module 4: genome function Flashcards
(T/F) The way proteins get translated (how, when, where) is dictated by the way DNA packs into the nucleus.
True!
Everything inside the nucleus has its place. The distribution of these various proteins and DNA is a direct consequence of the function of those particular compartments.
The chromosomes fold so that each DNA sequence is in an OPTIMAL AREA of the nucleus to carry out its function.
Where does most of our knowledge about how the nucleus is organized come from?
most of our knowledge about how the nucleus is organized came with the ADVENT OF FLUORESCENT MARKERS.
with a microscope, it becomes very hard to make any type of observation about the 3D structure of the nucleus.
The nucleus is _____ and ______ bound.
The nucleolus is _______-bound.
Specialized sub-domains inside the nucleus are called ______ _____ _______.
circular; membrane
non-membrane
Bio-Molecular Condensates
Briefly answer the questions regarding bio-molecular condensates:
1) what is the function of nucleolus?
2) what are cajal bodies?
3) what are speckles?
1) transcription and processing of ribosomal RNA (rRNA)
2) cajal bodies are sites of RNA processing. snRNAs are transcribed and processed here.
3) speckles are sites of mRNA splicing.
(T/F) The nucleus is a pretty static organelle.
False!
The nucleus is HIGHLY dynamic.
Briefly describe the FRAP technique.
Fluorescence Recovery After Photobleaching (FRAP) tracks the movement of fluorescently labelled proteins.
A protein that is present everywhere in the nucleus is labelled with a GFP. Then, a pulse from a high-energy laser is used to bleach a small area of the nucleus (PHOTOBLEACH).
Eventually, the photobleached area is RECOVERED with the protein. By measuring how long it takes for recovery, you can determine if it is a highly mobile (fast recovery) or relatively static protein (slow recovery).
(T/F) By using the FRAP technique, we know that the nucleus is extremely dynamic as a single protein can move around in a matter of minutes.
True!
Although the nucleus is jammed full of DNA (6 billion bp of DNA), everything is organized in a way that proteins can be very mobile.
Which one of the statements is true?
1) DAPI binds to dsDNA.
2) Even with the use of fluorescent dye that attaches to DNA, we are unable to tell the structure of the interphase, thus making it challenging for us to determine the typical packaging of DNA.
Both!
The 10nm chromatin fiber is also known as __________.
140-150bp DNA is wrapped around a core _____ of histone proteins (which proteins?), forming a ________.
The nucleosomes are separated by 50-70bp of ____ DNA.
Linker histone (___) binds to linker DNA and _______ DNA further.
beads-on-a-string
octamer (two of H2A, H2B, H3, H4); nucleosome
linker
H1; compacts
*role of H1 still unsure about
In human cells, there are ___ bp of DNA that wrap around the histone octamer.
147
Initially, what did we think happened to the 10nm fibre?
What were the two models we generated from this hypothesis?
We used to think the 10nm fibre would fold into itself, generating a 30nm fibre. 30nm fibre was thought to be the MAIN chromatin in the INTERPHASE nucleus.
The two models in which the 10nm fibre would coil onto itself:
1) The solenoid model
2) The helical ribbon model
The in-vivo existence of the 30 nm fiber was questioned by a paper. What did they suggest would happen instead?
Rather than having a uniform, coiled structure (30nm fiber), we have CLUTCHES of nucleosomes (think of eggs in a nest).
ACTIVE and SILENCED compartments arise from variations in packing densities.
The formation of the interphase DNA is LINKED to the CELL TYPE and the STAGE in DIFFERENTIATION. For example, compared to a stem cell, a somatic cell has more nucleosomes clustered together.
Briefly describe how STORM works.
STORM is a SUPER-RESOLUTION microscopy technique that makes use of PHOTO-SWITCHABLE FLUOROPHORES.
It took histones found in the nucleosomes and tagged them with the photo-switchable fluorophores. Only a small subset of the fluorophores are at the ON state at a given time.
It switched between fluorescent and dark states.
This limited the number of fluorophores active in each frame, preventing anything closer than 250nm from being on at the same time, generating distinct spots.
It combined pictures taken each round into a single molecule image.
1) What does resolution mean regarding microscopes?
2) What must the minimum separation of two fluorophores to be able to them apart? What happens over and below this number?
3) Can nucleosomes show up as distinct spots?
1) Seeing two separate objects as two separate objects.
2) The minimum separation must be 250nm (rayleigh criterion). If two fluorophores are closer than 250nm, we see them as one blurry spot (unresolved). If they are farther than 250nm, they are seen as two distinct spots (resolved).
3) No! Nucleosomes won’t show up as distinct spots because they are closer than 250nm. This limits our availability to investigate the organization of our nucleus.
1) How was STORM used to show that we have clutches of nucleosomes?
2) What were the results?
1) H2B was tagged with the photo-switchable fluorophores. Since this protein is part of the core histone octamer, localization should reflect the ARRANGEMENT OF NUCLEOSOMES within the chromatin fiber.
2) The histone protein appeared clustered in discrete and spatially separated nanodomains (no pattern). There was a higher density of it in the nuclear periphery compared to the interior.
When there are more clutches (nucleosomes are packed), genes are turned ___.
_________ tends to cluster on the periphery, while _______ tends to cluster on the inside.
Off
Heterochromatin (silenced); euchromatin
*the organization of genes is not random!
Differentiate facultative heterochromatin from constitutive heterochromatin.
Facultative: sometimes heterochromatin, sometimes euchromatin. depends on cues.
Constitutive: always heterochromatin (telomeres, centromeres, highly repetitive DNA).
1) what are the two types of networks of fibres present in the nucleus? briefly describe each.
2) why are these important?
1) nuclear lamina (just underneath the nuclear envelope) and nuclear matrix (extends towards the nucleoplasm)
2) these serve as ATTACHMENT POINTS FOR CHROMOSOMAL DNA. proteins bind to these fibres and the same proteins bind to DNA, anchoring sequences at specific areas in the nucleus. in part, the placement of DNA inside the nucleus is facilitated by the presence of the LAMINA and MATRIX.
(T/F) Packaging of DNA inside the nucleus has to support the function of the cell. Thus, it has to change and has to differ from cell type to cell type.
True!
Briefly answer the following questions regarding the nuclear lamina.
1) What is the thickness?
2) What kind of filament is it made of?
3) What is its role?
4) What are the types of proteins? How many genes encode the different lamins?
1) Its thickness can range from 10nm to 100nm, typically falling within the 10-30nm range.
2) Type V filament
3) Its role is analogous to the role of the cytoskeleton; for rigidity and structure (helps anchor the DNA in the nuclear environment)
4) Lamin A, Lamin C, Lamin B1, and Lamin B2. Lamin A and C are made from the same gene through alternative splicing, while the rest are made from different genes, thus 3 genes encode the different lamins.
Lamins are made of three domains. Describe their structure and function.
1) Head domain: N-terminal
2) Coiled-coil rod domain: made of four subdomains (1A/B, 2A/B) and mediates the interaction with OTHER PROTEINS IN THE NUCLEAR LAMINA.
3) Ig-like fold domain: mediates interaction with NON-LAMINS such as histones.
Which one of the statements is true regarding lamins?
1) For assembly, lamins start as monomers and dimerize into dimers that are organized into head-to-tail polymers which then polymerize into anti-parallel filament.
2) One anti-parallel filament is made of all the lamins.
3) All lamins are permanently farnesylated to the inner nuclear membrane.
1!
For 2, an anti-parallel filament is made of up either lamin A/C, lamin B1 or lamin B2. These can interact with each other but they don’t mix within a filament.
For 3, lamins A, B1, and B2 undergo post-translation modifications (farnesylation) that cause them to be retained in the nuclear envelope. However, only B-type lamins are permanently farnesylated as farnesyl (isoprene unit of the cholesterol pathway) is removed from lamin A. Lamin A is still close to the membrane.
1) What is the nuclear matrix (does it exist)?
2) What are matrons?
3) What are S/MARs?
1) The existence of nuclear matrix is controversial. There is some evidence that some filaments extend throughout the nucleoplasm and deeper into the nucleus. The nuclear matrix provides anchor points to help organize the chromatin in 3D space (each chromosome occupies a certain region in the nucleus; this must be how).
2) Matrons are proteins that bind to the matrix and could potentially bind to DNA.
3) S/MARs are scaffold/matrix attachment regions (100-1000bp in length) that are AT rich in the DNA that are thought to interact with matrons.
In the 1960s, the idea that each chromosome occupied its own space was abandoned.
1) Why was it abandoned?
2) How was this proved to be true around 1980s/1990s?
1) It was abandoned in the early 60s as all we had were electron microscopy. These do not tell us anything regarding the structure of the nucleus.
2) Using the FISH technique where each chromosome binds a probe with a distinct colour, it was found that chromosomes occupy distinct regions. It was also found that the two neighbouring chromosomes can INTERMINGLE ON THE BORDERS OF TERRITORY.
Chromosome conformation capture (3C) is a method commonly used for studying ______ ___________ in eukaryotic cells.
It relies on protein ___-______ and ________ ______ to detect chromatin interactions.
It uncovered general features of genome _______.
chromatin interactions
cross-linking; proximity ligation
organization
*any two regions of DNA close together can be cross-linked. we can use NGS to find what those pieces of DNA are.
*this tool revolutionized the study of chromatin folding inside the nucleus.
Match the steps of 3C to the right order:
1) Step 1
2) Step 2
3) Step 3
4) Step 4
5) Step 5
A) Reverse crosslink.
B) PCR amplification of all circular DNA using primers of two known (‘biased’) genomic regions of interest.
C) Formaldehyde (freezes the nucleus) crosslinks chromatin proteins (covalent bond) to their associated DNA.
D) Digested ends are re-ligated (conditions favour ligation of juxtaposed DNA fragments; intramolecular ligation of nearby ends).
E) Digestion of crosslinked DNA with restriction enzymes (non-specific endonuclease DNAse I) to find points where selected DNA regions are connected through a protein complex.
Step 1: Formaldehyde (freezes the nucleus) crosslinks chromatin proteins (covalent bond) to their associated DNA.
Step 2: Digestion of crosslinked DNA with restriction enzymes (non-specific endonuclease DNAse I) to find points where selected DNA regions are connected through a protein complex.
Step 3: Digested ends are re-ligated (conditions favour ligation of juxtaposed DNA fragments; intramolecular ligation of nearby ends).
Step 4: Reverse crosslink.
Step 5: PCR amplification of all circular DNA using primers of two known (‘biased’) genomic regions of interest.
In 3C, you end up with a genome-wide ligation product library in which each ligation product corresponds to …?
In 3C, you end up with a genome-wide ligation product library in which each ligation product corresponds to a specific interaction between the two corresponding loci.
1) Why is 3C considered a one vs one method?
2) Why is 3C considered a biased method?
1) 3C is considered a one vs one method because it can only test ONE PAIRWISE INTERACTION at a time. One half of the primer pair is specific for one gene segment, while the other half of the primer pair is specific for another gene segment. These primers are tested on ALL of the circular DNA. If there is a positive signal (amplification), the two gene segments are next to each other.
2) 3C is considered a biased method because it only probes what we are asking it to probe. It only detects gene segments X and Y if we design primers for X and Y.
Which statement is true?
1) 3C is a very efficient method with high throughput.
2) Non-specific DNase I cuts any accessible region. Linker DNA can be cut but DNA around nucleosome can’t be cut as it is more compact and less accessible.
2!
For 1) 3C is not a very efficient method. It has a low throughput as it is a one vs one method. It does allow to ask precise questions. Back when this method came out, sequencing was not common thus they had to use PCR.
Another chromatin conformation capture technique is ChIA-PET (chromatin interaction analysis by paired-end tag sequencing).
Match the steps of this method in the right order.
1) Step 1
2) Step 2
3) Step 3
4) Step 4
A) Sonication to break DNA into smaller pieces, immunoprecipitation (antibody precipitating a protein of interest), and ligation of adapters to fragment ends.
B) Digestion generating isolated paired-end tags (PETs) and massively parallel DNA sequencing to identify the binding sites of DNA-associated proteins.
C) Crosslinking with formaldehyde
D) Proximity ligation and reverse crosslinking
Step 1: Crosslinking with formaldehyde
Step 2: Sonication to break DNA into smaller pieces, immunoprecipitation (antibody precipitating a protein of interest), and ligation of adapters to fragment ends.
Step 3: Proximity ligation and reverse crosslinking
Step 4: Digestion generating isolated paired-end tags (PETs) and massively parallel DNA sequencing to identify the binding sites of DNA-associated proteins.
What is the specific question that is being asked by ChIA-PET?
What sequences are found next to each other but are mediated by a specific protein?
*this method looks at where a certain protein is bound in the genome and determines which two sequences are being brought together.
1) What is the composition of the paired-end tags (PETs) in ChIA-PET?
2) Which DNA sequencing method can be used in ChIA-PET?
1) The paired-end tags (PETs) of ChIA-PET are composed of two different DNA segments brought close together in 3D space by the mediator.
2) ChIA-PET can use NGS (Illumina paired-end sequencing), where the forward read is region X and the reverse read is region Y.
Compare and contrast 3C with ChIA-PET.
Differences:
3C ligates free ends while ChIA-PET adds adapters to sequence the DNA and ligates them.
3C is one vs one while ChIA-PET is all vs all but mediated by a protein.
Similarities:
Both do not give information on all of the interaction regions across the genome. 3C is very low throughput and ChIA-PET has higher throughput but it only gives information about all the interacting regions MEDIATED BY A PROTEIN.
Another chromatin conformation capture technique is Hi-C.
Match the steps of this method in the right order.
1) Step 1
2) Step 2
3) Step 3
4) Step 4
5) Step 5
6) Step 6
A) Restriction fragment ends labelled with biotin (VB7).
B) Sonication and STREPTAVIDIN pull-down. Sequencing using NGS to identify interacting fragments.
C) Reverse crosslinking (all ligation junctions are marked with BIOTIN).
D) Formaldehyde crosslinking of chromatin proteins to their associated DNA.
E) Digested ends are re-ligated (conditions favour ligation of juxtaposed DNA fragments).
F) Digestion of crosslinked DNA with restriction enzymes.
Step 1: Formaldehyde crosslinking of chromatin proteins to their associated DNA.
Step 2: Digestion of crosslinked DNA with restriction enzymes.
Step 3: Restriction fragment ends labelled with biotin (VB7).
Step 4: Digested ends are re-ligated (conditions favour ligation of juxtaposed DNA fragments).
Step 5: Reverse crosslinking (all ligation junctions are marked with BIOTIN).
Step 6: Sonication and STREPTAVIDIN pull-down. Sequencing using NGS to identify interacting fragments.
Briefly describe the step of streptavidin pull-down in Hi-C.
Biotin and streptavidin are the STRONGEST non-covalent interactions found in nature.
All ligated ends are marked with a biotin molecule. Sequences with biotin = two different sequences ligated together. This step picks out sequences with BIOTIN using immunoprecipitation.
Streptavidin binds to biotin and an antibody specific to streptavidin binds to the streptavidin-biotin complex.
Out of the three (3C, ChIA-PET, Hi-C), which one is the true all vs all method?
Hi-C!
Hi-C asks if sequence A is close to sequence B, C, D, E, etc. It is probing every single possible probe-wise interaction across the entire genome.
It is not biased like 3C.
1) What is Hi-C mostly used for?
2) What is the disadvantage of the Hi-C method?
1) Hi-C is mostly used for characterizing long-range chromatin interactions across an entire genome (how DNA folds inside the nucleus).
2) Hi-C requires MILLIONS of cells so it is revealing the architecture shared by an ENTIRE population of individual cells. Nuclear architecture is not the same for all cells; it is generating an aggregated image.
3C uses ________, ChiA pet uses ________, and Hi-C uses ________ enzymes to fragment the DNA into manageable pieces to analyze.
non-specific DNase I; sonication; restriction enzymes
(T/F) In 2009, the first genome-wide view of interactions between all sequences in the mammalian genome was published.
True!
From this, basic organizing principles of the nucleus emerged.
1) What is the Hi-C contact map/matrix? What kind of information can we get from them?
2) What do Hi-C contact points represent?
3) How is the linear genome partitioned for Hi-C?
1) Hi-C results are displayed in Hi-C contact map/matrix (chromosome-wide matrix of interaction frequencies). It visually represents all of the interactions across the entire genome (which regions are close together) and gives information on the FREQUENCY of these interactions (how often found).
2) Hi-C contact points represent pairs of genomic positions that were adjacent to each other in 3D space.
3) Linear genomes (chromosomes, series of chromosomes) are partitioned into bins (loci) of a fixed size.
Why were initial Hi-C studies giving low-resolution maps?
How could the resolution be increased?
Initial Hi-C studies used 1Mb “bins.” Each “square” on the Hi-C contact matrix represents 1 million base pairs of sequence. The low-resolution map is due to each bin being so large. While Hi-C with 1Mb bins can tell us that DNA from one bin interacts with DNA from another bin, it doesn’t reveal which specific sections within those large bins are involved in the interaction.
We can increase resolution by decreasing bin size. This can be done by sequencing more DNA and increasing sequencing runs.
(T/F) Hi-C plays a central role in mapping the 3D architecture of the interphase genome.
True!
Depending on the enzyme used, we generate more or less fragments.
If we use 6bp-cutting enzyme, the frequency of cutting is ______ (sequence will occur at random once every _____ bp). We generate ~10^6 restriction fragments; 10^12 possible pairwise interactions.
If we use 4bp-cutting enzyme, the frequency of cutting is _____ (sequence will occur at random once every ____ bp). We generate ~10^7 restriction fragments; 10^14 possible pairwise interactions.
4^6; 4096
4^4; 256
Why did researchers bin data?
There would be one million squares on each axis if using a 6bp-cutting enzyme.
If we only sequence 100 million fragments, our matrix will remain underpopulated.
When Hi-C was initially introduced, the capacity for sequencing DNA, particularly achieving 10^12 interactions, was beyond our capabilities then and remains so even now. Consequently, data binning was employed.
Rather than stating that our chromosome is divided into 4096 bp fragments, we grouped the fragments. Bins were set at one million bps, where any fragment corresponding to that one million receives a tick in the respective box, populating the matrix.
What is a significant challenge encountered by Hi-C?
Achieving sufficient sequencing coverage to support maximal resolution is a significant challenge.
____-chromosomal interactions are rare, while _____-chromosomal interactions are frequent.
inter (trans); intra
*chromosomes interact more amongst themselves than between
With multi-Mb large bins for Hi-C, they saw the appearance of checkerboard patterns. What did they conclude from this?
DNA segregates into two main compartments in the cells: Active (A) and inactive (B).
In the A compartment, transcription of genes is occurring and there are histone modifications that correlate with transcription (acylation).
In the B compartment, transcription of genes is repressed and there are histone modifications that correlate with transcription repression (methylation).
The checkerboard represents that the A compartment sequences tend to interact with A compartment sequences more frequently than with B.
(T/F) There is a continuous segregation of the DNA along the length of a chromosome.
False!
There is NOT a continuous segregation of the DNA along the length of a chromosome. It alternates between active and inactive. In 3D space, the active regions are close to each other.
When zooming into a chromosome, its A compartment is more towards the ______, while the B compartment is more towards the _______.
Central; periphery
Decreasing sequencing costs led to richer Hi-C data sets (200-300 million paired ends), partitioning into ~40kb bins.
What did this help to identify?
By increasing the resolution of the maps (going from 10 million paired-end reads for 1mbp bins), they identified TOPOLOGICALLY ASSOCIATING DOMAINS (TADs).
*triangles with very sharp boundaries at the bottom of the Hi-C maps.
TADs are smaller discontinuous ______ ______ within the ___ compartments. They are also known as __________ _______.
They are _________ segment of DNA folded into coils and loops (______ ______).
They are a _________ feature of the genome (2-3k TADs) that can range up to _____ bp in size.
Self-interaction is very ___, creating triangles with ____ boundaries in the Hi-C maps. Interaction between TADs is ____.
SELF-ASSOCIATING DOMAINS; A/B compartments; Self-Interaction Domains.
continuous; (chromatin globule).
CONSERVED; 1 million bp
high; sharp; low.
*chromatin fiber makes up TADs
*TADs help isolate certain sequences
Each chromosome occupies a distinct territory. Why must the chromosomes fold in a specific way in the territory?
- Make it fit in the interphase nucleus (decondensed)
- Has to fold to support gene expression (genes have to be accessible for transcriptional machinery)
(T/F) Gene sequences within a TAD are not regulated similarly.
False!
Gene sequences found within a TAD tend to be regulated similarly (co-ordinately).
Cohesions form _____, very large loops of DNA, via a _____-_____ ______ model.
Cohesions are made of three proteins: _____, _____ and _______.
Structural Maintenance of Chromosomes (SMCs) have a ___ and a ____ domain that is linked by a _________ domain.
Cohesions are one of the main players of ______ the structure of the nucleus.
TADs, cohesion-mediated extrusion (loop extrusion)
SMC1; SMC3; RAD21
head; hinge; coiled-coil
organizing
Match the steps of the loop extrusion model:
1) Step 1
2) Step 2
3) Step 3
4) Step 4
A) The cohesion ring entraps DNA and then moves along, extruding the DNA through its lumen until an obstacle (CTCF) is met.
B) Cohesion is released by WAPL and the loop falls apart. These loops are made and fall apart continuously.
C) Cohesion complex is loaded onto DNA with the help of a loading factor (NIPBL). Loading a cohesion onto a DNA is random.
D) Two convergent CTCFs prevent cohesion from moving along the DNA.
Step 1: Cohesion complex is loaded onto DNA with the help of a loading factor (NIPBL). Loading a cohesion onto a DNA is random.
Step 2: The cohesion ring entraps DNA and then moves along, extruding the DNA through its lumen until an obstacle (CTCF) is met.
Step 3: Two convergent CTCFs prevent cohesion from moving along the DNA.
Step 4: Cohesion is released by WAPL and the loop falls apart. These loops are made and fall apart continuously.
(T/F) A dot within TADs on a Hi-C map indicates a highly prevalent and consistently positioned interaction, likely present in every cell across our dataset.
True!
