V(D)J Recombination Flashcards

1
Q

Describe antibody structure?

A

2 heavy and 2 light chains, joined together by disulphide bonds at the hinge region

The light chain = 1 variable and 1 constant domain - each being around 10 amino acids
The constant region is either kappa or lambda (doesn’t matter which)
It only functions to keep the variable region attached to the heavy chain

The heavy chain = 1 variable and either 3 constant domain + hinge region or 4 constant domains
There are 5 different constant regions - alpha, theta, epsilon, gamma or micro

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2
Q

What does the antibody binding site comprise of?

A

Made up of the variable regions of the heavy and light chains
Both variable domains are fully variable as we need certain structural components

In the variable region there are 3 hypervariable regions (complementarity determining loops)
They form contact with the antigen

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3
Q

Give an overview of V(D)J recombination?

A

There are 30,000 genes in the genome and we need 10^13 different antibodies - theoretically we would need 10^13 genes

We assemble the exon that encodes the variable region, from scratch in individual B-cells
So we join individual gene segments in the V(D)J recombination
V - variable
D - diversity
J - joining

Recombining these in different ways allows many different antibody binding pockets
Variable (V) gene segments encode around 93 aa of 110 aa of the variable exon - the rest are encoded by D and J gene segments

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4
Q

Describe the V(D)J recombination mechanism?

A
  1. The heavy chain locus will recombine one of the D’s with one the J gene segments - this is a permenant somatic recombination
    The bit of DNA that was in between these segments is kicked out and lost from the genome
  2. Permanently join a DJ bit to a variable region and again lose the DNA in between (somatic recombination NOT splicing = alteration in the DNA)
  3. The Light chain locus recombines one V and J gene segment, to create the variable exon
  4. The genes are then transcribed as one bit of DNA - all the way to the constant region
  5. The introns in between the variable and constant region are then spliced out
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5
Q

What does V(D)J recombination mean for antibodies?

A

Immunoglobulin has multiple V, D and J gene segments
This allows for a lot of versatility - by mixing and matching the different segments

Number of functional gene segments in human immunoglobulin loci
Possible combinations = 1.9 x 10^6
= adaptive immune system

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6
Q

How is this recombination signalled?

A

V and J are signalled to undergo recombination - they have a recombination signal sequence (RSS) - often depicted by triangles
RSS consist of a conserved heptamer separated by either a 23 or 12 bp spacer from a conserved nonamer
This heptamer is always adjacent the coding region of the V or the J

For the lambda chain V’s RSS = 23 bp spacer and J’s RSS = 12 bp spacer
This ensures we always get recombination between a V and J (not V and V)
The RSS is recognised by RAG proteins

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7
Q

Describe RAG proteins?

A

RAGs - recombination activation gene proteins
They recognise RSSs

There are two types:
RAG1
1040 amino acids - it has no introns
Active region is at the C-terminal core - amino acids 384 – 1008
Active site DDE motif (aa 600, 708, 962) - this carries out the cleavage
This possess the nonamer binding domain - binds the nonamer of the RSS

RAG2
527 amino acids
Active region is N-terminal core – amino acids 1-387
C-terminus has PHD finger that is critical for chromatin binding
This directs the catalytically active site of RAG1 to the heptamer coding region boundary

Both proteins are needed for catalysis - RAG2 is an essential cofactor

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8
Q

Describe RAGs interaction during recombination?

A
  1. A tetramer of the RAG1-RAG2 complex binds to one of the gene segments’ RSS
  2. The tetramer complex with 1 RSS bound scans for the complimentary RSS
  3. The stable 12-23 RSS complex = synaptic complex (V and J gene has been combined)
  4. In a coupled cleavage reaction RAG 1 makes a nick on one strand of the DNA - on the heptamer RSS boundary - this releases a free 3’OH
  5. The 3’OH attacks the PO4- groups on the DNA on the opposite strand = direct transesterification reaction
    This results in a hairpin structure at the coding ends (continuous sugar phosphate backbone) and a blunt double strand break at the signal ends
  6. The 4 DNA ends are processed/joined by the non-homologous end joining machinery (NHEJ)
    Join the blunt ds ends to form an excised signal circle
    Open the hairpin ends, process them to give you the exon of the antibody gene
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9
Q

Describe the processing of the DNA ends after RAGs interaction further?

A

Once the RSSs have been cleaved and there are hairpin structures and blunt ends - Ku70:Ku80 binds the DNA ends with high affinity

At the blunt signal joints - recruit DNA ligase 4 and cofactor XRCC4 = direct joining

At the coding ends - Ku70:Ku80 recruits the DNA-PK Cs = Ku proteins and DNA-PK form the DNA PK complex
The DNA-PK complex then recruits Artemis - this opens the hairpins
Terminal deoxynucleotidyl transferase (TdT) processes the DNA ends
Then DNA ligase 4 and cofactor XRCC4 ligates the DNA ends

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10
Q

Describe an additional level of diversity?

A

The joining/processing of the coding ends by addition of N and P nucleotides
Artemis:DNA-PK complex opens the DNA hairpins generating palindromic P-nucleotides
It nicks the strand around 7 nucleotides from the apex of the hairpin
This nick allows the DNA to fold out, and the nucleotides on the lower strand are now on the top strand = palindromic P-nucleotides

N-nucleotides are added by terminal deoxynucleotide transferase (TdT) - up to 20
This has no template and adds random nucleotides (not complimentary)
TdT is expressed only in pro-B cells and early T cells

The ends then pair (not bp perfectly)
An exonuclease removes unpaired nucleotides and the gaps are filled by DNA polymerase and ligation to form the coding joint
Exonuclease and TdT can be interchangeable in order

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11
Q

What are the methods of diversity?

A
  1. Combinatorial joining of gene segments
  2. Junctional diversification during gene segment joining
  3. Combinatorial joining of L and H chains
  4. Somatic hypermutation

1-3 occurs during early B-cell development and 4 in later B-cell development

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12
Q

Describe the stages of B-cell development?

A

Heavy chain rearrangement - takes place at the pro-B cell type
When rearranging most times the antibody is out of frame = no functional protein (non-productive rearrangement)
Only rearrange one allele at a time - so if the rearrangement is non-productive, you can try rearrange the second allele
If one allele is productive the B-cell will survive - if not it will die

The productive heavy chain - it is expressed on the cell surface with a surrogate light chain
This signals the B-cell to replicate
This can then be paired with different light chains once that is rearranged in pre-B-cells

Immature naïve B-cells are released from the bone marrow and have IgM
Once they start to circulate then can express IgM and IgD
This differential expression is achieved via alternate splicing

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13
Q

What are the primary and secondary responses of B-cells?

A

Primary:
It undergoes proliferation and maturation - class switch recombination (which type of heavy chain needed) and hypermutation
Hypermutations - random mutations are introduced into the variable exon - some will make binding better = selection and they will survive (others will die)
We then make memory B-cells and plasma cells to fight the

Secondary:
We activate the memory cells - which can then go clonal expansion and further class switch recombination and hypermutation
This is because we could still manage to make improvements to the antibody response

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14
Q

Describe the determination of antibody classes?

A

They can be either membrane bound or secreted
This is determined by alternative splicing of polyA sites

Cell membrane - splice a down polyA site = addition of 25 hydrophobic amino acids which keep the antibody anchored in the membrane
Secreted - splice to an upstream polyA site = hydrophilic amino acids

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15
Q

What are the different antibody types of heavy chain?

A

IgG - 80% circulating in the blood and activates compliment

IgM- Pentameric - joined by the J chain (15 kDa protein) this recognises 18 amino acids in the tail of the IgM molecule
Activates compliment

IgA - Dimers - held together by the J chain protein
Found in secretions of the gut mucosa

IgD - Quite unknown - may help the efficiency of the immune response
Only a small amount found

IgE - Fights parasights but can give an allergic response

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16
Q

What is class switching recombination?

A

If we want a different immune response - we can switch the heavy chain to get a different antibody
IgG, IgE or IgA have a switch region (everything but IgD)
When you activate the switch region upstream - by T-cell signalling - allows transcription of these regions

We get ds breaks and repair of the NHEJ machinery = permenant recombination
We can’t then go back to expressing the previous antibody
But we can switch to another - with a constant region downstream

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17
Q

What is involved in class switch recombination?

A

This involves the deletion of intervening DNA
Upstream of the heavy chains have a switch region (not IgD) between 2-10 kbp long and G-rich with repeated elements
We activate the switch region of both heavy chains - the one you’ve got and the one you want
The enzyme needed for this requires ssDNA
The switch regions lie in introns = always results in genes in frame

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18
Q

Describe the switch regions and AID in class switch recombination?

A

The switch regions are transcribed and we form some R-loops (RNA/DNA hybrids) - however, they are stable for longer than normal
The stable R-loops give AID a longer time to act on the ssDNA on the non-template strand

ssDNA is attacked by AID - it catalyses the deamination
It is a zinc dependent enzyme which completes a nucleophilic attack on an amine bond
The amino group is replaced with a carbonyl group = mutation to uridine
This forms a GU mismatch

19
Q

Describe the mechanism of class switching?

A
  1. DNA in variable/switch region is transcribed and produces local ssDNA
    Initiated by activation of the upstream promotor
  2. In B-cells, AID attacks cytidine in ssDNA to produce uridine
  3. Uracil-DNA glycosylase (UNG) removes uracil to form apyrimidinic residue
    It removes the base and leaves the sugar phosphate backbone
  4. Apurinic/apyrimidinic endonuclease (APE1) excises ribose to form a single stranded nick in the DNA

All the enzymes - AID, UNG and APE1 introduce multiple clustered nicks on both strands of DNA

  1. DNA-PK and other repair proteins act to initiate double-strand break repair (DSBR)
  2. The NHEJ machinery aided by 53BP1 (helps pull the long range DNA in chromatin together) it repairs the broken DNA and re-joins the switch regions
    = a different constant region attached to the variable exon
20
Q

What is somatic hypermutation?

A

This takes place in mature B-cells
B-cells introduce mutations into only the variable exon - this increases antibody diversity
Predominantly takes place in the hypervariable regions
This increases the affinity of the antibody to the antigen

This is carried out by AID
It takes place in the variable region as it has more pause sites for the polymerase (stalling leads to more recruitment of AID) and AID can be recruited to the sites around promotors

21
Q

Describe the mechanism of somatic hypermutation?

A

AID deaminates cytosine to give uridine = GU mismatch
This mismatch can be processed in 3 different ways

  1. Replicate over the mismatch causing the U-G pair to transition the G to A or U to C, meaning there is variation of that base in the cell = mutation
  2. Uracil DNA-glycosylase removes the mismatch base and creates an abasic site
    An error prone polymerase TLS then inserts bases at ransom sites opposite the abasic site (lesion bypass) = introduces transversion mutation
  3. Via mismatch repair by excising bases around the U-G and an error prone polymerase (MMR) inserts mutations - preferably at A:T
22
Q

Give a comparison of class switch recombination and somatic hypermutation?

A

Somatic hypermutation - low density of AID-dependent lesions
This leads to a SSB which is more sufficient for mutation

Class switch recombination - high density of AID-dependent lesions
There is a greater tendency of the switch regions to form stable R-loops as AID is much more active 
This leads to more mutations and more opportunity of double stranded breaks, therefore, more chance of undergoing class switch mutation
23
Q

What can antibodies form if not secreted?

A

Membrane-bound antibodies = complexes
They are known as B-cell receptors
Immunoglobulin is in a complex with Iga and Igb
When it binds an antigen it can signal to the B-cell to proliferate

24
Q

List all the steps in generation of immunoglobulin genes?

A

V(D)J recombination
Junctional diversity - N-sequence insertion
Transcriptional activation - activation of promotor by proximity to the enhancer
Switch recombination
Somatic hypermutation

IgM, IgD expression on surface
Membrane v secreted form

25
Q

Describe T-cell receptors?

A

The are found in a complex in the membrane with CD3 - always membrane bound
They have only 1 antigen binding domain
The antigen binding pocket is made up of two dissimilar polypeptide chains - alpha and beta (held together by di-sulfide bonds)
It has a variable domain - responsible for antigen binding and a constant domain
They have the immunoglobulin fold in the variable regions

26
Q

Describe T-cell recognition?

A

T-cell receptor recognise their antigen when in complex with MHC class I or II
They then signal to the correct T-cell to proliferate and to activate B-cells
T-cell receptors actually recognise MHC I or II - so less of the pocket is available for the binding of the antigen

To compensate this hypervariable region 3 is a lot more diverse in T-cell receptors - it has to recognise a lot more antigens
There are a lot more J gene segments to facilitate this

27
Q

Describe the generation of T-cell receptor diversity?

A

This also occurs by V(D)J recombination
TCR beta - undergoes V, D and J recombination
TCR alpha - undergoes V and J recombination
There is a small portion of gamma T-cells which are responsible for immunity in the gut - similar rules to how they are generated

The mechanism is exactly the same as for B-cells - also following the 12/23 rule
They rely more on junctional diversity
There are few results that end up out of frame

Somatic hypermutation and class switch recombination do not occur at T-cell receptors as they are membrane bound (no AID)

28
Q

How is V(D)J recombination regulated?

A

Lineage specificity - RAG1 and RAG2 recognise the same RSSs in both B and T cells
Stage specificity - recombination occurs at distinct stages of lymphocyte development
Allelic exclusion - restricts expression to one allele per cell
Cell cycle regulation

29
Q

Describe cell cycle regulation?

A

We don’t want to introduce breaks in the DNA within mitosis
V(D)J recombination it strictly done in G1 phase
Therefore this is dependent on having both RAG 1 and RAG 2 - with RAG2 undergoing phosphorylation as a control mechanism (moving between stable and unstable)

  1. At the end of G1 cyclin dependent kinases phosphorylate RAG2 on phosphorylation of Thr 490
  2. This signals RAG2 for degradation by the ubiquitin phosphorylation pathway
  3. This allows suppression of V(D)J recombination until the cyclin dependent kinases are suppressed and RAG2 can build up again for the start of G1
30
Q

What does regulation of RAG cutting depend on?

A

The accessibility of RSSs
RSSs are only made accessible to RAGs in the correct cell type
RSSs are only made accessible to RAGs at the correct stage of lymphocyte development
RSSs are only made accessible (usually) on one of the two alleles

This is due to remodelling chromatin - allowing accessibility

31
Q

What features were looked at for influencing accessiblity of RSSs?

A
Chromatin
Histone modifications
Non-coding transcripts
Transcription
Enhancers
Allelic exclusion
32
Q

Describe experiements looking into chromatin and histone modifications - for accessiblity of RSSs?

A

Chromatin
If we take nucleosomes and package a RSS into it -> can RAG1 and RAG2 cut?
This showed the nucleosome completely blocked the cutting by RAG1 and RAG2

Histone modifications
Acetylation of the TCR alpha locus showed an increase in histone acetylation, correlating when the locus undergoes V(D)J recombination

Methylation (H3K4me3) - this increases at the immunoglobulin kappa locus during V(D)J recombination
RAG2 is recruited by the interaction of its PHD finger with tri-methylated H3K4 - evidently showed these are undergoing V(D)J recombination

33
Q

Describe non-coding transcription?

A

Non-coding transcripts traverse the RSSs of all antigen receptor loci
AKA sterile transcripts
They pass through the loci at the time they are undergoing recombination
Transcribed by RNA polymerase II and are spliced
Correlate with onset of recombination
Essential for regulation of the factors affecting accessibility of RSSs

Evidence:
Insertion of a transcription terminator showed it prevented sterile transcription, which prevented recombination
Termination of sterile transcription also substantially reduced histone acetylation and H3K4me3
This shows these non-coding regions help add the positive marks on the chromatin

However…
Increased acetylation and H3K4me3 alone are not enough for recombination

34
Q

How does transcription help in accessibility of RSSs?

A

During transcription the H2A/2B dimer is evicted transiently for around 6 minutes
When the dimer is evicted only 105 bp are protected = freed 40 bases of DNA
Which allows RNA polymerase II to transcribe the DNA as the euchromatin is in an open conformation

If the RSS is within the 40 freed bases of DNA, whilst the RNA polymerase is passing through - it can be cut by RAG
With limited RSS availability this limits RAG function - therefore not too many ds breaks

35
Q

How are enhancers useful in accessibility of RSSs?

A

They are essential to regulate V(D)J recombination
They are bound by stage specific transcription factors
Only active in the correct cell lineage
Activated at the correct stage of development
Can contact the promoters of sterile transcription to activate transcription and recombination

This explains lineage and stage specificity of activation

36
Q

Describe allelic exclusion?

A

This restricts expression to one allele per cell
During recombination - we only recombine one allele at a time, as we only want to express a single type of antigen receptor per cell

Inactive allele associates with pericentric heterochromatin
Active allele goes to active region of nucleus - so when loci want to undergo recombination they move towards the center of the nucleus
This happens one allele at a time
If rearrangement is non-productive, then other allele can be activated

37
Q

How can V(D)J recombination cause errors in cancer?

A

Errors in V(D)J recombination seem to contribute greatly to these cancers - as we introduce DNA breaks
This is one of the worst things for a genome (normally) = increased risk in cancer

There are different proposals to why there are translations of oncogenes into the antigen receptor loci:
Cryptic recombination
End donation
Transposition

38
Q

Describe cryptic recombination?

A

Recombination with a cryptic RSS outside of the antigen receptor loci
Similar looking sequences to an RSS but are elsewhere in the genome
Meaning RAGs may recognise the cryptic RSS - allowing them to recombine bits of other chromosome into the antigen receptor loci

There are 3500 active cryptic RSSs in the genome
Examples: LMO2, SIL/SCL, TCR/IgH inversion
We have found cryptic RSSs at both sides of the indel (insertion/deletion of bases)
Showing errors in RAG cutting is contributing to altering the chromosomes in B/T-cells that have gone on to become cancerous

39
Q

Describe end donation?

A

Broken end on other chromosome aberrantly recombined into antigen receptor loci
Meaning another random ds break is joined onto the break in the antigen receptor loci - due to ionising radiation, stalled replication forks etc… (no RSSs)
If the non-specific ds break is next to an oncogene then this can be incorporated

Examples: BCL-2/IgH – most common translocation in human cancer, accounting for most follicular lymphomas and Mantle lymphoma: BCL1-IgH translocation
Both of these prevent apoptosis - so if they are overexpressed they prevent the wrong B-cells dying

40
Q

Describe transposition?

A

It involves the excised circular DNA - that retain the RSSs, where RAGs can rebind
RAGs can insert and join the ESC DNA anywhere - but 5 bp apart = staggered insertion
This generates a duplication at the insertion point
This happens very often in vitro but in cells it hardly ever happens (only shown once)
The RAG2 C-terminus has a region that inhibits this = not really a cause of cancer

41
Q

What is similar to transposition in vivo?

A

Re-integration
This is similar to transposition except the target site is an RSS
So the ESC DNA is often re-integrated into B/T-cells
This is a reverse of the recombination reaction and always happens at a RSS or cryptic RSS

Estimated to occur once every 10,000 to 100,000 recombination reactions
Hard to detect cytogenetically but proposed to be a major cause of cancer
Most are proposed to be silent or integrated into an essential region meaning the B/T-cell dies

42
Q

How can oncogenes be activated by re-integration?

A

Often the ESC carry promotors = upregulation of oncogenes or inactivation of tumour suppressor genes
The ESC stimulates RAG cutting at an RSS but the ESC itself remains uncut

Cut and run - cutting breaks in the genome at cryptic RSSs and then running on to other targets
This was proved true in around 20% of the breaks in acute lymphoblastic leukaemia

43
Q

How can class switch recombination cause cancer?

A

Translocations involving class switch recombination can lead to cancer
One of the most common is where Myc is translocated into IgH switch region
Translocation critically requires AID, so it was knocked out to show there was no translocation
This contributes to oncogenic translocation