MCB 121 Final

Explain how immunity is acquired in vertebrates

acquired immunity: encounter infectiouis agent, you have a memory and strore this memory, so the next time there is an infectious agent, you can mount an immune response

1. For primary infection and memory, t cells first activate the b cell army. This leads to the formation of memory for b cells.

2. For recognition and amplification, B cells are activated upon antigen binding.

3.For the inactivation step upon secondary infection, b cells produce inactivating antibodies

explain how immunity is acquired in bacteria?

1.Involves insertion of bacteriophage sequences into CRISPR array. Bacteria have a CRISPR locus that is used to store information about a primary infection about a bacteriophage. Can cut bacteriophage into pieces and insert some of the segments into the CRISPR locus as spacer DNA, which stores information of primary infection. So, when the CRISPR RNAs are transcribed, they can be processed into guide RNAs that can go and cut the virus/bacteriophage again if it affects a second time

2.Bacteria express and process pre-crRNA

3.Bacteria have sequence-specific Cas9 cleavage

explain how immunity is acquired in plants

1.In plants, dsRNA viruses are processed by Dicer?RISC to form siRNAs. siRNAs that stored the information of the previous infection

2.RdRp can sustain pool of siRNAs; RdRp synthesizes dsRNA using siRNA as a primer

3.Spread siRNA to different plant tissues and destroy dsRNA with siRNAs

How are memory B cells formed? during first infection or vaccination

1.B cell army and will look for viruses for any information that might have been saved from a previous infection. Our B cells have immunoglobins are the outside, essentially antibodies. Will be on surveillance for virus if it comes back and reinfects

2.These antibodies are bound to an antigen that would be on the virus that they could recognize. Once they recognize an antigen, they can undergo differentiation and will amplify so that they have the same antibody on a population of b cells.

3.These would be plasma cells (short term) and then they will become memory B cells (long-lasting)

How does the mRNA vaccine work? Moderna and Pfizer

1.Corona virus has a spike protein. A mRNA vaccine will have mRNA that express the spike protein (is taken up by dendritic cells to receive and translate spike protein mRNA). The B cell army will be abele to recognize it and create memory B cells.

2.So, when exposed to the virus, immune response is now prepared. We take memory B cells that will be amplified and will antibodies to inactivate the virus

Explain in detail how the moderna and Pfizer-BioNTech vaccines work

1.The in vitro transcribed mRNA is delivered in the LNP which is taken up by dendritic cells that then translate the spike protein and present it on the cell surface.

2.T-cells recognize the spike protein and then stimulate B cells that have antibodies to the spike protein.

3.Resulting memory B cells will surveille the blood and are able to respond in the event of an infection

What were the challenges of making the Covid mRNA vaccine? (in detail)

Success of covid vaccine depended on the use of uridine analogs

1. in vitro transcribed mRNA is rapidly degraded due to 2’OH that would cause the hydrolysis of the phosphodiester backbone

2. injected mRNA leads to innate immunogenicity (inflammation)

3.negatively charged phosphodiester backbone makes it difficult to cross biological membranes

What was the first key innovation to make the covid mRNA vaccine

Uridine analogs: all Us are replaced with N1-methyl Psuedouridine

1.Evades innate immune sensors

2.Improves translational capacity

3.Increased protection against nucleases

4.Reduced cytotoxicity compared to unmodified bases

What was the second key innovation to make the covid mRNA vaccine *****

mRNA lipid nanoparticles: have a space inside the lipid membrane. Allows mRNA to pass through cell membrane. Protects mRNA from degradation

Use shell of lipids nanoparticles (LNPs) to protect RNA and are delivered to immune cells in the lymph nodes where they produce spike protein which is then displayed to other immune cells (e.g. T cells and B cells)

Describe B cells and antibody variants ****

In an antibody, there are 4 polypeptides: 2 light chains and 2 heavy chains. There is one antibody per cell. Antibody has a variable and constant region

The antigen binding site is composed of 3 components:

1.variable region

2.joint region

3.diversity (heavy chain only)

These make up V(D)J configuration of sequences. Each unique V(D)J combination has a constant region, forming a complete antibody chain. Each unique V(D)J combination creates a different antigen binding site

Immunoglobin light chain genomic locus has hundreds of possible V segments as well as multiple J segments

True

for light chains, 1 V and 1J are chosen

for heavy chains, 1V, 1D, and 1J

2 immunoglobin loci undergo VDJ recombination so only one of each is selected. This is a rearragment event, happens at the DNA level, and is irreversible

Describe how VDJ regions of immunoglobulin genes bind to antigens in detail

1.Antibodies have two light chains and two heavy chains. All the antibodies expressed on B cells has the same unique combination of V (Variable), J (Joint), and C (constant) regions. Heavy chains have an additional D (Diversity) region.

2.When an antigen binds to the antibody on the surface of a B cell, that B-cell will undergo an expansion and create memory B-cells. When the cell encounters the antibody again later it can ramp up production to fight off subsequent infections. A vaccine will also produce memory B cells.

Describe the DNA/RNA processing steps that encode the light chain polypeptide

1.DNA VJ Recombination of the V and J segments results in deletion of DNA between the V and J segments

2.This mature light chain gene is transcribed after VJ recombination to produce a primary transcript

3.This transcript undergoes RNA processing to remove extra J segments. This creates a mature processed RNA

4.The processed RNA undergoes translation to produce a light chain polypeptide.

5.After protein folding and assembly, the light chain polypeptide forms a light chain that is part of the antibody molecule

Describe how immunoglobin genes are organized in detail

1.There are 1,500 possible V-J light chain rearrangements and ~1,700 heavy chain rearrangements of the IgG class of genes. Combined, there are ~107 possible combinations.

2.More variation is introduced by a high mutation rate in the V regions

How do DNA and RNA processing steps lead to a unique antigen binding site? in detail

Following recombination, the primary transcript is further processed to splice out the remaining J sequences.

The final transcript after DNA and RNA processing expresses one V segment, one J segment and one C region. Heavy and Light chains are bound to produce the antibody

How does alternative 3’ processing affect immunoglobin

1.There are two types of immunoglobulin that can be made from the same gene, depends on alternative splicing on which 3’ cleavage site is used for adding the poly A tail

2.One variant contains the M region if 2nd cleavage site is used. When no antigen is detected, use of 2nd 3’ cleavage site joins exon 4 to M exons. M region is responsible for anchoring the antibody to the membrane of the cell. The IgM protein/antibody is anchored in the plasma membrane.

2.Another variant does not have M (membrane) region. When antigen is detected, CstF-64 represses splicing and promotes use of the 1st 3’ cleavage site. The IgM protein is secreted in the plasma membrane to inactivate antigen

How does alternative 3’ processing affect immunoglobin in detail

1.There are two flavors of the immunoglobulin that are controlled by alterative 3’ end cleavage sites.

2.If the first is used then the Ig is secreted to the plasma but if the second is used then the M exons are included and result in the Ig being anchored to the cell membrane. This is regulated by the presence or absence of the antigen, respectively

Describe how VDJ recombination is mediated by RAG proteins and NHEJ (non-homologous end joining)?

1.Recombination signal sequences (RSS) are used. The sequence will recongize two proteins (RAG1 and RAG2)

2.RAG1 and RAG2 bind to RSS and will create a nick. Then, the 3’OH will make a nucleophilic attack on the opposite strand. This will cleave on strand on each side. This is an intermolecular transesterification

3.A hairpin intermediate is formed, and intervening DNA will be discarded

4.Hairpins are processed to make free ends and there is double-strand break repair by end joining.

5.The gap will be sealed using ligation at the ends.

Describe how VDJ recombination is mediated by RAG proteins and NHEJ in detail

Rag1 and Rag2 bind to the recombination signal sequence and remove intervening sequences through a hairpin intermediate Each segment is adjacent to a recombination signal sequence (RSS) that binds to the RAG proteins. Each end forms a hairpin structure and the intervening sequences are removed. The hairpins must be processed so that the ends can be repaired.

The Tn5 transposon in bacteria excises through a similar mechanism. This led to the idea that the RAG proteins may have transpose activity and was later shown to be true. This leads to the interesting hypothesis that VDJ recombination evolved from a transposon, flanked by RSS repeats, that inserted into the genome in an ancestor of jawed vertebrates, like us and sharks.

The hairpin intermediate is similar to a mechanism of a transposon in bacteria called TN5 (bacterial transposon)

True, the transposon also undergoes a hairpin intermediate in its processing. The Tn5 transposon in bacteria excises through a similar mechanism. This led to the idea that the RAG proteins may have transpose activity and was later shown to be true. This leads to the interesting hypothesis that VDJ recombination evolved from a transposon, flanked by RSS repeats, that inserted into the genome in an ancestor of jawed vertebrates, like us and sharks

How do we choose 1V and 1J and remove the other sequences? Explain the 12-23 rule ****

The 12-23 rule ensures that only 1V, (1D) and 1 J segment is used in each cell

1.In front of the RSS, adjacent to the a V region, has 12 nucleotides

2.The RSS adjacent to the J region has 23 nucleotides

3.This ensures that recombination will involve joining of 12nt RSS and 23nt RSS. Since all the V regions have 12nt, they will not recombine with each other, same with the J region who all have 23 nt and will not recombine with each other, so recombination is only between the V and J segments

4.For VDJ recombination for the heavy chain, it will still follow the 12-23 rule (D segments have 23nt and J have 12nt)

5.Coding sequences will be joined together. Synapsis, cleavage, and coding end hairpin formation occurs

6. This forms a resolved genomic coding junction (Products are V-J fusion). Liberation of an excised circle will occur that has intervening sequences. The liberated excision circle contains the resolved signal junction

Explain the 12-23 rule in detail. What keeps the V, J and D regions recombining within the array?

12-23 RSS rule ensures V-J and J-D recombination.

1.There are two types of signal
sequences 12mers and 23 mers.

2.All V segments are followed by a 12mer and each J segment is preceded by a 23mer. In this way only 12/23 combinations will form a complex with the RAG proteins

How do our B cells express 1 version of 1 allele the of VDJ antibody on its cell surface despite our diploidy? (we have 2 alleles of the immunoglobin loci)

1.Due to allelic exclusion (replication timing can influence nuclear location of each allele and the ability to undergo VDJ recombination. By chance, one of the alleles will replicate

1.Two stages: early replication and late replication

2.Late replication allele will go to the nuclear periphery and will be silenced. Will be reinfocred by epigentic chromatin marks. The cytosines will be methylated to keep it silenced.

3.The other locus will have H3K4me3 which is associated with promoter regions or open DNA.

4.There will be different epigentic marks on the two alleles so that one will be active and be able to recombine and the other allele will be placed in a part of the nucleus that will be silenced (B compartment). The other will be placed in the A compartment

Explain allelic exclusion

Allelic exclusion: Replication timing influences nuclear location of each allele and the ability
to undergo V(D)J recombination.

Late replicating DNA will more likely be in silenced and at the nuclear periphery. If one homolog is
replicated early then it is more likely to undergo VDJ recombination and acquire active chromatin
marks such as H3Ac, H3K4me3 and unmethylated CpGs.

Explain what transposons or transposable elements are

A chromosome segment that can change position in the genome by the act of transposition (jumping genes). Can be DNA or RNA segments

define transponase

enzyme that catalyzes transposition

what are the different transposable element types?

1.DNA transposons: have a cut and paste mechanism. They encode their own transposase

2.Retrotransposons: transposase through RNA intermediate. Requires reverse transcriptase

3.LTR retroviruses: characterized by long terminal repeats at the ends. Come from retroviruses which are remnants of previous retroviral infections

4.Non-LTR: Two types, LINE (long interspersed nuclear elements) and SINE (short interspersed nuclear elements)

Describe transposons in detail

Transposable elements were discovered by Barbara McClintock using maize (corn)
• Transposition is the transfer of a chromosomal segment to a new position on the same or
another chromosome. Jargon: a transposon jumps or hops into the chromosome
• A transposon is considered a mobile element
• Transposase is the enzyme that catalyzes transposition.
• There are two general classes of transposons: DNA and retrotransposons, with the latter
going through and RNA intermediate.
• Endogenous retroviruses are retrotransposons that are remnants of past retroviral infections
(up to 400 million years ago in fish; ~70 million years ago in mammals)
• LINE, SINE elements (Alu elements in primates are examples of SINEs)
• Transposons make up almost 50% of our genome

Describe the three types of transposon mechanisms

1.Cut and paste: Transposon excises from DNA of the donor location and inserts into another
location in the genome. Leaves a DSB that needs to be repaired
2.Replicative: Leaves a copy of the transposon at the original site
3.RNA intermediate: Retrotransposon is transcribed from its position in the genome, which is
followed by reverse transcription to create a DNA fragment. This inserts into the genome by
integrase in a mechanism similar to “cut and paste” DNA transposons

ntrons of human genes are filled with transposons

true

CpGs of gene bodies tend to be methylated while the promoters are relatively
unmethylated. DNA methylation will suppress the transcription of these intragenic transposons

Difference between DNA transposon mechanim and that of LTR retrotransposon

DNA transposon

1.A cut and paste transposon will excise the transposon.

2.This creates free 3’OH that allow insertion

3.Insertion leaves leftover nicks that are repaired

LTR retrotransposon

1.Start from RNA. Have integrated in genome or are in RNA of retrovirus. They will be reverse transcribed (2 rounds). 1st round will use reverse transcriptase to make DNA using RNA template. 2nd round of synthesis will synthesize the second strand of DNA.

2.Undergoes cleavage

3.Undergoes insertion like the DNA transposon (cleavage leaves free 3’OH that allows insertion)

3.Insertion leaves leftover nicks that are repaired

Describe the overview of the “cut and paste” transposition

1.Transposase cleaves each side of the transposon to expose the 3’ ends. This is the excision

2.Liberated 3’OH attack the opposite strand, generating hairpin ends and excising the transposon. This forms a hairpin intermediate

3.Transpoase cleaves hairpins to reexpose the 3’ ends, which attack and target DNA

4.Transposon inserts into target site

5.Replication fills in the gaps at each end and ligation completes the process

Describe the overview of the “cut and paste” transposition

Transposition makes use of transposases, specialized enzymes that cleave phosphodiester bonds
and catalyze transesterification reactions in which the 3′ end of a cleaved DNA strand attacks a
phosphodiester bond.
Step 1: Transposase cleaves each side of the transposon to reveal 3’ ends. This reaction uses water
to induce hydrolytic nicking.
Step 2: Liberated 3’-OH groups attack the opposite strand to generate a hairpin. This is a type of
phosphoryl transfer reaction we have seen many times.
Step 3: Transposase cleaves the hairpins to reexpose 3’ ends which then attack the target DNA
Step 4: 3’ OH cleave target site and the transposon is pasted in
Step 5: there are nicks on each side of the transposon that need to be sealed by DNA ligase

Describe excision by transposase

1.hydrolytic nicking (involves water to make initial nicks)

2.3’OH will make a nucleophilic attack to make a concerted cleavage ligation. This forms a closed hairpin which is then processed in step 3

3.hydrolytic nicking creates free 3’OH

Describe insertion by transposase

1.The excised transposon will undergo two transesterification reactions, which is a concerted cleavage/ligation to be inserted in the two strands

2.This leaves leftover nicks that are repaired by DNA ligase after insertion

3.Nicks are sealed and repaired with DNA ligase and requires ATP to create bond

Describe the life cycle of a retrovirus and its integration

The mechanism of retroviral integration is similar to the cut-and-paste mechanism. Integrase is
encoded by the retroviral genome (e.g. HIV) and functions similarly to transposase.

1. Retroviruses have an RNA genome that is reverse transcribed by reverse transcriptase. It is
similar to telomerase since DNA is synthesized using an RNA template. The product of
RTase is cDNA (copied DNA)
2. RNase H degrades the RNA component of the RNA/DNA hybrid followed by another round
of DNA synthesis to create dsDNA
3. Integrase is similar to transposase and catalyzes the cut-and-paste transposition

Retroviruses and retrotransposons use integrase to insert their genome into genomic DNA. The
insertion mechanism is similar to cut and paste transposition except that the DNA product first goes
through an RNA intermediate.

Since the whole cDNA is inserted there is no need to excise through a
hairpin intermediate. Retroviruses replicate by transcribing the genes for translation of the viral
particles that then will assemble and package the viral RNA

Describe lifecycle of HIV virus

HIV is a retrovirus with an RNA genome in a capsid

When infection occurs, the capsid will interact with plasma membrane of the cell and will introduce RNA genome in the cell

Has rtase and integrase (transposase activty of retrovirus)

1.mRNA or RNA genome of the virus is introduced to the cell and infects

2.Its innate reverse transcriptase will make viral DNA copies

3.These copies will be integrated into the genome after moving to the nucleus from the cyotplasm. It will undergo transcription.

4.Transcript is transported out of the nucleus and and will be translated to make proteins that assemble the viral capsid, as well as components like RTase, protease, envelope protein

5.These components will be assembled and pacakaged to make viral capsid

6.This leads to lysis of the cytoplasmic membrane and the virus will infect other cells, leading to new retroviruses

Describe genome of retrovirus***

Describe retroviral integration

1.Undergoes reverse transcription (in 2 steps) to make cDNA

2.Circularization by integrase

3.The free transposon 3’ ends attack the target DNA (Does not go through hairpin intermediate since it has the 3’OH ends, did not need to be excised )

4.Integration into target DNA using transesterification and nicks are sealed with ligase

Describe endogenous retrovirus

We have many ERVs in our genome. The oldest ERV in the mammalian genome is ~70 million
years old and one in fish and amphibians is more than 400 million years old. These tend to have so
many mutations that they no longer move.

Dogs have some active ERVs that have ongoing retrotransposition (transcribed to RNA and udnergoes RTase to make cDNA after another round of replication. cDNA can insert in another place in genome) and can produce infectious particles.

Even in humans, there can be some transcription of ERVs that can even lead to
some viral particles they are not infectious. They are implicated in some diseases including cancer.

Describe Retrotransposons *****

Retrotransposons move through an RNA intermediate

Since LTR retrotransposons originate from ERVs, they can form a virus like particle after being exported in the nucleus that can result in integration of the cDNA just like in a normal LTR viral infection.

Integration still goes through and RNA intermediate. ERVs are generally at their original insertion sites during infection. LTR transposons are ERV sequences that have retained mobility through an RNA intermediate.

Describe non-LTR elements in LINE and SINEs

LINE gene is autonomous (has all the elements to direct the transposition): Has two open reading frames and encodes 3 different activites

integration of transpoase activity includes an endonuclease, so it will cut or nick a DNA in the genome and copy itself into the genome using reverse transcriptase.

SINE gene is nonautonomous: lacks open reading frames.Stil transpose through an RNA intermediate, so L1 or LINE proteins act on the SINE elements themselves. They require RTase encoded in LINE elements

Describe non-LTR elements in LINE and SINEs in detail

1.LINEs and SINEs make up the majority of the transposon sequences in the human genome
LINE: Long Interspersed Nuclear Elements;

2.SINE: Short Interspersed Nuclear Elements. The main difference is that SINEs do not encode reverse transcriptase and can use that expressed from LINEs. LINE elements are autonomous and encode all proteins needed to transpose. SINEs are nonautonomous require LINE proteins to transpose. A particular SINE found only in primates is called an Alu element. It has had a large role in shaping our genome.

SINEs may play a role in defining TAD boundaries

True. when a SINE element moves into a genome, it can sometimes take extra sequences with it. This can include bring a CTCF binding site (binds CTCF proteins which are at the basis of TADs). By having active SINE elements, the genome and transcription programs can be shaped. Often, CTCF sites are found near SINE sequences and there are even instances were new CTCF sites have arisen and these are found near SINE insertion.

Describe L1 insertion: target primed reverse transcription (TPRT)

Gene in chromosome is transcribed to become mRNA

2.mRNA is exported from the nucleus

3.ORF1 and ORF2 are translated in the cytoplasm

4.ORF1 will forma ribonucleoprotein complex that will bind to RNA. It is polyadenylated

5.Once ribonucleoprotein complex is formed, it will be imported back into the nucleus and there will be a target site. Target site often T-rich to bind to poly A tail (helps it enter nucleus)

6.Endonuclease will make a nick and RTase will be used. RNA will be copied into DNA. As it is copied into DNA, it will insert into the genome. We will have the transposition of a DNA sequence but it goes through a RNA intermediate.

SINE elements use ORF (ORF1p)

1.Will go back into nucleus and use L1 endonuclease and RTase to insert into genome

Describe L1 insertion: target primed reverse transcription (TPRT) in detail

L1 transcripts are polyadenylated and encode two ORFs. The gene product of ORF1 binds to L1
transcript to form an RNP, it is imported back to the nucleus where the product of ORF2 (reverse
transcriptase + endonuclease) will form a nick in an T-rich region. The polyA of L1 will base pair with
the repeat of Ts which will form a primer for reverse transcriptase to create a new DNA strand using
the RNA of the L1 as a template

TE insertions are associated with both noncancerous and cancer diseases

These events cans lead to exon skipping, formation of new polyA sites, formation of an exon or
creating new regulatory regions. They can also insert into exons causing frameshift mutations.
Similar events can lead to activation of oncogenes.

explain ATAC-Seq: Transposase-Accessible Chromatin with high-throughput sequencing *****

1.Isolate chromatin

2.Incubate with Tn5 DNA plus transposase

3.Tn5 cleaves DNA and adds “tags” at nucleosome free regions

4.Tags are used for illumina sequencing We can then see the open parts of chromatin

explain ATAC-Seq: Transposase-Accessible Chromatin with high-throughput sequencing. in detail

1.Faster and more sensitive method to evaluate chromatin structure in different cell types
compared to Mnase-seq and DNase-seq.
2.Relies on bacteria transposon Tn5 bound by short oligo tags.
3.They system is designed so that a modified transposase will cleave the target and add the
sequence tags excised DNA fragments that can then be sequenced

Genomic regions subject to ATAC-seq will be sequenced and peaks will be compared to other
known data about the region using a genome browser. Different cell types will have different
signature peaks. These can be used to identify candidate genes that function in the same cellular
process

Explain ATAC-Seq Network Analysis

What are the causes of DNA double strand breaks

1.Ionizing Radiation: X-rays, gamma rays outer-space, free radicals from metabolism

2.DNA replication: ssDNA nicks can form SSBs which become DSBs

3.CRISPR: guide RNA + Cas9. Cas9 cleaves bacteriophage DNA

4.Meiosis: Spol 11-induced to stimulate homologous recombination/chromosome segregation. This will form crossovers between chromosomes, essential for segregation in meiosis

5.Chemotherapy agents: Can inhibit DNA replication form and kill dividing cellsBulky adducts: topoisomerase inhibitors

What are the two ways to repair for DSBs *****

NHEJ (Nonhomologous End Joining)

1.The two ends are glued/stuck together, however this is error-prone and leads to small insertions or deletions. May result in frameshift mtuation

HDR (Homology Directed Repair)

1.Uses a donor template with a similar sequence. Uses a homologous DNA sequence as a template for repair. Not as error-prone as non-homologous end-joining. You will be able to replace one sequence with another sequence.

What are the two ways to repair DSBs? In detail

A. Nonhomologous end joining: Processes ends and seals them together. Joining can be imprecise
which can cause errors small insertions and deletions. Used in G1 of cell cycle (before DNA
replication). The process is referred to as “error prone”.

2.If changes result in the addition of a number of bases that is not a multiple of 3 you can get a frameshift mutation. This would cause the reading frame to change so that when translation occurs the ribosome may encounter a premature stop codon and form a truncated protein.

B. Homologous recombination uses a homologous DNA sequence (sister chromatid or homologous
chromosome).

2.The process relies on exonucleases to resect ends to reveal 3’ ssDNA tails. These
will invade a homologous DNA template using Rad51. DNA information is copied to repair the DNA.
This is also an important process in meiosis.

CRISPR-Cas9 creates DSBs that can be repaired by NHEJ or HDR ****

guide RNA binds to Cas9 protein

NHEJ:
creates deletions/insertions that results in frameshift mutations and premature stop codons

HDR:
replaces one sequence for another. Uses template for repair (allele replacement)

uses precise gene editing

what are the two ways CRISPR-Cas9 is used to make mutations?

The goal of CRISPR mutagenesis is to change the sequence of the DNA in the vicinity of the DSB it
creates. There are two DSB repair pathways that can be used. A DSB can be repaired using
nonhomologous end joining (NHEJ). This is a reaction that uses Ku and other proteins to reseal a
DSB. It is an error prone reaction since sometimes there can be insertions and deletions.

In insertion or deletion can cause a frameshift mutation and lead to a truncated protein product (or no protein product due to nonsense mediated decay). The second pathway uses homology directed repair
(HDR). This occurs if a DNA template with the mutation you want to create is added with the Cas9
and guide RNA. Enzymes involved in HDR will use the homologous DNA sequenced to be copied
into the DNA with the DSB

what is the difference between VDJ recombination and NHEJ recombination? ****

Both use Ku70/Ku80 (bind to gapped DNA, binds to ends of DSB), DNA-PKcs, Ligase IV/XRCC4 (joins ends together)

***In NHEJ, goes through hairpin intermediate. This is processed by Artemis to help direct coding joint to end frame. opens hairpin intermediates and coding joint uses a precise mechanism to maintian the open reading frame between VDJ segments. You cannot have insertions/deletions

V(D)J Recombination vs. Non-Homologous End Joining (NHEJ) in detail

V(D)J recombination and classical NHEJ are mechanistically related and share core DNA repair
proteins.

1. In both processes, the Ku70/Ku80 heterodimer binds to double-strand breaks (DSBs) and
recruits other components of the repair machinery. DNA-PKcs (DNA-dependent protein kinase
catalytic subunit) and XRCC4 are essential for aligning and processing DNA ends, while DNA ligase
IV completes the repair by ligating the DNA ends.

2.In V(D)J recombination, precision is critical at the coding joint, because insertions or deletions can
disrupt the reading frame of immunoglobulin genes. This precision is aided by the formation of
hairpin structures at coding ends, generated by RAG1 and RAG2. These hairpins are opened by the
Artemis nuclease, which allows for controlled processing of DNA ends.

3.In contrast, during general NHEJ (outside of antigen receptor loci), DNA end processing is often
more variable. Some nucleotides may be trimmed or added, which can result in small insertions or
deletions which are features commonly seen in NHEJ-mediated repair of random DSBs.

Explain the philadelphia chromosome NHEJ between different chromosomes ****

translocations can occur due to swapping of chromosome regions

ex:chromosome 22 has sequence that is the B cell receptor. on chromosome 9, we have ABL, which codes for a tyrosine kinase that is involved in cell growth control

In this translocation, a gene fusion of BCR-ABL is made. BCR-ABL becomes an oncogene, which is expressed in B cells. Now ABL is inappropriately expressed in B cells, which can result in leukuemia.

may due to RAG 1 and RAG 2 nucleases (found in B cells)

Explain the philadelphia chromosome NHEJ between different chromosome in detail

The Philadelphia chromosome is a result of a reciprocal translocation between human chromosomes
9 and 22, which causes chronic myeloid leukemia (CML). This translocation fuses the BCR gene on
chromosome 22 with the ABL tyrosine kinase gene on chromosome 9, creating the BCR-ABL fusion
gene. This fusion leads to constitutive ABL kinase activity, which promotes uncontrolled cell
proliferation and survival—hallmarks of cancer.
This chromosomal rearrangement is relatively rare, but its occurrence in a single hematopoietic stem
cell can be sufficient to initiate leukemia. There are a couple of reasons why this translocation may
occur more frequently in B cells:

1. RAG1 and RAG2 endonucleases, which are expressed in developing B cells to mediate
V(D)J recombination, may inadvertently introduce off-target double-strand breaks (DSBs),
making translocations more likely.
2. The loci on chromosomes 9 and 22 involved in the translocation may be spatially proximate
within the nucleus, increasing the chance of misrepair via non-homologous end joining
(NHEJ)

Double strand breaks in repeated DNA sequences can result in chromosome rearrangements by unequal crossing over involving homologous recombination ****

true

Homologous recombination between repeated DNA elements (e.g. transposons) can lead to genome rearrangements, such as deletions, inversions, and duplication

Unequal crossing over can lead to the creation of new genes “Exon shuffling”

Crossing over between LINE and SINEs lead to exon shuffling at the DNA level. This happens over evolutionary time scales (unlike VDJ recombination). Exons can be swapped between genes by HR between Alu elements flanking and exon and another two Alu’s flanking the exon of another gene.

Gene families have mix and matched modules

True, Exons tend to be short (e.g. 100-200 nt) and roughly similar in size to protein coding domains. Since many proteins have a modular form, it is conceivable that recombination/transposition that moves exon DNA sequences between genes (over evolutionary timescales) could lead to the expansion of gene families. Look at the splicing helicases and you can see that they have mix and match
domains

Exons tend to define protein domain modules, promotes exon-shuffling. Exon shuffling may have contributed to wide range of splicing factors

Describe meiosis

Involves one round of cell division with two segregation events

In Meiosis 1: homologs separate

Meiosis 2: sister chromatids separate

Crossovers are essential for proper segregation at Meiosis I and crossovers form as a result of the formation and repair of DSBs. The cell makes >200DSBs on purpose

Describe meiosis in detail

Meiosis is required for the formation of haploid gametes. The reduction in ploidy occurs through two
rounds of chromosome segregation following one round of DNA replication. In the first round,
homologous chromosome separate. In the second round, sister chromatids separate. Proper
segregation requires the formation of crossovers between homologous chromosomes. Crossing over
occurs through homologous recombination and is initiated by the formation of hundreds of DSBs

How does the meiotic chromosome undergo pairing and synapsis?

Crossover occurs during Prophase I and is divided in 4 stages:
Pre-leptotene, Leptotene, Zygotene, Pachytene

Pre-leptotene:

Leptotene: involves homolog pairing: homolog 1: 2 sister chromatids and homolog 2: 2 sister chromatids. Double-strand breaks form here. The homologous chromosomes pair through DSB repair. Homology search partner (this will be the homolog), leading to a crossover

Leptotene: Involves synapsis formation, with formation of synaptonemal complex
Pachytene: synapsis elongation, crossover maturation. Where crossing over will occur, so that when they separate in meiosis I, they will be able to line up together and spindles will split them apart

Pairing, Synapsis, and Recombination Are Essential for Accurate Chromosome Segregation

true

Homolog pairing, synapsis, and recombination are interdependent events that occur during
prophase I of meiosis and are crucial for ensuring the proper segregation of homologous
chromosomes.
Prophase I is subdivided into five stages, with Leptotene, Zygotene, and Pachytene being the key
phases for homologous recombination:

1. Pre-leptotene: Chromosomes begin to organize into a loop-axis structure, in which DNA
loops—each containing sister chromatids—are anchored to a linear proteinaceous axis.
2.Leptotene: Programmed double-strand breaks (DSBs) are introduced by the Spo11 protein,
initiating the search for homologous sequences.
3.Zygotene: Homologous chromosomes begin to pair through homology searching, and the
synaptonemal complex (SC) begins to assemble between them.

4.Pachytene: Full synapsis occurs. The synaptonemal complex, a conserved proteinaceous
structure, stabilizes the side-by-side alignment of homologs. Recombination intermediates
are processed at this stage, allowing for crossover formation.
The synaptonemal complex provides a scaffold that both maintains alignment and facilitates the
repair of DSBs into either crossovers or non-crossovers. These crossovers are essential for the
generation of chiasmata, which physically link homologs and ensure their accurate segregation
during anaphase I

what is the significance of Spo11?

it induces DSBS to initiate homologous recombination (1st step of homologous recombination)

Involves a transesterase

1.Spol II has a tyrosine with an OH, this OH will make a nucleophilic attack on the DNA sequence and will form covalent attachment to the 5’ end of DSB (both sides are attacked, forming DSB)

2.An endonuclease will remove Spol II oligos

3.Now we have double strand breaks with 3’OH overhangs. Endonuclease will cleave off DNA so leave longer stretches of ssDNA

4.Exonuclease will chew back DNA in 5 to 3’ direction and will be used in homology search to find homologous chromsome to allow pairing

concerted cleavage reaction

Describe the mechanism of Spol II

Spo11: A Transesterase that Initiates Meiotic DSBs
1. Spo11 initiates meiotic recombination by generating double-strand breaks (DSBs)—
potentially hundreds per cell—that are repaired by homologous recombination (HR).

2.Spo11 catalyzes DSB formation via a transesterification reaction. It becomes covalently
attached to the 5′ ends through a tyrosine residue in its active site. This reaction is
mechanistically similar to nucleotidyl transferases, except that the nucleophile is the hydroxyl
group of a tyrosine, rather than a 3′ hydroxyl from DNA or RNA.

3. After DSB formation, Spo11 remains covalently attached to short oligonucleotides at the
break sites. These Spo11-linked oligos are subsequently removed by specialized
endonucleases (such as MRN complex and Sae2 in yeast), exposing a free 5′ end.

4. The exonuclease Exo1 then degrades the 5′ strand in a 5′ to 3′ direction, creating 3′ single-
stranded DNA (ssDNA) tails.

5.These 3′ ssDNA overhangs are essential for the next steps of homologous recombination:
they undergo a homology search and initiate strand invasion of the homologous
chromosome to form recombination intermediates.

what is the 2nd step of homologous recombination? Repair of Spol II-induced DSBs result in crossover and noncrossover events

1.3’ overhangs will bind to RPA (single stranded binding protein, same RPA used in DNA replication)

2.Recombinases (Rad 51 and DMC1) will perform strand invasion (search for homology, dives into duplexes to find sequences that base pair with 3’ overhangs)

3.two different outcomes of homologous recombination: intermediate double holidy junction (will undergo resolution to form a crossover, cleaves inside strands, then outside strands, which is resolved to form a crossover. 2nd outcome a noncrossover. Strand invasion copies some DNA to fill in the gap, happens by single strand annealing

Describe how the Repair of Spo11-induced DSBs results in crossover and noncrossovers events in detail

The homology search is mediated by the recombinases Rad51 and Dmc1. These enzymes mediate
strand invasion so that the ssDNA overhang can invade a duplex to find homology. This is still a
mysterious process since it is a genome wide search. Resolution of a double-Holliday junction
intermediate to form a crossover. Synthesis Dependent Strand Annealing (SDSA) simply
synthesizes new DNA, is unwound by helicases and then anneals to the other end of the break. The
number of crossovers per chromosome is regulated by ZMM proteins

How are crossovers held together? What is the role of the chiasmata? Describe in detail

1.Homologs are held together by the combination of sister chromatid cohesion and the crossover
(called here the chiasmata) in humans, early prophase I events occur in the fetal ovary and
oocytes arrest prior to MI until ovulation so this could last decades.

2.The maternal age effect
demonstrates increased incidence of confirmed conceptions as trisomies. This increases
dramatically past age 35 . Since chromosomes are arrested in the paired configuration for decades, with only cohesins mediating cohesion, it is though that their loss over time could lead
to homologs no longer being held together.

chromsomes are physically held together by the combination of a crossover and cohesion between sister chromatids

true

In M1, the coheins (which hold sisters together) are removed

Aneuploidy is the leading cause of birth defects and pregnancy loss

true

Aneuploidy is when a cell or gamete has the wrong number of chromosomes
• Trisomy describes the chromosome state of the embryo after fertilization of a disomic
gamete.
• Rates of aneuploidy in eggs is very high (15-30% or higher). Aneuploid sperm are less
common but since they come from a stem cell population there is increased risk of point
mutations.
• The maternal age-effect describes the higher incidence of aneuploidy leading to trisomy or
pregnancy loss as women age. Trisomy 21 is the cause of Down’s syndrome.

Depending on when the DSBS form will determine how they will be repaired

true, DSBs in G1 are repaired by NHEJ and DSBS in G2 are repaired by HDR. Depends on whether sister chromatids are present or not

Describe the chemotherapuetic agents that target DNA replication in detail

Tumors divide rapidly so the DNA replication machinery is a good target. Many tumor cells have
mutations in DNA repair genes so drugs that create DNA damage can make effective cancer
therapies. There is a double-edged sword; cancer cells are dividing rapidly so unrepaired damage
can kill the cells. We have dividing cells in our gut and bone marrow, so these drugs can make us
very sick

Cisplatin: causes inter-strand and intra-strand DNA crosslinks that block DNA replication fork
treatment of metastatic testicular cancer
• Etoposide; inhibits ability of topoisomerase to reform DNA backbone after strand cleavage forms
DNA breaks that re especially lethal.

Describe the chemotherapuetic agents that target DNA replication

1.Cisplatin: causes intrastrand (2 bases of the same strand of DNA) and interstrand DNA crosslinks. These modifications can cause DSBs and cause replication to stall

2.Etoposide: topoisomerase inhibitor. Blocks transition from conformation 2 to conformation 3 Innhibits resealing of gap

Explain function of Etoposide

Etoposide blocks sealing Gapped DNA before topoisomerase reseals gap in G strand
1.Topoisomerase II is a AAA+-ATPase that uses ATP hydrolysis to make the reaction go
forward.
2.There are two segments of DNA; the G segment (green) is the strand that will be broken,
and the T segment (orange) is the strand that will pass through the gap; once it passes
through the gap the DNA gap will be resealed

3. Etoposide inhibits resealing the gap; double strand breaks are especially lethal to cells.
4.Topoisomerase inhibitors specific to bacteria are used as antibacterial drugs; one example is
ciprofloxacin (Cipro) is a broad-spectrum antibiotic

Which of the following is NOT a correct distinction between exon shuffling and alternative splicing?

Exon shuffling is an mRNA processing event that generates entirely novel proteins, while alternative splicing only produces different isoforms of the same protein.

This is false, as Exon shuffling is not an mRNA processing event, it happens at the DNA level and can lead to expansion of gene families. While alternative splicing does occurring mRNA processing and does produce different isoforms of the same proteins

which of the following are distinctions between exon shufflig and alternative splicing?

1.Exon shuffling occurs over evolutionary timescales, while alternative splicing occurs during normal gene expression on a daily basis.

2.Exon shuffling involves recombination between different genes at the genomic level, whereas alternative splicing modifies a single gene’s transcript.

3.Exon shuffling typically arises from unequal crossing over or transposition, whereas alternative splicing results from differential use of splice sites.

You hypothesize that a transposable element transposes through an RNA intermediate. Which of the following experiments would most definitively test this hypothesis?

You are designing an mRNA vaccine for a novel virus. Based on prior SARS-CoV-2 vaccines, you incorporate modified uridine residues. What is the effect will these modifications?

Help the mRNA evade host innate immune detection.

how???

The newly identified virus is a retrovirus. You aim to develop a nucleotide analog to inhibit viral replication. Which viral enzyme is the most appropriate drug target?

RNA-dependent DNA polymerase ***

1.Retroviruses convert from RNA to DNA (creates cDNA, complementary DNA strand using reverse transcriptase/RdDp which is able to use and read the RNA template to make DNA)

2.RNA-dependent DNA polymerase then synthesizes the second complementary strand to make dsDNA

3.The dsDNA, after integration, can then be transcribed to make viral proteins

CRISPR-Cas9 induces double-strand breaks that can be repaired by either nonhomologous end joining (NHEJ) or homologous recombination (HR). How can researchers influence which repair pathway is used

NHEJ is favored when a short RNA hairpin is co-delivered with Cas9 and the guide RNA.

NHEJ is favored in the absence of any DNA repair template.

HR is favored when antisense DNA is co-delivered with Cas9 and the guide RNA.

HR is favored when antisense DNA is co-delivered with Cas9 and the guide RNA.

HR is favored when no additional nucleic acid sequences are provided.

NHEJ is favored in the absence of any DNA repair template

NHEJ does not use a repair template, it simply joins the broken DNA ends back together

HR is favored when a DNA repair template, homologous to the region around the break, is delivered. This is not the antisense DNA, since the antisense DNA is complementary to the mRNA, not the genomic DNA.

HR is not favored with no additional nucleic acid sequences are provided, this favors NHEJ instead since it not need a template or nucleotides.

Which of the following genome rearrangements is NOT typically a product of homologous recombination?

Gene duplications

Deletions

Translocations

Inversions

Immunoglobulin V(D)J recombination

Immunoglobulin VDJ recombination

This is not a product of homologous recombination, since it involves site-specific recombination, not homology.

A patient presents with a blood-borne pathogenic yeast strain that is rapidly dividing and resistant to all current antifungal treatments. Considering chemotherapeutics used for cancer, which approach is LEAST likely to be effective in this case?

Lipid nanoparticles containing modified uridines

Cisplatin

Etoposide

Ionizing radiation

Lipid nanoparticles containing modified uridines

Lipid nanoparticles containing modified uridines are used in mRNA therapies, like mRNA vaccines, to allow the mRNA to be taken up effectively. This is used in mammalian cells, not fungal

How is genomic instability related to aging and cancer?

1.Exogenous sources include chemicals, x rays, gamma rays, chemotherapy

2.Endogenous sources include reactive oxygen species, single strand breaks, and mitochondria

The consequences include stalling of RNA/DNA polymerase, chromosome rearrangements, such as mutations, deletions, insertions, aneuploidy

How is genomic instability related to aging and cancer? in destail

DNA damage plays a critical role in the development of aging and cancer and can arise from both
endogenous and exogenous sources. Endogenous damage primarily results from reactive oxygen species (ROS), which are generated as byproducts of normal cellular metabolism, especially by mitochondria.

These ROS can inflict harm on DNA, proteins, and lipids. Exogenous sources of damage include environmental and certain chemotherapeutic drugs like etoposide; and ionizing radiation, including X-rays and gamma rays, which are known to induce double-strand breaks (DSBs).

The molecular consequences of such DNA damage can include the stalling of DNA or RNA polymerases, chromosomal rearrangements, and aneuploidy. At the cellular and systemic levels, accumulated DNA damage contributes to processes such as accelerated aging and the initiation and progression of cancer

How is unpaired DNA damage associated with age associated diseases

Genetic risk factors and environmental risk factors contribute to the balance of damage and repair. Excessive damage over time can lead to unrepaired damage accumulation, which has several effects at the cellular level. For instance, apoptosis may occur, senescence, and somatic oncogenic mutations. Apoptosis, which is cell death mediated by tumor suppressor genes (P53), and senescence (cell is arrested but still metabolically active) contribute to loss of regenerative capacity, inflammation, etc., which is associated with aging. Somatic oncogenic mutations occur due to the accumulation of mutations and clonal expansion, which lead to cancer.

what inherited syndromes are associated with defects in DNA repair

Several inherited syndromes are caused by defects in DNA repair pathways, leading to increased cancer risk. Earlier in the course, we discussed the role of mismatch repair (MMR) in correcting DNA replication errors, and its association with colorectal cancer. Mutations in MMR genes significantly elevate the risk of colorectal cancer, which is notably rising in incidence among younger adults (called Lynch syndrome). Another important example involves mutations in the BRCA1 and BRCA2 genes, which are key players in the homologous recombination pathway of DNA repair. Inherited mutations in either gene impair the cell’s ability to accurately repair double-strand breaks. Women who carry a heterozygous loss-of-function mutation in the tumor suppressor gene BRCA1 face a cumulative lifetime risk of 50% to 80% for developing breast cancer

what is the function of BRAC1 and BRAC2 in end processing and strand invasion of HDR?

typically, HDR involves a sister chromatid

BRAC1 and BRAC mediate the strand invasion step

BRAC1 is invovled in end processing

BRAC2 will facilitate loading of BRAC2

what is the function of BRAC1 and BRAC2 in end processing and strand invasion of HDR? In detail

BRCA1 and BRCA2 are essential components of the homologous recombination (HR) pathway, a high-
fidelity mechanism for repairing double-strand DNA breaks. These proteins play a critical role in the end resection and strand invasion steps of homology-directed repair (HDR). Specifically, BRCA1 is involved in the processing of DNA ends to generate single-stranded DNA overhangs, while BRCA2 facilitates the loading of RAD51 onto the single-stranded DNA, enabling strand invasion into the homologous template. When BRCA1 or BRCA2 is defective, this repair pathway is compromised, resulting in the accumulation of unrepaired double-strand breaks and increased genomic instability

Describe PARP inhibitors ****

Is used to kill BRAC ½ Mutant cells

PARP repairs SSBs, so inhibition of PARP will not allow repair of SSBs
During DNA replication, SSBs are converted to DSB, when fork counters SSB

A PARP inhibitor will create DSBs cell

PARP inhibitors will not cause many side effects if the cell is HR proficient (will repair DSB)

In a HR defieinct cell (cell is homozygous for mutation of BRAC/BRAC2), then the cell will die by apoptosis.

PARP inhibitors can kill cancer cells directly since they cannot perform HR

Describe PARP inhibitorsin detail

1.PARP inhibitors are an effective targeted therapy for breast cancers associated with BRCA1 or BRCA2
mutations. PARP (poly[ADP-ribose] polymerase) enzymes normally function in the repair of DNA single-strand breaks. Inhibition of PARP leads to the accumulation of unrepaired single-strand breaks, which cause replication forks to stall and collapse, resulting in double-strand breaks (DSBs). In cells with functional BRCA1 or BRCA2, these DSBs can be accurately repaired via homologous recombination.

2.However, in BRCA1- or BRCA2-deficient (BRCA–/–) cells, this repair pathway is disrupted, leading to persistent DNA damage and ultimately cell death through apoptosis. In contrast, BRCA+/– heterozygous cells retain one functional copy of the gene and can survive PARP inhibition. This therapeutic strategy offers a selective advantage over traditional DNA-damaging agents such as etoposide or cisplatin, which induce DSBs that are more difficult for all cells to repair, increasing toxicity to healthy tissue

How do cancer cell arise?

arise from multiple sequential mutations (multi-hit hypothesis)

1.1 mutation occurs when a protooncogene is converted to a oncogene. protooncogene normally regulate growth. However, in oncogenes, cell growth is no longer controlled, and these mutations are often sporadic and dominant (only one copy of the gene needs to be mutated)

2.2nd mutation is a mutation in a tumor suppressor gene: involved in DNA repair, apoptosis, or regulating cell cycle (BRAC1/BRAC2, p53, miRNA)

3.Usually telomerase is silent in most of our somatic cells, but a cancer cell will activate telomerase gene. Allows telomeres to be added to chromosomes and allows the cell to divide repeatedly. Caused by mutation (often in promoter region). Canalso cause hypomethylation and copy number gain of telomerase gene

describe the multi-hit model for carcinogenesis

Typically only a handful of mutations are required for a cell to become a cancer cell.
A. A mutation that causes the cell to lose control of division and differentiation (oncogene). An example of
an oncogene we saw previously was the BCR-ABL oncogene that causes leukemia. These mutations
typically have dominant phenotypes (predisposition to cancer) and only one allele needs to be mutated.
The unmutated version is called a protooncogene.
B. Loss of function mutations in genes involved in DNA repair/apoptosis/cell cycle regulation- tumor
suppressor genes (e.g. BRCA1/2); many TS miRNAs. Typically both copies of TS genes have to be
inactivated for them to have an effect, though for some just one copy is not enough.
C. Mutation that activates telomerase is often in metastatic carcinomas where cells undergo many divisions
and need to maintain telomere length-dominant. These can be promoter mutations, hypomethylation or
gain in copy number.

describe tumor suppressor gene in detail

1.Tumor suppressor (TS) genes play a critical role in maintaining genomic stability and preventing
uncontrolled cell growth. Their normal functions include repairing DNA damage, regulating the cell cycle, and triggering apoptosis when cellular damage is beyond repair.

2.Apoptosis, or programmed cell death,
serves as a fail-safe mechanism to eliminate potentially harmful cells. For a tumor suppressor gene to be fully inactivated, both alleles typically need to acquire loss-of-function mutations—a phenomenon known as the "two-hit hypothesis."

3.In cases of inherited cancer syndromes, individuals are born with one mutated
allele, which significantly increases the likelihood that the second allele will also become mutated during their lifetime, thereby accelerating tumor development.

what are tumor suppressor genes?

the normal function of TS genes is to protect cells, function as DNA damage repair genes, reglate the cell cycle, and initate apoptosis as the last resort

A loss of function mutation in both alleles is often required

At the cellular level, the genes are recessive

Describe tumor suppressor genes in detail

Tumor suppressor (TS) genes play a critical role in maintaining genomic stability and preventing
uncontrolled cell growth. Their normal functions include repairing DNA damage, regulating the cell cycle, and triggering apoptosis when cellular damage is beyond repair. Apoptosis, or programmed cell death, serves as a fail-safe mechanism to eliminate potentially harmful cells. For a tumor suppressor gene to be fully inactivated, both alleles typically need to acquire loss-of-function mutations—a phenomenon known as the "two-hit hypothesis." In cases of inherited cancer syndromes, individuals are born with one mutated allele, which significantly increases the likelihood that the second allele will also become mutated during their lifetime, thereby accelerating tumor development.

Describe how loss of heterozygosity can affect tumor suppressor genes

Loss of heterozygosity (LOH) is a key event in tumorigenesis, often affecting tumor suppressor genes. There are several mechanisms by which LOH can occur.

One pathway involves homologous recombination during G2: if a crossover occurs between the centromere
and a mutated gene, the resulting sister chromatids can segregate such that one daughter cell inherits two copies of the mutant allele, losing the wild-type allele entirely.
A second pathway involves chromosome missegregation, or nondisjunction, which results in aneuploidy—daughter cells receiving an abnormal number of chromosomes, one too many or one too few. In the case of an extra chromosome the chromosome with the “good” allele is lost in a subsequent division, restoring diploidy but potentially leaving the cell with two copies of the mutant allele.
Additionally, LOH can also result from epigenetic silencing of the wild-type allele. For example, changes in
higher-order chromatin structure, such as aberrant topologically associating domain (TAD) formation, can lead to transcriptional silencing of the functional copy of a gene, functionally mimicking a genetic mutation.