PSwk7key - Problem
Set
Week
7
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Unformatted text preview: Problem
Set
Week
7
 MCB110,
Spring
‘11
 
 1.

True
or
false:
 ___F____
TBPs
and
TAFs
are
part
of
TFIIA

 ___
F
___
Eukaryotic
core
promoters
contain
a
‐35
and
‐10
region
 ___
T___RNAP
II
is
responsible
for
transcribing
nearly
all
genes
 ____F___Eukaryotic
core
promoter
recognition
requires
3
proteins
 ____F___There
are
2
phosphorylation
steps
in
bacterial
transcription
initiation
 ___
T
___CpG
islands
are
commonly
found
upstream
of
housekeeping
genes
 ___F___
The
gel
used
in
Gel
Shift
experiments
(EMSA)
is
denaturing
 ___
F
__
DNaseI
footprinting
assay
is
an
in
vitro
way
to
test
whether
a
protein
 activates
transcription
It
only
tells
you
whether
a
protein
binds
to
DNA
 ___
F
___
Most
DNA
binding
domains
of
transcription
factors
are
devoid
of
arginines
 and
lysines
 ___
T___
High
salt
washes
in
affinity
chromatography
get
rid
of
nonspecifically
or
 weakly
interacting
proteins.
 
 2.
List
the
functions
of
the
three
different
eukaryotic
RNA
polymerases.
 
 RNAP
I

makes
pre‐rRNA
for
18,
5.8,
and
28S
rRNAs,
makes
1
gene
 RNAP
II
makes
pre‐mRNA
and
some
snRNAs,
transcribes
nearly
all
genes
 RNAP
III
makes
pre‐tRNAs,
5S
rRNA,
some
snRNAs,
makes
30‐50
genes
 
 
 3.

If
you
are
given
a
candidate
RNA
Polymerase
protein
how
might
you
determine
 which
type
of
RNA
Polymerase
it
is?
 
 Test
the
protein
for
RNAP
activity
in
the
presence
or
absence
of
alpha‐amanatin.


 
 If
the
protein
is
not
sensitive
then
it
is
RNAP
I.

 If
the
protein
is
sensitive
to
alpha‐amanatin
at
1
ug/ml
then
it
is
RNAP
II.
 If
the
protein
is
sensitive
at
10
ug/ml
then
it
is
RNAP
III.
 
 
 
 
 
 
 4.

Describe
how
Eukaryotic
RNAP
II
is
recruited
to
the
core
promoter.

How
is
this
 different
from
the
recruitment
RNAP
in
bacteria.
 In
Euks,

RNAP
II
is
recruited
to
the
core
promoter
by
multiple
different
 transcription
factors.
 In
bacteria,
only
sigma
factor
is
used
to
help
recruit
RNAP
to
the
promoter.
 
 5.

What
marks
the
1st
step
of
eukaryotic
promoter
recognition?
 TFIID
recognizing
and
binding
to
the
TATA‐box
 
 6.

Imagine
you
are
conducting
an
in
vitro
step‐wise
assembly
of
the
RNAP
II
Pre‐ initiation
complex
at
a
core
promoter.

 A.

What
would
be
the
specific
consequence
of
leaving
TFIID
out
of
the
assembly
 reaction?
 Failure
to
do
first
step
of
promoter
recognition.

TFIID
is
required
to
recognize
the
 core
promoter
or
TATA
box.

Therefore,
the
rest
of
the
PIC
can’t
assemble.
 
 
 B.
What
would
be
the
specific
consequence
of
leaving
TFIIE
out
of
the
assembly
 reaction?
 Couldn’t
assemble
TFIIH
into
the
PIC.

This
would
prevent
switching
from
the
closed
 complex
to
the
open
complex.
 
 C.
What
would
be
the
specific
consequence
of
leaving
TFIIH
out
of
the
assembly
 reaction?
 Couldn’t
switch
from
the
closed
complex
to
the
open
complex.

TFIIH
contains
 helicase
activities
that
facilitate
DNA
unwinding
to
produce
the
open
complex.


 
 
 
 

 7.

Describe
how
RNAP
II
is
regulated
during
transcription,
especially
during
 promoter
clearance
and
elongation.
 Phosphorylation
of
the
CTD
of
RNAP
II
regulates
RNAP
II
during
transcription.
 TFIIH
first
phosphorylates
Ser5
of
the
CTD
after
PIC
assembly.

This
allows
for
 promoter
clearance
and
pausing
to
allow
for
5’capping.
 In
a
second
phosphorylation
step,
pTEFb
phosphorylates
Ser2
of
the
CTD
of
RNAPII.

 This
results
in
release
from
pausing
and
therefore
allows
for
productive
elongation.
 
 
 8.

We
have
learned
that
a
gel
shift
assay
is
one
method
that
can
be
used
to
study
a
 protein‐DNA
interaction.

What
type
of
gel
is
used
in
this
assay?

What
are
the
 advantages
and
disadvantages
to
using
this
assay?

 Native
gel.

Protein
needs
to
be
folded
in
order
to
bind
the
probe
 Advantage:

Highly
sensitive
assay
whereby
only
0.2%
of
your
probe
needs
to
be
 bound
by
your
DNA‐binding
protein.
 Disadvantage:

You
need
a
highly
stable
protein‐DNA
complex.

You
don’t
learn
 precisely
which
DNA
bases
are
contacted/bound
by
your
protein
of
interest.
 9.

When
doing
an
in
vitro
transcription
assay,
what
components
must
you
include
 in
the
reaction?
 Reaction
buffer
must
contain:
rNTPs,
DNA
template,
RNAP
II

 
 
 
 
 
 
 
 10.

Imagine
you
are
doing
an
in
vitro
transcription
assay
to
look
for
the
ability
of
a
 particular
transcription
activator
to
activate
transcription
of
a
gene.

What
can
be
 used
as
a
positive
control
for
this
assay?
 For
the
positive
control
use
a
DNA
template
that
RNAP
II
can
transcribe
RNA
from.

 You
need
to
make
sure
your
purified
RNAP
II
is
active
for
RNA
synthesis.

So,
for
 example,
you
could
use
Sp1
and
a
DNA
template
with
a
GC‐rich
proximal
promoter.
 
 11.
If
you
want
to
test
whether
a
protein
is
a
transcription
repressor,
what
 technique
we’ve
learned
so
far
in
class
would
help
you
accomplish
that?
 In
vivo
transcription
assay;
in
vitro
techniques,
gel
shift
and
DNaseI
footpringint
can
 only
tell
you
whether
a
protein
binds
to
a
piece
of
DNA.
In
this
experiment,
you
 could
use
a
plasmid
containing
the
gene
encoding
the
potential
transcription
 repressor,
and
co‐transfect
it
(introduce
it
into
cells)
with
a
plasmid
containing
a
 reporter
gene
(often
a
luciferase).
One
of
your
negative
controls
should
be
reporter
 plasmid‐alone
transfection.
This
way,
if
your
luciferase
activities
(i.e.
your
 transcription
activities)
in
the
co‐transfected
cells
were
lower
than
reporter
alone
 cells,
then
your
protein
of
interest
is
potentially
a
transcription
repressor.
 
 12.
How
would
you
tell
whether
a
DNA
sequence
could
be
a
bona
fide,
true
enhancer
 sequence
without
doing
any
bench
work?
 Assuming
the
gene
for
which
this
DNA
sequence
is
an
enhancer
is
conserved
 evolutionarily,
you
can
align
the
DNA
sequence
of
interest
from
whatever
organism
 that
you
are
testing
(e.g.
humans)
with
similar
regions
in
other
animals.
By
 comparing
whether
this
sequence
is
present
in
the
genomic
sequences
of
other
 organisms,
you
can
tell
whether
it’s
a
true
enhancer
or
not.
Because
when
a
gene
is
 evolutionarily
conserved,
its
regulatory
regions
often
are
as
well.
 
 13.
You
discovered
a
transcription
factor
that
prefers
to
bind
promoters
thathave
a
 consensus
AT‐rich
binding
site
 a.
How
would
you
purify
this
protein?
 You
could
make
a
DNA
affinity
purification
column
where
many
copies
of
the
 consensus
AT‐rich
binding
sites
are
linked
to
a
matrix.
Your
transcription
factor
 should
preferably
bind
to
the
consensus
sequence
in
the
column,
thus
being
 enriched.
Meanwhile,
the
weakly
interacting
proteins
should
be
washed
away
by
 increasing
concentration
of
salt
solution.
 
 b.
If
there
are
three
AT‐rich
similar
but
slightly
different
binding
sites
in
the
 promoter
of
a
particular
target
gene
of
this
transcription
factor,
then
describe
two
 experiments,
one
in
vitro
and
one
in
vivo,
that
you
can
use
to
tell
exactly
which
site
 is
the
true
binding
site
 For
both
in
vitro
and
in
vivo
experiments,
you
need
to
generate
mutant
promoters
 containing
only
one
of
the
three
consensus
binding
sites
at
a
time
(by
deletion
of
 different
DNA
pieces).
 For
in
vitro
assay,
you
could
use
gel
shift
with
each
of
your
three
mutant
promoter
 fragments
plus
the
purified
transcription
factor
(DNA
alone
would
be
your
 controls).
If
your
protein
binds
to
only
one
of
the
three
regions,
it
would
produce
a
 shift
in
mobility
of
that
DNA
fragment.
(This
can
be
done
with
DNase
I
footprinting
 as
well,
but
gel
shift
is
more
sensitive
and
easier
to
read/perform)
 For
in
vivo
assay,
you
could
use
the
same
three
promoter
fragments
and
put
them
 into
a
plasmid
and
co‐transfecting
them
(one
at
a
time)
with
a
luciferase
reporter
 plasmid.
Then
test
for
luciferease
activity
as
in
problem
11.

 
 c.
What
initial
(bench)
experiment
could
you
do
to
search
for
new
direct
targets
of
 this
protein,
knowing
that
cycloheximide
is
a
commonly
used
small
molecule
that
 inhibits
translation
of
new
protein
products
(by
inhibiting
translocation
of
 ribosome)
 Direct
targets
mean
transcripts
that
are
specifically
upregulated
by
your
 transcription
factor
of
interest,
and
not
by
secondary
effects.
So
by
transfecting
cells
 with
your
protein
of
interest
(with
untransfected
cells
as
a
negative
control)
and
 immediately
inhibit
translation
of
new
protein
products
with
cycloheximide,
you
 ensure
all
the
transcripts
that
are
upregulated
are
direct
targets
of
your
 transcription
factor.
So
you
would
isolate
RNA
from
both
transfected
and
negative
 control
cells
after
cycloheximide
treatment,
make
cDNA
with
different
fluorophores
 attached,
mix
the
samples,
and
hybridize
them
on
to
DNA
microarrays.
Any
over‐ represented
gene
products
from
the
transfected
cells
comparing
to
untransfected
 would
be
tentative
direct
targets
of
your
transcription
factor.
 
 14.
Describe
the
three
major
DNA
binding
motifs
of
transcription
factors
that
we’ve
 covered
in
class.
What
are
some
of
the
common
features
shared
by
two,
or
among
all
 three?
 1.
C2H2
zinc
finger
where
2
His
and
2
Cys
residues
coordinate
a
zinc
atom;
2.
basic
 helix‐loop‐helix;
3.
leucine
zipper
(b‐Zip).
2
and
3
both
contain
many
basic
residues
 such
as
arginine
and
lysines
to
interact
with
the
negatively
charged
DNA
backbone.
 All
three
cannot
usually
act
on
their
own:
Zn
fingers
usually
work
with
other
Zn
 finger
domains
on
the
same
protein
(for
sequence
specificity
and
tighter
 interactions),
whereas
bHLH
and
b‐Zip
domain
containing
proteins
usually
form
 dimers
 15.
How
can
you
map
(determine
the
location
of)
the
DNA
binding
module
of
a
 known
transcription
factor?
What
about
the
dimerization
module?
What
assay
 would
you
choose
and
describe
the
experimental
set
up.

Hint:
for
both
you
can
use
 an
in
vitro
assay
we
talked
about,
and
for
the
former,
you
can
also
use
an
in
vivo
 technique
that
we’ve
mentioned.
 Again,
you
can
use
deletion
studies
to
truncate
the
gene
encoding
the
proteins
into
 different
mutants:
some
won’t
contain
the
DNA
binding
module,
some
will
contain
 neither,
and
some
won’t
contain
the
dimerization
module
(you
want
smaller
and
 smaller
pieces
to
pinpoint
the
minimal
domain
required
for
either
activity).
For
a
 protein
without
either,
you
probably
won’t
get
any
transcription
activation,
so
you
 could
probably
use
the
co‐transfection
of
mutants
+
luciferase
reporter
assays
that
 we
talked
about.
But
this
won’t
tell
you
whether
you
just
deleted
the
activation
 domains
of
the
protein.
 For
in
vitro
assays,
you
can
use
either
gel
shift
or
DNaseI
footprinting
to
analyze
 whether
a
specific
type
of
truncation
has
lost
ability
to
bind
DNA.
For
the
 dimerization
module,
you
could
use
gel
shift.
Since
dimers‐bound
DNA
and
 monomer‐bound
DNA
probably
shift
differently,
you
could
have
several
controls:
 naked
DNA,
DNA
with
WT
proteins
that
form
a
dimer
in
solution,
DNA
with
your
 truncated
mutants.
If
one
of
the
mutants
shifts
compared
to
naked
DNA,
but
not
as
 much
as
WT
protein
(i.e.
going
faster
than
the
WT
protein),
then
this
mutant
 probably
affects
the
dimerization
module
of
the
transcription
factor.
 ...
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