Immunoglobulins. Generation of Diversity

Transcription

1 Immunoglobulins Generation of Diversity

2 Unfortunately, for this theory to be true the number of antibody genes would need to be fold greater than the entire human genome Introduction Immunologist estimate that each person has the ability to produce a range of individual antibodies capable of binding to a total of well over epitopes According to the germline theory, a unique gene encodes each antibody

3 Theories An alternative theory, the somatic mutation theory, holds that a single germline immunoglobulin gene undergoes multiple mutations that generate immunoglobulin diversity. This scheme, however, requires an unimaginable mutation rate The immune system has developed a much more elegant solution- the chromosomal rearrangement of separate gene segments, which employs some elements of the germline and somatic mutation theories

4 Gene Rearrangement Each light and heavy chain is encoded by a series of genes occurring in clusters along the chromosome In humans, the series of genes encoding the k light chain, λ light chain, and the heavy chain are located on chromosomes 2, 22, and 14 respectively When a cell becomes committed to the B lymphocyte lineage, it rearranges the DNA, encoding its light and heavy chains by cutting and splicing together some of the DNA sequences, thus modifying the sequence of the variable region gene

5 Tonegawa s demonstration 1976 used restriction enzymes and DNA probes to show that germ cell DNA contained several smaller DNA segments compared to DNA taken from developed lymphocytes (myeloma cells)

6 k H l VH1 VH2 VH65 DH JH Cg 1 gene 1 transcript 1 protein Antibody specificities more than 1,000,000,000,000 Human genome about 30,000 genes Human Antibody genes H: chromosome 14 k: chromosome 2 l: chromosome 22

7 Ig gene sequencing complicated the model Structures of germline V L genes were similar for Vk, and Vl, However there was an anomaly between germline and rearranged DNA: L V L C L L V L J L C L ~ 95aa ~ 100aa ~ 95aa ~ 100aa L V L C L ~ 208aa Where do the extra 13 amino acids come from? Extra amino acids provided by one of a small set of J or JOINING regions

8 Further diversity in the Ig heavy chain L V H J H D H C H Heavy chain: between up to 8 additional amino acids between J H and C H The D or DIVERSITY region Each heavy chain requires three recombination events: J H to D H, J H D H to V H and J H D H V H to C H L V L J L C L Each light chain requires two recombination events: V L to J L and V L J L to C L

9 Problems? 1. How is an infinite diversity of specificity generated from finite amounts of DNA? 2. How can the same specificity of antibody be on the cell surface and secreted? 3. How do V region find J regions and why don t they join to C regions? 4. How does the DNA break and rejoin?

10 Vk & Jk Loci: Vl & Jl Loci: Diversity: Multiple germline genes 132 Vk genes on the short arm of chromosome 2 29 functional Vk genes with products identified 87 pseudo Vk genes 16 functional Vk genes - with no products identified 25 orphans Vk genes on the long arm of chromosome 2 5 Jk regions 105 Vl genes on the short arm of chromosome functional genes with products identified 56 pseudogenes 6 functional genes - with no products identified 13 relics (<200bp of Vl sequence) 25 orphans on the long arm of chromosome 22 4 Jl regions

11 Diversity: Multiple Germline Genes VH Locus: JH Locus: DH Locus: 123 V H genes on chromosome functional V H genes with products identified 79 pseudo V H genes 4 functional V H genes - with no products identified 24 non-functional, orphan V H sequences on chromosomes 15 & 16 9 J H genes 6 functional J H genes with products identified 3 pseudo J H genes 27 D H genes 23 functional D H genes with products identified 4 pseudo D H genes Additional non-functional D H sequences on the chromosome 15 orphan locus reading D H regions in 3 frames functionally increases number of D H regions

12 Reading D segment in 3 frames Analysis of D regions from different antibodies One D region can be used in any of three frames Different protein sequences lead to antibody diversity GGGACAGGGGGC GlyThrGlyGly Frame 1 GGGACAGGGGGC GlyGlnGly GGGACAGGGGGC AspArgGly Frame 2 Frame 3

13 Estimates of combinatorial diversity Using functional V, D and J genes: 40 VH x 27 DH x 6JH = 5,520 combinations D can be read in 3 frames: 5,520 x 3 = 16,560 combinations 29 Vk x 5 Jk = 145 combinations 30 Vl x 4 Jl = 120 combinations = 265 different light chains If H and L chains pair randomly as H 2 L 2 i.e. 16,560 x 265 = 4,388,400 possibilities Due only to COMBINATORIAL diversity In practice, some H + L combinations are unstable. Certain V and J genes are also used more frequently than others. Other mechanisms add diversity at the junctions between genes JUNCTIONAL diversity

14 Problems? 1. How is an infinite diversity of specificity generated from finite amounts of DNA? Mathematically, Combinatorial Diversity can account for some diversity how do the elements rearrange? 2. How can the same specificity of antibody be on the cell surface and secreted? 3. How do V region find J regions and why don t they join to C regions? 4. How does the DNA break and rejoin?

15 Genomic organisation of Ig genes (Numbers include pseudogenes etc.) L H V H D H 1-27 J H 1-9 Cm Lk1-132 Vk1-132 Jk 1-5 Ck Ll1-105 Vl1-105 Cl1 Jl1 Cl2 Jl2 Cl3 Jl3 Cl4 Jl4

16 Ig light chain gene rearrangement by somatic recombination Germline Vk Jk Ck Rearranged 1 transcript Spliced mrna

17 Ig light chain rearrangement: Rescue pathway There is only a 1:3 chance of the join between the V and J region being in frame Vk Jk Ck Non-productive rearrangement Light chain has a second chance to make a productive join using new V and J elements Spliced mrna transcript

18 Ig heavy chain gene rearrangement V H D H 1-27 J H 1-9 Cm Somatic recombination occurs at the level of DNA which can now be transcribed

19 RNA processing Primary transcript RNA V D J 8 J 9 Cm1 h Cm2 Cm3 Cm4 pas AAAAA mrna V D J 8 Cm1 h Cm2 Cm3 Cm4 AAAAA The Heavy chain mrna is completed by splicing the VDJ region to the C region The H and L chain mrna are now ready for translation V L J L C L AAAAA V H D H J H h C H AAAAA

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21 Problems? 1. How is an infinite diversity of specificity generated from finite amounts of DNA? Combinatorial Diversity and genomic organisation can account for some diversity 2. How can the same specificity of antibody be on the cell surface and secreted? 3. How do V region find J regions and why don t they join to C regions? 4. How does the DNA break and rejoin?

22 Remember These Facts? Cell surface antigen receptor on B cells Allows B cells to sense their antigenic environment Connects extracellular space with intracellular signalling machinery Secreted antibody functions Neutralisation Arming/recruiting effector cells Complement fixation How does the model of recombination allow for two different forms of the same protein?

23 The constant region has additional, optional exons Primary transcript RNA Cm AAAAA Each H chain domain (& the hinge) encoded by separate exons Secretion coding sequence Polyadenylation site (secreted) pas Polyadenylation site (membrane) pam Cm1 h Cm2 Cm3 Cm4 Membrane coding sequence

24 Membrane IgM constant region DNA Cm1 h Cm2 Cm3 Cm4 Transcription 1 transcript Cleavage & polyadenylation at pam and RNA splicing mrna Cm1 h Cm2 Cm3 Cm4 Cm1 h Cm2 Cm3 Cm4 AAAAA pam AAAAA Protein Fc Membrane coding sequence encodes transmembrane region that retains IgM in the cell membrane

25 Secreted IgM constant region DNA Cm1 h Cm2 Cm3 Cm4 Transcription 1 transcript Cleavage polyadenylation at pas and RNA splicing mrna Cm1 h Cm2 Cm3 Cm4 Cm1 h Cm2 Cm3 Cm4 AAAAA pas AAAAA Protein Secretion coding sequence encodes the C terminus of soluble, secreted IgM Fc

26 Alternative RNA processing generates transmembrane or secreted Ig

27 (a) Secreted & membrane forms of the heavy chain by alternative ( differential ) RNA processing of primary transcript.

28 Synthesis, assembly, and secretion of the immunoglobulin molecule.

29 Problems? 1. How is an infinite diversity of specificity generated from finite amounts of DNA? Combinatorial Diversity and genomic organisation accounts for some diversity 2. How can the same specificity of antibody be on the cell surface and secreted? Use of alternate polyadenylation sites 3. How do V region find J regions and why don t they join to C regions? 4. How does the DNA break and rejoin?

30 V, D, J flanking sequences Sequencing up and down stream of V, D and J elements Conserved sequences of 7, 23, 9 and 12 nucleotides in an arrangement that depended upon the locus Vl Jl Vk Jk D V H J H

31 Gene rearrangements are made at recombination signal sequences (RSS). RSSs are heptamer-nonamer sequences Each RSS contains a conserved heptamer, a conserved nonamer and a spacer of either 12 or 23 base pairs.

32 There is a RSS downstream of every V gene segment, upstream of every J gene segment and flanking every D gene segment Generic light chain locus Generic heavy chain locus V D J

33 Recombination signal sequences (RSS) HEPTAMER - Always contiguous with coding sequence D NONAMER - Separated from the heptamer by a 12 or 23 nucleotide spacer V H J H D V H J H RULE A gene segment flanked by a 23mer RSS can only be linked to a segment flanked by a 12mer RSS

34 1. Rearrangements only occur between segments on the same chromosome. 2. A heptamer must pair with a complementary heptamer; a nonamer must pair with a complementary nonamer. 3. One of the RSSs must have a spacer with 12 base pairs and the other must be 23 base pairs (the 12/23 rule).

35 - RSS having a one-turn spacer can join only with RSS having a two-turn spacer : one-turn / two-turn joining rule - This ensures that V,D,J segments join in proper order & that segments of the same type do not join each other. - The enzymes recognizing RSS : recombination-activating genes. ( RAG-1, -2), lymphoid-specific gene products

36 Molecular explanation of the rule 23-mer = two turns 12-mer = one turn 23 Intervening DNA of any length 12 V D J

37 7 Molecular explanation of the rule V1 V2 V3 V4 V7 V6 V5 V8 V9 D J Heptamers and nonamers align back-to-back The shape generated by the RSS s acts as a target for recombinases 23-mer V V2 12-mer D J V3 V4 Loop of intervening DNA is excised V9 V5 V8 V6 V7 An appropriate shape can not be formed if two 23-mer flanked elements attempted to join (i.e. the rule)

38 7 12 Junctional diversity 9 Mini-circle of DNA is permanently lost from the genome V 7 D J Coding joint V D J Signal joint Imprecise and random events that occur when the DNA breaks and rejoins allows new nucleotides to be inserted or lost from the sequence at and around the coding joint.

39 Non-deletional recombination V1 V2 V3 V4 V9 D J Looping out works if all V genes are in the same transcriptional orientation V D 9 J V1 V2 V3 V4 V9 D J How does recombination occur when a V gene is in opposite orientation to the DJ region? V D J

40 Non-deletional recombination V D J V4 and DJ in opposite transcriptional orientations V V V V D J

41 V V D J D J Heptamer ligation - signal joint formation V4 D J V to DJ ligation - coding joint formation V4 D J Fully recombined VDJ regions in same transcriptional orientation No DNA is deleted

42 Problems? 1. How is an infinite diversity of specificity generated from finite amounts of DNA? Combinatorial Diversity and genomic organisation accounts for some diversity 2. How can the same specificity of antibody be on the cell surface and secreted? Use of alternative polyadenylation sites 3. How do V region find J regions and why don t they join to C regions? The rule 4. How does the DNA break and rejoin?

43 V V D J V D J Steps of Ig gene recombination D J Recombination activating gene products, (RAG1 & RAG 2) and high mobility group proteins bind to the RSS The two RAG1/RAG 2 complexes bind to each other and bring the V region adjacent to the DJ region The recombinase complex makes single stranded nicks in the DNA. The free OH on the 3 end hydrolyses the phosphodiester bond on the other strand. This seals the nicks to form a hairpin structure at the end of the V and D regions and a flush double strand break at the ends of the heptamers. The recombinase complex remains associated with the break

44 9 12 Steps of Ig gene recombination V D J A number of other proteins, (Ku70:Ku80, XRCC4 and DNA dependent protein kinases) bind to the hairpins and the heptamer ends. V D J V D J The hairpins at the end of the V and D regions are opened, and exonucleases and transferases remove or add random nucleotides to the gap between the V and D region DNA ligase IV joins the ends of the V and D region to form the coding joint and the two heptamers to form the signal joint.

45 Junctional diversity: P nucleotide additions V D J J V TC AG JD J TC CACAGTG V AG GTGTCAC AT GTGACAC D TA CACTGTG The recombinase complex makes single stranded nicks at random sites close to the ends of the V and D region DNA. TC CACAGTG V AT TA AG GTGTCAC AT GTGACAC D TA CACTGTG The 2nd strand is cleaved and hairpins form between the complimentary bases at ends of the V and D region.

46 CACAGTG GTGTCAC V2 V3 V4 GTGACAC CACTGTG Heptamers are ligated by DNA ligase IV V9 V8 V7 V5 V6 J V TC AG D AT TA V TC AG AT TAD J V and D regions juxtaposed

47 Generation of the palindromic sequence V TC AG AT TAD J Regions to be joined are juxtaposed V TC AG AT TAD J Endonuclease cleaves single strand at random sites in V and D segment The nicked strand flips out V TC~GA AG AT TA~TA D J The nucleotides that flip out, become part of the complementary DNA strand In terms of G to C and T to A pairing, the new nucleotides are palindromic. The nucleotides GA and TA were not in the genomic sequence and introduce diversity of sequence at the V to D join. (Palindrome - A Santa at NASA)

48 Junctional Diversity N nucleotide additions V TC~GA CACTCCTTA AT TTCTTGCAA AG TA~TA V TC~GA CACACCTTA AT TTCTTGCAA AG TA~TA D J D J Terminal deoxynucleotidyl transferase (TdT) adds nucleotides randomly to the P nucleotide ends of the singlestranded V and D segment DNA Complementary bases anneal V TC~GACACACCTTA TTCTTGCAA TA~TAD J Exonucleases nibble back free ends TC~GACACACCTTA AT V GTT AT AT AG TTCTTGCAA AGC TA~TA D J DNA polymerases fill in the gaps with complementary nucleotides and DNA ligase IV joins the strands

49 Generation of Antibody Diversity P-nucleotide and N-nucleotide addition during joining.

50 P and N region nucleotide alteration adds to diversity of V region During recombination some nucleotide bases are cut from or add to the coding regions (p nucleotides) Up to 15 or so randomly inserted nucleotide bases are added at the cut sites of the V, D and J regions (n nucleotides_ TdT (terminal deoxynucleotidyl transferase) a unique enzyme found only in lymphocytes Since these bases are random, the amino acid sequence generated by these bases will also be random

51 Junctional Diversity V TCGACGTTATAT D J AGCTGCAATATA TTTTT TTTTT TTTTT Germline-encoded nucleotides Palindromic (P) nucleotides - not in the germline Non-template (N) encoded nucleotides - not in the germline Creates an essentially random sequence between the V region, D region and J region in heavy chains and the V region and J region in light chains.

52 Problems? 1. How is an infinite diversity of specificity generated from finite amounts of DNA? Combinatorial Diversity, genomic organisation and Junctional Diversity 2. How can the same specificity of antibody be on the cell surface and secreted? Use of alternative polyadenylation sites 3. How do V region find J regions and why don t they join to C regions? The rule 4. How does the DNA break and rejoin? Imprecisely to allow Junctional Diversity

53 Why do V regions not join to J or C regions? V H D H J H C IF the elements of Ig did not assemble in the correct order, diversity of specificity would be severely compromised 2x DIVERSITY Full potential of the H chain for diversity needs V- D-J-C joining - in the correct order 1x DIVERSITY Were V-J joins allowed in the heavy chain, diversity would be reduced due to loss of the imprecise join between the V and D regions

54 Additional Degrees of Variation Somatic hypermutation: Stimulated memory B cells accumulate small mutations on the V L or V H leading to affinity maturation to antigens that are frequently or chronically present Isotype switching

55 Somatic hypermutation FR1 CDR1 FR2 CDR2 FR3 CDR3 FR4 100 Variability Wu - Kabat analysis compares point mutations in Ig of different specificity. Amino acid No. What about mutation throughout an immune response to a single epitope? How does this affect the specificity and affinity of the antibody?

56 CDR1 CDR2 CDR3 CDR1 CDR2 CDR3 CDR1 CDR2 Hypermutation is T cell dependent Mutations focussed on hot spots (i.e. the CDRs) due to double stranded breaks repaired by an error prone DNA repair enzyme. CDR3 CDR1 CDR2 CDR3 Somatic hypermutation leads to affinity maturation Clone 1 Clone 2 Clone 3 Clone 4 Clone 5 Clone 6 Clone 7 Clone 8 Clone 9 Clone 10 Day 6 Day 8 Day 12 Day 18 Cells with accumulated mutations in the CDR are selected for high antigen binding capacity thus the affinity matures throughout the course of the response Deleterious mutation Beneficial mutation Neutral mutation Lower affinity - Not clonally selected Higher affinity - Clonally selected Identical affinity - No influence on clonal selection

57 Allelic Exclusion A single B cell can express only one V L and one V H allele to the exclusion of all others Both must be on the same member of the chromosome pair-either maternal or paternal The restriction of V L and V H expression to a single member of the chromosome pair is termed allelic exclusion The presence of both maternal and paternal allotypes in the serum reflects the expression of different alleles by different population of B cells

58 Allelic exclusion: only one chromosome is active in any one lymphocyte

59 Model to account for allelic exclusion: If one allele arranges nonproductively, a B cell still can rearrange the other allele productively; once a productive rearrangement( 33%) have occurred, the recombination machinery is turned off. ( the protein product acts as a signal to prevent further gene rearrangement)

60 Antibody isotype switching Throughout an immune response the specificity of an antibody will remain the same (notwithstanding affinity maturation) The effector function of antibodies throughout a response needs to change drastically as the response progresses. Antibodies are able to retain variable regions whilst exchanging constant regions that contain the structures that interact with cells. Organisation of the functional human heavy chain C region genes J regions Cm Cd Cg3 Cg1 Ca1 Cg2 Cg4 Ce Ca2

61 Switch regions Cm Cd Cg3 Cg1 Ca1 Cg2 Cg4 Ce Ca2 Sm Sg3 Sg1 Sa1 Sg2 Sg4 Se Sa2 Upstream of C regions are repetitive regions of DNA called switch regions. (The exception is the Cd region that has no switch region). The Sm consists of 150 repeats of [(GAGCT)n(GGGGGT)] where n is between 3 and 7. Switching is mechanistically similar in may ways to V(D)J recombination. Isotype switching does not take place in the bone marrow, however, and it will only occur after B cell activation by antigen and interactions with T cells.

62 7 means of generating antibody diversity

63 Generation of Antibody Diversity Germ line diversity. Combinatorial diversity. Junctional diversity. Somatic hypermutation ( affinity maturation)