Chapter 13 Chromatin Structure and its Effects on Transcription - PowerPoint PPT Presentation

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Title: Chapter 13 Chromatin Structure and its Effects on Transcription


1
Chapter 13Chromatin Structure and its Effects on
Transcription
2
Students must be positive that they
understand standard PCR.
  • There is a resource on the web for this purpose.
  • Warn them before this class.

3
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).

4
Eukaryotes five different Histone Classes
Histone 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
5
Compaction
700 nm fiber
Beads On A string
solenoid
Snake The Solenoid PR 25
Double helix
Probably involve SARS
6
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7
Nucleosome
DNA
Protein
Core nucleosome contains 2X H2A, 2X H2B, 2X H3
2X H4
8
Beads on a string
9
Histone H1
  • Outside of the core
  • Easiest to remove by high salt extraction

10
Solenoid
30 nm solenoid
6 per turn
Packing ratio of 8
11
Tetranucleosome Fig 13.7
12
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13
45-80 kb loops
14
The movie was deleted to save space. You can
download it separately.
15
Evidence that histones help to regulate gene
expression
  • Xenopus 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

16
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.

17
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.
18
Nucleosome compete with transcription factors
  • This is too simple toallow one to
    modulateexpression or is it?

19
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.

20
Laybourne Kadonaga 1991
  • Chromatin form of the Drosophila Krüppel gene
  • nuclear extract transcribes it at 25 of the
    maximum rate. This is 75 repression.
  • Interpretation1) 100 of the genes are
    transcribed at a 75 rate of transcription2) 25
    of the genes are transcribed at 100 of the rate.
  • 25 of promoters unoccupied. How was this
    determined?

pg 370 Weaver 4th ed
21
Histones can act as repressors
  • Thus, via competition for binding sites,
    transcription factors can de-repress (your book
    calls it antirepression.

Histone H1 Nucleosome core
22
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.

23
De-repression or Anti-repression is the amount of
transcription that you get when the histone is
not interfering by hiding the promoter.
24
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.

25
Yaniv saw that some transcriptionally active SV40
virus DNAs had nucleosome free zones.
Fig. 13.21 Weaver 3rd ed.
Figure 13.17
26
Is the promoter region of an active gene a
nucleosome free zone?
Figure 13. 18
27
BamHI
BamHI
BamHI
BglII
BglII
BglII
28
DNase I hypersensitivity
Transcribing SV40 virus isolated from infected
monkey cell tissue culture.
29
Two fragments
Figure 13. 20
Gel electrophoresis Southern Blotting
This is a Southern Blot.
30
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?

31
Histone acetylation
  • amino groups of lysine side chains
  • unacetylated histones tend to repress
    transcription
  • acetylated histones tend to activate
    transcription
  • Histone acetyl transferase (HAT)- see Figure
    13.23 for how to detect them.
  • Histone deacetylase

32
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 acetyl-CoA. Wash.
    Fluorography.

You dont have this slide.
33
How to purify a HAT
heated chemical inactivation
BSA no histones no
protein
Acetylates histones
A way to assay for its presence
Demonstrates that one is detecting the presence
of a single enzyme. Tells you the relative size
of the enzyme.
34
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.

35
Degenerate PCR
mRNA
mRNA
mRNA
DNA
36
Degenerate PCR
mRNA
DNA
37
Degenerate PCR
mRNA
AAAAAAA
38
Degenerate PCR
39
Degenerate PCR
UPPER STRAND Primer cocktail K G W
M D I M 5' AA(AG) GG(N) TGG
ATG GA(TC) AT(TCA) ATG 3' 2 X 4 X
1 X 1 X 2 X 3 X 1 48 A 21 mer
primer with 48 fold degeneracy.
40
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 CAT (N)GT (N)AC CAT (AG)TT 3'
4 X 1 X 1 X 4 X 4 X 1 X 2 128
41
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 CAT (N)GT (N)AC CAT (AG)TT 3'
4 X 1 X 1 X 4 X 4 X 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 (N)GT (N)AC CAT
(AG)TT 3' Degeneracy is 1 X 1 X 1 X 4 X
4 X 1 X 2 32 fold degeneracy.
42
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 GAYAAYTTYAAYMGNCARAARCARAARYTNGGNGGNGARGAYY
TNTTYATGACNGARGARCARAARAARTAYTAYAAYGCNATGAARAARYTN
GGNWSNAARAARG 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-uracilT V-not T W-A or T Y-C or T -
gap
43
Screen a cDNA library
  • DNA copies of every mRNA made by a cell are
    produced and cloned into a bacterial vector.
  • Screening.
  • Google.

44
5-RACE
Figure 4.16
45
HAT type A
  • Have bromodomain
  • Binds aceylated lysines. So HAT As can recoginze
    partially acetylated histone tails.
  • Examples p55, Gcn5p CBP/p300, TAF250

46
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

47
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.

48
END
49
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50
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 de-repression or antirepression
    in your book.
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