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 Problem
Set
Week
9
 
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Unformatted text preview: 
 Problem
Set
Week
9
 
 MCB110,
Spring‘11
 
 1. True
or
False
 
 ___T____
The
Mediator
complex
contacts
RNAP
II,
general
transcription
factors,
and
 specific
transcription
factors
 
 ___F___

Cell‐Type
Specific
transcription
factors
bind
only
within
proximal
promoter
 elements.


 
 ___F____
Glucocorticoid
receptor
GR
can
enter
the
nucleus
without
its
LBD
(ligand
 binding
domain)
 
 ___F____

mature
eukaryotic
mRNAs
can
be
exported
out
of
the
nucleus
without
a
 5’Cap
 
 ___F____

The
5’cap
of
eukaryotic
mRNA
contains
only
1
phosphate.

5’Cap
is
linked
 via
a
5’—5’
linkage
with
3
phosphates.
 
 ___T____

Splicing
reaction
occurs
through
2
transesterification
reactions
 
 ___T___

During
5’cap
addition,
Guanylyltransferase
is
recruited
and
activated
 through
binding
to
the
Ser5‐
phosphorylated
Pol
II
CTD
 
 ___T____

During,
addition
of
the
poly
A
tail,

PAP
(PolyA
polymerase)
binds
to
the
 complex
prior
to
cleavage
of
the
Poly
signal
sequence
 
 ___T_____

The
CTD
of
RNA
Pol
II
serves
as
a
scaffold
for
coupling
transcription
to
 mRNA
capping,
polyadenylation,
and
splicing
 
 ___
T
____
Splicing
factors
can
also
act
like
elongation
factors
by
facilitating
 elongation
 
 ___F____

All
of
the
steps
in
E.
coli
protein
synthesis
utilize
GTP
as
an
energy
source
 
 ___
F____
Aminoacyl‐tRNA
synthetase
recognizes
their
specific
tRNA
by
a
single
 interaction
at
the
anti‐codon
loop.

tRNA
Synthetases
recognize
tRNAs
by
tRNA
 shape
and
sequence
of
anticodon
loop.
 
 ___T_____
Mistakes
in
decoding
mRNA
into
amino
acid
sequences
usually
occur
and
 are
fixed
during
matching
an
amino
acid
to
the
appropriate
tRNA.
 
 ___F____
Eukaryotic
and
prokaryotic
translation
are
both
directed
by
a
consensus
 Shine‐Dalgarno
sequence.

Bacteria
recognize
the
Shine‐Dalgarno
Sequence
using
 16S
rRNA
in
the
30S
subunit.

Eukaryotic
translation
initiation
begins
with
the
small
 40S
subunit
+eIF‐2‐Met‐tRNA
recognizing
the
5’Cap
of
mRNA
followed
by
scanning
 of
the
mRNA
in
the
5’—3’
direction
for
the
first
AUG
start
codon.


 
 
 2. Why
are
insulator/boundary
elements
important
in
transcriptional
 regulation?

 
 Insulator
and
boundary
elements
assist
the
cell
in
differentiating
between
genes
 that
need
to
be
actively
transcribed
and
those
that
should
be
repressed.
These
 elements
are
particularly
important
for
preventing
enhancers
from
acting
on
the
 wrong
promoters
and
preventing
spreading
of
silenced
chromatin.

 
 3. How
is
the
5’
G7
methyl
cap
added
to
nascent
mRNAs?

 
 The
cap
is
added
after
20‐30
nucleotides
of
the
mRNA
is
made.
Phosphohydrolase
 and
guanylyltransferase
form
a
complex.
Phosphohydrolase
hydrolyzes
a
 phosphate
group
on
the
nascent
strand
that
allows
for
binding
of
the
guanine
by
 guanyltransferase.
Guanyltransferase
adds
the
guanine
in
5’‐5’
linkage.
PPi
is
 released.
Then
guanine‐7‐methyltransferase
uses
S­adenosylmethionine
to
add
 a
methyl
group
to
the
7’
N
of
guanine.

 
 4. In
class
we
discussed
the
importance
of
the
Poly
A
Signal
sequence
in
Poly
A
 tail
addition.

Describe
an
experiment
that
you
could
do
to
verify
whether
the
 entire
polyA
signal
sequence
is
necessary
for
cleavage.
 
 Mutate
each
base
of
the
poly
A
signal
sequence.

Then
perform
in
vitro
transcription
 +
polyadenylation
assays.

Add
all
of
the
proteins
required
for
polyadenylation
 would
would
be
added
to
your
reaction
buffer
(CPSF,
CstF(Cleavage
stimulatory
 factor)
CF1,
CF2,PAP,
PABPII).


 
 Run
mRNA
on
gel,
northern
blot
to
check
for
polyA
tail
addition.

If
you
observe
no
 polyA
tail
addition
then
mutant
results
in
no
cleavage
of
polyA
signal
sequence
and
 therefore
no
subsequence
polyA
tail
addition
by
Poly
A
Polymerase
(PAP).

 
 5.

If
you
are
given
the
DNA
sequence
for
a
gene.

How
can
you
identify
the
introns
of
 this
sequence?

Hint:

Remember
there
are
5
conserved
elements.


 5’SS,
3’SS,
BPS,
Pyrimidine‐rich
tract,

Conserved
distance
between
BPS
A
and
3’SS
 (20‐50
bp)
 
 
 6.
Why
is
a
gene
so
much
bigger
than
an
mRNA?
 A
gene
and
its
corresponding
pre‐mRNA
contain
introns
that
are
spliced
out
to
make
 a
shorter
mature
mRNA.
 
 
7.
What
are
two
benefits
of
splicing?
 Alternative
splicing
generates
different
proteins
from
a
single
gene,
therefore
more
 diverse
protein
function.
 Introns
increase
the
chance
for
exon
shuffling
through
recombination,
therefore
 providing
evolutionary
diversity.

 
 8.
Name
two
RNA
modifications,
where
they
occur
on
the
RNA,
where
in
the
cell
the
 RNA
is
modified
and
what
role
the
modifications
serve
in
protein
synthesis.
 5’
cap
and
polyA
tail
are
added
to
the
mRNA
in
the
nucleus.

Both
modifications
are
 required
for
nuclear
export
and
both
enhance
RNA
stability
and
translation.
 9.

Describe
the
2
steps
of
the
Spliceosome‐mediated
splicing
reaction?
 Step1:

The
2’OH
of
the
BPS
A
attacks
the
phosphodiester
bond
of
the
5’Splice
site.

 This
results
in
cleavage
at
the
5’SS
and
covalent
linkage
of
the
branch
point
A
to
the
 5’end
of
the
intron.

 Step2:

3’OH
of
upstream
exon
attacks
the
3’Splice
site.

This
results
in
excision
of
 the
intron
as
a
lariat‐like
structure
and
ligation
of
the
exons.
 
 10.

How
are
the
5’Splice
Site
and
the
BPS
(Branch
point
site)
recognized
by
the
 spliceosome?
 Both
site
are
recognized
through
RNA‐RNA
base‐pairing
interactions.
 U1
snRNA
(of
U1
snRNP)
recognizes
the
5’SS
and
U2
snRNA
(of
U2
snRNP)
 recognizes
the
BPS.
 
 11.
You
are
studying
a
gene
called
TREE.

You
discover
that
a
mutation
within
TREE
 residing
within
the
5’Splice
site
of
exon
1.

This
mutation
results
in
failure
to
 undergo
splicing.

You
observe
that
the
mutation
is
not
within
the
GU
sequence
of
 the
5’SS.
 What
would
you
hypothesize
is
the
reason
this
mutation
result
in
a
failure
to
do
 splicing?
 This
mutation
likely
prevents
base‐pairing
with
U1
snRNA.
 
 What
experiment
would
you
do
to
support
your
hypothesis?

 Perform
a
compensatory
mutation
in
the
U1
snRNA
that
restores
complementarity
 of
the
U1
snRNA
with
the
mutant
5’SS
of
TREE.
 
 12.

You
are
working
with
a
gene
called
PANDA
containing
6
Exons.

Heart
cells
 express
a
version
of
PANDA
which
includes
all
exons
whereas
muscle
cells
express
a
 version
of
PANDA
that
doesn’t
contain
Exon
4.


 Describe
an
experiment
which
would
allow
you
to
determine
whether
mRNA
 transcripts
made
from
either
heart
or
muscle
cells
contain
Exon
4.
 Perform
a
northern
blot
using
a
probe
complementary
to
Exon
4.

On
your
northern
 blot,
mRNA
from
Heart
cells
will
show
up
whereas
mRNA
from
muscle
cells
will
not.
 
 13.

What
role
do
SR
proteins
play
in
splicing?
Explain
in
both
general
and
specific
 terms.
 Generally,
SR
proteins
serve
a
role
in
constitutive
and
alternative
splicing
by
 defining
an
exon
to
be
included
into
the
final
mRNA
transcript.
 Specifically,
SR
proteins
bind
to
exonic
splicing
enhancers
(ESEs)
located
within
an
 exon.

SR
proteins
bound
to
ESEs
serve
to
promote
cooperative
binding
of
U1
snRNP
 to
the
5’SS
of
the
downstream
intron,the
65‐
and
35
kDa
subunits
of
U2AF
to
the
 pyrimidine‐rich
region,
and
the
3’SS
of
the
upstream
intron.
This
complex
promotes
 RNA
splicing
allowing
the
exon
to
be
included
into
the
mature
mRNA
transcript.
 
 14.
Describe
the
four
sequence
components
of
the
tRNA
that
all
of
them
share.
 CCA‐3’
end
in
the
acceptor
stem
 TψCG
loop
that
has
ribothymidine
and
pseudouridine
 D
loop
which
contains
dihydrouridine
nucleotides
 Anti‐codon
loop
that
base
pairs
with
the
mRNA
sense
codon.
 
 15.
How
can
an
aminoacyl
tRNA
synthetase
fix
a
mistake
in
attachment
of
the
wrong
 amino
acid
to
a
tRNA?
 The
aminoacyl‐tRNA
synthetase
has
many
contacts
that
recognize
the
shape
and
 sequence
of
the
anticodon
loop
and
the
whole
tRNA
molecule
itself.
If
a
wrong
tRNA
 binds
to
this
enzyme
and
receives
an
amino
acid,
the
enzyme
has
the
ability
to
 hydrolyze
that
bond
again
once
the
correct
tRNA
molecule
displaces
the
incorrect
 one.
Displacement
and
hydrolysis
are
driven
by
the
more
favorable
extensive
 interactions
with
the
correct
tRNA.
 
 16.
Consider
prokaryotic
translation
and
answer
the
following
questions:
 a.
Which
subunit
contains
the
E,
P,
A
sites?
 The
smaller
30S
subunit
 b.
What’s
special
about
the
first
amino
acid
to
a
polypeptide?
What
does
the
 modification
prevent?
 fMet
instead
of
regular
Methionine.
It
cannot
form
a
peptide
bond
at
its
amino
group
 because
of
the
formyl
functional
group
that
is
added
by
transformylase
from
a
 formyl
donor
N10‐formyl
tetrahydrofolate.
This
way,
all
further
amino
acids
have
to
 be
added
to
its
carboxy
end
only,
elongating
the
polypeptide.
 c.
Since
formylmethione
and
methione
are
both
conjugated
to
the
same
tRNA,
how
is
 it
possible
for
cells
to
ensure
a
fMet
is
placed
at
the
AUG
start
codon
and
that
Met
is
 placed
elsewhere?
 tRNA
is
bound
to
IF‐2,
an
initiator
factor,
and
thus
selectively
guided
to
the
 ribosome
during
translation
initiation
 
 d.

At
the
initiation
of
translation,
what
are
the
roles
of
IF‐1
and
IF‐3?
 IF‐1
binds
to
the
A
site
to
prevent
premature
binding
of
tRNAs
to
the
A
site
 IF‐3
binds
to
30S
and
prevents
premature
binding
of
the
50S
subunit,
also
hels
IF2
 for
specific
binding
of
fMet‐tRNAfMet.
 
 17.
Howdoes
the
ribosome
know
where
to
start
in
both
prokaryotes
and
 eukaryotes?
What
part
of
the
ribosome
is
involved
in
recognition?
 3’
end
of
the
16S
RNA
component
of
the
prokaryotic
ribosome
needs
to
base
pair
 with
a
sequence
~10nt
away
from
AUG
start
codon,
called
the
Shine‐Dalgarno.
This
 tells
the
ribosome
where
to
start
translation.
In
eukaryotes,
the
40S
subunit

(with
 eIF‐2‐Met‐tRNA
bound)
recognizes
the
5’
cap,
then
the
ribosome
scans
for
the
first
 AUG
codon
to
begin
translation.
 ...
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This document was uploaded on 09/12/2011.

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