Compare and contrast operon vs gene fusions.

1. Draw a simple diagram showing for each type of fusion: the transcription and translation start and stop sites; the mRNA transcript expected; the protein products expected.

   ANSWER: See figure below.

   

2. For each type of fusion, indicate how frequently you would obtain insertions that express the reporter gene product if a transposon derivative (such as Mud) was randomly inserted into a gene.

   ANSWER:

   • Operon = 1/2 or 50%
   • Gene = 1/2 x 1/3 = 1/6 or 16.7%

3. List 3 different uses of operon or gene fusions.

   ANSWER: There are many possible answers. A few examples include:

   • Study gene expression in vivo
   • Purify fusion proteins
   • Study cell localization
   • In vivo cloning
   • Portable region of homology for constructing duplications, Hfrs, etc
   • Use as linked marker with selectable phenotype
   • To isolate mutations in genes

Kenyon and Walker used Mud(lac, Ap) operon fusions to identify genes whose expression is increased by DNA damage (Kenyon, C., and G. Walker. 1980. DNA-damaging agents stimulate gene expression at specific loci in Escherichia coli. Proc. Natl. Acad. Sci. USA 77: 2819-2823).

1. What was the selection they used?

2. How did they screen for din fusions?

3. How did they confirm that the din fusions were controlled by the SOS response?

4. If the polA gene was regulated by the SOS response would they have found polA::Mud (lac,Ap) fusions? Why or why not?

The iclR gene encodes a repressor that turns off transcription of the aceA gene in the absence of acetate. Given a Salmonella typhimurium strain with an aceA::MudJ operon fusion that expresses LacZ under control of the aceA promoter, how would you do localized mutagenesis to obtain iclR mutants? [Draw a diagram showing how you would do the experiment, including the donor and recipient used and how you would identify the desired mutants. When looking for recombinants indicate whether the approach is a selection or a screen.]

ANSWER: The desired mutants would be expected to express LacZ at high levels in the presence or absence of acetate ("Lac^c"). Such mutants could be obtained by localized mutagenesis as shown in the diagram below.

Roof and Roth (1988. J. Bacteriol. 170: 3855-3863) isolated a large number of MudA (lac⁺, Amp) and MudJ(lac⁺, Kan) operon fusions in the ethanolamine utilization operon (eut) in S. typhimurium. They also isolated a large number of Tn10 (Tet^R) insertion mutations in the eut operon. Using the Tn10 insertions and Mud operon fusions in the eut genes, they constructed double mutants that carried one of the Tn10 insertions and one of the MudJ(lac, Kan) insertions. The results are shown below.

^a The numbers shown are allele numbers.

^b + indicates the mutant expresses B-galactosidase and - indicates the mutant does not express B-galactosidase.

Based upon these results, draw a simple diagram showing the order of the Tn10 insertions relative to the MudJ insertions and indicate the direction of transcription. Briefly explain your rationale for this order and orientation.

ANSWER: Based upon the rationale that an upstream Tn10 insertion will be polar on a downstream Mud insertion. Thus, the order of each of the mutations (with transcription from left to right) is shown below:

Operon and gene fusions to a target gene can be used to distinguish transcriptional regulation from translational or post-translational regulation of the target gene. For example, to determine how the aceA gene from Salmonella typhimurium is regulated, aceA::MudJ and aceA::MudK fusions were isolated.

1. If the two types of ace::Mud fusions were constructed by transitory cis-transposition, what would be the expected frequency of X-gal⁺ operon fusions vs X-gal⁺ gene fusions? [Briefly explain your answer.]

   ANSWER: MudJ forms operon fusions. Thus, of the total MudJ insertions 1/2 would be in the correct orientation for transcription of lacZ, and thus would be X-gal⁺. [Some will be strongly X-gal⁺ because they are inserted into highly expressed genes and some will be weakly X-gal+ because they are inserted into poorly expressed genes.]

   MudK forms gene fusions. Thus, of the total MudK insertions 1/2 would be in the correct orientation for transcription of lacZ, and 1/3 of these would result in insertions in the correct reading frame for translation of B-galactosidase, thus only 1/6 would be X-gal⁺.

2. Expression of the lacZ reporter gene in the ace::Mud fusions gave the results shown in the following table. Based upon the results presented in this table, is the aceA gene regulated at the transcriptional, translational, or post-translational level? [Briefly explain your answer.]

   ANSWER: Note that there is no significant difference between the amount of B-galactosidase expressed from the MudJ vs MudK fusions. Because the MudJ carries its own translational start sites for lacZ, regulation of MudJ is strictly due to the transcriptional control of the promoter for aceA. Thus, the regulation must be at the transcriptional level. [If regulation was at the translational level, you would see regulation of Beta-galactosidase expression in the MudK fusion but not in the MudJ fusion.]

3. The crp gene encodes a DNA-binding protein that activates gene expression under poor growth conditions but not under good growth conditions. The crp* allele makes a altered Crp protein that is locked into the activating conformation. Given a P22 transducing lysate grown on a strain with a Tn10 insertion that is 80% linked to a crp* gene, how could you determine whether the aceA gene is regulated by the Crp protein? [Draw a diagram showing the recipient strain you would use, what media you would use, and how you would interpret the results.]

   ANSWER: This experiment simply requires a two factor cross to replace the crp⁺ gene with the crp* gene in a recipient with the aceA::MudJ fusion, then examining the expression of B-galactosidase in cells grown on poor vs rich growth conditions. You do not have a way of directly selecting for the exchange of these genes, but you can take advantage of the linked Tn10 insertion by selecting for Tet^R.

Lactose permease is an integral membrane protein encoded by the lacY gene. Computer predictions based upon the hydrophobicity of the amino acid sequence of lactose permease suggested that it spans the membrane 12 times. TnphoA fusions were used to experimentally determine the membrane topology of lactose permease. The results are shown below. [The shaded area represents the membrane bilayer, the curved line that wraps back and forth across the membrane represents the predicted topology of lactose permease, the lines with filled in triangles represent the position of PhoA⁺ fusions in the LacY protein, and the lines with open triangles represent the position of PhoA^- fusions in the LacY protein.]

1. Do these results support the conclusion that there are 12 membrane spanning domains? [Briefly explain your answer.]

   ANSWER: No. Based upon the evidence shown here, it is possible that the entire protein is secreted outside the cell. It is possible to show that the Lac^- fusions on the inside (as shown in the figure) yield hybrid protein using immunological methods, which would support the model.

2. Would you expect to find any PhoA⁺ fusions in internal domains of LacY? [Briefly explain your answer.]

   ANSWER: No. Alkaline phosphatase must be outside the cell to allow proper formation of S-S bonds required for the enzymatic activity.

3. Would you expect to find any PhoA^- fusions in external domains of LacY? [Briefly explain your answer.]

   ANSWER: Yes. This is a gene fusion so 2/3 of the insertions would be expected to be in the wrong reading frame.

To determine how the conjugal genes from IncP plasmids are regulated, two types of fusion mutations were constructed: operon fusions with the galK gene (encoding galactose kinase) and gene fusions with the lacZ gene (encoding §-galactosidase). The results are shown in the following table.

1. What do the galK fusion results indicate about the regulation of the trbF and trbG genes?

   ANSWER:
   The trbF gene is not regulated at the transcriptional level.
   The trbG gene is regulated at the transcriptional level.

2. What do the lacZ fusion results indicate about the regulation of the trbF and trbG genes?

   ANSWER:
   The trbF gene is regulated at the translational level.
   The trbG gene is regulated to the same extent in the gene fusion as in the operon fusion.

3. Combining the galK and lacZ results, briefly compare and contrast the regulation of trbF and trbG genes.

   ANSWER:
   The trbF gene is regulated at the translational level.
   The trbG gene is primarily regulated at the transcriptional level.

E. coli can use maltose as a carbon source. The malE and malF genes are required for transport of maltose into the cell. The malE gene encodes a periplasmic protein and the malF gene encodes a cytoplasmic membrane protein.

TnphoA insertions were isolated in the malE and malF genes in a host deleted for the chromosomal phoA gene. Some TnphoA insertions in malE and malF express alkaline phosphatase activity and some do not express alkaline phosphatase activity.

1. List 2 reasons why some malE::TnphoA insertions would not express alkaline phosphatase activity.

   ANSWER:
   Insertions may be in the wrong orientation
   Insertions may be in the wrong reading frame

2. List 3 reasons why some malF::TnphoA insertions would not express alkaline phosphatase activity.

   ANSWER:
   Insertions may be in the wrong orientation
   Insertions may be in the wrong reading frame
   Insertions may lack export signals

3. One of the malF::TnphoA insertions does not express alkaline phosphatase enzyme activity but reacts with antibody against alkaline phosphatase. What is a likely explanation for this phenotype? ANSWER: The gene fusion probably results in a hybrid protein but the PhoA domain is not exported so it remains inactive. Antibody can detect the presence of the inactive protein in the cell.

In addition to fusions that express a reporter gene from a target promoter, it is possible to isolate fusions that express a target gene from a promoter provided by a transposon. For example, recently John Roth's lab constructed a tetracycline resistant derivative of Tn10 called Tpop that has promoters pointing outward from both sides of the transposon. These "outward" promoters are induced by tetracycline. If T-pop insertions were isolated in the lac operon, would you expect to find T-pop insertions that would result in both a Lac^- phenotype and regulation of lacZ expression by tetracycline?

ANSWER: For the promoter within T-pop to drive lacZ expression, T-pop must be inserted between the promoter and the very beginning of the lacZ gene. Because the T-pop promoter is induced by tetracycline, these insertions would be polar on lac operon expression in the absence of tetracycline (and thus Lac^-), but Lac+ in the presence of tetracycline.

Fusions to the chloramphenicol transacetylase (cat) gene make the host cell resistant to chlorampenicol. Given a lacZ::cat gene fusion, how could you select for mutations that disrupt the lacI repressor?

ANSWER: Because transcription of the lacZ gene is repressed by the LacI repressor, a strain with the lacZ::cat gene fusion would be make very little chloramphenicol transacetylase and will be sensitive to chloramphenicol on medium without lactose or IPTG. However, mutations in the lacI gene will result in constitutive, high-level expression of the cat gene and will be resistant to chloramphenicol. Thus, it is possible to select for lacI mutants by simply selecting for resistance to chloramphenicol on medium lacking inducer.

Reporter genes are commonly used to facilitate the purification of specific proteins. A variety of reporter genes kits for this purpose are sold by different biotech companies. For example, New England Biolabs (NEB) sells a kit for making fusions of the N-terminus of the maltose binding protein from E. coli (malE) to the C-terminus of any target protein. The target gene is cloned into the proper reading frame near the end of the malE gene on a multicopy plasmid. The MalE-fusion protein can be purified by affinity chromatography, then the MalE domain can be removed using a highly specific protease. Why are such fusions usually constructed by cloning instead of by transposition?

ANSWER: In this case you usually want an insertion at the very end of a gene and it must be placed in the correct reading frame. Because this is a small target relative to the rest of the gene, the probability of finding a transposon insertion precisely at this position is expected to be very low.

Jim Hu and colleagues developed a system for assaying protein dimerization in bacteria. The system involves construction of gene fusions that result in the fusion of a protein to the N-terminus of the Lambda cI protein as shown in the diagram below. The Lambda cI protein must dimerize to bind to its operator site and repress transcription (A). The N-terminus of the Lambda cI protein retains the DNA binding domain but lacks the dimerization, so it cannot repress transcription (B). If another protein fused to the N-terminus the Lambda cI protein can dimerize, it will substitute for the missing dimerization domain of the Lambda cI protein (C).

Mark Surber isolated fusions that join the N-terminus of the Lambda cI protein and portions of the PutA protein. In addition to the fusion protein, in some cells he also expressed portions of the PutA protein separately. The results are shown in the following table.

	Fusion	Separate PutA	B-galactosidase expression
1	Lambda cI N-terminus	--	100%
2	Lambda cI N-terminus-PutA (1-end)	--	5%
3	Lambda cI N-terminus-	PutA (AA 1-end)	100%
4	Lambda cI N-terminus-PutA (1-end)	PutA (AA 1-end)	80%
5	Lambda cI N-terminus	PutA (AA 1-250)	100%
6	Lambda cI N-terminus-PutA (1-end)	PutA (AA 1-250)	80%
7	Lambda cI N-terminus	PutA (AA 575-885)	100%
8	Lambda cI N-terminus-PutA (1-end)	PutA (AA 575-885)	5%

*The fragment of PutA protein is indicated from 1, for the first amino acid, through end, for the last amino acid (i.e., AA 1-end indicates the full length PutA protein).

1. Suggest a simple explanation for the results in rows 1-2.
2. Suggest a simple explanation for the results in rows 3-8.

Please send comments, suggestions, or questions to smaloy@sciences.sdsu.edu
Last modified October 30, 2003