1. Give an example of the type of mutation in DNA that would be repaired by
A. base excision repair (4 pts)
point mutation caused by deamination . Other possibilities also accepted if accurate
B. nucleotide excision repair (4 pts)
bulky adducts such as thymine-thymine dimers
C. Choose either A or B and outline the steps that would occur in the repair process. (8 pts)
Consult the textbook for the two mechanisms.
2. Describe the location of the three major control regions that regulate the efficiency of transcriptional initiation, relative to the startpoint of transcription. (6 pts)
TATA box, 25-35 nucleotides upstream
upstream promoter elements (promoter proximal elements), 100-500 bases upstream
enhancers, 1 or more kilobases upstream, downstream, or even within an intron.
3. How does a histone acetylase alter chromatin structure? (5 pts)
It attaches acetyl groups to the amino terminals of the histone proteins, causing a charge repulsion between the histones and the DNA, thus loosening the chromatin structure.
4. What two roles are played by TFIIH in the initiation of transcription? (6 pts)
It acts as a helicase, melting H bonds in the vicinity of the general transcription factors at the TATA box so that RNA polymerase has access to the template strand for transcription.
It acts as a protein kinase, attaching phosphates to the carboxy terminal domain (CTD) of RNA polymerase II so that it can clear away from the TATA region and begin transcription.
5. Transciption of the hsp70 gene is controlled by initiation events involving the upstream promoter region as well as by anti-termination events. Explain this. (8 pts)
Transcription of the gene begins but is paused very soon after initiation. If a heat shock occurs, the HSTF (heat shock transcription factor) changes its shape from an inactive to an active form and binds to regions rich in G and U residues next to promoter proximal elements upstream from the startpoint of transcription. This induces the hyperphosphorylation of the CTD of RNA Polymerase II and the stall is removed and it completes transcription. The binding also increases the efficiency of additional transcription initiation events.
6. Describe the chemical modifications that occur at the 5’ end of eukaryotic primary transcripts and to the first two nucleotides of the transcript in vertebrates. You do not need to describe the mechanism by which these modifications occur. (6 pts)
The Cap is a G nucleotide, methylated at the N7 position, attached to the 5’ end of the transcript by a 5’ – 5’ triphosphate bridge. The first two nucleotides of the transcript are also methylated at the 2’OH of the ribose.
7. How does the choice of cleavage and polyadenylation site influence the type of antibody molecule that forms in developing B cells as they mature. (7 pts)
Prior to activation of a B cell by antigen, antibody molecules are held as membrane proteins. To do so requires the antibody protein contain a hydrophobic region near the carboxy terminal. After antigen activation, the antibody molecules are secreted out of the cell (now a plasma cell) and therefore they no longer have the hydrophobic region but instead have a hydrophilic region at their carboxy terminal. This is achieved by differential choice of cleavage and polyadenylation site. If an early site is prevented from being used, a later site is used instead, producing a longer transcript that becomes the mRNA for the hydrophobic containing antibody molecule. Later, the first site becomes accessible, producing a shorter transcript that produces a mRNA for the hydrophilic antibody molecule.
8. If a gene contained 4 introns, what would a hybrid between its template strand and mRNA transcribed from that template strand look like? Illustrate it. (8 pts)
The picture should contain the hybrid ds regions between the exons in mRNA and their corresponding DNA sequences in the template strand. It should also contain 4 single stranded looped out regions of template DNA (the introns). It should also show the poly A tail attached to the mRNA. The polarity of the interacting strands should also be indicated.
9. Illustrate or describe in words the two transesterification reactions that remove introns. Only show the chemistry, not the interactions of the spliceosome. (8 pts)
The 2’OH of the branch point A attacks the intron-exon junction between the end of the first exon and the intron. This breaks that bond and the 5’ phosphate of the intron’s first nucleotide (G) is linked in a phosphodiester bond to the 2’OH of A. Then the 3’OH of the last nucleotide of exon 1 attacks the 5’ phosphate of the first nucleotide of the second exon, breaking the bond between it and the last nucleotide of the intron. Then a phosphodiester bond links the two exons into a continuous piece. The intron leaves as a lariat and is degraded.
10. How do the U1 and U2 snirps interact with intron/exon junctions? (8 pts)
U1 binds to the 100% conserved GU region of the intron at the first exon/intron junction by complementary, anti-parallel hydrogen bonding between that region and the RNA component of U1. U2 binds to the region surrounding the branch point A in the intron through similar hydrogen bonding with U2’s RNA component and the region of the intron. However, the A itself does not bind but bulges out from the hybrid, facilitating the first transesterification reaction.
11. Describe the complete pathway by which the female phenotype is determined during the embryonic development of the fruit-fly. (10 pts)
A. An early promoter is used by the female embryo and transcribes the early sex-lethal gene. The protein is made.
B. Later, a late promoter transcribes the late sex-lethal gene. The early sex-lethal protein binds to a 3’ intron/exon junction in the transcript and prevents normal splicing of an intron. The result is that the following exon is removed as if it was part of the intron. It contained a premature stop codon which is no longer part of the mRNA. Thus, functional late sex-lethal protein is made.
C. The late sex-lethal protein continues to autoregulate its own transcription as described in step B. It also regulates the alternative splicing of another transcript, the transformer gene (tra). By an analogous mechanism, sex-lethal binds to a 3’ intron/exon junction and causes the following exon to be removed as part of the intron, removing a stop codon. The resulting tra mRNA is used to make functional tra protein.
D. The tra protein binds with tra2 protein and RBP1 protein, forming a complex that binds to an exon in the transcript of the double-sex gene, enhancing the use of a weak intron/exon junction and thus allowing a splice to occur and the removal of an intron. The resulting mRNA produces the female form of the double-sex protein. That protein acts as a transcriptional repressor to prevent expression of male-specific genes, resulting in the female phenotype.
12. What phenomenon explains the need for third base degeneracy in the genetic code? Explain your answer. (6 pts)
The phenomenon of wobble explains this. Wobble is the term used to describe hydrogen bonding between anticodon position 1 and codon position 3 that does not follow normal Watson/Crick base pairing rules, yet is stable. Therefore, if used during translation, the tRNA would remain at the ribosome A site long enough to be used in the formation of a peptide bond. Since the amino acid attached to the ribosome would then become part of the protein being made, it is essential that it be the correct, expected amino acid, coded for by that codon in the mRNA and not some other amino acid that might be coded for if the third nucleotide of the codon were important in determining the amino acid. To protect against error, for those proteins carried by tRNAs that can participate in wobble relationships, the third base of their respective codons have evolved to be irrelevant and the amino acid is essentially coded for by the first two nucleotides of the codon in those cases.
13. If no radioactivity appeared on a filter during the Nirenberg experiment, what conclusion could you draw? Assume he did not make any mistakes. (6 pts)
If not found on ANY filter, the trinucleotide being tested is most likely a stop codon.
Some of the class interpreted the question to mean that there was no radioactivity on one filter, rather than on any filter. The correct answer then would be that the trinucleotide being tested coded for an amino acid that was not the radioactive amino acid in that particular assay.
I gave credit for either answer.
Extra Credit: Write two sentences for each debate we have had thus far that summarizes the major controversy being debated. (4 pts)
A. 2 sentences about the pros and cons of posting genomic sequences for dangerous viruses on-line.
B. 2 sentences about federal funding of adult versus embryonic stem cell research.