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Immunology > Antibody genetics and function > Flashcards

Antibody genetics and function Flashcards

1
Q

What are the main molecules for antigen recognition?

A

PRRs, MHC, TCR and BCR. Antibodies which are glycoproteins, not main.

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2
Q

Describe the structure of antibody molecules.

A

There are 2 identical heavy and 2 identical light chains which have disulphide bonds between light chains and heavy chains to bond them.

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3
Q

What is the molecular weight of heavy chains?

A

50 kDa

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4
Q

What is the molecular weight of light chains?

A

25 kDa

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5
Q

What is the molecular weight of an antibody?

A

~150 kDa

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6
Q

What enables antigen specificity of antibodies?

A

the variable regions of the antibody, which vary between one antibody and another. There are hypervariable regions on complementarity determining regions (CDR1/2/3). The amino acids at the CDR region are complementary to the amino acids on the epitope of the antigen.

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7
Q

How many subclasses of IgG are there?

A

4

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8
Q

How many subclasses of IgA, E, D and M?

A

Ig A has 2, however the others do not have subclasses.

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9
Q

How many possible constant regions of antibodies are there?

A

9

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10
Q

How does antibody variability arise?

A

By genetic recombination of Ig genes for the variability regions. They only recombine in B cells and have a germ line configuration

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11
Q

Describe heavy chain genetics

A

They are syntenic and there are 4 groups of genes: Variable, Diversity, Joining and constant.

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12
Q

How many variable genes are there?

A

~40

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13
Q

How many diversity genes are there?

A

~27

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14
Q

How many joining genes are there?

A

~6

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15
Q

How many constant genes are there?

A

9

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16
Q

Describe light chain genetics

A

They are not syntenic and do not have diversity genes, made up of 3 gene segments.

17
Q

Describe B cell development

A

They start in the bone marrow as haematopoietic stem cells and develop down the lymphoid pathway. B cells then recombine their Ig genes, T cells recombine TCR genes. For B cells, in germ line configuration, they recombine the VDJ gene segments in a semi-random way:
1/27 D genes are positioned next to 1/6 joining genes. Chromosomal inversion loops enable recombination.
2ndary recombination occurs, where 1/40 V segments are recombined with the DJ genes. Primary RNA transcript transcribed, introns removed from pre-mRNA (splicing). The mRNA then has VDJC mRNA. Recombination of kappa and lambda light chains also occurs; if kappa fails then lambda occurs.

18
Q

Which part of the antibody has the most variable structure?

A

The Ig heavy chains (1/40 x 1/27 x 1/6 = 1/6480)

19
Q

What is the variability of Igkappa?

A

40 x 5 = 200

20
Q

What is the variability of Iglambda?

A

30 x 5 = 150

21
Q

What is the Ig variability value?

A

200 (with kappa) x 6480 = 1.3 x 10^6

22
Q

How many kappa light chain genes are there?

A

50

23
Q

How many lambda light chain genes are there?

A

30

24
Q

How many constant genes are there for light chains?

A

5 for both kappa and lambda

25
Q

How is V-D-J recombination mediated?

A

By recombinase enzymes encoded by recombination-activating genes; RAG-1 and RAG-2 which recognise the recombination signal sequence (RSS) nucleotides flanking the V,D and J gene segments.

26
Q

Where do V segments have their RSS

A

3’ downstream

27
Q

Where do D segments have their RSS?

A

at 5’ and 3’ flanking sites?

28
Q

Where do J segments have their RSS?

A

upstream 5’

29
Q

How many types of RSS are there?

A

2; heptamer (7) - nonamer (9) sequences.

30
Q

What is the sequence of RSS?

A

5’-heptamer-unconserved spacer - nonamer - 3’ sequences.

31
Q

How long are unconserved spacers?

A

12 or 23 nucleotides long. The length and sequence of RSS is conserved at heptamer and nonamer sequences. This 12-23 rule means that a sequence can only be recombined that has an RSS of 12bp spacer to a sequence with a 23bp spacer. There is no recombiantion between V and a J not with 2 Ds. This 12-23 rule ensures that the sequence is V-D-J.

32
Q

How does additional diversity of antibodies arise?

A

recombinatorial inaccuracies, N-nucleotide addition using terminal deoxynucleotidyl transferase (TdT), chain combinations, somatic hypermutation

33
Q

What does class switching enable?

A

Allows expression of different antibody isotopes, IgM to IgE.

34
Q

Are RAG-1/2 used in the recombination of heavy chains?

A

No, different enzymes involved.

35
Q

Are there switch sequences between Cmu and Cdelta?

A

No, this results in IgM and IgD being constitutively expressed on naive B cells.

36
Q

How is TCR diversity generated?

A

Multiple germ line genes. VJ and VDJ recombination, recombinatorial inaccuracies, N-nucleotide addition and chain combinations.

37
Q

Can TCR diversity be generated through somatic hypermutation?

A

No, nor through class switch recombination.

38
Q

Describe antibody structure

A

Fab regions; the fragment of antigen binding region and an Fc region which can be split by papain. Fc region binds to FcR. Ab heavy and light chains are folded into domains. Domains are stabilised through intra-chain disulphide bonds. The hinge region enables flexibility of two Fab regions. The Ig domain structure becomes stable through folding. 4 beta strands make 2 beta sheets which are hydrophobic and intra=chain disulphide bonds stabilises. the hypervariable regions are exposed to the exterior. In the V domain, the antigen-binding CDRs are located in Ag-binding loops at the tip of the Ab.

39
Q

Can antibodies form polymers?

A

Yes, particularly secretory IgA; dimeric using J-chain protein. Circulating IgM can form a pentameric polymer with J chains linking.

Immunology (18 decks)

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