Chapter 13Chromatin Structure

and its Effects on Transcription

There is a resource on the web for this purpose.

Warn them before this class.

Students must be positive that they understandstandard PCR.

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 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 five different Histone Classes

Eukaryotes contain many copies of each histone gene.10-20 in mice, 100X in Drosophila

Compaction

Doublehelix

BeadsOnAstring

solenoid SnakeTheSolenoidPR 25

700 nm fiber

Probably involve SARS

Nucleosome

Core nucleosome contains2X H2A, 2X H2B, 2X H3 & 2X H4

DNAProtein

Beads on a string

Beads on a stringpacking ratio 5

Histone H1

• Outside of the core

• Easiest to remove by high salt extraction

H2B

H4

H3

H1

H2B

H2A

DNA

27Å

110Å

57Å

Solenoid

H1 histone

DNA

Nucleosomecore

30 nm solenoid

6 per turn

Packing ratio of 8

Tetranucleosome Fig 13.7

45-80 kb loops

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%

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 toallow one to modulateexpression 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 & 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

Histones can act as repressors

Histone H1Nucleosome core

• Thus, via competition for binding sites, transcription factors can de-repress (your book calls it antirepression.

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.

Figure 13.17

Fig. 13.21Weaver 3rd ed.

Is the promoter region of an active gene a nucleosome free zone?

Figure 13. 18

BglI

EcoRI BamHI

BamHI BamHI BamHI BglII BglII BglII

DNase I hypersensitivity

EcoRI

BglI

EcoRI BamHI

DNAse I

BglI

EcoRI BamHI

EcoRI

Transcribing SV40 virus isolated from infected monkey cell tissue culture.

Two fragments

Gelelectrophoresis

& Southern Blotting

Figure 13. 20

This is a Southern Blot.

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?

DNase I hypersensitivity in Chromatin lectures

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

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 don’t have this slide.

How to purify a HAT

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.

What do you know about it?•

What does this knowledge provide you?

•

What is the advantage of using a gel?

•

heated chemical inactivation BSA no histones no protein

• 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.

Now purify it.

KGWMDIM NMVTMMV

mRNA

Degenerate PCR

mRNAmRNA

DNA

Degenerate PCR

mRNA

DNA

KGWMDIM NMVTMMV

KGWMDIM NMVTMMV

Degenerate PCR

mRNAAAAAAAA

REVERSE TRANSCRIBE TO DNA

KGWMDIM NMVTMMV

Degenerate PCR

Codon Usage by amino acid.F L S Y C W P H Q R

TTT TTA TCT TAT TGT TGG CCT CAT CAA CGTTTC TTG TCC TAC TGC CCC CAC CAG CGC

CTT TCA CCA CGACTC TCG CCG CGGCTA AGT AGACTG AGC AGG

I M T N K V A D E GATT ATG ACT AAT AAA GTT GCT GAT GAA GGTATC ACC AAC AAG GTC GCC GAC GAG GGCATA ACA GTA GCA GGA

ACG GTG GCG GGG

KGWMDIM NMVTMMV

UPPER STRAND Primer cocktail K G W M D I M5' 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.

Degenerate PCR

KGWMDIM NMVTMMV

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

Degenerate PCRLOWER 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.

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

A-adenineB-not AC-cytosineG-guanineH-not G

T-thymineU-uracil=TV-not TW-A or TY-C or T- = gap

K-G or TM-A or CN-A, C, G or TR-A or GS-C or G

Ambiguous Bases

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 de-repression or antirepression in your book.