Humoral Immunity; Antibodies and the life cycle of B cells

Question 1

Define antibody

Answer 1

Antibodies – Class of proteins called Immunoglobulins that Directly Bind to Specific Antigen. They are produced by Plasma Cells (Activated B-Cells) in response to infection/immunisation.

Question 2

Describe the structure of an antibody

Answer 2

On document

Question 3

What is the nomenclature of antibodies?

Answer 3

• Fab = Antigen Binding Site
• FC = Constant Region
• VH = Variable Region of a Heavy Chain
• VL = Variable Region of a Light Chain
ƒ CH = Constant Region of a Heavy Chain
• CL = Constant Region of a Light Chain
• CDR = Complementarity Determining Region

Question 4

What is the CDR?

Answer 4

• These are 3 finger like protrusions in the variable region and they are the ones interacting with the antigen
• They are located in the variable light and heavy regions

Question 5

What are the 4 functions of antibodies?

Answer 5

They bind specifically to their corresponding antigens, leading to:

1. Neutralisation of Pathogens/Toxins -> Phagocytosed.
2. Opsinisation of Pathogen -> Marks them for destruction by Phagocytes.
3. Activation of Complement -> Lysis of Extracellular Bacteria -> Phagocytosed.
4. Ab-Dependent Cellular Cytoxicity (ADCC) -> Lysis of a Target Cell that has been bound by Specific Antibodies.
   a. NK-Cells -> Lysis of a Pathogen-Infected Cell.
   b. Eosinophils (Via IgE) -> Kills Parasites that are too big for phagocytosis

Question 6

What heavy chain does IgM/D/G/A/E have?

What is the function of them

Answer 6

On table

Question 7

What is heavy chain class switching?

Answer 7

• This only affects the heavy chain constant region

* It produces different effector functions to deal with different pathogens

Question 8

What are the two types of heavy chain class switching?

Answer 8

• Minor (does not affect DNA of the B-cell): this occurs between IgM and IgD at the Mrna level
• Major: DNA recombination: IgM ->G/A/E and IgG ->A/E

Question 9

How does the B-cell know what class to switch to?

Answer 9

They do this by sensing chemicals around released from the T-cells

Question 10

What does Major class switching (class switching recombination) require?

Answer 10

• Cytokine signals
• Switch regions
• AID and DSB repair proteins

Question 11

Compare and contrast membrane vs secreted Ig

Answer 11

Same constant and variable chains but secreted ig has tail piece, whilst membrane IgG has a transmembrane cytoplasmic tail.

Question 12

How do we code for either a secreted or membrane Ig?

Answer 12

1. The DNA has genes (for example for IgM), which are u1,2,3,4,5 and then a polyA tail and a stop codon.
2. Then M1 and M2- transmembrane region and cytoplasmic tail.
3. Then a stop codon and a polyA tail.
4. For secretory Ig- this will be transcribed into mRNA (up to first stop codon) and splicing too.
5. For membrane bound Ig, the whole region up to the second polyA tail will be transcribed into RNA.
6. 8 regions, including genes encoding tail piece and stop signal, will be spliced out.

Question 13

What is somatic recombination and give some examples?

Answer 13

• Somatic occurs at the DNA level

Examples:
•	V(D)J recombination
•	Tdt nucleotide addition
•	Somatic hypermutation
•	Class switching
•	Any changed at the DNA level during somatic recombination, it will affect mRNA and the protein

Question 14

What is differential splicing and give some examples?

Answer 14

• Differential splicing occurs at the Mrna level

Examples:
• Is the formation of IgM and IgD
• Membrane bound and secreted Ig

Question 15

Describe the life cycle of the B-cell (independent)

Answer 15

1. Stem cell in bone marrow differentiates to pro-B cell.
2. DNA in pro-b undergoes D-J and V-DJ recombination to permanently code in heavy chain variable region.
3. Variable region is expressed with mu, default first heavy chain expressed by b cell. Called Pre-B cell- only heavy chain but placeholder light chain (psuedoantibody).
4. V-J recombination to code in (k or l) light chain and constant to become immature B cell. This will express IgM and will mature over time. During the V(D)J recombination, can also give additional diversity through junctional flexibility and P/N nucleotide addition. Though random, can produce millions of unique B cells.
5. Can also express IgD through differential splicing. Once both expressed- it is mature and can circulate in blood stream, spleen and lymph nodes. Naïve b cell.

Question 16

Describe the life cycle of the B-cell (dependant)

Answer 16

1. Activation- B cell will complete ability maturation in germinal centre (GC). Only most suited to pathogen will survive.
2. B cell will receive information about the type of pathogen and undergo class switching to the appropriate effector regions. BCR becomes IgG or IgA.
3. Majority of B cell will further develop onto plasma cells, secreting more Ig.
4. Some B cells coding for IgM will differentiate into plasma cells secreting IgM as first defence.
5. After infection, some B cells remain as memory B cells.

Question 17

What is VDJ and VJ recombination for variable fragment diversity?

How is this done in the light and heavy chains?

What does the J gene code for?

Answer 17

Create diversity in the heavy chain and light chain variable regions.
• Complete Antibody genes are not inherited, only gene segments are.
• Arranging these gene segments in different combinations generate many Ig sequences

Light chain:
• Consists of VJC segments
• The gene segments under recombination (one V segment is selected along with one J segment)
• The DNA will now encode a complete antibody with the V/J AND VDJ segments encoding for the light and heavy chain variables
• THE C SEGMENTS CODE FOR THE CONSTANT DOMAINS

Heavy chain:
• Consists of VDJC segments
• The gene segments under recombination (one V segment is selected along with one J segment and d segment)
• The DNA will now encode a complete antibody with the V/J AND VDJ segments encoding for the light and heavy chain variables
• THE C SEGMENTS CODE FOR THE CONSTANT DOMAINS

THE J GENE CODES FOR CDR3 – INTERACTS WITH ANTIGEN

Question 18

Describe VJ recombination of kappa light chain chromosome 2

Describe VDJ recombination of gamma heavy chain genes chromosome 14

Answer 18

Gene consists of:
• 40 V segments (30 in lambda).
• 5 J segments (4 in lambda).
• C segment.

V and J segments farther from each other than J and C.

1. In front of each V segment is a leader sequence that tell cell where protein will end up.
2. V and J segments are randomly chosen to form a leader and an extra j segment and C are added.
3. Transcribed and spliced (extra J removed, as well as other parts) to form mature RNA, with only leader, V, J and C with poly-A tail.
4. Translation into amino acid chain and then folded into protein.
5. Leader cleaved off once protein reached its destination.
6. Lambda chain recombination is similar but more complex.

Genes consist of:
•	51 V segment.
•	27 diversity (D) segments.
•	6 J segments.
•	Constant region.

1. First recombination is D to J joining.
2. V segment is also be joined.
3. Transcription and differential splicing.
4. Begin with leading sequence, then VDJ and Cu.
5. Eventually translated into IgM heavy chain.
6. Or begin with leading sequence, then VDJ and Cd.
7. Eventually translated into IgD heavy chain, signalling maturity

Question 19

Describe the mechanism of VDJ recombination

Answer 19

• Recombination signal sequences (RSS) are conserved sequences up/downstream of gene segments.
• RSS made up of turns consisting of a heptamer (conserved 7 base pair DNA sequence- always the same) and then a 12 or 23 bp spacer and then a nonamer (conserved 9 base pair DNA sequence).
• 2 types of turns: two turn (23 spacer) and one turn (12 spacer).
• DNA recombination only occurs between segment with a 12bp spacer or a 23bp spacer. One-turn/two-turn or 12/23 rule.
• 2 turns are downstream of the V segment and upstream of J segment in both the heavy and light chain.
• The 1 turns are on both sides of the heavy chain D segments, upstream of the lambda light chain J segment and downstream of the kappa light chain V segment.
• Heavy chain- 12/23 or one-turn/two-turn join between D and J and then D and V, preventing other types of recombinations. Same principle for the light chains.
• RAG1 and RAG2 enzymes bind to the turns and pulls them together to form a major hairpin.
• DNA is then cleaved at both RSS turns and retains hairpin shape at ends of gene segment. This is the minor hair pin.
• Many enzymes repair and process joints- forming coding joint with the V and J next to each other in light chain or D and J next to each other in heavy chain.
• Signal joint produced with turns and other DNA between the gene segments.

Question 20

What is Combinatorial diversity?

What other mechanisms produce diversity?

Answer 20

• How many possible combinations from heavy chain V,D,J and light chain V,J?
• Heavy chain= V(51), D (27) and J (6)= 51 x 27 x 6= 8262.
• Light chain kappa= V (40) and J (5)= 40 x 5= 200.
• Together= 8262 x 200 x 120= 198,288,000= 1.98 x 108 possible combinations.
• Need 1 billion but this is less than 200 million so therefore there are additional mechanisms to generate diversity.
• Multiple germline V, D and J gene segments
• Combination V-J and V-D-J joining
• Junctional flexibility
• P-nucleotide addition
• N-nucleotide addition
• Combinatorial association of heavy and light chains  Somatic hypermutation during affinity maturation (next lecture)

Question 21

Why is junctional diversity good and bad?

Describe the mechanism for it?

Answer 21

• Through junctional flexibility during VDJ recombination, P and N nucleotide additions.
• Good= antibody diversity.
• Bad= non-productive rearrangements (incorrect reading frame due to loss of stop codon)- which makes this a wasteful process.

Mechanism:
• Minor hairpin- opened by enzyme Artemis. Artemis cleaves one strand of the double stranded DNA randomly. Nicked ends will linearize forming overhanging ends.
• Repair enzyme will fill in gaps leading to P-nucleotide addition.
• Terminal deoxynucleotidyl transferase (tDt) then adds N-nucleotides between the two ends before the ends are ligated together again.
• N-nucleotide addition occurs almost exclusively in heavy chain.
• Addition of P and N-nucleotides before joining the segments together causes addition of amino acids which may shift reading frame- this produces even more diversity.

Question 22

What is junctional flexibility?

What enzyme does it require?

Answer 22

• Removal of nucleotides between gene segments during V(D)J recombination.
• Involves exonucleases.
• Once Artemis cleaves strand, there are overhanging ends.
• If there is enough complementarity, these ends can overlap (between the two DNA segments).
• There will be mismatches that will be removed by exonucleases before repair enzyme can work.
• Exonuclease can sometimes over-trim the end, removing a number of nucleotides of varying amounts. Even if same V and J segments are chosen for a kappa light chain, this process will still create diversity as different lengths will be removed, leading to changes in amino acids, in frameshift so brand-new unique proteins produced.
• Joining of signal joints (RSS/RSS) is always precise.

Question 23

What is Allelic exclusion?

Answer 23

• Two copies of Ig gene (one from each parent).
• Only one heavy chain allele and one light chain kappa and light chain lambda are expressed. ½ of the possible ones.
• Order of loci rearrangement- heavy chain first allele (if successful move on), then kappa, then lambda. If 1st chain does not produce a functional chain, the 2nd allele is used.
• These ensures that each B cell only makes one type of antibody.

Question 24

Describe what happens when the body encounters a pathogen

Answer 24

1. When the body encounters a pathogen a subset of naïve B-cells will become activated.
2. The B-cells will then make clones of itself through clonal expansion
3. Some of these clones will secrete IgM to initially fight the pathogen
4. The rest of the B-cells will migrate to the lymph nodes to wait for T-cell activation
5. The b-cells will engulf the pathogen and present the antigen to the T helper via the MHC 2 receptor
6. Dendritic cells will also engulf and present to the T-helper cells
7. CD40/L allows the B-cell to recognise the T-cell, cytokines are also produced by the T-helper cell.
8. The B-cells will now undergo affinity maturation and class switching, and differentiate into plasma cells where they secrete antibodies.

Question 25

Describe Clonal expansion and affinity maturation

Answer 25

Naïve B-cells (not exposed to antigen) has a unique receptor, the b-cell which has the best binding affinity will be selected for, it will proliferate.
Affinity maturation will then occur:
• This aims to improve the affinity of the antibody for the antigen
• Mutations in the V segment allow this so the antibody binds better.
• This process occurs in the germinal centre of the lymph node of the GC
• T-follicular helper cells (T-cells that can only enter the GC) and follicular dendritic cells (FDC) assist in affinity maturation