Antibody genetics I - B cell development and generation of diversity

Question 1

Where do B cells develop in birds and in mammals?

Answer 1

Birds = Bursa of Fabricius
Mammals = fetal liver and postnatal bone marrow

Question 2

Describe B cell development in fetal liver (humans)

Answer 2

• B cells develop directly from lymphoid stem cells in the hematopoietic tissue of the fetal liver
• occurs at 8–9 weeks of gestation
• Later, the site of B cell production moves from the liver to the bone marrow, where it continues through adult life.
• This fetal liver-bone marrow migration of stem
  cells is also true for cells of other hematopoietic lineages such as erythrocytes, granulocytes, monocytes, and platelets.

Question 3

Describe the process B cell development in the bone marrow

Answer 3

B cell progenitors line the bone endosteum.

In the bone marrow, B cells mature in close association with stromal reticular cells termed as adventitial reticular cells based on their location.
-Where the B cells differentiate, the reticular cells have mixed phenotypic features with some similarities to fibroblasts, endothelial cells, and myofibroblasts.
-The reticular cells produce type IV collagen, laminin and the smooth muscle form of actin.
-Experiments in vitro have shown that reticular cells sustain B cell differentiation, possibly by producing the cytokine IL-7.
- Adventitial reticular cells may be important for the release of mature B cells into the central sinus.

Question 4

Describe the B cell selection process

Answer 4

• Most B cells maturing in the bone marrow do not reach the circulation, but undergo apoptosis and are phagocytosed by bone marrow macrophages.
  -B cell–stromal cell interactions enhance the survival of developing B cells and mediate a form of selection that rescues a minority of B cells with productive rearrangements of their immunoglobulin genes from apoptosis
  -Many self-reactive/autoreactive B cells are also eliminated through negative selection in the bone marrow.

Question 5

when do immunoglobulin gene rearrangements and phenotypic changes take place?

Answer 5

During B cell ontogeny

Question 6

What is Erlich’s side chain hypothesis?

Answer 6

He proposed antigen-induced selection. His model is close to our present view of clonal selection, except that he placed receptors of several different specificities on the same cell and did not address the question of how the diverse receptors were generated.

Question 7

What is Dreyer and Bennet hypothesis?

Answer 7

This led Dreyer and Bennett to propose the ‘two genes one polypeptide chain’ hypothesis, i.e. the light chain sequence is encoded by at least two different genes, contrary to a central dogma of molecular biology at the time (‘one gene, one protein’)

Question 8

How did Tonegawa prove the Dryer and Bennet hypothesis?

Answer 8

by demonstrating that the DNA encoding the variable and constant region were located at distant sites, on the same chromosome.

Question 9

What are the mechanisms of diversity

Answer 9

Somatic recombination: this allows the joining of one segment of the gene to another.

Somatic mutation: this allows for ‘sloppy’ joining of those segments.

combinatorial pairing: the pairing of a unique
VL domain with a unique VH domain generates a unique antigen-binding site

Question 10

Describe the order of diversity generation

Answer 10

Heavy-chain gene rearrangements occur in B cell progenitors and represent the earliest indication of B lineage commitment.

This is followed by light-chain gene rearrangements, which occur at later pre-B cell stages.If successful get k light chain. If no k then lambda chain rearranges

Once a B cell has synthesized light chains, it becomes committed to the antigen-binding specificity of its surface IgM (sIgM) antigen receptor.

Question 11

Describe the heavy chain VDJ

Answer 11

*On human chromosome 14
* These are the componants of the Ig diversity region
* Compose 3 genetic ‘libraries’
– V region 80 gene segments, approx 50 are functional
– D region 23 DH segments
– J region 6 segments
* Recombination of any one V with any one D and any one J forms the functional heavy chain VDJ region
* Each V region codes for a signal peptide which directs the polypeptide to the RER, Golgi and then out of the cell.

Question 12

Describe Heavy chain recombination

Answer 12

The first event is recombination between a JH gene segment and DH segments. DH segments:
* are highly variable, both in number of codons and sequence
* may be read in three possible reading frames without generating stop codons;
* Can be used singly or in combination!

*Productive recombination between DH and JH gene segments signals recombination of this DJ sequence to a VH gene segment, forming a contiguous DNA sequence encoding the entire VH protein sequence
* The recombined VH, DH, and JH gene segments generate widely diverse hypervariable HV3 (CDR3) sequences, which contribute greatly to the diversity
* VDJ recombines to Cμ if naïve B-cell or one of the other 8 C regions after antigen experience

Question 13

Describe light chain rearrangement (k light chain)

Answer 13

• Human k light chain locus is on Chromosome 2
• has 31–35 functional Vk gene segments that encode the N terminal 95 residues of the Vk region
• No Dk segments
  *The C terminal residues of the Vk region are encoded within five Jk gene segments
  *A leader or signal sequence ( responsible for targeting the chain to the endoplasmic reticulum) precedes each Vk segment. It is cleaved in the endoplasmic reticulum, and the antibody molecule is then processed through the intracellular secretory pathway.
• If a non-productive VJ rearrangement is created then the other k allele is used.

Question 14

What happens if k light chain rearrangement is non-productive?

Describe this

Answer 14

If this is also non-productive then the cell
moves to the……..

• λ locus on Chromosome 22
• Has 29-33 functional Vλ gene segments that encode the N terminal 95 residues
• Has 7-11 Jλ segments each linked to a Cλ region (the number of J l C l sequences is dependent on the haplotype).
• Extra diversity is generated by imprecise joining.

Question 15

Recombination involves recognition of signal sequences by the V(D)J recombinase.

Describe the recombination Signal sequence (RSS).

Answer 15

• Each V, D, and J segment is flanked by recombination signal sequences (RSS)
• CACAGTG heptamer is downstream (3’) of VH, VL and DH (or analogue)
• Followed by a spacer of 12 or 23 non-conserved bases
• Then a ACAAAAACC nonamer (or analogue)
• Upstream (5’) of JH, DH and JL is a corresponding nonamer, spacer (12 or 23) and heptamer sequence
• The 12 and 23 spacers correspond to one or two turns of the double helix.
• Only 12 will combine with 23, not with another 12 (12/23 rule)

Question 16

What provides N-region junctional diversity?

Answer 16

the enzyme terminal deoxynucleotidyl transferase (TdT), may add random nucleotides to the exposed cut ends of the DNA.
Nucleotides may therefore be inserted between DH and JH, and between VH and DH, without need of a template

Question 17

Mechanism of recombination (Lymphoid specific)

Answer 17

• Firstly the portions of the gene are made available to the recombination machinery. Two selected coding segments and their RSSs are brought together. Chromosomal loop.
• Double stranded breaks are generated at the RSS-sequence junctions by RAG1/RAG2 complex (VDJ recombinase). This creates a hairpin end on the coding region.
• Opening of the hairpins (Artemis) and addition or removal of bases (TdT) to add extra diversity.
• The coding ends are then rejoined by a number of factors (inc DNA ligase.

Question 18

How is recombination mediated?

Answer 18

mediated by the protein products of the two recombination-activating genes (RAG-1 and RAG-2):
* a RAG-1–RAG-2 complex recognizes the RSS, bringing a 12-RSS and a 23-RSS together into a synaptic complex
* the RAG proteins initiate cleavage by introducing a nick in the area bordering the 50 end of the signal heptamer and the coding region;
* the RAG proteins then convert this nick into a double strand break, generating a hairpin at the coding end and a blunt cut at the signal sequence, resulting in a blunt signal end;
* the hairpinned coding end must be opened (Artemis) before the joining step, and usually undergoes further processing (the addition or deletion of nucleotides), resulting in an imprecise junction.
*The loss or addition of bases (TdT) during coding joint formation (‘junctional diversity’) creates additional diversity that is not
encoded by the V, D, or J segments, and also contributes to the generation of nonproductive joints, since some of the modification create stop codons.
*Signal ends, in contrast, are usually joined precisely to form circular signal joints that have no known immunological function and are lost from the cell

Question 19

Describe Juctional diversity

Answer 19

• When RAG1/RAG2 breaks the DNA strands and Artemis
  breaks open the hairpin loop, it does so asymmetrically.
• One DNA loop is longer than the other
• The shorter strand is extended with complimentary nucleotides
  to the longer strand before ligation. Called P nucleotides
• Then TdT adds a few more, called N nucleotides
• Adds non-germline sequences to the mix
• TdT mediated N-region diversity is more common in heavy
  chains than light.
• These junctional sequences form the CDR3 (3rd hypervariable
  region)

– Addition and removal of bases at the V,D,J junctions
– Largest contribution to diversity

Question 20

What are idotypes?

Answer 20

• An idiotype is that region which is specific for an antibody.
• This could be it’s binding site
• So all antibodies binding the same epitope may share an idiotype
• This would be a novel structure and may be recognised as non-self by the rest of the immune system.
• May represent a control system. Id anti-Id

Question 21

What if the mechanisms of genetic diversity go wrong?

Answer 21

• SCID- Severe Combined Immunodeficiency
  Syndrome- ‘Bubble boy’
• Approx 1/50,000. Mostly male.
• Most forms affect T-cells but some affect T
  and B cells
• If mutations in RAG1 and/or RAG2