Antibody biology

Question 1

What are B cells and what do they do?

Answer 1

B cells are lymphocytes central to our humoralresponse - they are responsible for production of antibodies.B cells originate and mature in the bone marrow.On activation with antigen: B cells can become antibody secreting cells (plasma cells)- differentiation, or memory B cells- basis of immunological memory.Antigen receptor on B cells = antibody the B cell is programmed to make (B cell receptor- form of antibody that the cell is programmed to make.)

Question 2

What do plasma cells do?

Answer 2

Plasma cells secrete antibody of the same specificity as themembrane-bound immunoglobulin expressed by their B cell precursor.Membrane bound Ig has specific binding to antigen, stimulates B cell to undergo further differentiation and make daughter cells. Stimulated B cell gives rise to plasma cells- have same specificity as B cell. Programmed to affect antigen that was originally detected.

Question 3

What is the structure of immunoglobulins?

Answer 3

All immunoglobulins are structurally similar. Made up of heavy chains (inner) and light chains (outer). Antibodies can express one of two possible light chains, λ or κ.Variable region at tips of arms. N terminal. Named VH and VL. Constant region makes up the rest of the molecule. C terminal. Named CL, CH1, CH2, CH3 etc.FAB regions are arms. Fc region is stem. Disulphide bridges between the two regions, at hinge.

Question 4

What are the products of enzymatic digestion of antibodies?

Answer 4

Proteolytic cleavage by papain yields:FAB- Fragment antigen binding - responsible for binding antigen through tip of FAB arms.Fc- Fragment crystallisable (tends to crystallise) - responsible for effector function.

Question 5

What does the Ig flexible hinge do?

Answer 5

Flexible hinge of IgG allows both arms to bind to many arrangements of antigens on pathogen surfaces.True for all antibodies to an extent.Flexibility at hinge region.Greater occupancy rates due to bivalent binding- greater effector function happens as a result.

Question 6

What are light chains and heavy chains?

Answer 6

Two types of light chain - lambda (λ) and kappa (κ).All antibody classes use these same light chains (same type of chains in each antibody molecule).In any individual antibody molecule both lightchains are the same.Heavy chain determines class of antibody:IgG- γ heavy chain.IgA- α heavy chain.IgM- μ heavy chain.IgE- ε heavy chain.IgD- δ heavy chain.Domains fold up into globular domains.

Question 7

How to antibodies fold into globular domains?

Answer 7

Antibodies fold into globular domainswith a distinctive 3D structure.Each domain comprises 2 β sheets that lie over each other, which are linked by a disulphidebridge to form a β barrel.Each sheet made up of a number of B strands. Slight difference in V and C domains- slightly more folds in V domain.Loops that join lie at either end of domain.Folded structure is known as the‘immunoglobulin fold’.

Question 8

What are hypervariable loops?

Answer 8

Variable domains in both light and heavychains have hypervariable loops which varyextensively between different antibodies.Hypervariability- very different sequences from one antibody to the next.Framework regions-These are still variable, but much less so.Hypervariable loops form antigen binding site.Hypervariable loops are more commonly termed the complementarity-determining regions (CDRs).When VH and VL domains are paired the CDRs create theantigen-binding site.Antigen binding site hascomplementary 3D structure to that of antigenic epitopeproviding high specificity and high affinity.Loops lie at one end of domain.6 loops brought close together in binding- form binding site.CDR- bind to antigen through complementarity of structure eg. hand in glove, key in lock etc.High specificity binding- they fit together in a specific way- one one type of antigen will bind into one type of site.Adjust loop sequence give subtle structural changes- changes which antigen can fit in site.

Question 9

What are linear and discontinuous epitopes?

Answer 9

Epitopes (Regions recognised by antigen.AKA antigenic determinants) in protein antigens may comprise a single stretch ofpolypeptide chain = linear or continuous epitope. Binds in a linear sequence.Or may depend on 3D folded structure = conformational ordiscontinuous epitope. Doesn’t bind in a linear way.

Question 10

How do epitopes bind to antibodies?

Answer 10

Antibody recognises structural ‘epitopes’ or ‘antigenicdeterminants’ on the antigen.Epitopes can bind to pockets, grooves, extended surfaces orprojections in antigen binding sites.

Question 11

Where are epitopes located?

Answer 11

Epitopes for antibodies are exposed on pathogen surfaces.Sites need to be accessible for antibody binding.

Question 12

How is antibody diversity achieved?

Answer 12

Almost any substance can elicit an antibody response.Responses to even simple substances are diverse.Each individual has an antibody repertoire (i.e. available antigenspecificities) of approximately 10^11- Need to recognise 10^11 different antibody structures.To achieve this, Ig genes are organised differently to other genes.In all cells except B cells, Ig genes are in fragmented form thatcannot be expressed.Ig heavy and light chain loci consist of families of gene segments,arrayed sequentially along the chromosome- This is why they are in fragmented form.They are inherited in this form through egg and sperm (germline)Therefore their arrangement is termed “germline configuration”.In developing B cells Ig gene segments must be rearranged toassemble functional light and heavy chain gene.

Question 13

What are the gene segments and how are they arranged?

Answer 13

V = V gene segment encodes most of the V domain (95-101 aa)J = ‘Joining’ segment (encodes up to 13 aa)D = ‘Diversity’ segmentVH domain is encoded by V, D and J gene segments.VL domain is encoded by V and J gene segments.Lambda light chain locus- J segments upstream of corresponding constant region.Kappa light chain locus- just Vk and Jk genes- no D genes.Heavy chain locus- Have around 40 VH segments arranged one after other. DH (diversity segment). 23 joining segments.

Question 14

When do Ig gene rearrangements occur?

Answer 14

Ig gene rearrangements occur at definedpoints in B cell development.Heavy chain Ig rearranges first- D-J rearranging at early pro-B cell stage. V-DJ rearranging at late pro-B cell stage.Light chain Ig reararnges later- V-J rearranging at small pre-B cell stage.

Question 15

How to pro-B cells rearrange heavy chain gene segments?

Answer 15

Pro-B cells randomly recombine heavy chain gene segments.Works through alternative splicing.D gene segment is joined to a J gene segment.V gene segment joined to DJ gene segment.These events are called somatic recombination because they occur in cells other than germ cells.DNA that is spliced out is lost- irreversible step.VH-CH1-CH2-CH3-CH4: Transcript to produce IgM- first antibody that is usually produced.Codes for IgM heavy polypeptide chain (VH domain).

Question 16

How is the kappa light chain rearranged?

Answer 16

Works through alternative splicing again.V kappa domain polypeptide is: Vk-Ck.

Question 17

What are repeated light chain rearrangements?

Answer 17

If non productive joint is made (eg. out of frame), can redo it one or two times until we get a light chain that can function with heavy chain.

Question 18

How is rearrangement guided?

Answer 18

Rearrangement guided by conserved non-coding sequences known as Recombination Signal Sequences (RSS).These flank gene segments.Heptamer/nonamer and either 23 or 12 spacer.Have one or either of spacer at end.Only get combos of 23 and 12 (not 12 and 12 or 23 and 23).

Question 19

What is the 12/23 rule?

Answer 19

RSS flank recombination sites.RSS consist of conserved heptamer, nonconserved spacer (12 or 23 bp),conserved nonamer.Genes with a 12bp spacer can recombine with one with a 23bp spacer = 12/23 rule.

Question 20

Whar are the enzymatic requirements for recombination?

Answer 20

Set of enzymes needed to recombine V, D, and J = V(D)J recombinase. (Only VJ recombinase in light chain).Two components are made only in lymphocytes - Products of the recombination-activating genes RAG-1 and RAG-2 (key elements that allow recombination in B cells).RAG-1 and RAG-2 associate with each other and other proteins known as high mobility group of proteins- RAG complex.Other components are present in all nucleated cells have activities that:-repair double-stranded DNA.-bend DNA.-modify the ends of broken DNA strands.They include:-DNA ligase IV.-DNA-dependent protein kinase (DNA-PK).-Artemis (a nuclease).-Ku protein.

Question 21

How do RAGs work?

Answer 21

1. One RAG complex binds 23 spacer RSS, another bind 12 spacer RSS.2. RAG complex interaction aligns RSSs and cleaves DNA at ends of gene segments.3. DNA hairpin formed at end of each gene segment, and clean break at heptamer ends.4. RAG complexes hold DNA in place while broken ends rejoined by DNA repairenzymes = nonhomologous end-joining.5. Coding joint formed in chromosome, signal joint in removed piece of circular DNA (This has been excised and will be lost).

Question 22

What is happening in the RAG hairpins?

Answer 22

Enzymes that open hairpins and form codingjoint introduce additional sequence diversityinto CDR3 (This is one of hypervariable loops) in several ways.Nick (by RAG) that opens hairpin can occur at several different positions.Forms single-stranded end in whichcomplementary bases that were originally ontwo DNA strands are now on same strandCreates sequence that would form palindromein dsDNA – i.e. identical when read from either end. These nucleotides are termed P nucleotides.Ends variably modified by exonucleases (removenucleotides) or terminal nucleotidyl transferase(TdT) which randomly adds nucleotides= N nucleotides since they are non-templated.Once sequence complementarity enables thestrands to pair, single-stranded gaps are filled in with complementary nucleotides.Contribution of P and N nucleotides toresulting amino acid sequence in CDR3is called junctional diversity.Can increase diversity of antibody repertoire by up to 3 x 10^7.(Fill in the extra ones.Some are P and some are N.New piece of sequence between D and J parts. Can gets lots of variability here.)

Question 23

What processes generate the diversity of Ig repertoire?

Answer 23

1. Combinatorial diversity - many copies of each gene segment type which can combine in many different ways.2. Junctional diversity - nucleotides added or subtracted at joins between gene segments during enzymatic steps.3. Many different combinations of light and heavy chains V regions that pair to form antigen binding site. (Can combine any light and heavy chains.)(1-3 occur in naïve B cells).4. Somatic hypermutation introduces point mutations into rearranged V region genes of activated B cells.

Question 24

What is happening at the Ig heavy chain locus?

Answer 24

CH genes (encoding Ig heavy chain constant regions) form a cluster 3’ to JH.Each CH gene has multiple exons (not shown), each encoding an individual Ig domain.Cμ closest to JH gene segments and first complete transcript produced.Subsequently additional isotypes generated by an irreversible change in DNA by process known as isotype switching.Switching from one antibody class to next.

Question 25

What do B cells use alternative splicing for?

Answer 25

Developing and naïve B cells use alternative splicing to express both IgM and IgD.Circulating B cells that haven’t yet encountered antigen = naïve B cells.Express both IgM and IgD on surface.Accomplished through differential splicing of same primary RNA transcript.IgD- Keep same VH part, but different constant region part to IgM. IgD has 3 constant regions, IgM has 4.

Question 26

How does each B cell produce Ig of a single antigen specificity?

Answer 26

Ig gene rearrangement is tightly controlled.Ensures only one heavy chain and one light chain expressed.Known as allelic exclusion - only one copy or allele of heavy chain locus and one of the light chain locus are rearranged to form functional genes.Thus each B cell produces Ig of a single antigen specificity.(due to loss of other genes permanently after splicing).

Question 27

How is Ig first made in a B cell?

Answer 27

Ig is first made in a membrane-bound formthat is present on B cell surface.B cell receptor (BCR) is made up of cell surface IgM with invariant proteins Igα and Igβ.Invariant chains are important for signaling once antigen binds to IgM.Ig heavy chains have hydrophobic sequence near C terminus that anchors them in membrane.IgM is the first Ig class that is secreted.

Question 28

How are membrane and secreted antibody forms generated?

Answer 28

Membrane and secreted forms are generated by alternative RNA processing of heavy chain.Transmembrane Ig has a hydrophobic transmembrane domain of around 25aa thatacts as an anchor in plasma membrane.Secretory Ig has a hydrophilic secretory tail.The 2 carboxy termini are encoded in separate exons and selected by alternative RNA processing.Secreted IgM- Splice slightly different region at C terminus that membrane IgM- hydrophilic makes secretory form, hydrophobic makes membrane bound form.

Question 29

What is needed for B cells to respond to antigens?

Answer 29

Help from T cells.

Question 30

How do T cells help B cells?

Answer 30

B and T recognise same antigen.1. Co-stimulatory signal through ligation of CD40 and CD40L.2. Secretion of cytokines which aid in the induction of isotype switching.

Question 31

What are primary and secondary immune responses?

Answer 31

Primary- large amount of infective agent. Medium levels of antibodies produced (IgM)- peak at 10 days. Slow response to infection. Disease.Secondary- see infection for a second time. Low levels of infective agent. High levels of antibody (IgG)- peak at 4 days. Much faster response. No disease.

Question 32

What are the different immunoglobulin classes?

Answer 32

IgG, IgM, IgD, IgA1, IgE.Heavy chain determines the class of antibody - denoted by greek letter γ, μ, δ, α, ε.5 main classes of antibody exist.IgG and IgA also have subclasses.

Question 33

How are IgG, IgA and IgE isotypes generated?

Answer 33

IgG, IgA, and IgE isotypes are generated by an irreversible change in DNA in a process known as isotype switching or class switching.Association of VH exon with different CH genes – Ag specificity remains the same.Occurs through mechanism of nonhomologous DNA recombination guided by stretches of repetitive DNA called switch regions, upstream of CH gene.Switch regions contain repeats of GAGCT and GGGGGT sequences.

Question 34

What is AID and what does it do?

Answer 34

Recombination to excise previously expressed C gene and juxtapose another one with the VH.Depends on action of enzyme named activation-induced cytidine deaminase (AID), only found in proliferating B cells.AID targets switch regions, causing nicks in both DNA strands.Nicks facilitate recombination between switch regions, leading to excision of intervening DNA.