Antibody Diversity and Gene Rearrangement

Question 1

Describe the life cycle of a B cell

Answer 1

1. A stem cell (pluripotent) differentiates into a Pro-B cell.
2. This Pro-B cell will undergo the first DNA recombination, which is the VDJ recombination that codes in the heavy chain and variable region.
3. It is now a Pre-B cell which will undergo VJ recombination for the light chain’s variable and constant regions = Immature B cell
4. The B cell will continue to mature until it expresses both IgM and IgD through differential mRNA splicing until it becomes a mature recirculating B cell. They will patrol the blood and spleen, on the lookout for pathogens.
5. A B cell will be activated in the spleen or lymphoid tissues when it encounters a foreign antigen
6. The B cells will migrate to the germinal centre (GC) and will undergo a special selection to hone its variable region to that particular pathogen.
7. They will then undergo affinity maturation and class switching
8. They then differentiates either into plasma cells (which secrete antibodies) and memory B cells.

Question 2

How do B cells develop in the bone marrow and how are they selected?

Answer 2

Progenitor B cells line the bone endosteum – internal bit of bone marrow next to bony lamella

They are then transported to stromal reticular cells, which have a mixed phenotype (fibroblast; endothelial; myofibroblasts) - this is to maintain cell viability and they produce IL-7 (a haematopoietic growth factor) to sustain B cell differentiation

Each progenitor can produce up to 64 progeny which migrate to the centre of spongy bone and then are selected by adventitial reticular cells.

75% of B cells don’t make it out of the bone as they are selected and undergo apoptosis. These include auto reactive B cells, which produce seld-antibodies are removed by negative selection. The dead B cells are then phagocytosed by macrophages. The B cells that survive and are successful have rearranged their Ig gene produce surface IgG

These successful cells then move from the reticular cells through the central sinus and then to the venous sinusoid and then out to the circulation

Question 3

What is the difference between somatic recombination and differential splicing (and examples of each)?

Answer 3

SOMATIC RECOMBINATION 
This is any change at the DNA level. This is called somatic change. Examples include:
- V(D)J recombination 
- Tdt nucleotide addition 
- Somatic hypermutation 
- Class switching 

DIFFERENTIAL SPLICING
This is any change at the mRNA level (at protein level) and examples include
- IgM and IgD formation
- Membrane bound and secreted Ig

If there is a change at the mRNA level, the DNA is not affected. However, if the change is at the DNA level, then both the mRNA and the protein will have changes.

Question 4

What is the antigen independent stage and how does it lead to diversity of B cells

Answer 4

The Antigen-independent stage is where the antibody diversity comes from. It occurs inside the bone marrow and before the B cell encounters a foreign antigen. The body doesn’t produce specific antibodies for specific antigens so they have to make resting B cells which contain ‘random’ BCRs.

1. The pro-B cell in the BM firstly undergoes V->DJ recombination that codes in the heavy chain and variable region.
2. The D first recombines with the J region, then the V recombines with the recombined DJ region.
3. the 3rd recombination V->J codes the light chain, both variable and constant regions
4. It is now an immature B cell with a functional IgM attached, the default before encountering pathogens
5. We also get junctional diversity via P and N nucleotide addition which adds another level of diversity.
6. The naive B cell then expresses both IgM and IgD so is then capable of differential splicing at the mRNA level
7. After that, it becomes a mature recirculating B cell. Those will patrol the blood and spleen, on the lookout for pathogens.

Question 5

Where are the three genetic loci encoding Ig?

Answer 5

Two are for the light chains: kappa (κ) and lambda (λ) locus, and one for the heavy chain.

They are located on different chromosomes.

• λ light chain: chromosome 22
• κ light chain: chromosome 2
• heavy chain: chromosome 14

Question 6

Describe VDJ recombination.

Answer 6

VDJ combination of gamma heavy chain on chromosome 14.

It has 80 variable gene segments (51 are functional), 23 diversity segments, 6 joining segments and a constant region. The VDJ- region constitutes the highly diverse 3rd hypervariable region (CDR3) of the Ig molecule

1. First, the DJ recombination occurs.
2. Then, a random V segment is selected to join with the DJ recombined segment.
3. This will be transcribed into mRNA. Only the first two constant regions, Cμ and Cδ will be transcribed.
4. Due to the two constant regions, differential splicing allows the B cell to make two different classes of B cell receptor/ antibody, which is why is can express both IgM and IgD.

Question 7

Describe VJ (light chain) recombination in the kappa chain

Answer 7

The light chain can be made from kappa of lamda genes. Kappa is preferential so is the first set of segments to try successful recombination

The kappa gene has 30-35 functional V segments and 5 J gene sequences and a constant region. In front of each V segment, there is a postcode sequence specific to the V segment.

1. One of the V segments will be selected at random, and the same with the J segments, and they will be recombined together.
2. That recombination, along with an additional J segment and the constant region are transcribed into mRNA.
3. The postcode sequence is spliced out, as well as the bit between the J and C segment and the extra J sequence
4. A poly-A tail is added and a stop codon, ending up with mature mRNA

It will be translated into protein, and once that chain is correctly folded, the leader sequence will be cleaved off as it is no longer needed. This end product will be the light chain of your receptor or antibody.

Question 8

Describe VJ recombination in the lamda chain

Answer 8

The lamda locus on Chromosome 22 and the recombination works very similarly to the kappa chain, which if unsuccessful will turn to the lamda recombination. There can only be 1 type of light chain

Lamda has 29-33 functional V gene segments and 7-11 J segments each linked to a C region (the number of J->C sequences is dependent on the haplotype).

Extra diversity is generated by imprecise joining.

1. one V and one J combine together and a constant region is added.
2. Intron between them is VJ and C is cleaved and the J region is removed, leaving a mature mRNA sequence

Question 9

Describe how ‘turns’ help in recombination.

Answer 9

VDJ recombination requires Recombination Signal Sequences (RSS) – these are conserved sequences upstream or downstream of gene segments (also known as ‘turns’).

‘Turns’ consist of heptamer and nonamer with a 12 or 23 basepair spacer.

Heptamer(7) = Sequence of 7 nucleotides e.g. CACAGTG (7) - downstream (3’) of VH, VL and DH

Followed by a 12 or 23 spacer made of non-conserved bases which correspond to one or two turns (loops) of the double helix - 12/23 rule

• Heavy chain (V = 23 down) (J = 23 up) (D = 12 up & down)
• Light chain (K) - (V = 12 down) (J = 23 up)
• Light chain (lamda) - (V = 23 up) (J = 12 down)

Nonomer (9) = sequence of 9 nucleotides e.g. ACAAAAACC -> Upstream (5’) of JH, DH and JL

Question 10

Describe the mechanism of recombination at a molecular level.

Answer 10

• Two selected coding segments and their RSSs are brought together in a chromosomal loop
• RAG1 and RAG2 will climb onto the RSS at both ends, forming a major hairpin (the structure just folds).
• It will then create nicks in the DNA, so there are now minor hairpins.
• Enzyme called Artemis opens the hairpins and there is addition/removal of P and N bases by exonucleases and TdT before DNA ligase joins these ends together, adding extra diversity - Causes frameshift which will also contribute to the diversity of the antibody produced.
• WE now have the coding join with imprecise joining and the signal joint (including the turns) with precise joining

Question 11

How do the enzymes cleaving cause junctional diversity?

Answer 11

The site at which Artemis cleaves the hairpins is random. When they unfurl, they may not be equal in length of nucleotides, resulting in overhangs.

The DNA has repair enzymes to add (P) nucleotides to fix these overhangs.

Almost exclusively in heavy chains, we have an additional enzyme called TdT (Terminal deoxynucleotidyl Transferase) which also adds (N) nucleotides in between the two segments, adding even more diversity.

Question 12

What is an idiotype and give example of a disease caused by an idiotype

Answer 12

In immunology, an idiotype is a shared characteristic between a group of Igs or TCR molecules based upon the antigen binding specificity and therefore structure of their variable region.
These idiotypes can become targets for the immune system so all antibodies binding the same epitope may share an idiotype > a novel structure that may be recognised as non-self by the rest of the immune system.

Example = Systemic lupus erythematosus (SLE)
- Anti-Id (idiotype 16/6) causing apoptosis and production of nuclear antigens (Ab against DS DNA in SLE)

• Some people will have genes which are not good at clearing these antigens away and gene which recongise these anitgens as foreign
• inflammatory response and Type 3 hypersensitivity reaction
• Presentation – fever, rash, joint pain in women of child-bearing age