characterize their mechanism of integration. To try to gain insight into the mechanisms they use, you infect two of the phage into several different E. coli strains with mutations in genes effecting
DNA processing. Below is a table of the results.
E. coli strain
Wild-type
Phage A
Integrates
No integration
Phage B
Integrates
Integrates polA- (gene for DNA polymerase I) lig- (gene for DNA ligase) recB- (gene for RecB)
No integration
Integrates
Integrates
Integrates
Based on these data, which type of recombination: homologous, transposition, or sitespecific, is most likely to be used for integration by Phage A? Explain your answer.
Transposition. Phage A does not require recB, so it is not homologous recombination. However, it does require polymerase and ligase, which are necessary for transposition but not for site specific recombination.

7.28 Spring ’01
Problem Set #2 Answers
3b) Which type of recombination: homologous, transposition, or site-specific, is most likely to be used for integration by Phage B? Explain your answer.
Site Specific Recombination. Phage B does not require recB either, and so it does not use homologous recombination. Site specific recombination does not require any synthesis of new DNA, and so does not require polymerase. The recombinase joins the broken strands, so ligase is not necessary either.
3c) You decide that this method of determining integration mechanism is rather labor intensive and indirect. Suggest a more direct assay that should allow you to determine whether the phage you isolate use the integration pathway used by phage A or that used by phage B. You have at your disposal a DNA probe to the antibiotic resistance genes carried by both types of phage. Explain the assay, and what you would expect to see with phage of each class.
One simple assay would be to do a genomic southern blot on E. coli strains infected by each phage. Isolate the DNA from E.coli infected with each phage, digest it with a restriction enzyme, and run a Southern blot (run on a nondenaturing gel, denature the
DNA, transfer to a membrane, incubate with a probe, and expose to film). E. coli infected by phage B should always give a band of the same size when probed with phage B DNA, since site-specific recombination always causes integration into the same place. E. coli infected with phage A will give lots of different bands when probed with phage A DNA, since transposons transpose into random spots in the genome.
3d) To stop the spread of these antibiotic resistance phage you want to isolate an inhibitor of the integration process. Based on studies with other enzyme inhibitors, you reason that the best target for the inhibitor would be the active site of the recombination enzyme responsible for integration. To identify the active site in the recombinase for
Phage B you use a consensus sequence approach. First you identify a region from the
Phage B genome that is weakly homologous to the phage λ integrase gene. Then you sequence this region from five phage that are all close relatives of Phage B. Below is a region of amino acid sequence (1 letter code) from these six B-like phage.
Phage B IFGYARVSYSQE
B-001 LFGYARVSYSQE
B-002 LFGYARVSTSQQ
B-003 IAGWIRVSTFDQ
B-004 IAGYIRVSTFDQ
B-005 IAGYIRVSTFDQ
Based on this alignment data, which amino acid in this region is most likely to form part of the active site? Give two reasons to justify your answer.
The conserved serine is most likely part of the active site. It is present in every phage
B relative, and so it must be important. Also, serine contains an OH group which is necessary to attack the DNA phosphate and form a covalent DNA-protein intermediate.

7.28 Spring ’01
Problem Set #2 Answers
4a) A colleague is studying a dominantly inherited disease and comes across a patient with no history of the disease in his family. After ascertaining that the patient was indeed correct as to the identity of his parents, your colleague contacted you, an expert in transposons. You sequence the disease gene from the patient, and find a large insertion in the gene.
What sequence characteristics might you expect this insertion to have if it is a transposon?
Short direct repeats (target site duplication), inverted repeats, genes with homology to transposases.
4b) You decide to test whether you have indeed found a new transposon by checking whether the transposon can transfer between different DNAs in a bacterial cell.
(Assume that the human elements can transpose in bacteria.) Your UROP inserts an
Amp R gene into the suspected transposon sequence and clones it into a bacterial plasmid containing a Tet R gene. The UROP transforms this plasmid into a bacterial strain carrying an F’ plasmid with a Kan R gene. Then she grows these donor cells up, mates them to an Amp S Kan S Tet S recipient strain, kills the donor cells, and plates the surviving bacteria onto various antibiotic plates to test the phenotype of the recipient cells. 0.05% of the Kan R cells she recovers are Kan R Amp R Tet S , while the remaining
99.5% are Kan R Amp S Tet S . What can you conclude from this data?
Cut and paste transposon, transposing at a low rate.
4c) You accuse your UROP of messing up the assay, as you expected a higher percentage of Amp R cells. Offended, your UROP repeats the experiment in duplicate, this time transforming into a number of bacterial strains with different F’ plasmids, all containing Kan R genes. This time, she finds that one of the strains results in recipient cells in which 40% of the Kan R genes are Kan R Amp R . You tell her to sequence an F’ plasmid that has gained a transposon insertion and discover that it has a very high GC content. Given that regions of the human genome that contain coding sequences have a much higher GC content than regions that don’t, why is it strange that this transposon appears to prefer to integrate into GC-rich DNA? (Why might most transposons integrate into non-coding regions?)
Most transposons have a preference for noncoding DNA, presumably so that they can be retained in the genome without accidentally killing the host.
4d) What’s an explanation for why this transposon might prefer to integrate into actively transcribed regions with open chromatin structures? (What kind of transposon might it be?)
Perhaps it is a retrotransposon, and needs to be transcribed into RNA before transposing. Retrotransposons usually have their own promoters, but they could presumably transpose more frequently if they were in active regions of the genome with open chromatin structures.

7.28 Spring ’01
Problem Set #2 Answers
4e) How does this hypothesis change your conclusion about the results in part b?
Retrotransposons transpose replicatively (they leave a copy behind in the genome from which the RNA was transcribed), but they do not form a cointegrant. Therefore, in the bacterial assay they would not bring the entire plasmid containing the TetR gene with them into the recipient cell. What looked like cut and paste transposition in your assay was actually replicative transposition.
4f) How might this transposon achieve this preference for GC-rich DNA?
The transposase could specifically recognize a short GC rich sequence in the target
DNA.
5) You are interested in studying the regulation of genes in mice important during early development. Your professor asks you to continue the work of a former graduate student. This student was studying two novel genes that he named Hip and Hop . Both
Hip and Hop are necessary in early embryos but not afterwards. The graduate student was able to generate the following restriction map for Hip and Hop :
You decide to perform a Southern blot on Hip and Hop . Knowing that these genes are important during early development, you isolate genomic DNA from early embryos and also from young adults. You digest with HindIII and find the following results:

7.28 Spring ’01
Problem Set #2 Answers
5a) You are surprised by the results. How can you explain the result seen when probing your Southern blot with Hop? Given your explanation, what can you conclude about the regulation of Hop during mouse development? Is Hop necessary in adult mice, or only in embryonic development?
The HOP gene is important only during embryonic development. The lack of a band when probing adult DNA indicates that the HOP gene is no longer present in adult mice. It is likely that site specific recombination is used to excise this gene after it is no longer necessary for development. This type of regulation is only possible in cells that are terminally differentiated.
5b) You would like to further investigate regulation of Hip. In an attempt to do so, you digest the DNA you isolated above with two enzymes, HindIII and EcoRI. This time, you see the following results:
Given this new data, how do you think Hip is regulated during mouse development?
Provide an explanation from the data above.
In this case, site specific recombination is also being used to regulate the gene HIP.
However, rather than an excision event, an inversion has occurred. This can be seen by the change in restriction fragment sizes on the Southern blot. The asymmetric location of the EcoRI site yields two distinct patterns before and after the inversion event.
5c) Using the same strains of mice you are studying, a fellow labmate bred mutant mice that fail to develop properly. One of her mutants shows multiple embryonic characteristics, even when the mouse has reached an adult stage. She asks you to perform your Southern blot characterization on her mutant mice to see if either Hip or
Hop is mutant in her strain. You are happy to do so and digest the genomic DNA from her mice with HindIII and EcoRI. You find the following results:

7.28 Spring ’01
Problem Set #2 Answers
Given the data above, which gene do you think is mutated in your labmate's mice? How do you think the regulation of this gene is mutated to give the mutant characteristics described above?
The HIP gene appears to be mutated in your labmate
’ s strains. The gene is normally regulated via an inversion event. The restriction pattern on the gel above suggests that some cells fail to undergo this inversion event. The middle two bands are generated from cells where HIP is correctly regulated, whereas the outer two bands are generated from the cells who fail to invert the HIP gene.
6) You have developed a transcription assay that recapitulates the in vivo regulation of the promoter for the RNA Pol II-transcribed Ratchet gene. By fractionating this extract, you have purified and cloned two specific transcription factors that, in addition to the general RNA Pol II transcription factors, are required for correct regulation of the promoter.
As a first step to understand the function of these factors you make a series of deletion mutants of the Ratchet promoter and test their effect on transcription in the presence and absence of the Click and Clack factors. You find the following results.
6a) Based on this data, what would you predict the function of elements 3, 6, and 7 are?

7.28 Spring ’01
Problem Set #2 Answers
Element 3 is most likely an activator binding site. Since this element is required for transcriptional activation by Click and Clack, it is probably the binding site for one of these proteins.
Elements 6 and 7 are probably part of the core promoter element, possibly the TATA box, and the INR sequence. This can be determined from their locations relative to the start of transcription and by the fact that eliminating these sequences leads to the loss of basal levels of transcription.
To further characterize the factors that you have identified, you perform a gel shift assay to look at the ability of the different factors to bind to the Ratchet promoter. You obtain the following results.
6b) Based on this data, what can you conclude concerning the function of Click and
Clack? How is this finding consistent with the data obtained in part 3a?
This shows that Clack is a DNA binding protein, and that Click depends on Clack for association with DNA.
This is consistent with the fact that there is only one activator binding site, as determined by your analysis in part 3a. Since Click depends on Clack for DNA association, the only binding site required is the one that Clack binds to. If both proteins bound DNA, one would probably observe a binding site for each protein.
6c) How would you determine the region of the Clack factor required for activation of the
Ratchet promoter?
To map the Clack activation domain:
1. Fuse Clack to the DNA binding domain of a known DNA binding protein (i.e. LexA). This allows you to map activation domains independent of effects on DNA binding.
2. Make a series of deletions of the end of Clack, and fuse each to the LexA DNA binding domain
3. Test the ability of each fusion to activate transcription at a test promoter. This promoter must contain LexA binding sites in the region upstream of the trancriptional start site and a reporter gene
(like LacZ) to measure transcription.
This technique allows you to determine which regions of the Clack protein are necessary and sufficient for transcriptional activation.
** IMPORTANT: If you answered the question by explaining how to map the DNA binding domain of Clack, this would lose points on an exam. Many proteins have *both* DNA binding domains *and* transcriptional activation domains. Be sure you read and answer the question that is asked.

7.28 Spring ’01
Problem Set #2 Answers
Throughout your analysis of the Ratchet promoter you have been frustrated by the lack of effect of a third transcription factor, Clock, that is known to activate this promoter in vivo. All of your previous assays for transcription were performed using S1 assays and a ssDNA probe of 300 bases in length that overlaps the ratchet start site by 100 bases.
Thinking you may have missed something, you analyze the products of a set of transcription reactions performed in the presence of radio-labeled UTP on a denaturing agarose gel. To cause termination of RNA Pol II artificially, you cut the template DNA 2 kb downstream of the promoter with a restriction enzyme. You obtain the following results (hint: promoter clearance is complete after 100 bases of RNA synthesis).
6d) What aspect of RNA Pol II function is effected by addition of Clock? Why was this effect missed by the S1 assay and detected by the transcription run off assay?
Clock serves as a processivity factor for RNA Pol II. Notice that when Clock is added, the transcripts increase in length from ~300 bp to 2000 bp. This is not an effect on promoter clearance, as promoter clearance (and thus initiation of transcription) is complete by 100 bp. This is an effect on elongation of the new transcript.
Another plausible answer is that Clock acts as a negative regulator of transcription termination. Thus, when Clock is present, transcripts are longer than in the absence of
Clock.

7.28 Spring ’01
Problem Set #2 Answers
In the S1 assay, a labeled 200 bp DNA probe that is hybridized to the RNA transcript is your readout for transcription. A band corresponding to this 200 bp probe will be observed on the autoradiogram whether the RNA that it is hybridized to is 200 bp long or 2000 bp long. Thus, in an S1 protection assay, no RNA length information is provided, and the results with and without Clock protein would be the same.
In the run-off transcription assay, one measures incorporation of rNTPs. rNTPs will be incorporated from the start site until termination occurs (which will depend on the intrinsic processivity of RNA Pol II without Clock, and on the RNA Pol II
“ running off
”
the DNA with Clock). This assay allows you to determine RNA length information, and thus will allow you to see different products with and without the
Clock protein.
7) Your lab studies a specialized strain of E. coli that contains unusual genes that aid in its growth under conditions of low nutrients. You are further examining a gene that is induced under starvation conditions, the funE gene. You have used S1 nuclease protection assays to determine the start site of transcription for the funE mRNA. The sequence upstream of the start site includes -10 and -35 regions that differ substantially from the standard consensus sequences for E. coli. Using a convenient restriction site, you fuse the funE promoter with a β-galactosidase reporter gene in a plasmid to allow detailed study of expression from this promoter.
7a) You mutagenize this strain and transform it with your reporter plasmid to find mutants that affect expression of β-galactosidase. One mutant, the craZ mutant, has a complete loss of induction of β-gal activity under starvation conditions. You examine expression of other genes that your lab studies that are induced in starvation conditions, and find that these genes are also no longer induced in the craZ mutant. Given that these genes have a similar -35 region to the funE gene, propose an explanation for the nature of the craZ mutant.
Recall that the -35 region is primarily involved in the specificity of RNA Polymerase holoenzyme binding. The craZ mutation is likely to be in a specialized sigma factor
(σ) that recognizes the unusual -35 regions of the funE gene and related genes.
Expression or activity of this σ subunit must be induced under conditions of starvation.

7.28 Spring ’01
Problem Set #2 Answers
7b) galactosidase fusion in wild type cells. You find that adding calcium significantly
-gal signal. To examine this effect at the transcriptional level, you first
- perform incorporation assays in wildtype cellular extracts, adding the reporter plasmid and [H3]UTP. You do the assay in the presence and absence of calcium, then run the reactions on a gel and expose the gel to film. The resulting film is shown below:
This result suggests a difference in the rate limiting step of transcriptional initiation in the presence and absence of calcium. What is the changed rate limiting step?
For the incorporation assay in the absence of calcium, you observe that the majority of products are short (~5 bp) fragments. This indicates that ternary complexes have formed but promoter clearance has not occurred. For the incorporation assay in the presence of calcium, you observe that the majority of products are large (>300 bp), presumably full-length transcripts. Therefore in the absence of calcium, promoter clearance must be rate limiting, but in the presence of calcium this step is made more efficient and no longer limits transcription.
7c) Using biochemical fractionation and a gel shift assay, your labmate isolates a protein that binds upstream of the funE -35 region. He determines the sequence of the gene and notes that the protein has two domains, a DNA binding domain and a calcium binding domain. The two of you conclude that this protein is likely to be mediating the effects you studied in (b). Suggest TWO possible mechanisms for how this factor could result in Ca++- dependent transcription. What genetic experiment could you perform to distinguish between them?
The protein could either be acting as a repressor of transcription that is inactivated by calcium binding, or as an activator of transcription that is activated by calcium binding.
To distinguish between these possibilities, you could knockout the gene. If the protein is a repressor, the knockout will be calcium insensitive and constitutively express
FunE at a high level (the level caused by calcium in wild type). If the protein is an activator, the knockout will be calcium insensitive and express FunE at a lower level
(the level without calcium in wild type).

7.28 Spring ’01
Problem Set #2 Answers
7d) To further study transcription from this promoter, you mutagenize the promoter region of your reporter plasmid and look for effects on
β
-gal expression in the presence of calcium. You obtain a mutation in the -10 region, called silE, that increases β -gal expression in the presence of calcium. However, this mutation has no effect on β -gal expression in the absence of calcium.
You use KMnO4 to examine DNA unwinding in the transcription reactions performed in vitro with a linear, end-labelled, double-stranded DNA fragment carrying the funE promoter / bgal fusion. The substrate DNA and resulting film are shown below.
What do these results indicate about the nature of the silE mutation? Why does the silE mutation only affect the expression of
β
-gal in the presence of calcium (Hint: consider the data from part (b) as well)?
The silE mutation increases the extent of DNA unwinding and open complex formation, perhaps by bringing the unusual funE -10 sequence closer to the E. coli consensus -10 region.
Expression of β-gal from the fusion gene is determined by the rate limiting step of initiation. In the absence of calcium, promoter clearance is rate limiting. Therefore increasing DNA unwinding and open complex formation in the absence of calcium will have no effect on expression of β-gal. In the presence of calcium, promoter clearance is no longer rate limiting. Open complex formation appears to be rate limiting under these conditions, so that the silE mutation which increases the extent of
DNA unwinding (and therefore open complex formation) also increases expression of
β-gal.

7.28 Spring ’01
Problem Set #2 Answers
8) You are studying a pair of transcription factors that regulate the Pumped promoter
(involved in the production of a muscle-specific protein). You have purified both factors and test their ability to regulate the transcription of the Pumped promoter either without or with mutations in several key elements of the promoter.
8a) Based on the transcription data above propose roles for the A and B elements of the
Pumped promoter. Be brief.
The A element is required for activated transcription only. It is likely to contain a sequence (UAS) recognized by a sequence-specific DNA-binding transcriptional activator.
The B element is required for basal and activated transcription. It is likely to contain the TATA-box and/or the initiator elements recognized by the general transcription factors.
8b) You are intrigued by the results of the transcription assay in which both PTF1 and
PTF2 were added at the same time. You hypothesize that PTF2 might inhibit PTF1 function by altering its DNA binding properties. To address this you perform gel shift experiments using the the Pumped promoter.
Previous experiments argued that both PTF1 and PTF2 are dimers when bound to DNA.
With that in mind, how can you explain the results of this gel shift assay?
PTF2 is a larger protein than PTF1, and the DNA-protein complex containing a PTF2 dimer runs more slowly than the one containing a PTF1 dimer. When both PTF1 and
PTF2 are present, a heterodimer containing one molecule of PTF1 and one molecule of PTF2 is formed. This DNA-protein complex runs at an intermediate size between the two homodimers.

7.28 Spring ’01
Problem Set #2 Answers
8c) Having mapped the activation domain of PTF1, you now ask what part of PTF2 is necessary to inhibit PTF1 function. You identify three regions as important. Two of these regions are involved in the DNA binding and dimerization functions of PTF2. The third is a separate domain that is not required for either DNA binding or dimerization. Propose a hypothesis for the function of this third domain and describe a model for the transcription results in part (a).
The third domain is likely responsible for inhibiting basal transcription in conjunction with the PTF1 activation domain (perhaps via a physical proteinprotein interaction).
(Note that the PTF2 homodimer does not inhibit basal transcription)
MODEL: Normally, both PTF1 and PTF2 are present in equal concentrations in the cell. They form predominantly heterodimers at these concentrations, and these heterodimers bind to the A-box at the promoter. The PTF2 inhibition and PTF1 activation domains interact to inhibit even basal transcription at the Pumped promoter.
Under activating conditions, the ratio of PTF1 to PTF2 in the cell changes so that
PTF1 predominates (e.g. either more PTF1 is produced, less PTF2 is produced, or
PTF2 is degraded). This forces the formation of PTF1 homodimers (rather than the heterodimers), which bind to the A-box and activate transcription at the Pumped promoter through the activation domains.
[PTF1] = [PTF2] OFF
[PTF1] >> [PTF2] ON

7.28 Spring ’01
Problem Set #2 Answers
9) A sketch of the lac promoter region is shown below. Please fill in the information missing on the chart below.
Growth
Conditions
Glucose +
IPTG -
Glucose +
IPTG + activity
-
Lac Z Explanation, if less than maximal activity repressed by LacI
Proteins bound to promoter and the position of each.
LacI bound to operator
Glucose -
IPTG -
-
- not activated
CAPcAMP by
CAP-cAMP closed complex promotes formation and LacI promotes closed complex formation but blocks open complex formation.
~ nothing
CAP-cAMP bound to CAP site, RNAP-sigma70 bound to -35 and -10, and
LacI bound to operator.
Glucose -
IPTG +
+ go!
CAP-cAMP binding at
CAP sites activates promoter, RNAP sigma 70 will bind to and rapidly clear promoter region.
cya-
IPTG +
-
No cAMP to bind to CAP, so CAP will not bind CAP site and promoter is not activated.
~ nothing cya- = a mutant in the adenylate cyclase gene.
CAP-cAMP activation of the lac promoter increases transcription ~50 fold, so assume that LacZ activity refers to this highly activated activity (activated by CAP and not repressed by LacI).

7.28 Spring ’01
Problem Set #2 Answers
10) You are studying a newly identified bacterial operon ygo. It appears to be transcriptionally regulated by YGP and a small molecule product, YGO, of the operon. It is transcribed by the RNAP plus sigma 70. a) The results of genetic analysis of promoter regulation using a promoter fusion to lacZ are shown below. Interpret each set of results by filling in the molecular defect column, and whether the mutation is cis or trans. Assume in all cases that the reporter lacZ is not mutated.
Note mutants 1 and 4 have the same phenotypes, but there are two possible mutations that could cause this.

7.28 Spring ’01
Problem Set #2 Answers b) You isolate a sixth mutation that has a phenotype like mutant 3 except that if you overexpress the mutant 6 protein (it is a trans- mutation) repression is now observed in the presence of YGO. How can you explain this finding? What assay could you do to test your hypothesis and what result would you expect.
YGPwt when bound by YGO could bind the operator in a cooperative manner and mutation 6 abolishes this cooperative binding. Overexpression could then suppress the effect of loss of cooperativity and restore repression. To test this hypothesis use a gel shift assay of the ygo operon promoter with increasing concentrations of YGPwt or mutant 6 in the presence of YGO. The result you expect if YGPwt binds the operator cooperatively when activated by YGO,and mutant 6 does not, is shown below.