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SEM 4: Immunology > Humoural Immunity - Part 2 > Flashcards

Humoural Immunity - Part 2 Flashcards

1
Q

What are the two stages of B-cell development?

A
  1. The B cell is first in the bone marrow, the blood and secondary lymph node organs and spleen.
  2. It is divided into antigen-independent stage and antigen-dependent stage which happens in the bone marrow, the blood and the lymph node.
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2
Q

Describe the life cycle of a B-cell

A
  1. Starts as a stem cell and differentiates into a pro-B cell.
  2. The pro-B cell will undergo DNA recombination (V-DJ) and encodes the heavy chain variable region. It then becomes a pre-B cell when the heavy chain is functional.
  3. The pre-B cell undergoes another V-J recombination to code in the light chain variable and constant region. This is an immature B cell.
  4. The B-cell will continue to mature until it expresses IgM and IgD through differential mRNA splicing; then it becomes a mature recirculating B cell.
  5. At this stage, the mature recirculating B cells are patrolling the blood the spleen.
  6. This cell will be activated when the body encounters a pathogen and migrates into the germinal centre (GC) and undergo special forces selection to hone the variable region to the particular pathogen.
  7. It will undergo affinity maturation and class switching and then it will differentiate into plasma cells where it will secrete antibodies and memory B cells.
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3
Q

What is the difference between somatic recombination and differential splicing?

A

Any changes in the DNA level is called somatic recombination. Any changes in the mRNA level is called differential splicing.
The differential splicing changes will produce a protein that will look different but does not change the DNA level. However, changes at the DNA level will change the entire protein completely.

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4
Q

Give examples of somatic recombination

A

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

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5
Q

Describe the life cycle of the antigen independent stage

A
  1. Stem cell differentiates into the pro-B cell.
  2. Then, the pro-B cell undergoes DNA recombination which changes in the DNA level -> first one is D-J and then the V region combines with DJ.
  3. Then, this codes in for the heavy chain variable region which is co-expressed with the Mu constant region. This constant region is for the IgM antibody class and this is the default for all B cells before it encounters pathogens.
  4. This is the pre-B cell. It expresses the place holder light chain as the normal light chain is not generated yet.
  5. It undergoes a 3rd DNA recombination which is the V-J recombination to code in the light chain variable and constant region. There are additional mechanisms which contribute more to the iversity of the antibody (junctional flexibility and P/N nucleotide addtions).
  6. Then, once the B-cell can express full IgM and IgD. They are capable of alternative splicing wihich is the mRNA level. This is when it becomes an immature B-cell or naive B-cell.
  7. It becomes a mature B-cell when it has the capacity to produce both IgM and IgD through alternative splicing.
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6
Q

What are the 3 different loci for encoding Ig?

A
  • Chromosome 22 for the lambda light chain
  • Chromosome 2 for the kappa light chain
  • Chromosome 14 for the heavy chain
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7
Q

How are different Ig sequences generated?

A

Through gene segments in different combinations which is unique for each B-cell.

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8
Q

What are the different segments in the germ line for an antibody?

A
  • Light chain segments: V,J and C

- Heavy chain segments: V, D, J and Cu

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9
Q

What happens when the Ab gene segments are rearranged at the DNA level?

A

They code for the CDRs.
VDJ genes code for the heavy and light variable regions
C genes code for the constant regions
J/DJ genes code for CDR3 which interacts with antigens and is the most variable region.

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10
Q

Example VJ recombination of kappa light chain genes

A
  1. In the 40 variable V segments, one of the segments is selected at random e.g. V23 - J4 and is recombined in the DNA.
  2. This is transcribed to mRNA and in the primary transcript, there are extra bits that are spliced out.
  3. This forms the mature mRNA with a stop codon and a polyA tail - containing the leader, V, J and Cmu.
  4. It is translated to polypeptides and folds to the area and the leader sequence cleaves off when the proteins reaches the specific area.
  5. This will form the kappa light chain.
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11
Q

Explain VDJ recombination of gamma heavy chain genes

A
  1. The first recombination that takes place is the D/J joining. The D7/J3 are selected at random to join.
  2. Then, the V20 is recombined at random with this section (D7/J3) that has formed. This is transcribed into mRNA transcript. Then, differential splicing occurs.
  3. Only the first two constant region segments will be transcribed with the rest of the L-VDJ segment.
  4. The B-cell expresses IgM and IgD, so alternative splicing occurs. It becomes a new heavy chain. The B-cell can make two different classes of the B receptor without changing the DNA but just through splicing.
  5. L-VDJ-Cmu sequence is eventually translated into the IgM heavy chain and the other sequence is L-VDJ-Cdelta, the Cm and extra J segment is spliced out and is eventurally translated into the IgD chain.
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12
Q

What is the sequence of genes in the light chain germline DNA?

A

40 variable (V) segments, 5 joining (J) segments and constant (C) region segments in the germline DNA. In front of each V segment, there is a leader sequence which directs the amino acid to a specific region in the cell.

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13
Q

What is the sequence of genes in the heavy chain germline DNA?

A

51 variable (V) segments, 27 diversity (D) segments, 6 joining (J) segments and a constant (C) region segment.

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14
Q

How many VDJ combinations per mutations can the B segment make?

A
  1. 98 x 10(^8)
    - 8262 from the 94 gene segments in the heavy chain
    - 200 from the 45 segments in the light kappa chain
    - 120 from the 30 segments in the light lambda chain
    - In total = 8262 x 200 x 120 = 1.98 x 10 (^8)
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15
Q

What are other things that contribute to the antibody diversity?

A
  • Multiple germline V, D and J gene segments - different genes from parentals
  • Combination V-J and V-D-J joining
  • Junctional flexibility
  • P-nucleotide addition
  • N-nucleotide addition
  • Combinational association of heavy and light chains
  • Somatic hypermutation during affinity maturation
  • So 1.98 x 10(^8) will become 3 x 10(^11).
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16
Q

What are recombination signal sequences (RSS)?

A

They are conserved sequences upstream or downstream of genes segments required by VDJ recombination.

17
Q

What is the sequence of RSS’?

A

These are turns consisting of a conserved heptamer (7 DNA base pair - same in all RSS) and conserved nonamer (9 DNA base pair) with a 12 or 23 bp spacer.

18
Q

What are the two types of turns in RSS’?

A

One with the 12 bp spacer

One is with the 23 bp spacer

19
Q

Why are there two types of turns? What is the one-turn/two-turn rule?

A

This is because recombination only occurs between a segment with a 12bp and a 23 bp spacer. This is the one-turn/two-turn rule.

20
Q

Where are the two turns and one turns located?

A

The two turns are located downstream of every J segment whereas the one turn is located upstream and downstream of the D segment. This ensures that the correct recombinations occur.

21
Q

Describe the mechanism of recombination of V17 and J3 with their respective RSS

A
  1. RAG1 and RAG2 clamp on top of the RSS on both sides which forms a major hairpin (entire DNA folded in half) fold.
  2. This will create nits in the DNA and form minor hairpins (small folds).
  3. Artemis enzyme opens the hairpins by randomly nicking between base pairs (e.g. A-A or A-T) and the ends will unfurl. This leaves overhangs.
  4. In some cases, the overhangs are repaired by enzymes with added nucleotides - these nucleotides added are called P nucleotides.
  5. On the other end, in the heavy chain, almost exclusively, the Tdt enzyme adds N nucleotides before the joining of the end. The enzymes which carry out these processes are called exonucleases and Tdt perform end processing.
  6. There are P and N nucleotides between the V and J joint due to end-processing occuring. This generates antibody diversity as well.
  7. This results in V and J joining (coding joint) and a circular signal joint.
22
Q

How does junctional flexibility occur?

A

The precise mechanisms is unknown but it involves the removal of DNA base pairs. It involves exonucleases which remove the mismatched nucleotides for example in the DNA overhangs left by the artemis enzyme. It sometimes removes DNA off the segments. However, junctional flexibility means the coding joints will lose base pairs. Whereas, the signal joints are always the same.

23
Q

What is junctional diversity?

A

It is formed by junctional flexibility during V(D)J recombination, P and N nucleotide additions.

24
Q

What are the positivies of junctional diversity?

A

It generates antibody diversity

25
Q

What are the negatives of junctional diversity?

A

It can produce non-productive rearrangments (incorrect reading frame) therefore is a wasteful process.

26
Q

How are antibody genes different to normal alleles?

A

Two copies of each Ig gene from one mother to one father. For example, the islets of langerhan, both genes will be expressed. But antibody genes are different in B cells, that means only one heavy chain allele and one light chain allele is expressed.

27
Q

What is the order of hierachy in the antibody genes?

A

Heavy > Kappa > Lambda

  • 1st allele than 2nd.
  • If it produces a successful heavy chain then it will proceed but if not successful, it will not continue making the other genes.
28
Q

Why is junctional flexibility and allelic exclusion important?

A

These mechanisms ensure that each B cell makes one type of antibody.

SEM 4: Immunology (15 decks)

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