Adaptive immunity - Generation of diversity Flashcards
1
Q
What is the immune repertoire
A
-
“Immune Repertoire”
- the amount of antibody types we intheory can make in our lifetime
- ~10^14 antibody receptors
- ~10^18 T cell receptors
2
Q
What are the 2 basis of antibody diversity
A
Structural and genetic basis of antibody diversity
3
Q
Describe the structural basis of antibody diversity
A
- Variations in the sequence and length of CDRs are the main determinants of antibody diversity.
- One CDR (hypervariable domain) CDR3 tends to be the most variable in length and sequence
- true for both H and L chains
- Heavy chain generally contributes more to antigen binding and is more variable than the light chain
4
Q
Describe the genetic basis of antibody diversity
A
- Antibody repertoire > 10^14 in humans.
- How are these proteins encoded by DNA? (only 20 - 25,000 genes in genome!)
-
Diversity:
- Arises by somatic recombination and mutation of a limited number of inherited gene segments, which make up the V regions.
-
V(D)J recombination is the mechanism of Somatic recombination
-Somatic recombination, somatic hypermutation, affinity maturation?? revise this card!
5
Q
How are immunoglobulins encoded in genes? what do genes consist of
A
- Heavy (H) chains - Chromosome 14
- Kappa chains - Chromosome 2
- Lambda chains - chromosome 22
- Each locus (Chromosome) has multiple VARIABLE region genes and one CONSTANT region genes.
- The variable regions (each) are encoded by 2 or more exons
-
Light chain V regions are encoded by 2 segments of DNA:
- Kappa chainsas example
- single constant region gene, multiple downstream genes making variable region
- most variable region encoded by Vkappa exon, some made by Jkappa exon
- in intact kappa protein we will find most of sequence is made of constant region, variable region made mostly up of Vkappa gene(exon) and a little bit of J kappa gene(exon)
-
Heavy chains are encoded by 3 segments of DNA
- 3 exons/segments of DNA
- V D and J exon. (D for diversity)
- most made up by Constant region exon, rest made up of variable, mainly V exon
6
Q
Describe the V D and J segments of the genome and how many copies of each
A
-
The V, J and D segments are present in multiple copies in the genome:
- Kappa light chain
- single 1 constant region gene, 5 J segments, and further downstream we have 38 V genes (exons)
- 1C + 38V + 5J
- Kappa heavy chain
- single 1 constant region exon coding for IgM, 6 J segments, 23 D segments and much further downstream we have 40 V genes
- 1C + 40 V + 23D + 6J
- So when the B cell differentiates how do we get these genes so that the V segments are close to the constant region genes? → next lecture slide
- Kappa light chain
7
Q
Whats the difference between somatic recombination and V(D)J recombination
A
V(D)J recombination is the mechanism of Somatic recombination
they are the same thing… used interchangeably
8
Q
Descsribe the process of somatic V(D)J recombination
A
- B cell differentiation triggered:
-
Rearrangement of light and heavy chain genes occurs during B cell during differentiation from Lymphoid SC into B cell (before antigen contact) → permanent changes in the DNA
- Using Kappa light chain as example:
- DNA is permanently rearranged: a V gene is spliced to a J gene, and intervening DNA is excised.
- V and J segments are spliced together and lie very close to the constant region
- Each V gene has its own promoter → an enhancer element allows the gene to be transcribed: This occurs because now the V gene (including its promoter) are now close to the enhancer element after splicing (gene rearrangement occurred)
- Intervening sequences are removed by RNA processing → mRNA produced for variable and constant regions → kappa light chain made
- Using Kappa light chain as example:
9
Q
Describe somatic V(D)J recombination of heavy chains
A
- Particular segment aligned with a particular D and J segment → somatic recombination → intervening DNA lost:
- First: D segment rearranges to lie next to J segment (D-J joining)
- a particular V gene somatic recombined with the D-J joined segment → V-D-J joining
- RNA processing removes unwanted RNA →
- IgM heavy chain made.
- Notice very large constant region, large V, and small D and J segments
- NB!
- CDR1 and CDR2 are encoded by the V segments (i.e. germline)
-
CDR3 corresponds to the VDJ (or VJ for light chain) join
- i.e. where VDJ /VJ join
10
Q
Explain the mechanisms of somatic V(D)J recombination
A
- Involves lymphocyte specific recombinases (several enzymes collectively called this)
- How do the recombinases know which bits of DNA to join?
- they recognise specific signals called recognition signal sequences (RSSs) that lie adjacent to the genes that need to be joined together
-
RSS = conserved heptamer (7 b.p.) + nonamer (9 b.p.) separated by 12 or 23 random nucleotides (spacers of any sequence)
- RSSs are found directly adjacent to the coding sequence of V, D or J gene segments.
- These guide rearrangement of the V, D and J segments.
-
12-23 base pair rule: a gene segment with a 12 bp spacer only joins with a gene segment with a 23 bp spacer.
- 12 and 23 BP corresponds to half a turn / full respectively turn of DNA helix
- Ensures correct V-D-J joining (instead of e.g. V-J-D joining
11
Q
Describe the V(D)J recombinase
A
-
Complex of several enzymes required for somatic V-region gene recombination:
- some are normal DNA cleavage/repair enzymes
- most important complex:
-
RAG1-RAG2 protein complex (encoded by Recombination Activation Genes*)
- specialised endonuclease (cut DNA) expressed only in developing lymphocytes
-
RAG1-RAG2 protein complex (encoded by Recombination Activation Genes*)
- Other important enzyme:
- Terminal deoxynucleotide transferase (TdT)
- Mutations in these genes* (RAG1 or RAG2 genes) result in severe combined immunodeficiency (scid) → they can’t produce antibody receptors or T cell receptors.
12
Q
Describe the action of RAG1-RAG2 driven recombination
A
- In image: want to join V1 with J
- During recombination, the RAG1-RAG2 complex recognises and aligns the RSSs adjacent to the gene segments to be joined
- so RAG1-RAG2 complex aligns the RSS sequences
- The RAG1-RAG2 complex has endonuclease activity and cleaves the DNA
- The cleaved DNA is repaired to form the coding joint (V and J segments now next to one another) and the signal joint (intervening DNA excised)
- Signal join repaired (joined) to be excised
13
Q
Describe the structure and function and importance of RAG1-RAG2
A
- RAG1-RAG2 complex consists of:
- dimer of RAG1 (recognises DNA and cleaves DNA) associated with dimer of RAG2 (acts as a cofactor)
- Function:
- RAG1 has 2 paired domains (NBD - nonamer binding domain) binds the nonamer of RSS sequence
- When not binding DNA → open configuration. When DNA bound → RAG1 closes (like a nutcracker)
- RAG1-RAG2 acts as a transposase (enzyme found in retrovirus that cleave bits of DNA and remove them. though that RAG1-RAG2 gene originated from retrovirus transposase
- Essential in the development of adaptive immunity
- Structure of RAG1-RAG2 complexed with DNA explains the 12-23bp rule.
- when molecule binds DNA → causes a tilt (bends) towards RSS sequence with 12 bp spacer → other side of molecule is forced to bind to 23-RSS bp.
- ensures proper recombination
- RAG1-RAG2 associated with HMGB (high mobility group box 1 protein) helps bend the DNA facilitating the recobination process
14
Q
Describe how junctional diversity occurs
A
- RSS recognised by RAG1-RAG2 complex → RAG1-RAG2 make single-strand break, which exposes hydroxyl group (OH) which is very reactive and reacts with the other strand to form a hairpin → clean double stranded break : signal joint created and removed.
- Reparation of hairpins carried out by normal DNA repair enzymes
- asymmetric or symmetric opening of hairpin
- asymmetric → palindromic overhand → reparation (nucleotides added or removed)
- symmetric → TdT can add DNA
- DNA is “patched up” imprecisely creating variability
- RAG-1-RAG-2 complex recognises and aligns the RSSs adjacent to the gene segments to be joined (germline DNA folded)
- Two ssDNA breaks are made close to the RSSs.
- Free 3’-OH attacks phosphodiester bond on other strand of DNA to create a hairpin at the segments to be joined and a flush ds break at RSS boundary.
- 4 – 7. Other proteins bind to repair the joints, but this process is imprecise, with nucleotides added or subtracted.
- DNA hairpins are cleaved at random, symmetrically (4) or asymmetrically (5). For V-D-J joining of the H chain, nucleotides can be added by terminal deoxynucleotide transferase (TdT) (6).
- Unpaired overhangs are filled in by DNA polymerase (7) or may be excised by an exonuclease
- DNA ligase joins the nicked and repaired hairpins to form the “coding joint”. (The blunt ends formed at Stage 3 are ligated to form the “signal joint” and this DNA is typically excised).
15
Q
What are the 4 ways in which B cell receptor and antibody diversity is generated
A
-
Multiple copies of each V region gene segments
- [Vn x Jn or Vn x Dn x Jn]
-
Heavy x light chain combination
- [Vkappa * Jkappa] + [Vlambda * Jlambda] * V_H * D_H * J
- Recombination is imprecise → junctional diversity
- nucleotides may lost or added variable addition of nucleotides at junctions contributes to diversity of CDR3
-
Terminal deoxynucleotide transferase (TdT) randomly adds up to 12 nucleotides to V-D-J joined (heavy chain)
- Junctional diversity increases overall diversity by a factor of $3*10^7$
- with junctional diversity + points 1 and 2 = 3x10^13 so where is the rest of variability coming from? → all these processes occur when B cells are developing in the bone marrow, completely independently of B cell interacting with antigen. When B cells encounter antigen →
-
Somatic mutation of V regions following antigen activation
- only occurs when B cell encounters antigen → point mutations → most changes occur in CDR loops
- point mutations
- base changes tend to be clustered in CDRs
- NB! only B cells undergo somatic hypermutation - T cells don’t
23
Q
Describe the mechanism of class switching
A
- Shown below at the top is the arrangement of all the heavy chain genes in the genome
- Mature B cell VDJ + IgM (Constant region mu) closest → IgM
- Adjacent to the heavy chain genes is a gene sequence know as a switch sequence.
- this allows recombination of the VDJ gene with another class constant region gene e.g. IgA heavy constant region.
- Looks similar to somatic recombination but enzymes and process is different
- DNA inbetween is lost (irreversible). This means that the B cell cannot switch back to e.g. IgM
- switch regions aren’t conserved but tend to be rich in G which AID likes to bind to.
- How is the enzyme acting → next
-
Class switching by DNA recombination between switch regions:
- intervening DNA lost
- irreversible
- NB. Different mechanism to V-D-J joining. Initiated by AID acting at switch regions (G –rich tandem repeated DNA sequences found close to C region genes).
