Humoral Immunity: Generation of Antibody Diversity Flashcards

1
Q

What do antibodies do?

A

Antibodies are Y-shaped molecules expressed by immune cells (B cells).

They can work by preventing bacteria from entering cells, by binding to their docking sites, or they can neutralise them via toxins, etc.

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2
Q

Revise the structure of an antibody.

A

An antibody is made up of 2 heavy and 2 light chains. The types of heavy chains are μ, δ, γ, α or ε chains; these can be divided into further classes. The types of light chains are κ or λ chains.

The variable region of the antibody binds to the antigen, CH1 support.

The constant region executes effector functions (such as activating complement, binding phagocytes, etc.).

Disulphide bonds stabilize the structure.

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3
Q

Describe the two forms of antibodies.

A

Antibodies have two forms: the final form will be the secreted form. During its development, it is anchored to the plasma membranes of B cells. At that stage, it is called a B cell receptor.

Throughout its development, it will incrementally secrete antibodies until the last stage of development, where it becomes a plasma cell and exclusively secretes antibodies.

The antibodies are secreted as monomers, but when they come together they can form polymeric structures.

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4
Q

Describe the life cycle of B-cells, from stem cells to plasma and memory cells.

A

We start with a stem cell that differentiates into a Pro-B cell. This Pro-B cell will undergo the first DNA recombination, which is the VDJ recombination that codes in the heavy chain and variable region.
It will then undergo another recombination, this time VJ which codes for the light chain and both variable and constants regions. Now, it is an immature B cell.

The B cell will continue to mature until it expresses both IgM and IgD through differential mRNA splicing. Then, it becomes a mature recirculating B cell.
They will patrol the blood and spleen, on the lookout for pathogens.

When the body encounters a pathogen, those circulating B cells will be activated. They will migrate to the germinal centre (GC) and will undergo a special force of selection to hone its variable region to that particular pathogen. It will undergo affinity maturation and class switching, and will then differentiate either into plasma cells (which secrete antibodies) and memory B cells.

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5
Q

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

A

SOMATIC RECOMBINATION (at DNA level):

  • V(D)J recombination
  • Tdt nucleotide addition
  • Somatic hypermutation
  • Class switching

DIFFERENTIAL SPLICING (at protein level):

  • IgM and IgD made
  • Membrane bound and secreted Ig

DNA is the code for everything you need in your body, and the mRNA is a snippet of the code which will be translated to proteins.

Any change at the DNA level is called somatic change.
Any change at the mRNA level is called differential splicing.

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.

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6
Q

Describe the chain structure of antibodies.

A

Each chain is made up of one single polypeptide.

There is a hinge region between the CH1 and CH2, a stretch of polypeptides that makes the antibody flex.

There’s a sugar group decorating one of the amino acids in the CH2 region; this forms part of the active site.

The glycosylation pocket is where receptors of other immune cells are attracted to.

The variable fragment (Fv) is the variable region of the light and heavy chains working together. The antibody fragment (Fab) is the Fv and the first constant region. The factor regions (Fc) will be the heavy CH2 and CH3 domains.

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7
Q

What are CDRs?

A

CDRs are Complementarity Determining Regions.

The CDRs are three finger-like protrusions in the variable region, and they are what interact with antigens.

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8
Q

Go into detail on the antigen independent stage of the life cycle of B cells.

A

We start with a stem cell that differentiates into a Pro-B cell. This Pro-B cell will undergo the first DNA recombination, which is the VDJ recombination that codes in the heavy chain and variable region. The D first recombines with the J region, then the V recombines with the recombined DJ region.
This variable region will be co-expressed with a new constant region. It is the constant for the IgM antibody+, and is a default for all B cells before they encounter pathogens.

It will then undergo another recombination, this time VJ which codes for the light chain and both variable and constants regions. Now, it is an immature B cell.

The Pre-B that follows will already express a new, functional, valid heavy chain. It will also express a placeholder light chain, just to hold it together as at this stage the light chain is not generated yet.
It will now undergo a third recombination of VJ in the light chain,

During the VDJ recombination, there are additional mechanisms that come into play to contribute more to the diversity of the antibody: junctional flexibility and P & N nucleotide addition. And then, when we get an immature B cells that expresses both IgM and IgD, the cell is capable of alternate splicing at the mRNA level.

After that, it becomes a mature recirculating B cell. Those will patrol the blood and spleen, on the lookout for pathogens.

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9
Q

Where are the three genetic loci encoding Ig?

A

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
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10
Q

How many antibody genes are inherited?

A

NONE! No complete genes are inherited, only gene segments.

Arranging these gene segments in different combinations generate many Ig sequences.

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11
Q

Describe VJ recombination.

A

In front of each V segment, there is an leader sequence, which is like a postcode, with a sequence specific to the V segment it is in front.

One of the V segments will be selected at random, and the same with the J segments, and they will be recombined together. That recombination, along with an additional J segment and the constant region are transcribed into mRNA. Any extra bits will be spliced out, such as the space between the J and the C segment, including the extra J segment. Thus, we end up with mature mRNA (with a stop codon and a PolyA tail).

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.

Lambda light chain production is similar, but has a few more steps.

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12
Q

Describe VDJ recombination.

A

First, the DJ recombination occurs. Then, a random V segment is selected to join with the DJ recombined segment.
This will be transcribed into mRNA. Only the first two constant regions, Cμ and Cδ will be transcribed.

Due to the two constant regions, alternative 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.

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13
Q

Describe how ‘turns’ help in recombination.

A

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.

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14
Q

Describe how the one-turn/ two-turn rule aids VDJ recombination.

A

Recombination only occurs between a segment with a 12 basepair spacer and a 23 basepair spacer.

A two turn will be located at the end of every V segment, as well as upstream of every J segment.
One turns will be located upstream and downstream of the D segments.

The 12/23 rules explains the order in which recombination occurs in heavy chains: first we get recombination between the D and J segments, as they have a one turn and two turn, and then we get recombination between the D and V segments.

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15
Q

Describe the mechanism of recombination at a molecular level.

A

First, 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.

Then, all the enzymes will come into play, which will result in V and J joining, creating a coding joint, and the creation of a signal joint, so the extra bit gets cut out.

First, Artemis will come and nick open the hairpins. Exonucleases and TdT will then preform end processing (so they will mess about with the free ends, adding and deleting base pairs) before joining these ends together.

So, we now have V and J combined, and we have N and P nucleotides in between because of end processing. These additions of nucleotides cause a frameshift, which will also contribute to the diversity of the antibody produced.

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16
Q

How do the enzymes cleaving cause junctional diversity?

A

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.

17
Q

Describe junctional flexibility.

A

It’s precise mechanisms are unknown, but it involves Exonuclease, which removes mismatched nucleotides.

Previously, with junctional diversity, we were looking at addition of base pairs. Here, with junctional flexibility, we are looking at deletions.

The coding joints may be missing base pairs from the original segments, which can contribute to frame shift. The signal joints are always precise.

18
Q

What is the advantage and disadvantage of junctional diversity?

A

GOOD: It generates a huge amount of antibody diversity - about 100-fold.

BAD: It produces non-productive rearrangements (due to the incorrect reading frame) – so it can be a wasteful process.

19
Q

Describe allelic exclusion.

A

We have two copies of each Ig gene – one from the mother and one from the father.
In most cases (such as in the Islets of Langerhan), both genes are expressed.

Antibody genes are different – only one heavy chain allele and one light chain allele is expressed in an antibody.

Their order of rearrangement is: Heavy > kappa > lambda; 1st allele (for all) then 2nd.

This ensures that each B cell makes one type of antibody.