Download Chapter 13 Chromatin Structure and its Effects on

Chapter 13
Chromatin Structure
and its Effects on
Transcription
Students must be positive that they understand
standard PCR.
There is a resource on the web for this
purpose.
Warn them before this class.
Notes
• Do not attempt to interpret figure 13.14.
• Figure 13.18. The concentration of
nucleosomes does not prevent the
restriction enzyme from finding its cut
sites in many molecules (in some the
nucleosomes block the cut site).
Histone
Eukaryotes five
different Histone
Classes
size amino acids
Molecular weight
H2A
150
14,000
H2B
150
13,770
H3
150
15,400
H4
100
11,340
H1 (H5)
200
21,500
Eukaryotes contain many copies of each histone gene.
10-20 in mice, 100X in Drosophila
Compaction
Double
helix
Beads
On
A
string
solenoid
Snake
The
Solenoid
PR 25
700 nm fiber
Probably involve SARS
Nucleosome
DNA
Protein
Core nucleosome contains
2X H2A, 2X H2B, 2X H3 & 2X H4
Beads on a string
Beads on a string
packing ratio 5
Histone H1
• Outside of the core
• Easiest to remove by high salt
H2B
extraction
H2 A H3
H1
H4
H2 B
DNA
Solenoid
27Å
DNA
H1 hist one
Nucleosome
cor e
30 nm solenoid
6 per turn
Packing ratio of 8
57Å
Tetranucleosome Fig 13.7
45-80 kb loops
The movie was deleted to save space. You
can download it separately.
Evidence that
histones help to
regulate gene
• Xenopusexpression
laevis 5s rRNA (20,000
copies)
• oocyte 5S rRNA genes, expressed only
in oocytes - 98%
• somatic 5S rRNA genes, expressed in
both oocytes and somatic cells - 2%
Chromatin is required
for specificity
• With DNA, RNA polymerase III
transcribes both well
• With oocyte chromatin, both expressed
• With somatic cell chromatin only
somatic 5S rRNA genes expressed.
Somatic cell
chromatin
• Inactive oocyte genes contained all 5
histones*
• Active somatic genes contain only core
histones
• Remove H1 & oocyte genes turn on.
Add it back and they turn off.
pg 369 Weaver 4th ed.
Nucleosome
compete with
transcription
factors
• This is too simple to
allow one to modulate
expression or is it?
mRNA encoding
genes Class II genes transcribed by RNA
polymerase II
• Core histones cause mild (4X)
repression of gene expression
• Activators do not affect this
• H1 increases repression (25 to 100X)
•
Activators can prevent this - similar to the 5S genes.
Laybourne &
• Chromatin
form of the Drosophila
Kadonaga
1991
Krüppel gene
• nuclear extract transcribes it at 25% of
the maximum rate. This is 75%
repression.
• Interpretation
1) 100% of the genes are transcribed at
a 75% rate of transcription
2) 25% of the genes are transcribed at
100% of the rate.
• 25% of promoters unoccupied. How
pg 370 Weaver 4th ed
Histones can act as
repressors
• Thus, via
competition for
binding sites,
transcription
factors can derepress (your
book calls it
antirepression.
Histone H1
Nucleosome core
•
•
•
Histone
antirepression
With respect to
histone repression,
Gal4 acts as an
antirepressor. In
addition, it acts as
an activator.
SP1 acts as both a
histone antirepressor
and as an activator.
GAGA factor seems
to only act as a
histone
antirepressor.
De-repression or Anti-repression is the
amount of transcription that you get
when the histone is not interfering by
hiding the promoter.
•
•
•
Histone
antirepression
With respect to
histone repression,
Gal4 acts as an
antirepressor. In
addition, it acts as
an activator.
SP1 acts as both a
histone antirepressor
and as an activator.
GAGA factor seems
to only act as a
histone
antirepressor.
Yaniv saw that some transcriptionally
active SV40 virus DNAs had
nucleosome free zones.
Fig. 13.21
Weaver 3rd ed.
Figure 13.17
Is the promoter region of an active
gene a nucleosome free zone?
Figure 13. 18
BamHI
BamHI
BamHI
BglII
BglII
BglII
DNase I hypersensitivity
Transcribing SV40 virus isolated from infected monkey cell tissue culture.
Two fragments
Figure
13. 20
Gel
electrophoresi
s
& Southern
Blotting
This is a Southern Blot.
DNase I hypersensitivity in Chromatin lectures
Is this caused by the promoter or something else near the promoter? Can this occur with the promoter at a
different location? Try a modified SV40 that has a second promoter inserted.
Does transcription cause this or is it caused by something else that binds the promoter? Do we need to have
transcription for this to occur? Make a nuclear lysate that supports transcription & chromatin assembly. Then
deplete it for RNA polymerase II. Or starve for nucleotides.
Is this peculiar to the SV40 promoter. Try a plasmid with a completely different type of promoter.
Does the promoter have to be active for this to happen? Try this with a promoter that can be turned on and off.
eg. Tet-on.
Why does the smaller band disappear?
*Less intense because amount of probe that it can collect is smaller. Test w/2 probes of the same size.
*Less abundant, because transcription is going that way and polymerarse might expand the nucleosome clear
zone to the left. Test by reversing the direction of the plasmid.
What is the origin of the 100% band. Does it confound our interpretation?
Histone acetylation
• amino groups of lysine side chains
• unacetylated histones tend to repress
transcription
• acetylated histones tend to activate
transcription
• Histone acetyl transferase (HAT)
• Histone deacetylase
- see Figure 13.23 for how to detect them.
Histone acetylation
• First discovered by Vincent Allfrey in
1964 takes ‘till 1996 (Brownell & Allis) to
purify a HAT
• Tetrahymena has heavily acetylated
histones.
• Take macronuclei extracts. SDS gel
electrophoresis in gel impregnated with
histones. Soak in radiolabeled acetylCoA. Wash. Fluorography.
You don’t have this slide.
How to purify a HAT
• What do you know about it?
Acetylates histones
• What does this
knowledge provide you?
A way to assay for its presence
• What is the advantage of
using a gel?
Demonstrates that one is detecting the
presence of a single enzyme. Tells
you the relative size of the enzyme.
heated
chemical inactivation
BSA no histones
no protein
Now purify it.
• Standard biochemical techniques to
purify p55.
• Can use the assay to tell where it is.
• Once pure get partial amino acid
sequence. Very small number of amino
acids.
Degenerate PCR
KGWMDIM
NMVTMMV
mRNA
DNA
Degenerate PCR
KGWMDIM
NMVTMMV
mRNA
DNA
Degenerate PCR
KGWMDIM
NMVTMMV
mRNA
AAAAAAA
REVERSE TRANSCRIBE TO DNA
Degenerate PCR
KGWMDIM
NMVTMMV
Codon Usage by amino acid.
F
L
S
Y
TTT
TTA
TCT
TAT
TTC
TTG
TCC
TAC
CTT
TCA
CTC
TCG
CTA
AGT
CTG
AGC
I
ATT
ATC
ATA
M
ATG
T
ACT
ACC
ACA
ACG
N
AAT
AAC
C
TGT
TGC
W
TGG
P
CCT
CCC
CCA
CCG
H
CAT
CAC
Q
CAA
CAG
R
CGT
CGC
CGA
CGG
AGA
AGG
K
AAA
AAG
V
GTT
GTC
GTA
GTG
A
GCT
GCC
GCA
GCG
D
GAT
GAC
E
GAA
GAG
G
GGT
GGC
GGA
GGG
Degenerate PCR
KGWMDIM
NMVTMMV
UPPER STRAND Primer cocktail
K
G
W
M
5' AA(AG) GG(N) TGG ATG
2 X
4 X
1 X
1 X
D
I
GA(TC)
2 X
M
AT(TCA) ATG 3'
3 X
A 21 mer primer with 48 fold degeneracy.
1 = 48
Degenerate PCR
KGWMDIM
NMVTMMV
LOWER STRAND Primer cocktail
N
M
V
T
M
M
V
5' AA(TC) ATG GT(N) AC(N) ATG ATG GT(N) 3'
Take the reverse complement for the lower strand.
The reverse complement is merely the opposite strand in a DNA
helix also written in the 5' to 3 direction.
5' (N)AC
4 X
CAT
1 X
CAT
1 X
(N)GT
(N)AC
4 X
4 X
CAT
1 X
(AG)TT 3'
2 = 128
Degenerate PCR
LOWER STRAND Primer cocktail
N
M
V
T
M
M
V
5' AA(TC) ATG GT(N) AC(N) ATG ATG GT(N) 3'
Take the reverse complement for the lower strand.
The reverse complement is merely the opposite strand in a DNA
helix also written in the 5' to 3 direction.
5' (N)AC
CAT
4 X
CAT
1 X
1 X
(N)GT
(N)AC
4 X
4 X
CAT
(AG)TT 3'
1 X
2 = 128
If we wish to have a less degenerate cocktail what can we do?
Let's shave off the 5' end by one base to reduce the degeneracy
to 32 fold.
5'
AC
CAT
CAT
Degeneracy is
1 X 1 X 1 X
(N)GT
4 X
(N)AC
4 X
CAT
1 X
(AG)TT 3'
2 = 32 fold degeneracy.
Degenerate PCR
D N F N R Q K Q K L G G E D L F M T E E Q K K Y Y N A M K K L G S K K
GAYAAYTTYAAYMGNCARAARCARAARYTNGGNGGNGARGAYYTNTTYATGACNGARGARCARAARAARTAYTAYAAYGCNATGAARAARYTNGGNWSNAARAARG
2 2 2 2 6 2 2 2 2 6 4 4 2 2 6 2 1 4 2 2 2 2 2 2 2 2 4 1 2 2 6 4 6 2 2
L G G E D L F
YTNGGNGGNGARGAYYTNTTY
6 4 4 2 2 6 2 = 4608
D N F N R Q K ...........................................K K Y Y N A M
GAYAAYTTYAAYMGNCARAAR..........................................AARAARTAYTAYAAYGCNATG
2 2 2 2 6 2 2 = 384.....................................2 2 2 2 2 4 1 = 128
Ambiguous Bases
A-adenine
B-not A
C-cytosine
G-guanine
H-not G
K-G or T
M-A or C
N-A, C, G or T
R-A or G
S-C or G
T-thymine
U-uracil=T
V-not T
W-A or T
Y-C or T
- = gap
Screen a cDNA
library
• DNA copies of every mRNA made by a
cell are produced and cloned into a
bacterial vector.
• Screening.
• Google.
5’-RACE
Figure 4.16
HAT type A
• Have bromodomain
• Binds aceylated lysines. So HAT A’s
can recoginze partially acetylated
histone tails.
• Examples p55, Gcn5p CBP/p300,
TAF250
Acetylation continued
• Acetylation of histone tails neutralizes
some of the positive charge, causing
them to relax their grip on the DNA.
• Reduces nucleosome cross-linking.
That is; the interaction between
histones in neighboring nucleosome.
eg. basic n-terminal tail of H4 in one
nucleosome and an acidic pocket in
H2A-H2B dimer in the next nucleosome
Acetylation continued
• Also some TFs recognize acetylated
histones. eg. TAFII250 has a double
bromodomain and recognizes low level
acetylated histones. Once bound it is a
HAT and increases acetylation.
• low level acetylation of histones occurs
in inactive chromatin.
END
Take home
• The presence of nucleosomes can interfere
with the binding of TFs to enhancers and
with the preinitiation complex to the
promoter. = Repression
• When other proteins (simple TFs) are
bound to the DNA they can prevent the
histones from binding. This is competitive
inhibition of histone binding. Called derepression or antirepression in your book.