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Immunoglobulins

Immunoglobulins. Generation of Diversity. 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 10 10 epitopes According to the germline theory , a unique gene encodes each antibody

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Immunoglobulins

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  1. Immunoglobulins Generation of Diversity

  2. 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 1010epitopes • According to the germline theory, a unique gene encodes each antibody • Unfortunately, for this theory to be true the number of antibody genes would need to be 100-1000-fold greater than the entire human genome

  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 klight 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. H k l Cg VH1 VH2 VH65 DH1-------27 JH1-----6 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. L VL JL CL L VL CL ~ 95aa ~ 100aa ~ 95aa ~ 100aa VL CL L Extra amino acids provided by one of a small set of J or JOINING regions ~ 208aa Ig gene sequencing complicated the model Structures of germline VL genes were similar for Vk, and Vl, However there was an anomaly between germline and rearranged DNA: Where do the extra 13 amino acids come from?

  8. L VL JL CL L VH JH DH CH Further diversity in the Ig heavy chain Heavy chain: between up to 8 additional amino acids between JH and CH The D or DIVERSITY region Each heavy chain requires three recombination events: JH to DH, JHDH to VH and JHDH VHto CH Each light chain requires two recombination events: VL to JL and VLJL to CL

  9. Problems? • How is an infinite diversity of specificity generated from finite amounts of DNA? • How can the same specificity of antibody be on the cell surface and secreted? • How do V region find J regions and why don’t they join to C regions? • How does the DNA break and rejoin?

  10. Vk & Jk Loci: • 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 Vl & Jl Loci: • 105 Vl genes on the short arm of chromosome 22 • 30 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 Diversity: Multiple germline genes

  11. VH Locus: • 123 VH genes on chromosome 14 • 40 functional VH genes with products identified • 79 pseudo VH genes • 4 functional VH genes - with no products identified • 24 non-functional, orphan VH sequences on chromosomes 15 & 16 JH Locus: • 9 JH genes • 6 functional JH genes with products identified • 3 pseudo JH genes DH Locus: • 27 DH genes • 23 functional DH genes with products identified • 4 pseudo DH genes • Additional non-functional DH sequences on the chromosome 15 orphan locus • reading DH regions in 3 frames functionally increases number of DH regions Diversity: Multiple Germline Genes

  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 GGGACAGGGGGC GlyGlnGly GGGACAGGGGGC AspArgGly Frame 1 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 H2L2 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 genesJUNCTIONAL diversity

  14. Problems? • 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? • How can the same specificity of antibody be on the cell surface and secreted? • How do V region find J regions and why don’t they join to C regions? • How does the DNA break and rejoin?

  15. Genomic organisation of Ig genes (Numbers include pseudogenes etc.) LH1-123 VH 1-123 DH1-27 JH 1-9 Cm Lk1-132 Vk1-132 Jk 1-5 Ck Ll1-105 Vl1-105 Cl1 Jl1 Cl2 Jl2 Cl3 Jl3 Cl4 Jl4

  16. Vk Jk Ck Germline Rearranged 1° transcript SplicedmRNA Ig light chain gene rearrangement by somatic recombination

  17. 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 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

  18. VH 1-123 DH1-27 JH 1-9 Cm Ig heavy chain gene rearrangement Somatic recombination occurs at the level of DNA which can now be transcribed

  19. Primary transcript RNA pAs D J8 J9 Cm1 h Cm2 Cm3 Cm4 V AAAAA D J8 Cm1 h Cm2 Cm3 Cm4 V mRNA AAAAA The H and L chain mRNA are now ready for translation AAAAA JL VL CL DH JH CH h VH AAAAA RNA processing The Heavy chain mRNA is completed by splicing the VDJ region to the C region

  20. Problems? • How is an infinite diversity of specificity generated from finite amounts of DNA?Combinatorial Diversity and genomic organisation can account for some diversity • How can the same specificity of antibody be on the cell surface and secreted? • How do V region find J regions and why don’t they join to C regions? • How does the DNA break and rejoin?

  21. 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?

  22. Cm Primary transcript RNA AAAAA Each H chain domain (& the hinge) encoded by separate exons Secretioncodingsequence Polyadenylation site (secreted) pAs Polyadenylation site (membrane) pAm Cm1 Cm2 Cm3 Cm4 Membranecodingsequence The constant region has additional, optional exons h

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

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

  25. Alternative RNA processing generates transmembrane or secreted Ig

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

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

  28. Problems? • How is an infinite diversity of specificity generated from finite amounts of DNA?Combinatorial Diversity and genomic organisation accounts for some diversity • How can the same specificity of antibody be on the cell surface and secreted?Use of alternate polyadenylation sites • How do V region find J regions and why don’t they join to C regions? • How does the DNA break and rejoin?

  29. Vl Jl 7 23 12 7 9 9 Vk JH Jk 9 9 7 12 23 23 7 7 9 D 12 7 7 12 9 9 VH 9 7 23 V, D, J flanking sequences Sequencing up and down stream of V, D and J elementsConserved sequences of 7, 23, 9 and 12 nucleotides in an arrangement that depended upon the locus

  30. 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.

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

  32. HEPTAMER - Always contiguous with coding sequence NONAMER - Separated fromthe heptamer by a 12 or 23 nucleotide spacer √ √ JH JH 9 9 23 23 7 7 D D 12 12 7 7 7 7 12 12 9 9 9 9 VH VH 9 9 7 7 23 23 Recombination signal sequences (RSS) 12-23 RULE – A gene segment flanked by a 23mer RSS can only be linked to a segment flanked by a 12mer RSS

  33. 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).

  34. 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

  35. 12-mer = one turn 23-mer = two turns Intervening DNA of any length 23 12 V 7 9 7 D J 9 Molecular explanation of the 12-23 rule

  36. V4 V5 V3 V1 V3 V4 V2 V6 V2 V5 V6 V7 V8 V7 9 V9 D J V8 V9 9 23-mer • Heptamers and nonamers align back-to-back • The shape generated by the RSS’s acts as a target for recombinases 12-mer 7 7 D J V1 Molecular explanation of the 12-23 rule Loop of intervening DNA is excised • An appropriate shape can not be formed if two 23-mer flanked elements attempted to join (i.e. the 12-23 rule)

  37. 9 23 7 7 12 9 9 9 23 Coding joint Signal joint 12 V D J 7 7 V D J Junctional diversity Mini-circle of DNA is permanently lost from the genome 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.

  38. Looping out works if all V genes are in the same transcriptional orientation V1 V3 V4 V9 V2 D D J J V1 D J 9 7 23 9 23 7 How does recombination occur when a V gene is in opposite orientation to the DJ region? 12 7 9 V1 V3 V4 V9 V2 V4 D J 12 7 9 Non-deletional recombination

  39. V4 and DJ in opposite transcriptional orientations 1. 2. D D D D J J J J 12 12 12 12 7 7 7 7 9 9 9 9 9 9 9 9 9 23 23 23 23 23 7 7 7 7 7 V4 V4 V4 V4 V4 3. 4. D J 12 7 9 Non-deletional recombination

  40. 1. 2. V4 Heptamer ligation - signal joint formation D J D J 12 12 12 7 7 7 9 9 9 12 7 3. 9 9 9 9 9 23 23 23 23 7 7 7 7 V4 V to DJ ligation - coding joint formation D J V4 4. V4 D J Fully recombined VDJ regions in same transcriptional orientation No DNA is deleted

  41. Problems? • How is an infinite diversity of specificity generated from finite amounts of DNA?Combinatorial Diversity and genomic organisation accounts for some diversity • How can the same specificity of antibody be on the cell surface and secreted?Use of alternative polyadenylation sites • How do V region find J regions and why don’t they join to C regions?The 12-23 rule • How does the DNA break and rejoin?

  42. V D J 9 7 23 12 7 9 V 9 7 23 J D 7 12 9 V 9 9 7 7 23 23 J D 7 7 12 12 9 9 Steps of Ig gene recombination 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

  43. V 9 7 23 J D 7 12 9 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. V V 9 7 23 J J D D 7 12 9 Steps of Ig gene recombination A number of other proteins, (Ku70:Ku80, XRCC4 and DNA dependent protein kinases) bind to the hairpins and the heptamer ends.

  44. 7 9 TCCACAGTG AG GTGTCAC V 23 V 9 7 23 J D 12 9 7 AT GTGACAC TA CACTGTG J D 7 12 9 TC AG V U J D AT TA U 7 9 TC AG CACAGTG GTGTCAC V 23 J D 12 9 7 AT TA GTGACAC CACTGTG Junctional diversity: P nucleotide additions The recombinase complex makes single stranded nicks at random sites close to the ends of the V and D region DNA. The 2nd strand is cleaved and hairpins form between the complimentary bases at ends of the V and D region.

  45. V3 7 9 CACAGTG GTGTCAC 23 V2 V4 12 9 7 GTGACAC CACTGTG V5 V9 V8 V6 V7 U U D J AT TA U TC AG TC AG V V U J D AT TA Heptamers are ligated by DNA ligase IV V and D regions juxtaposed

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

  47. CACACCTTA Complementary bases anneal TTCTTGCAA CACACCTTA TC~GA V D J TA~TA Exonucleases nibble back free ends TTCTTGCAA D D J J AT TA~TA AT TA~TA TC~GA AG TC~GA AG V V DNA polymerases fill in the gaps with complementary nucleotides and DNA ligase IV joins the strands TC CACACCTTA TC~GA AG V V D D J AT TA~TA AG C TTCTTGCAA TA GTTAT AT Junctional Diversity – N nucleotide additions Terminal deoxynucleotidyl transferase (TdT) adds nucleotides randomly to the P nucleotide ends of the single-stranded V and D segment DNA CACTCCTTA TTCTTGCAA

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

  49. 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

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