1.                  Describe the role in transcription initiation played by the following regulatory regions in DNA. Include in your answer the location of each relative to the +1 transcription start point (4 pts each)

A.                 TATA box

Regulatory sequences in DNA, around 25-35 basepairs upstream from the startpoint of transcription that bind the general transcription factors that load on RNA polymerase II.

B.                 upstream promoter proximal elements

Regulatory sequences in DNA located between 200-500 base pairs upstream of the startpoint of transcription that bind to specific transcription factors and regulate the binding of the general transcription factors at the TATA box

C.                 enhancers

Regulatory sequences located far upstream or downstream from the startpoint of transcription, sometimes a kilobase or more away. They bind specific transcription factors and chromatin remodeling complexes as well as mediator and thus regulate access to the chromatin and the binding of specific and general transcription factors.

2.                  What is the CTD of RNA polymerase II and how is it important to the transcription initiation and elongation process?  (6 pts)

It is the carboxy terminal domain of the RNA polymerase molecule. It has amino acid side chains that can become phosphorylated by TFIIH. When phosphorylated, it helps loosen the RNA polymerase II from interaction with the general transcription factors at the TATA box and this promoter clearance allows RNA polymerase II to continue transcription up until it pauses. Then, other protein kinases can hyperphosphorylate the CTD and cause the RNA polymerease II to continue elongation if appropriate.

3.                  Chromatin remodeling complexes as well as histone acetylase and deacetylase enzymes are important in the control of transcription initiation. Explain how. (8 pts)

Chromatin remodeling complexes are able to push the DNA wrapped around the histones in a nucleosome in to the interior region of the nucleosome. This allows the nucleosome to “slide”, decondensing the chromatin structure and allowing specific transcription factors access to their binding sites on DNA. Histone acetylases, part of activation domains on specific transcription factors, acetylate lysines in the amino termini of histones, loosening their interaction with the negatively charged DNA, thus allowing easier access by specific and general transcription factors to their binding regions. Deacetylases can remove acetyl groups from the histone amino termini, thus relieveing the charge repulsion, helping deactivate the transcription initiation.

4.                  Since globin genes are not expressed in skin cells but are in red blood cell precursors, what result would you predict from a DNAase sensitivity assay using DNA extracted from either cell type. Be sure to explain both how the result would look and also why it would look that way. (8 pts)

The chromatin structure around and including the gene from the skin cells would be resistant to degredation by DNAse because it is more compact in a cell type that does not express the gene and the DNAse would be less able to access the DNA to degrade it. That from the red blood cell precursors would have a looser chromatin structure because the gene was expressed in that cell. Therefore when subjected to increasing DNAse concentrations in the sensitivity assay, the DNA of that gene would increasingly be degraded, resulting in loss of the DNA fragment that represents that gene. But even at the highest concentration, the fragment from the skin cells would not be degraded and would still be seen in the result.

5.                  Illustrate or describe in words what the cap at the 5’ end of eukaryotic mRNAS looks like in vertebrates. (5 pts)

It is a guanine nucleotide, methylated at the N7 position, that is attached to the first nucleotide in the mRNA by way of a 5’-5’ triphosphate bridge linkage. The 2’OH of the ribose on both the first and second nucleotides of the mRNA would also be methylated.

6.                  Describe the events that occur as RNA polymerase II transcribes the poly-A signal up to the addition of the poly-A tail on eukaryotic mRNAs. (8 pts)

It transcribes an AAUAA sequence and beyond, including a GU-rich region of the transcript. CPSF binds to the AAUAA sequence and CStF to the GU-rich region. This bends the DNA and binds CF1 and CFII, marking the cleavage site for formation of the 3’ end of the transcript. PAP binds here also. CF1 and II cleave the phosphodiester bond at the cleavage site, releasing the rest of the transcript beyond it. The PAP uses ATP to attach about 12 As to the 3’ end. Then PABII binds the A polymer, stimulating PAP to attach an additional 12 or so As again. More PABII binds to that stimulating PAP to act again. This continues until a polyA tail is completed.

7.                  If you analyzed a eukaryotic gene that contained 4 introns, using an R-looping experiment, what would you see in the result? You may illustrate the result or explain it in words. (6 pts)

The diagram would show the DNA of the template strand of the gene hybridized to complementary antiparallel sequences in the mRNA transcript of the gene. But the intron regions in the DNA would be looped out as single stranded loops because there would be no complementary bases in the mRNA. All hybridized regions would be double stranded containing one DNA and one RNA component.

8.                  Describe the location in introns that contain nucleotides that are 100% conserved. (6 pts)

At the 5’end of the intron is a 100% conserved GU sequence. At the 3’ end is a 100% conserved AG sequence. There is also a 100% conserved A (branchpoint A) nucleotide approximately 20-30 nucleotides upstream from the 3’ end, next to a pyrimidine-rich sequence.

9.                  Two transesterification reactions allow the removal of introns during splicing. Describe these chemical reactions. (6 pts)

First the 2’OH of the branchpoint A attacks the 5’phosphate of the G at the 5’ end of the intron. This breaks the bond between the G and the last nucleotide of the exon (exon 1 in our explanation). A new phosphodiester bond forms preferentially between the branchpoint A 2’ carbon and the 5’ phosphate of the G. Then the now free 3’OH at the end of exon 1 attacks the 5’ phosphate attached to the first nucleotide of the next exon (exon 2). This breaks the phosphodiester bond between the 3’ end of the intron and that first nucleotide of exon 2. A new phosphodiester bond forms between the 3’ end of exon 1 and the 5’ end of exon 2. The intron is released in lariat form and degraded. This completes the splicing reaction.

10.              How do the following influence the splicing reaction.? ( 4 pts each)

A.                 U1

U1 uses its RNA component to base-pair to sequences surrounding and including the 100% conserved GU at the 5’ end of the intron. It then interacts of the other components of the spliceosome, orienting the intron to facilitate the splicing reaction.

B.                 U2

U2, with the help of U2AF, base pairs to sequences surrounding the branchpoint A region in the intron, but not to the A itself. This causes the A to “bulge” out from the hybridized region. It then interacts with other spliceosome components, orienting the branchpoint A into the vicinity of the 3’ end of the exon (exon 1 in our explanation). This allows the first transesterification to happen.

C.                 SR proteins

These are proteins that bind to recognition sequences within exons called exonic slicing enhancers (ESEs). When bound there, the SR proteins facilitate the binding of U2AF to the pyrimidine-rich region in the intron to the 3’ side of the exon. That binding helps U1 bind (explained above). They also facilitate the binding of U1 to its binding site at the 5’ end of the intron (explained above) on the 3’ side of the exon.

11.              The sex-lethal protein (sxl) in Drosophila begins a cascade of events that determines the sex of a fruit-fly. Answer these questions about this system. 

A.                 How does sxl control the ability of tra to be made or not? (5 pts)

In females, the sxl protein binds to the U2AF binding site (see above) in one of the introns of the tra transcript. This prevents the 3’ end of the intron from interacting with U2AF and U2. Instead, the next U2AF and U2 sites are used to represent a 3’ end of an intron and all intervening sequences are removed from the transcript as one large intron. Subsequent joining of the exons gives an mRNA containing sequences that do not have a stop codon in frame during translation because it was part of an exon that got removed with the large intron. This tra protein when made is functional. In males, normal splicing occurs and the exon containing the stop codon remains in the mRNA. This truncates the protein during translation, resulting in nonfunctional tra.

B.                 How does tra determine whether the fruit-fly will be male or female? (7 pts)

In females, tra binds to tra2 and RB1 proteins and that complex binds to ESE regions in an exon, activating the 5’ end of an intron following that exon. The junction is recognized by the U1 and a spliceosome will form there. The resulting removal of this intron produces a transcript that contains a poly A signal within the next exon. When cleaved and polyadenylated, the resulting mRNA, when translated, produces the female form of the double-sex (dsx) protein. That protein is a specific transcription factor that represses genes needed for the male phenotype and the fly will be female. In the absence of functional tra in males, no complex forms to activate the splice junction. Therefore, the exon and the next intron are removed with the preceeding intron as one large intron. This removes the poly-A signal in the resulting transcript and a longer, different form of the dsx protein is made during translation. This dsx represses genes needed for the female phenotype and produces a male.

12.              The genetic code is degenerate. This is due to a phenomenon known as wobble. Describe what is meant by wobble. (5 pts)

Wobble refers to a non-Watson-Crick basepair that can form between the nucleotide at position 1 of an anticodon in some tRNAs with position 3 of a codon. This base pair will be stable. Therefore, during translation, when a codon-anticodon interaction occurs, the association will be stable and the amino acid attached to the tRNA will be used to make a new peptide bond. Therefore, the code for the amino acid carried in the mRNA codon must be such that the amino acid is coded for by a codon that does not need a specific nucleotide at position 3 to be used as long as the first two positions are correct. In other words, for these codons the third base is irrelevant to specifying the amino acid. So, if a wobble interaction occurs, the amino acid will still be accurately coded for.

13.              Debate Question:

Use one of the debates that you have heard and write 3 or 4 sentences that describe the major issues it addressed. (6 pts) 

      There are many possible answers to this question. The answer would be about either the suitablility of mandating HPV vaccinations for young girls to protect against cervical cancers OR about the suitability of allowing people to sell their organs.