back to Molecular Biology Homepage

back to Homework Page

1. Would you expect modulation of chromatin structure to be important in the regulation of bacterial gene expression? Why or why not? (You may want to look back at chapter 9 for more information on the structure of bacterial chromosomes in order to answer this question.)

[In assessing answers to this question, I am much more interested in whether you demonstrated an understanding of chromatin structure and how it affects transcription (and the similarities and differences in eukaryotic and prokaryotic chromosomes) than in whether you said "yes" or "no." Here are two full-credit student answers...]

#1) I would expect a method of chromosome regulation, however this regulation would vary from that of eukaryotic chromatin regulation. Bacterial DNA, like that of eukaryotes, needs to be packaged due to size constraints. [In bacteria,] this packaging begins with the binding of positively charged polyamines, to reduce the charge repulsion of the [DNA] phosphates. In addition, numberous small proteins (NOT histones) associate with the DNA to further compress the chromosomal structure. Supercoiling completes the condensation of the bacterial chromosome. If a gene is to be transcribed, transcription factors and polymerase must be able to attach. To accomplish this binding, the strucutre most likely has to be made accessible. Eukaryotes use chromatin regulation to decompress such specific regions of the chromosome for transcription [e.g. via histone acetylation]; bacteria probably use a similar mechanism. This method of bacterial regulation, however, would not be chromatin specific, simply beccause chromatin is defined as a eukaryotic protein (histone)/DNA complex. Since bacteria lack chromatin, they cannot use chromatin structure as a means of transcriptional regulation.

#2) I suppose that modulation of chromatin structure might be important to regulation of gene expression. It is known that in bacteria chromosomal DNA forms associations with numerous small proteins, which facilitates compacting of the chromosomal structure. Knowing this, it seems possible that bacteria could have a form of gene regulation that mimics that seen with the association of histones with DNA in a eukaryotic cell. Material from the book further suggests this idea by mentioning there are numerous different proteins that associate with the chromosomal DNA, with HNS dimers being the most prevalent. This suggests that other proteins associating with the chromosome could form complexes that repress or activate transcription.

2. Would you say that the lac repressor gene, lac i, is part of the lac operon? Why or why not? (What constitutes an operon?)

[full-credit student answer, with slight changes]

The lac i gene is not part of the lac operon. An operon is made up of a certain set of genes all found next to each other and under the control of a specific promoter. These genes are transcribed together to form polycistronic mRNA. The lac i gene is located upstream of the operator and its expression is therefore independent of the operator. The protein encoded by the lac i gene is a repressor protein and is not incorporated into the polycistronic mRNA encoded by the true lac operon genes.

[Note that the lac repressor is expressed all the time. Its activity is not controlled via transcription initiation, but rather via binding of an allosteric regulator, lactose.]

3. A bacterial strain has a mutation in the gene encoding adenyl cyclase, the enzyme that makes cAMP from ATP. The mutation results in a constitutively active enzyme, and thus constitutively high levels of cAMP. What levels of beta-galactosidase expression would you expect to see in this strain in the presence of glucose only? Glucose plus lactose? Lactose only? (Assume that all other genes and DNA elements important to the regulation of the lac operon are wild-type in this strain.)

[my answer]

In this strain, the expression of the lac operon will be effectively CAP-independent, because CAP will be constitutively active due to the high levels of cAMP. Transcription of the lac operon will thus depend only on the activity of the lac repressor, which in turn depends on the absence of lactose. If lactose is high, the repressor will be inactivated and transcription will be high. If lactose is low, transcription will be negligible. So...

Glucose only: negligible expression (lac repressor is active)

Glucose plus lactose: high expression (repressor is bound to lactose, and is inactive)

Lactose only: high expression (the same as glucose plus lactose)

4. Your textbook briefly describes the isolation of CAP as a cAMP-binding protein that is absent in mutant cells unable to activate alternate metabolic pathways (such as lac) in response to an increase in cAMP. We discussed this isolation method in class as well. Propose an alternate method that could have been used to isolate and identify this protein.

[First, note that the isolation method used for CAP, as we discussed it in class, relied on cAMP columns and not on DNA footprinting. It has been shown that CAP does footprint the lac control region, but that is not the assay that was used in CAP's isolation.]

[full-credit student answer...]

The gene for CAP could be isolated using complementation analysis. A [genomic] DNA library of wild-type bacteria could be transformed into mutant bacteria that were defective for CAP function (could not subsist on sugar media other than glucose and did not respond to increasing cAMP levels). The transformed bacterial colony that can grow on alternate sugars, such as lactose and galactose, is the colony that was transformed with the plasmid carrying the CAP gene. This colony should also respond to high levels of cAMP, even in the presence of glucose. This plasmid can then be used to direct synthesis of pure CAP protein using an inducible promoter.

Note 1: As noted on a previous key (exam 1, I believe) it does not make sense to make a cDNA library from bacteria. Bacteria have very small genomes, so a genomic library need not have an unmanageable number of different plasmids. The bacterial genome is also very compact--there is very little noncoding DNA--so using cDNA does not help you. Further, bacterial mRNA is notoriously short-lived--it is very, very difficult to isolate and work with, so making a library from it would be a bit of a nightmare. So this experiment should use a genomic library.

Note 2: There are many, many possible answers to this question. You could have proposed a biochemical assay such as footprinting or gel shift. However, it is not sufficient to propose such an assay. You must also describe how you could use that assay to isolate this particular protein. Two complications must be addressed. First, CAP is not the only protein to bind to the lac control region. You would need to devise a method to tell the difference between CAP and other proteins such as the lac repressor and RNA polymerase. Second, identifying an activity in an assay is not equivalent to isolating the reponsible protein. Such an assay would need to be combined with a biochemical purification. The assay could then be used to assess which biochemical fraction contained the protein of interest. Figure 10-35 in your textbook describes an example of such an experimental protocol. (A third complication is that CAP does not bind very tightly to the lac control region, and it does not bind in the absence of cAMP. You would not necessarily know either of these limitations when setting out to design your experiment, though.)

5. You've isolated and identified a eukaryotic transcriptional activator that is important in the regulated expression of your-favorite-gene. To further characterize this activator, you conduct an experiment designed to detect protein-protein interactions. You determine that your activator interacts with TBP, a subunit of the eukaryotic general transcription factor TFIID. Propose a mechanism by which your activator might stimulate transcription of your-favorite-gene. (Your answer should demonstrate that you know what TBP is and what role it plays in eukaryotic transcriptional activation.)

[full-credit student answer...]

TBP is the first protein to bind to the TATA box. TFIIB then binds to the TBP-promoter DNA complex, followed by TFIIF-PolII, TFIIE, and TFIIH. When the transcription-initiation complex has assembled the DNA double strand can unwind in the presence of ATP and transcription can begin. My activator might stimulate transcription of my favorite gene by interacting with TBP in a variety of ways. Interaction with my activator may stabilize the saddle-shape structure of TBP or expose TBP's active site and allow or increase binding to the minor groove of the DNA strand. Or, when bound to my activator, TBP may have a conformation that then allows the binding of TFIIB or increases the affinity of TBP for TFIIB and the following complex.

6. (extra credit, up to two points) Propose an experiment or series of experiments that you could use to explore whether the mechanism you proposed in #5 is in fact correct.

[full-credit student answer...]

I could do footprinting analysis to see if TBP binds to the TATA box [of my favorite gene] without my activator. If it does not bind to the DNA or does so only to a small degree, my activator probably stimulates transcription by helping TBP to bind to the promoter sequence. If the presence of my activator in the footprint analysis creates no difference in the binding of TBP to DNA, then my activator most likely works through a different mechanism. If my activator allows for higher affinity binding of TFIIB, an experiment that detects protein-protein interactions could compare interaction between TBP and TFIIB with or without my activator. My activator might also act as a bridge between TBP and TFIIB and bind to both these proteins to form the complex, so I could also test for an interaction between my activator and TFIIB. The protein-protein interaction analysis could be done using a technique similar to what we are doing in lab; one of the proteins could have a GST-tail and the other could be synthesized in the presence of labeled amino acids.

top of page

back to Homework Page

back to Molecular Biology Homepage