Chromatin and Transcription Initiation and elongation Flashcards

You may prefer our related Brainscape-certified flashcards:
1
Q

Histone Acetylation

A
  1. Importance of post-translational modifications of the histone tails.
  2. Changes in charge (Acetyl group is negatively charged) will promote nucleosome dissociation. It also may change the potential protein partner that could bind to the histone tails.
  3. The acetylation is reversible (HAT vs. HDAC). At least 10 different HDACs have been identified. HAT exists in two forms HAT-A (nuclear) and HAT-B (cytoplasmic)

N-terminal tail

histone-fold domain

Lysine modified by addition of acetyl group (HAT) or removal (HDAC)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Histone Acetylation

A

HAT can form a complex with an activator

HDAC (1 to 12) can form a complex with a repressor

Coactivator (HAT)+ Activator + acetyl group =Acetylated histones-active chromatin

Co-repressor(HDAC) + repressor=deacetu;ate histones - inactive chromatin

Hat is a co-activator (often needs to be brought in contact with histone tails). HDAC is a co-repressor requiring a partner to find histone tails

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

histone methylation and phosphorylation

A

Active chromatin - H3 N-terminal tail

Inactive chromatin

the histone N-termini are the target of the modification

Multiple methyl groups can be present (tri-methylation of Lys 4,9 or histone H3, Lys 20 on H4, play a role in regulation

Other histon modifications: Phosphorylation and methylation. Different charges and conformations will dictate the function

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Summary of location and type of histone tail modifications

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What is the role of histone in post-translational modifications

A
  1. Repression/silencing
  2. Inprinting (silent allele)
  3. Activation

Sumo: Small Ubiquitin-like modifier 18% identity with Ub, similar 3D structure

TSS: transcription start site

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Nucleosome remodeling factors

A

Binding of transcriptional activatio and chromatin remodeling faction

Chromatin remodeling factor + activator

Nucleosome displacement

Binding of general transcription factors and RNA polymerase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What is the role of nucleosome remodeling complexes

A

sliding, changes in DNA path partial dissociation

ATP dependence (through SWI2 or ISWI sub-unit)

a. translocation-dependent remodeling = SWI/SNF and ATP is converted to ADP considered Nucleosome sliding

b. remodeling of DNA contacts
ATP to ADP SWI/SNF Changes in DNA path

c. remodeling of octamer structure
Loss of H2A-H2B dimer

Swi/Snf, NURF, CHRAC, NURD
All complexes are ATP-dependent

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Epigenetic inheritance of histone modifications

A
  1. Parental chromatin
  2. DNA replication
  3. Parental nucleosomes distributed to progeny strands
  4. Incorporation of new nucleosomes
  5. Parental modified histones direct modification of newly incorporated histones
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Regulation of transcription by siRNAs

A
  1. siRNA-association with RITS unwinding of siRNA
  2. Pairing with mRNA transcript at target gene
  3. mRNA + siRNA + RITS
  4. methylation of H3 lysine-9 heterochromatin formation
  5. transcription repressed

RITS: RNA-induced initiation and transcriptional Gene silencing

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Chromatin Structure and Enhancers

A

Role of nucleosomes as a structural element that will redirect the DNA path and allow long distance interactions

here is an example of how the enhancer region can be brought to close proximity with the promoter region

the protein binding the enhancer region can now interact with TF or mediators

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Insulator

A

Euchromatin

Insulator

Heterochromatin

Role: Prevent spreading of heterochromatin to transcriptionally activ regions

chromatin domain are separated by DNA sequences called insulators

Trans acting factors are maintaining the structure of the insulator region in the proper configuration avoiding loss of cohesion

the euchromatin is organized in large loops that can undergo torsonial stress (changes in topology) that can be important for chromatin remodeling

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Chicken B-globin gene cluster

A

Proteins such as CTCF (CCCTC-binding factor, a zinc-finger protein) can bind to the 5’ hyper=sensitive region HS4 preventing “chromatin invasion”

Concept of Boundary elements where a specific region prevents heterochromatin from invading the region that needs to be transcribed

Boundary elements are cis elements (sequences) that can be bound by specific DNA-binding proteins (example for beta globin: CTCF is a cCCTC-binding factor (zinc finger protein)

DNase insenstive = Low histone acetylation

DNase senstive = high histone acetylation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

regulation of transcription in bacteria

A

Metabolism of lactose

Lac Operon: lactose is a precursor of glucose

regulation of transcription based on need for metabolism intermediates

Glucose requred will necessitate the lactose to be broken into glactose plus glucose

then enzyme required is the B-galactosidase

Lac Operon

  1. In the absence of lactose with a repressor = no lac nRNA

I, p, repressor, z, y, a

  1. Presence of lactose

I, p, o, z, y, z

z=galactosidase, y=permease, a=transacetylase

Transition for actively binding repressor vs inactiv binding

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Positive Control of the lac Operon by glucose

A

POSITIVE CONTROL:

  • The production of B-Galactosidase is controlled at the level of the Promoter-Operator region (operon).
  • The lac z gene codes for ß-galactosidase
  • Transduction of the signal for LOW GLUCOSE is transduced to an enzyme named ADENYLATE CYCLASE which role is to hydrolyze ATP to cAMP plus P~Pi (pyrophosphate).
  • cAMP will form a complex with the protein CAP (not to be confused with the cap structure at the end of mRNA 7 methyl guanosine).
  • The complex will bind DNA and the bacterial RNA polymerase at the Promoter-operator site.

CAP: Catabolic Activator Portein

P: promoter region

O: operator

Z: B-galactosidase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Negative control of the lac Operon

A
  • NEGATIVE CONTROL: Repression.
  • In presence of Lactose, no need for
  • The expression of the structural genes is not only influenced by the presence or absence of the inducer, it is also controlled by a specific regulatory gene. The regulatory gene may be next to or far from the genes that are being regulated.
  • The regulatory gene codes for a specific protein product called a REPRESSOR.
  • The repressor acts by binding to a specific region of the DNA called the operator which is adjacent to the structural genes being regulated.
  • The structural genes together with the operator region and the promoter is called an OPERON.
  • However, the binding of the repressor to the operator is prevented by the inducer and the inducer can also remove repressor that has already bound to the operator.
  • Thus, in the presence of the inducer the repressor is inactive and does not bind to the operator, resulting in transcription of the structural genes.
  • In contrast, in the absence of inducer the repressor is active and binds to the operator, resulting in inhibition of transcription of the structural genes.
  • This kind of control is referred to a NEGATIVE CONTROL since the function of the regulatory gene product (repressor) is to turn off transcription of the structural genes.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Regulation of transcription in Eukaryotes

A

Regulation of transcription in eukaryotes: is more complex

chromatin de-condensation may be a requiremnt for proper expression of genes

Decondensed Chromosome regions in Drosophila also referred to as : Puffs

17
Q

Identification of Eukaryotic Regulatory Sequences

A
  • The DNA sequences required for proper activation and/or repression have been determined by deletion or addition of DNA within or adjacent to the promoter region(s).
  • The regulatory sequence is cloned upstream of a reporter gene, the expression of which can be monitored by colorimetric assay (CAT or Chloramphenicol Acetyl Transferase assay), Northern blot of fluorescence measurements (GFP for example)
  • The DNA was cloned into specific vectors (plasmid that can be expressed in Eukaryotic cells).
  • Note that the plasmid is not integrated into the chromosome (transient expression will be lost when the plasmid is lost).

By changing the regulatory sequence one can determine the necessary region for proper regulation (ie the proteins that will bind to those sequences)

18
Q

DNA Methylation

A

In females the X-chromosome exist in duplicate (XX vs XY)

If both copies of the X-chromosome were active, there would be an imbalance in expression of the genes coded by the X

One of the two X-chromosome is therefore inactivated

The inactivation mechanism involves DNA modification and protein binding to various regions of the X-chromosome

DNA methylation of Cytosine (in CpG dinucleotides) acts as a flag for gene inactivation

Note that specific protein can bind to methylated C (ex. MeCP2 or other MDBs)

Role is to maintain repression of the methylated DNMA regions

19
Q

Maintenance of Methylation Patterns

A

Methylated parental DNA

DNA replication

Methylation

Methylated daughter DNAs

How is methylation maintained?

During replication only one strand will have a methyl group at the CpG

The second strand will be methylated by specific DNA methylase enzyme(s)

20
Q

Maintenance of repressive state: genomic imprinting

A

the X-chromosome is maintained in a repressed state (phenomenon of gene imprinting)

One copy of the X-chromosome is already methylate (sperm)

After fusion of Egg and Sperm one allele will be active and one will be inactive

If both X-chromosome are un-methylated one of the two alleles will be inactivated

21
Q

Overview processing eukaryotic mRNAs

A
22
Q

modifications of 3’ ends of eukaryotic mRNAs

A
23
Q

Overview of splicing: removal of introns in the nucleus

A

Splicing requires specific sequences to be recognized as intron-exon boundaries, RNA needs to be cleaved and then re-ligated at the exon-exon interface

3 steps:

recognition of intron and exon site

cleavage

joining

24
Q

Consensus sequence for intron splicing in higer eukaryotes

A

Consensus sequences at donor and acceptor sites plus presence of the branching point (A).
Donor obeys the GT-AG rule: GT at the 5’end absolutely required) same for AG at the 3’end. The rest of the requirement for sequences is more loose.
Distance bewteen branching point and acceptor site is 20-40 ntdes.

25
Q

splcing

A

not linked to the transcription machinery

can do in vitro

does not require mRNA to be capped or polyA

Splicing is independent of transcription machinery and does not require either capping of the 5’end or poly-A tail at the 3’end.
How does removal proceed?

26
Q

Splicing reaction: generation of Lariat

A

Splicing requires the formation of a specific structure and sequential cleavage.
The 5’ site (acceptor is cleaved first and will be transiently ligated to the A at the branching point (forming a lariat structure). After that the 3’ site will be cleaved resulting in the removal of the intron. The Exons will be re-ligated.

27
Q

Branch Poiint

A

The branching point displays a specific ligation motif. The nucleotide at the 5’ end will form a 5’-2’ bond between the G (5’) and the A(2’ at the branching point).

28
Q

Nuclear splicing is a 2 reaction process in which a free OH end attacks a phosphodiester bond

A

The free OH attacks the 5’ phosphate of the G at the acceptor site. Then Exon 1 OH will attack the Phosphate at the Donor site of Exon 2

29
Q

RNA splicing is Catalyzed by a spliceosome

A

After U1 sn RNA plus proteins forms U1 sn RNP binding at the 5’site, U2 sn RNA will bind to the branching point (through specific base-pairing at the Py tract). The complex U4/6/5 will
bind to the 5’ site and displace U1. SR splicing factors bind to specific sequence within exons, help recruit U1, U2AF and U2.

30
Q

How is the lariat-like intermediate formed?

A
  1. U4/U6 complex will interact with U2, U5 is brought in contact with the 3’ site (AG sequence).
  2. The RNA is cleaved and the exons 1 and 2 are ligated.
  • Note that Splicing Factors or SR proteins are present at the boundary of exons 1 and 2.
  • U5 binds exon #1 at 5’ splice site
  • U4/U6 binds U2
  • 1st OH attack by U2/U6
  • formation of lariat like intermediate
  • U5 shifts, binds exon #2 at 3’ splice site
  • U2/U6 catalyze 2’-5’ bond
  • excision of intron
  • ligation of exons
31
Q

small nuclear Ribonucleoproteins (snRNPs) catalyze splicing

A

snRNPs = snRNA + 6-10 proteins

Splicing requires catalytic activity for the actual cutting of the phosphate backbone of the RNA. It is achieved by Small Nuclear Ribonucleoproteins or sn RNPs.
They are composed of a snRNA and 6-10 proteins that will hold the complex together (example here is U1 snRNA).
The U1 sn RNA has a complementary sequence to the 5’ splicing site and will base-pair with the site.Other components will be joining U1 snRNA to form a spliceosome (ribozyme: complex of RNA and protein required for RNA modification). The catalytic activity is brought to the cutting site and the mRNA is cleaved.

32
Q

Group I and Group II introns are self-splicing

Self-splicing RNA (mitochondria, chloroplast, bacteria)

A

In bacteria and some organelles, the splicing mechanism exists in two different forms.
Group I: No lariat structure is formed, no A at the branching point. A guanosine substitute for the need of the A branching point.
Group II are similar to what was described in Eukaryotes.

33
Q

Alternative Splicing may generate multiple protein products from one gene

A

Alternative splicing allows for various RNA to generated from the same starting un-processed mRNA. Here are several examples

SV40 T/t antigens splice two 5’ sites to a common 3’ site

adenovirus E1A splices variable 5’ sites to a common 3’ site

D. melanogaster tre splices a 5’ site to alternative 3’ sites

D. melanogaster dsx skips and exon

Troponin T splices alternative exons

P elements splice out an extra intron

34
Q

Processing of rRNA

A
35
Q

processing of tRNA

A
36
Q

mRNA degradation

A
37
Q

RNA editing of Apolipoprotein B mRNA

A

Editing converts CAA to UAA in mRNA

UAA is a stop codon resulting in an obortive product

The process is cell=type-specific

RNA sequence can be modified after transcription: RNA editing

38
Q

RNA editing can affect multiple nucleotides

A
  1. mRNA for coxII in trypanmosoma is edited by insertion of 4 uridines
  2. one case of A to G editing (mammal glutamate receptor mRNA)
  3. Pyrimidine editing in tRNA
  4. Other genes (protozoan) require a guide RNA to base pair and will serve as a template for insertion of nucleotides
  5. Deletion of nucleotides have been reported but are infrequent
  6. editing is catalyzed by a complex of endonuclease, terminal uridyl transferase activity and RNA ligase

What sequences are affected?
Editing requires the presence of several enzymatic activities: endonuclease, terminal Uridyl Transferase and RNA ligase