Humoral Immunity - Generation of B-cells and Antibody Diversity

Question 1

What is the difference between somatic recombination and differential splicing?

Use examples from the B-cell life cycle

Answer 1

Somatic recombination = Alters genetic information at the DNA level

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

Differential Splicing = Changes made at the mRNA level

(Two identical copies of mRNA will be altered differently to produce two different protein products)

• IgM and IgD
• Membrane bound and secreted Ig

Question 2

What are the two stages of B-cell development?

Answer 2

• Antigen dependant
  
  • D→J (Heavy chain variable)
  • V→DJ (Heavy Chain variable)
  • V→J (light chain variable and constant)
• Antigen independant
  
  • Affinity maturation + Class Switching

Question 3

Describe the Life Cycle of B cells

Answer 3

1. The B cell will start in bone marrow
   
   • It starts as a stem cell and differentiate to a pro-B cell
   • The DNA of the pro B cell undergoes
     
     1. D→ J recombination
     2. And V→DJ recombination
        
        • This permanently codes in the heavy chain variable region
2. The variable region will be expressed with the μ heavy constant chain
   
   • This is the first default heavy chain expressed by the B-cell
3. The pro-B cell will become a pre-B cell when it expresses a heavy chain with a light chain placeholder
4. The pre B cell will undergo
   
   1. V→J recombination of either kappa or lambda to permanently code the light chain variable + constant region to become immature B-cell
5. The pre-B cell will become an immature B-cell once it can express both heavy and light chain IgM
   
   1. Note during V(D)J or VJ recombination we have junctional flexibillity + P/N nucleotide addition to generate diversity between B-cells with same comb of gene segments
6. Once they express both IgM and IgD on their surface (via alternative splicing of mRNA) = mature B-cells + circulate between blood stream, lymph nodes + spleen
7. They will patrol resting until they encounter a pathogen
8. The B-cell is activated once it encounters a pathogen and undergoes affinity maturation in the germinal centre (GC)
9. B-cells will recieve information on the type of pathogen and undergo major class switching for the appropriate effector function
   
   1. Note: B cell receptor can become IgG or IgA
10. Majority of B-cells will develop into professional plasma cells to secrete the antibody that they code for + some B-cells coding for IgM will differentiate into plasma cells secreting first defence IgM
11. After infection some B-cells will remain as memory B-cells

Question 4

Further describe the antigen independant phase

Answer 4

• This stage occurs in our bone marrow, makes 1, 000, 000, 000 resting B cells which each contain a unique random BCR
  
  • Lymphoid progenitor will differentiate into pro-B-cells
    
    • Pro-B undergo D to J recombination in its DNA
    • V segment will recombine with DJ segment to hard code the variable region of the heavy chain
  • This will be expressed with the μ constant region
  • The cell will become a pre-B cell when it can express a full heavy chain with unique variable region
    
    • Pre-B cells will express a light chain placeholder which forms a pseudoantibody with the heavy chain
  • The pre-B cell will undergo V – J recombination on the light chain regions, this will determine the light chain variable and constant region
  • The immature B-cell will express a full IgM (μ heavy chain, κ/λ light chain)
  • It becomes a mature B cell (also known as resting B cell/ naïve B cells) when it has the capacity to produce both IgM and IgD through differential splicing of the mRNA

Question 5

What processes occur during V→ DJ and V→J recombination which allow for additonal diversity?

Answer 5

1. Junctional flexibillity
2. P + N nucleotide addition

Question 6

How do we know that the immature B cell has become a mature B-cell?

What acts like the quality control?

Answer 6

• We know the immature B-cell has become mature when it has the capacity to produce both IgM and IgD through differential splicing
  
  • IgD = like quality control

Question 7

How are antibody genes inherited?

Answer 7

NONE are inherited

• No complete genes are inherited, only gene segments
• These gene segments are arranged in different combinations to generate many Ig sequences

Question 8

What different gene segments is the light chain and heavy chain loci made up of?

Answer 8

• Light chain is made up of V, J and C gene segments
• Heavy chain is made up of V, D, J and Cμ
  
  • In the light chain one V segment will be selected at random to recombine with one J segment
  • In the heavy chain random V(D)J segments will recombine
    
    • The DNA will now encode a complete antibody with the VJ of VDJ segments encoding for the light or heavy chain variable fragments and the Cμ segment coding for the constant domains

Question 9

What does the J or the J/D portion code for?

Answer 9

• The J or the D/J portion will code for the CDR3 (complementarity determining regions)
  
  • These are the finger like protrusions interacting with the antigen

Question 10

How many genetic loci are there coding Ig chains?

Answer 10

3 Genetic Loci

• 2 for light chain = Kappa and lambda locus
• 1 for heavy chain

Question 11

Explain the example of VJ recombination of kappa light chain genes on chromosome 2

Answer 11

In humans the Kappa chain contains: 40 (V) variable segments, 5 (J) segments and a constant region (C) segment

• In front of each segement there is a leader sequence which is like a postcode so the cells know where the protein will end up
• The V and J segments are randomly chosen
• This is then transcribed in to mRNA
• The extra J segment will be spliced out to form the leader with V, J and C segments
• The mRNA will be translated into amino acid sequences of the light chain
• The amino acid will be folded, and the leader sequence will be cleaved off when the protein reaches where it needs to go
  
  • This will result in a unique kappa light chain

Question 12

Describe VDJ recombination of heavy chain genes on chromosome 16

Answer 12

In humans the H chain will contain: 51 (V) variable segments, 27 (D) Diversity segments, 6 (J) joining segments, Constant region (C) segments

• The VDJ segments are rearranged on the chromosome below
  
  • The Cμ and Cδ code for the IgM and IgD heavy chain respectively
• First recombination which will take place is the is the D →J joining
  
  • In this D7/J3
• Then there will be V→ DJ joining
  
  • In this case it will be the V20 joining to D7 and J3
• The recombined hard coded DNA will be transcribed into mRNA transcripts
• The mRNA transcript will then undergo differential splicing
• In a population of mRNA transcripts some will express the leader sequence followed by VDJ and finally Cμ sequence
  
  • This will be translated into the IgM heavy chain
• The other mRNA transcript will express the leader, VDJ and Cδ
  
  • This will be translated later IgD heavy chain to signal maturity of B-cells

Question 13

How can V(D)J recombination take place?

Answer 13

It involves the use of recombination signal sequences (RSS)

• They are made up of turns containing a conserved heptamer and nonamer with a 12/23 bp spacer
  
  • These are located upstream or downstream of gene segments

Question 14

What are the two types of turns?

Answer 14

• Two turn contains a 23 base pair spacer
• One turn contains a 12 base pair spacer

Question 15

Why do we have these two types of different turns located in the recombination signal sequences (RSS)?

Answer 15

One turn/ Two turn rule (12/23)

• DNA recombination only occurs between a segment with a 12bp spacer and a segment with a 23bp spacer
  
  • Essentially a one turn can only recombine with a two turn
    
    • This is known as the one-turn/two-turn rule (12/23 rule)

Question 16

How can we generate antibody diversity

Answer 16

• Multiple germline V, D and J gene segments
• Combination of V-J and V-D-J joining
• Junctional flexibility
  
  • P-nucleotide addition
  • N-nucleotide addition (Tdt nucleotide addition)
• Combinatorial association of heavy and light chains (5 classes of heavy chain and 2 classes of light chain)
• Somatic hypermutation during affinity maturation

Question 17

What are the pros and cons of generating junctional diversity through V(D)J recombination and P and N nucleotide additions?

Answer 17

Pros = Antibody diversity

Cons = Can get disadvantageous combinations for example autoantibodies

Question 18

Explain the mechanism of recombination

Answer 18

• If you want to recombine the V17 segment with J3 the dark and light blue triangles will represent the turns
  
  • The enzymes RAG1 and 2 will bind to the turn and pull them together to form a hairpin
• The DNA will be nicked and will automatically form a minor hair pin at the ends of the gene segment
  
  • RSS ends will be joined forming a signal joint which will play no role in recombination process
• The hairpin will be opened by the enzyme artemis
• The DNA ends will be processed by Tdt (P/N nucleotide addition) and exonucleases (junctional flexibillity)
  
  • These will add or remove DNA nucleotides respectively
• The ends will then be joined together by another series of enzymes known as Ku70, DNA-PK, XRCC4

Question 19

Explain how P/N nucleotide addition occurs

Answer 19

• Once hairpin is formed artemis will randomly nick one end of dsDNA
• Nicked ends will linearise and form overhanging ends
• Repair enzyme will fill these gaps = P-nucleotide addition
• Terminal deoxynucleotidyl transferase (Tdt) will add the N nucleotides into the two ends before they are ligated again
  
  • The addition of P and N nucleotides before joining of segments together will cause addition of extra amino acids or shifting of the whole reading frame
    • So if by chance the B-cell selects the same VDJ or VJ segment they will end up with different Abs due to P/N nucleotide addition

Question 20

What is junctional flexibillity?

Answer 20

• This involved the removal of mismatched nucleotides during V(D)J recombination
• Involves EXONUCLEASES

Question 21

What is the mechanism of junctional flexibillity?

Answer 21

• As mentioned before artemis will create the overhanging ends
• In some these ends will overlap if there are enough complementary sequences
• However, if there are mismatched nucleotides these will have to be removed by the exonucleases before the repair enzyme can work
  
  • Note the joining of the signal joins are always precise

Question 22

What is allelic exclusion?

Answer 22

• Even though we have two copies of each Ig gene (one from each parent) and we end up with 2 x kappa and lambda VJC segments and 2 x VDJC heavy chain segments
  
  • In most genes both genes will be expressed but in antibody genes in B cells
    
    • Only one heavy chain allele and one light chain allele out of 6 possible alleles is expressed

Question 23

How does allelic exclusion take place?

Answer 23

• There is a particular order which the loci will undergo rearrangement
  
  • HEAVY > KAPPA > LAMDA, 1st allele then 2nd
• Essential the B cell will try the first allele of the heavy chain if it successfully produces a full heavy chain it will move onto the kappa chain allele, if not then it will try the second heavy chain allele
• Again for the light chain the first kappa allele will be recombined if successful then all recombination will stop immediately if it is not successful in producing a light chain it will move onto the second kappa allele etc
  
  • Heavy or light chain recombination is not successful the cell will undergo apoptosis
• This allelic exclusion mechanisms ensures that each B-cell will only make one type of antibody!!