These factors are used in the mechanism of nucleotide excision repair
(removal of thymine dimers). Consult pages 966 and 967 in the text.
2. What role is played by the following in eukaryotic transcription initiation? (3 pts each)
A. TATA box
Binds the general transcription factors.
B. TFIIH
Uses helicase activity to open the ds DNA during promoter clearance.
Uses kinase activity to phosphorylate the CTD of RNA pol II during promoter
clearance.
C. TBP
Binds to the TATA box directly, beginning the assembly of the general
transcription factors.
D. CTD of RNA pol II
Exposed domain of RNA pol II that must be phosphorylated before transcription
can continue after initiation. Also becomes hyperphosphorylated in some
systems to relieve premature stalling of transcription.
E. histone acetylase
Enzyme that attaches acetyl groups to the amino terminal region of
histones to facilitate loosening of the chromatin structure during transcription
activation.
F. chromatin remodeling complex
Transiently dissociates DNA from the nucleosome to permit nucleosome sliding, facilitating the loosening of the chromatin structure during activation of transcription.
G. mediator
Large multi protein complex that serves as a bridge between specific
transcription factors (activators) and to the general transcription factors
at the TATA box and to RNA pol II.
H. proximal promoter elements
Regulatory sequences located 100-200 base pairs from the start site
of transcription that bind to specific transcription factors that regulate
how efficiently the general transcription factors bind and initiate transcription.
I. enhancer
Regulatory sequence located far away (over 1 kilobase) from the startpoint
of transcription and sometimes on either side of it or even within the
coding area. Binds specific transcription factors to influence the efficiency
of transcription activation.
3. How does the HIV tat protein promote the transcription of HIV genes?
(6 pts)
Consult pages 485 and 486 for the mechanism.
4. Draw the result you would see in an electron microscope picture after a hybrid was formed between the template DNA strand for a gene and its mRNA if that gene had 3 introns. (8 pts)
The diagram would illustrate three single-stranded DNA loops representing
the introns, 4 hybrid regions (ssDNA paired with mRNA) representing the
4 exons. The poly A tail would be at the 3’ end of the diagram attached
as a single stranded region to the mRNA. Some single-stranded DNA would
also extend from the 5’ side representing extra DNA nucleotides in the
template strand not found in the mRNA.
NOTE: I did not remove any points for not including the poly A tail
or the additional DNA nucleotides in your answer.
5. Illustrate or describe in words the structure of the 5’ cap on eukaryotic
mRNAs. (6 pts)
Figure 4-13 on page 113 shows the structure.
6. Below is a diagram of a gene that produces one form of mRNA in skeletal
muscle cells and a different form in cardiac muscle cells. The arrows indicate
potential cleavage and polyadenylation sequences and the boxes are possible
introns. Describe how such a gene could produce two different mRNAs. (8
pts)
In one cell type the first cleavage and polyadenylation signal could be masked and the transcription would cleave and polyadenylate at the second site. This would produce a transcript with two normal introns which would be removed and the exons connected. This would give rise to one form of protein.
In the second cell the first cleavage and polyadenylation signal is not obscured and is therefore used, eliminating any further transcription. Therefore there would be no normal introns left and the transcript would include all sequences up to the first cleavage signal (nucleotides that would otherwise be part of the first intron). When translated, this would make a different form of protein.
NOTE: Some of you considered that the splicing apparatus would work
first to remove the first intron and with it the first cleavage signal
causing the second to be used. That would result in the same protein as
I described above in the first paragraph. Either explanation was considered
correct. The second protein would not be affected by splicing in any circumstance
since no complete intron would be present when the first signal is used.
7. Describe the two trans-esterification reactions that allow an intron to be removed from a eukaryotic primary transcript. (8 pts)
Consult pages 497-498.
8. The late sex-lethal protein is functional in female fruit-flies but not male fruit-flies. Outline the consequences of this for both types of flies showing how, ultimately, this will determine the sexual characteristics of these flies. (10 pts)
Consult pages 505-506. To gain complete credit you had to explain both the regulation of the tra gene by sxl and the dsx gene by the tra complex. The complex plays the role of an SR protein in females and this needed to be part of your answer. Also, the different forms of dsx protein repress transcription of the phenotype-specific genes of the opposite sex. This also needed to be part of your answer.
9. If Nirenberg was living in a world where the genetic code consisted of quadruplet sequences coding for 15 amino acids, but everything else was done identically to our world,
A. How would he set up his experiments? (6 pts)
Into 15 tubes he would put cell lysates that had been treated with nucleases to remove the endogenous RNAs. They would contain all factors needed for translation including all tRNAS and all amino acids etc.
He would then synthesize all possible 4 nucleotide combinations. Individually, he would test each combination by adding it to each of the 15 tubes.
He would then add one radioactive amino acid to each of the 15 tubes.
He would filter the mixture and test for radioactivity on the filter.
If a filter was radioactive he could conclude that it contained a ribosome
complexed to the tetranucleotide and to a tRNA carrying the amino acid
(radioactive) that the tetranucleotide coded for.
B. If the following table resulted from such an experiment, what could he conclude? (8 pts)
Sequence used labeled amino acid filter radioactive?
AGCU 3 no
AGCU 9 no
AGCU 11 yes
AGCU 15 no
CGAU 2 no
CGAU 6 yes
CGAU 8 no
CGAU 14 no
UCAG 1 yes
UCAG 4 no
UCAG 10 no
UCAG 14 no
GCUU 5 no
GCUU 7 no
GCUU 12 no
GCUU 13 no
AGCC 3 no
AGCC 8 no
AGCC 11 yes
AGCC 15 no
Amino acid 1 is coded by UCAG. Amino acid 6 is coded by CGAU. Amino
acid 11 is coded by both AGCU and AGCC and therefore the genetic code is
degenerate. Also, GCUU code for none of the 15 amino acids and therefore
may possibly serve the role of stop codons.
10. What was the topic of our first debate? (5 pts)
Should scientists awarding a limited amount of available federal
dollars for research choose human “adult” stem cell projects showing promise
already (umbilical cord; bone marrow, etc.) over projects involving human
embryonic stem cells (controversial, untested, but potentially very powerful)?