1. Below are three activities associated with RNA polymerase II. Describe what role is played by each during transcription elongation in eukaryotes. (4 pts each)

A. unwinding activity

Melts the double-stranded DNA in the transcribed region to allow RNA polymerase II to gain access to the template strand.
 

B. polymerizing activity

Creates the RNA polymer by catalyzing the formation of the phosphodiester bonds connecting nucleotides together.
 

C. rewinding activity

After 10-15 nucleotides of the transcript are made, this activity releases the short region of transcript associated with the template strand and allows the reestablishment of the double stranded DNA polymer.

2. The diagram below illustrates the organization of a “typical” eukaryotic gene and its regulatory regions. Indicate the following regions on the (show ALL regions if more than one applies) and indicate if they are bound by general or specific transcription factors. (12 pts)

· Location of the TATA box
· Promoter proximal elements
· Enhancer/silencer regions
 

See diagram in glass case near 021 McKinly laboratory.

TATA box is located about 30 nucleotides upstream of +1.
Promoter proximal elements are the boxes located from -50 to -200
Enhancer/silencer regions are boxes located far upstream or downstream on the diagram as well as potentially within the intron shown in the diagram.
 
 

3. Describe the role of the following in transcription initiation. (3 pts each)

TBP
Is the first protein to enter in the formation of the initiation complex. It binds directly to the TATA box. Called the TATA binding protein.
 

TFIIH
Plays two roles. It is the protein kinase that phosphorylates the carboxy terminal domain of RNA polymerase II, allowing it to release from the initiation complex and it also is a helicase that performs the initial melting of the double-stranded DNA allowing RNA polymerase to gain access to the template strand.
 

CTD of RNA polymerase II
As described above, it is the carboxy terminal domain of RNA polymerase II that has the amino acid side chains available for phosphorylation. This phosphorylation is required to clear RNA pol II’s association with the initiation complex, allowing it to begin elongation.
 

Acetylase enzymes

These are usually parts of activating complexes associated with the promoter proximal regions and/or the tafs which are part of TFIID. They are enzymes that catalyze the attachment of acetyl groups to the amino termini of histone proteins in chromatin in the region of the promoter. This loosens the chromatin structure allowing easier access to the TATA box by the transcription initiation complex .
 
 

4. Outline all steps that generate the 3’ end of a eukaryotic messenger RNA molecule. (8 pts)

See page 414, Figure 11-2. All steps needed to be included in the answer
 
 

5. Answer the following questions about the Cap that is attached to eukaryotic mRNAs. (3 pts each)

A. Describe the linkage between the Cap and the RNA transcript.

The linkage is a triphosphate bridge attaching the modified guanine in a 5’ to 5’ linkage, meaning the first phosphate, which is attached to the 5’ carbon of the modified guanine of the CAP is attached to diphophate on the end of the RNA transcript through the terminal phosphate’s linkage to the 5’ carbon of the first nucleotide. See diagram 4-18, page 115.
 

B. What type of modified nucleotide base is part of the CAP?
 

A modified guanine nucleotide, methylated at the N7 position of the base.
 

C. In vertebrates, what additional modifications are seen on the RNA transcript itself?

A methyl group is attached to the 2’OH of the ribose of both nucleotides number one and two of the primary transcript.
 

D. What molecule is the source of the nucleotide that will become the CAP?

GTP
 

6. Below is a result obtained from an R-looping experiment, using a cellular extract expressing a gene, X. Illustrate the organization of gene X. (8 pts)

See diagram in cabinet. In summary, the answer should include the location of the promoter regions (singe-stranded DNA at the 5’ side of the diagram), the location of the 4 exons and the 3 introns and their relative sizes, based on the sizes of the thicker hybrid regions in the diagram and size size of R loops, and some region of untranscribed double-stranded DNA at the 3’ end of the gene, represented by single-stranded DNA in the diagram. The poly-A tail seen in the diagram should not appear in the gene picture since this is a post-transcriptional modification.
 

7. You are teaching a course in Molecular Biology to a class of undergraduate Biology majors. The topic of splicing of eukaryotic genes is next on your syllabus. Summarize the lecture/s you will give on this topic, knowing that the University requires that the students learn the intron/exon junction sequences, the chemical reactions that remove the introns and reattach exons, and what types of cellular particles are needed to catalyze these reactions. (12 pts)

See pages 416, 417, and 419.

A. Describe the conserved sequences at the ends of the introns: GU at 5’ end and AG at 3’ end. Also the A residue approximately 20-50 base pairs near the 3’ end of the intron that is involved in the chemical reactions of splicing.

B. Two transesterification reactions occur. First the 2’OH of the branchpoint A (see above) attacks the phosphodiester linkage attaching the last nucleotide of exon 1 to first nucleotide of the intron (the G). A covalent phosphodiester linkage is made between the G in the intron and the A nucleotide (a 2’-5’ linkage instead of the usual 3’ to 5’ linkage). Then a second transesterification reaction occurs where the now available 3’ OH of the last nucleotide of exon 1 attacks the phosphodiester bond between the terminal nucleotide of the intron  and the first nucleotide of exon 2. The intron is released in its lariat form and degraded.

C. These reactions are assisted by small nuclear RNA-protein particles called snurps, consisting of small nuclear RNAs and proteins. These assemble during splicing into a particle known as the spliceosome. The major snurps we discussed were the U1 and U2. U1 associates with the conserved region at the first exon/intron junction using base-pairing between  the junction sequences and the small nuclear RNA component of U!. U2 assembles near the sequences surrounding the branchoint A, again through base-pairing. The branchpoint A itself does not base-pair and therefore bulges out slightly, facilitating its use in the beginning of the first transesterification reaction. Following these initial steps, the spatial changes bring the the 3’ end of the first exon in the vicinity of the 5’ end of the second exon. The remaining snurps assemble, additional spatial changes occur, and the two transesterification reactions are catalyzed by enzymatic activities contained in the spliceosomes.
 

8. The following is an illustration of the organization of the template strand of a eukaryotic gene that can give rise to two different forms of a protein from one primary RNA transcript. The different forms arise due to the use of alternative poly-A sites, indicated here as either #1 or #2. Shown also are potential intron/exon junctions and potential stop codons for translation that could possibly remain in the final mRNA that is made. Illustrate the two final mRNA forms that would be produced and compare/contrast (in general terms) the proteins that could be generated from each. (10 pts)
 

See diagram in cabinet. Two primary transcripts are possible, stopping either at poly A site 1 or poly A site 2. If poly A site 2 is used, a longer transcript is produced that includes exon/intron junctions at both ends of the intron and that can therefore be used in a successful splicing event, producing a mRNA that has the second stop codon since the first has been removed. If poly-A site 1 is used, a shorter transcript is made that does not have a functional acceptor sequence and no splicing would occur. This would produce a mRNA with a stop codon at stop codon one position. The protein products would contain amino acids coded for in these mRNAs, terminating at these differing stop codons. The sequence in the first mRNA would not be derived from any of the nucleotides contained in the removed intron.
 
 

9. Stem-loops structures in mRNAs play important regulatory roles. Answer the following questions about them.
 

A. When cells regulate iron concentration, they use the iron-response element binding protein. The production of what two proteins is regulated by this molecule? (4 pts)
 

Ferritin and the transferin receptor
 
 

B. Describe how each of the two proteins in question B is regulated when cells have high iron concentration. (10 pts)

 
Under high iron concentrations, the Iron-response element binding protein cannot bind to its binding site, called the iron response element (IREs). These IREs are located in the loops of stem-loop structures located in either the 5’ UTR of the ferritin mRNA or the 3’ UTR of the transferin receptor RNA. For ferritin, binding to the IRE prevents translation initiation of the iron storage protein, ferritin. For the transferin receptor RNA, binding does not allow mRNA degredation to occur. Therefore, when these IREs cannot be bound during high iron concentrations, ferritin CAN be translated and store the excess iron and the transferin receptor will be degraded resulting in less or no transferin receptor being made, since its role is to bring more iron into the cell and under high iron concentrations, this is not needed.
 
 

     Problem 2 Question

Choose one system that you discussed in your report (yeast, mammals, or whatever you prefer) and describe the major regulatory mechanisms that control cell-cycle progression in that system.
 

Include in the answer as much detail as possible, including stages of the cell cycle and the primary regulators of progression into each stage.