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100 points possible.

1. In the late 1960s, Roy Britten and David Kohne designed and carried out an experiment to assess the complexity of mammalian vs. bacterial genomic DNA (we discussed this experiment in class). They purified equal mass amounts of calf and bacterial genomic DNA, fragmented each preparation into small pieces, heated each to denature the double strands, and then lowered the temperature and monitored the rate of reannealing.

a. (1 pt) The following is a rough estimation of their data. One curve represents calf DNA and one represents E. coli DNA. Label the two curves&emdash;which is which?

The curve that was on top at first (reannealing slower) and then on bottom (reannealing faster) was E. coli. The other was calf.

b. (4 pts) How do you interpret the part of the graph labeled I? What do these data suggest?

[full-credit student answer] A subset of the colf DNA has a high # of repeated regions in its genome. For a denatured piece of DNA to find its base-pair-buddy [complement] it can take a while--but if it has a lot of base-pair-buddies with the same sequence it will be much, much faster. Calf DNA has a lot of repeats in its genome and thus has this subset of its DNA that reanneals quickly. That's what these data suggest.

c. (4 pts) How do you interpret the part of the graph labeled II? What do these data suggest?

Here the E. coli DNA is reannealing much more quickly than the calf DNA. Enough time has passed to give all of those repeats time to reanneal, so the calf DNA pieces aren't finding many base-pair-buddies [complements]. The E. coli, on the other hand, is present in many many more copies than the calf because the genome is so much smaller (and the same mass of DNA was used). It is faster. That's what the data suggest.

2. (3 pts) What is "satellite" DNA, and where in the human genome can such DNA be found?

[full-credit student answer] Satellite DNA is simple repeats of DNA where a short sequence is repeated. It is concentrated at the telomeres and the centromeres, but it is also found scattered throughout the genome.

3. (6 pts) How is a transposon different from an IS element?

[full-credit student answer] A transposon has an additional protein-encoding region that IS elements do not--it is usually drug resistance. The drug-resistance gene in the transposon is flanked by two IS elements. These IS elements have a protein coding region that includes transposase flanked by inverted repeats. The inverted repeat regions nearest the center of the transposon, however, are inactive for transposition, so the entire transposon gets transposed into target DNA. [The best answers were accompanied by diagrams, which I won't attempt to reproduce here...]

4. (2 pts) What biological function have "PEST" sequences been found to serve?

[full-credit student answer] PEST sequences target a protein for degradation by proteases. Such is the fate of IkappaB when a cellular reponse changes its conformation, prompting its destruction and preventing its binding to NFkappaB.

5. (3 pts) Under what biological conditions is CAP active, and how is this activity regulated?

[full-credit student answer] CAP is active when cAMP is upregulated in response to low glucose levels in the bacteria's environment. CAP/cAMP facilitates the binding of RNA polymerase to the lac operon, an action which in the presence of lactose will result in the expression of genes needed to use lactose as a food source.

6. (18 pts) Complete the table, giving the level of expression of the lac operon [i.e. beta-galactosidase activity] you would expect to observe in each case. Use qualitative words such as "high," "moderate," "low," and "very low." The first row has been completed for you.

Strain	lactose only	glucose + lactose	glucose only
wild-type E coli	High	Low	Very Low
a haploid strain carrying an oc mutation	High	Low	Low
a haploid strain carrying a loss-of-function lac i mutation	High	Low	Low
a haploid strain carrying a mutation in the gene encoding CAP that renders CAP constitutively active	High	High	Very Low
a "diploid" strain (for the lac region) with the following genotype (and WT for the gene encoding CAP): lac i- oc lac z+/lac i+ o+ lac z-	High	Low	Low
a "diploid" strain (for the lac region) with the following genotype (and WT for the gene encoding CAP): lac i- oc lac z-/ lac i+ o+ lac z+	High	Low	Very low
a haploid strain carrying a loss-of-function mutation in the gene encoding CAP	Low	Low	Very low

NOTE: in every case marked "high," the lac repressor is functionally inactive (and/or is unable to bind because of an oc mutation) and CAP is active. In every case marked "low," both CAP and the repressor are inactive. Cases where the lac repressor is active are marked "very low." In other words, every instance is directly analogous to one of the three situations provided for you in the first row.

Each cell in the table was worth one point. If your answer was within one category of the correct one (e.g. "moderate" instead of "low") you earned half a point for that cell.

7. You have identified a new gene in yeast. Analysis of its DNA sequence suggests that it may be involved in the metabolism of galactose&emdash;it has homology to other genes (in other organisms) that serve this function. Sure enough, when you create mutant strains that lack your-favorite-gene (YFG), they cannot grow on galactose. As a next step, you decide to explore whether expression of your gene is regulated by galactose.

a. (8 pts) Describe how you could determine whether mRNA levels of YFG are different in the presence or absence of galactose.

[full-credit student answer] A northern blot would be the most effective way to test this. To test this, we could grow two different colonies [cultures] of yeast, one on an agarose plate [in a liquid culture] with galactose, and one with an alternative food source. Next, we would separate out all the mRNA transcripts form each group of cells using a process similar to creating a cDNA ligrary, using oligo dT's complementary to the poly-A tails of cellular mRNA to separate it from other cellular parts. Then run the mRNA on a gel and transfer to a membrane. Then probe the membrane with probe complementary to YFG. The two lanes of the gel should look different, with the colony [culture] grown in galactose showing higher levels of mRNA for YFG if YFG is indeed induced by galactose.

[This is analogous to the experiment done in Lycan et al to assess expression of YAR1 under various conditions.]

b. (8 pts) Sure enough, your experiment in a demonstrated that mRNA levels of your-favorite-gene are much higher when galactose is present. Now you want to determine what regions of the YFG promoter are involved in this effect. Describe an experiment you could conduct to address this question.

[full-credit student answer] You could design a promoter "bashing" experiment to isolate the crucial sequences in your promoter that are necessary for the effect. You would need to make a number of different constructs with YFG promoter on a plasmid with a reporter gene (beta-galactosidase) attached to your differing promoter lengths. Then in the presence of galactose test for levels of your reporter gene. You can the whittle down to crucial sequences for galactose regulation.

[NOTE that the experimental question here has to do with galactose regulation of YFG, not simply with YFG expression, just as the experimental question in Hoshijima et al had to do with da/sis-b activation of Sxl, not simply Sxl expression. Your experiment should therefore include galactose, preferably comparing expression in the presence and absence of galactose.]

c. (8 pts) Your experiment defines a region of the YFG promoter that is necessary for galactose activation. You hypothesize that this cis regulatory element works because of an interaction with a trans element&emdash;a regulatory protein. Describe an experiment you could conduct to determine whether this region is in fact bound by a cellular protein. (You need not determine what cellular protein(s) yet&emdash;that's part d).

[full-credit student answer] DNA footprinting will show you whether or not this region is bound by a protein.

1) label one end of the DNA of region of the YFG promoter involved in galactose activation

2) mix the region of DNA with cellular proteins from cells

a) exposed to galactose (galactose may activate a DNA-binding protein) and

b) cells not exposed to galactose (interaction with galactose may inactivate a DNA-binding protein)

3) partially digest each mixture with DNase I

4) run gels and see if a section of the DNA was protected by the protein from DNase I and therefore missing bands on a gel (ompare to a control of the digested region w/out adding any protein)

[you could also use an EMSA to address this question...]

d. (8 pts) Your experiment in c demonstrates that the YFG promoter region you've identified is in fact bound by a cellular protein. Describe how you could go about isolating and identifying this protein. ("Isolate" means to purify this protein away from other cellular components, and "identify" means to obtain the sequence of its gene.)

[Student answer, with comments] I would use complementation analysis. Generate a series of mutants who do not repress YFG. Then [mate them to wild-type haploid cells] and find ones that are complemented. Those haploid mutants that can be complemented likely have mutations in the repressor and not the promoter, since they aren't dominant mutations. Make cDNA library and introduce the library using vectors into haploid cells characterized as having recessive mutations and look for restoration of wild-type phenotype in a colony. [You'd need an easy screen for this, since you'd have to screen through thousands of transformants.] The colony shouldhave a plasmid with the [gene encoding the] repressor of YFG. Now isolate the plasmid and use sequencing to determine the sequence of the gene, and therefore the sequence of the protein. [If you need purified protein, you can clone this gene into an expression vector and either induce expression in E. coli or perform in vitro transcription/translation.]

[You could also have proposed a biochemical approach, purifying the protein from other cellular components via a series of columns, and identifying which column fractions contained your protein via a footprint or EMSA analysis.]

8. Having completed your doctoral dissertation entitled "Examining the galactose regulation of my favorite gene in yeast," you've moved on to a post-doc. In your new lab, you're studying a bacterial operon that encodes genes necessary for the synthesis of the amino acid lysine. You notice that the length of the transcript of this operon varies with environmental conditions. When lysine is abundant, only very short transcripts are produced. When lysine is limiting, full-length transcripts are synthesized.

a. (8 pts) What regulatory mechanism would you hypothesize is producing this effect? Name and describe it.

[full-credit student answer.] Attenuation. This generally describes the mechanism by which RNA polymerase "decides" where to end transcription. Specifically: transcription and translation occur concurrently in bacteria. Transcription is terminanted if RNApol pauses at a seqeucne of U's (because weak A-U bonds dissociate easily). It pauses due to interaction with a step loop structure generated by complementary RNA sequences just transcribed. (seq. 3 and 4 below.) [A drawing follows which I won't attempt to replicate here, but it parallels the structure of the beginning of the trp operon.] However, if a translating ribosome happens to pause while translating region 1, regions 2 and 3 will base pair because they are also somewhat complementary. This in effect breaks the stem loop between 3 and 4 and RNA polymerase doesn't pause, instead transcribing past the U's. In this case, the absence of Lysine causes a ribosomal pause because lysine tRNA's are not charged (i.e. they are without lysine). On the other hand, the presence of lysine doesn't cause a pause and transcription is terminated at the U's [i.e. at the attenuator.]

b. (6 pts) If your hypothesis is correct, describe a protein-coding region that you might expect to find near the beginning of the transcript of this operon. What characteristics might this region exhibit? Why? What is the function of this region?

[my answer] I'd expect to find a leader peptide region near the beginning of the transcript, that could be translated from both the short and long transcripts. This region would include at least a few lysine codons, placed so that a ribosome pausing at them would cause a disruption in the "default" RNA structure that would otherwise allow termination at the attenuator. The function of this region is to make attenuation dependent on the concentration of lysine in the cell. If lysine is abundant, ribosomes will not pause while translating the leader peptide, the default RNA structure will form, and transcription will terminate. However, if lysine is limiting, the ribosomes will pause while translating the leader, and this pause will prevent termination of transcription at the attenuator.

This is the "L" region of the trp operon. It does not encode a protein that functions in the biosynthesis of trp--just a short leader peptide. Its stop codon is between regions 1 and 2.

9. (4 pts) A mutation has arisen which blocks the binding of sex-lethal protein to the transformer mRNA in Drosophila. Describe the phenotypic effect that you'd expect to see in a female fruit fly homozygous for this mutation. Explain your prediction.

[full-credit student answer] I would expect to see a male phenotype. If sxl cannot bind to the tra transciprt it cannot direct splicing of the preRNA to produce a functional Tra protein. If there is not Tra protein then no Tra/Tra2 complexes can form. This causes the male-specific double-sex protein to be expressed, causing a male phenotype [because the male-specific form of double-sex represses expression of female-specific genes.]

10. Below is figure 6 from Hoshijima et al. This figure addresses the mechanism of action of deadpan protein. The sequence shown in part A is part of the sex-lethal early promoter control region.

(9 pts) Describe the experiment presented in parts A and B. What is this technique called, how does it work, and what do these data show?

The technique is called DNaseI footprinting. It works by first incubating your [radiolabeled] hypothesized DNA binding stretch with YFP [your favorite protein] and a DNase. If your protein does bind DNA the stretch where it binds will be impervious to cleavage. If you then run the reaction out on a gel you will see a blank spot which corresponds to where your protein bound to DNA. If you run sequencing lanes opposite this you can tell where exactly this protein binds. The extra 4 or so nucleotides in fig A that depicts where dpn binds to DNA have to do with DNA's helical nature. The data show that dpn binds DNA at the specific region around -90 to -120.

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