MBB 206 Exam 2 May 19, 2004 30 points

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MBB 206 Exam 2May 19, 2004 30 points

• Please put your Name and Student I.D. Number in
  the upper right corner of this page.
• Please put your Student I.D. Number on every page
  of this exam
• There are 3 questions. Each question is worth 10
  points.

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Answer key

• I have entered the answers in blue under each
  question. In several instances, there were
  alternate explanations or possible pathways
  given. As long as the answer was scientifically
  correct and you could justify your model, I
  accepted the answer.

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Functional interactions of RNA-capping enzyme
withfactors that positively and negatively
regulatepromoter escape by RNA polymerase IIS.
S. Mandal, C. Chu, T. Wada, H. Handa, A. J.
Shatkin, and D. Reinberg PNAS May 18, 2004
vol. 101, 75727577
Question 1 Background Information

• There is increasing evidence that capping occurs
  cotranscriptionally and is facilitated by the
  association of capping enzyme (CE) with
  transcribing RNAPII complexes through interaction
  with the phosphorylated carboxyl-terminal domain
  (CTD) of RNAP II. The CTD consists of a tandemly
  repeated YSPTSPS motif that undergoes extensive
  serine phosphorylation and dephosphorylation
  during the transcription cycle. The CTD serves as
  a landing platform for, and regulator of, many
  transcription, splicing, polyadenylation, and
  termination factors. Several kinase activities
  have been implicated in CTD phosphorylation, and
  primarily one CTD-specific phosphatase, FCP1,
  serves to recycle RNAPII.
• Transcription begins with CTD phosphorylation at
  Ser-5 by the kinase activity of the Cdk7 subunit
  of transcription factor TFIIH.
• It is believed that during the formation of the
  transcription initiation complex, or soon after
  initiation, DRB sensitivity-inducing factor
  (DSIF) is recruited to the transcription complex.
  Additionally, after initiation of transcription,
  the negative elongation factor (NELF) is
  recruited through interaction with DSIF. This
  results in the arrest of the transcription
  complex before it enters into productive
  elongation, giving ample time for the recruitment
  and catalytic actions of the CE on nascent
  transcripts of 1722 nt.
• DSIF-NELF mediated arrest is then relieved by
  means of phosphorylation of the CTD at Ser-2 by
  positive transcription elongation factor b
  (P-TEFb), and the transcription complex resumes
  elongation.

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Question 1
Fig. 4. (A) Ternary 24 complexes alone (lane 1)
or incubated with 0.6 ng of CE (capping enzyme)
and GTP for 5 min at 30C (lane 2). Ternary
complexes were incubated with 60 ng of
FCP1(CTD-specific phosphatase) for 30 min at
30C, washed to remove FCP1, resuspended in TB60,
and treated with CE and GTP in the absence (lane
3) or presence (lanes 46) of varying amounts of
DSIF (DRB-sensitivity inducing factor) . As
control, complexes not treated with FCP1 were
incubated with DSIF and CE (lanes 78). (D)
Quantification of (A). (E) Ternary 24 complexes
were treated with FCP1 for 30 min, and then
treated and untreated complexes were incubated
with 0.6 ng CE in the absence and presence of 30
ng of DSIF. Reactions were stopped at the
indicated times, and the percentage of capped RNA
was determined by SDS-PAGE and autoradiography.
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Question 1- 10 points

• Please answer the following questions regarding
  Fig. 4 on the previous page
• What would conclude from Figure 4A (which is also
  shown quantified in Fig. 4D)? Why would you
  conclude this?
• Figure 4A demonstrates that dephosphorylation of
  the CTD by FCP1 inhibits capping because CE is
  recruited to the phosphorylated CTD. Both Ser-5
  and Ser-2 phosphorylated CTD can bind to CE but
  Ser-5 stimulates the activity of CE. DSIF appears
  to recruit CE to the CTD regardless of the
  phosphorylation state so it can overcome the
  effect of FCP1.
• What would you conclude from Fig. 4E? Why?
• DSIF can stimulate the activity of CE, possibly
  by recruitment, over and above the native
  activity of CE alone even when FCP1 is not
  present so presumably the CTD is appropriately
  phosphorylated.
• If after treating with FCP1 and extensively
  washing away the FCP1 , you then added P-TEFb,
  which phosphorylates Ser-2 on the CTD, what
  effect would you expect to see on capping in the
  absence of DSIF? In the presence of DSIF? Why
  would you expect to see these effects?
• P-TEFb phosphorylates on Ser-2, which binds CE
  but in vivo, CE binds to Ser-5 phosphorylated CTD
  and disassociates when elongation (Ser-2)
  phosphorylation begins. DSIF can stimulate CE so
  in the presence of DSIF, capping would occur and
  in the absence of DSIF, capping would be
  inefficient.

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Analysis of the requirement for RNA polymerase II
CTDheptapeptide repeats in pre-mRNA splicing
and3-end cleavageE. ROSONINA and B. J.
BLENCOWERNA (2004), 10581589.
Question 2

• Here, we demonstrate that 22 tandem repeats, from
  either the conserved or divergent halves of the
  CTD, are sufficient for approximate wild-type
  levels of transcription, splicing, and 3-end
  cleavage of two different pre-mRNAs, one
  containing a constitutively spliced intron, and
  the other containing an intron that depends on an
  exon enhancer for efficient splicing.
• In contrast, each block of 22 repeats is not
  sufficient for efficient inclusion of an
  alternatively spliced exon in another pre-mRNA.
  In this case, a longer CTD is important for
  counteracting the negative effect of a splicing
  silencer element located within the alternative
  exon.
• Our results indicate that the length, rather than
  the composition of CTD repeats, can be the major
  determinant in efficient processing of different
  pre-mRNA substrates.

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FIGURE 3. Truncation of the CTD results in
reduced inclusion levels of the alternatively
spliced CD44 v5 exon. (A) Diagram of the pETv5 wt
reporter minigene used to monitor alternative
splicing. The variant v5 exon, from the human
CD44 gene, is flanked by native CD44 intron
sequences (indicated by a bold line) and
intron/exon sequences from the constitutively
spliced exon 2-intron 2-exon 3 region of the
human insulin gene. Inclusion of v5 was detected
by RT-PCR using primers specific for the
constitutively spliced insulin exons (B) Human
293 cells were transfected with expression
plasmids for the pol II deletion mutants
indicated (shown above Fig. 3) and with pETv5 wt.
Recovered RNAs were analyzed by RT-PCR. Spliced
RNAs containing (included) or lacking v5
(excluded) are indicated. The level of actin RNA
was determined by RT-PCR as a control for RNA
recovery. (C) The percent v5 inclusion measured
from three independent experiments, with the
average level and standard deviations shown.
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Question 2- 10 points

• Please answer the following questions regarding
  Fig. 3 on the previous page
• The data shown in Fig. 3 indicate that the full
  length CTD is required for efficient inclusion of
  exon v5 in the mRNA and that this exon is skipped
  when the pre-mRNA is transcribed by the two
  shorter CTDs. Further, there is an ESS, exonic
  splicing silencer, in the v5 exon, which binds an
  hnRNP protein that interferes with binding of
  U2AF to the 3 splice site upstream of v5,
  thereby negatively regulating inclusion of this
  exon thus, making it a weak 3 splice site.
  Propose a model for why transcription by the
  longer CTD results in v5 exon being included in
  the mRNA, whereas this exon is skipped with the
  shorter polymerases. In making this model, assume
  the full length (longer) CTD has a higher
  processivity (it is faster) than the two shorter
  CTDs. You may also assume that the shorter CTDs
  are slower because they pause at different sites
  during transcription. Finally, assume that the
  hnRNP protein will bind to the ESS in exon v5
  within a few minutes after this part of the
  pre-mRNA is extruded from the polymerase. Propose
  a model that incorporates all of these points.
• What would happen if you mutated the ESS such
  that the hnRNP protein cannot recognize the site
  to bind to it? Would you expect to see a
  difference in exon inclusion/exclusion between
  the shorter and longer polymerases? Why or why
  not? (Make sure your answer fits your model).

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Question 2

• 1. The RNA polymerase with the longer CTD
  transcribes faster. Therefore, RNA pol II would
  transcribe the v5 exon region and move to exon 3.
  There would be a competition between hnRNP
  binding to the ESS in exon v5 and U2AF binding to
  the 3splice site upstream of v5. If hnRNP gets
  there first, splicing at the 3 upstream site is
  prevented and v5 is included in the mRNA. With a
  slower polymerase, as seen with the CTD
  truncations, there is pausing by the polymerase
  as the pre-mRNA is synthesized. Therefore, RNA
  pol II may be paused in v5 preventing the binding
  of hnRNP because the ESS site has not yet been
  synthesized or because it has not yet been
  extruded from the elongating polymerase. Splicing
  would occur at the upstream 3 splice site and v5
  would be spliced out of the mRNA, thus this
  alternate exon would be skipped.
• 2. If you mutate the ESS, hnRNP would not
  recognize it to bind so there would be little
  difference between the slower and faster
  polymerases. The alternate exon would be excluded
  because the upstream 3 splice site is no longer
  weak in the absence of the ESS.

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Pre-mRNA splicing and mRNA export linked by
direct interactions between UAP56 and AlyM. Luo,
Z. Zhou, K. Magni, C. Christoforides, J.
Rappsilber, M. Mann R. ReedNATURE VOL 413
11 OCTOBER 2001
Question 3- 10 points
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Question 3
What does the data in Figure 2a tell you about
where UAP56 is found? Why? UAP56 binds to
pre-mRNA, suggesting it is part of the splicing
complex and it binds to mRNA suggesting that it
remains bound after splicing and may have a role
in export. What does the data in Figure 2b
tell you about where UAP56 is found as splicing
progresses? Why? UAP56 associates with the mRNA
as splicing progresses, further suggesting a role
beyond splicing, such as export.
Figure 2. a) AdML pre-mRNA was incubated in HeLa
nuclear extract under splicing conditions for 45
min. After gel filtration, the fraction
containing spliceosomes and spliced mRNP was
incubated with protein A beads coupled to
control, UAP56 or Aly antibodies. b) Spliceosomal
complexes assembled on AdML pre-mRNA for the
indicated times were purified. RNA (top panel) or
western blot of purified complexes were probed
with UAP56 antibody. NE, nuclear extract
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Question 3 continued
Figure 3. a) Buffer or His-UAP56 (1x 140 fmol)
was injected into Xenopus oocyte cytoplasm. AdML
pre-mRNA, pre-tRNAs and U1 and U6 snRNAs were
then injected into nuclei. Oocytes were incubated
for 3 h. b) Buffer or 4x His-UAP56 was injected
as in a. ftz pre-mRNA, pre-tRNA and U1 and U6
snRNAs were then injected into nuclei and
incubated for 2 h. c) Same as b except that CTE
RNA and U6 snRNA were injected into nuclei and
incubated for 75 min. RNA from the nucleus (N) or
cytoplasm (C) was analyzed.
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Question 3 continued

• 1. What do these results tell you about export of
  AdML mRNA in the presence of excess UAP56?
• AdML export is inhibited by excess UAP56.
• 2. Why would excess UAP56 have an effect on AdML
  mRNA export?
• UAP56 binds Aly/REF and recruits it to the EJC.
  Excess UAP56 probably binds all available Aly and
  sequesters it so it is not available for RNA
  binding and export.
• 3. Why does excess UAP56 not have an effect on
  CTE export?
• CTE binds directly to TAP and doesnt require
  Aly or UAP56 as bridging or adapter proteins.
• 4. Why does excess UAP56 not have an effect tRNA
  export?
• tRNA is exported by a nuclear export receptor
  from the karyopherin-exportin family termed
  Exportin-t (Exp-t). It does not use Tap and does
  not require Aly or UAP56, which are involved in
  mRNA export only.

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