1. The Molecular Genetics of Immunoglobulins ©Dr. Colin R.A. Hewitt

2. The molecular genetics of immunoglobulins A single C region gene encoded in the GERMLINE and separate from the V region genes Multiple choices of V region genes available A mechanism to rearrange V and C genes in the genome so that they can fuse to form a complete Immunoglobulin gene. Dreyer & Bennett (1965) For a single isotype of antibody there may be: How can the bifunctional nature of antibodies be explained genetically? This was genetic heresy as it violated the then accepted notion that DNA was identical in every cell of an individual

3. Genetic models of the 1960’s were also unable to explain: How B cells shut down the Ig genes on just one of their chromosomes. All other genes known at the time were expressed co-dominantly. B cells expressed a light chain from one parent only and a heavy chain from one parent only (evidence from allotypes). A genetic mechanism to account for increased antibody affinity in an immune response How a single specificity of antibody sequentially switched isotype. How the same specificity of antibody was secreted and simultaneously expressed on the cell surface of a B cell.

4. Proof of the Dreyer - Bennett hypothesis V V V V V V V V V V V V V Rearranging V and C genes C V C Single germline C gene separate from multiple V genes Aim: to show multiple V genes and rearrangement to the C gene

5. Proof of the Dreyer - Bennett hypothesis Tools: cDNA probes to distinguish V from C regions C V V V V V V V V V Germline DNA Germline (e.g. placenta) and rearranged B cell DNA (e.g. from a myeloma B cell) DNA restriction enzymes to fragment DNA C V V V V V Rearranged DNA

6. VVV C VV VV VV Size fractionate by gel electrophoresis C V V V V V V V V V C V V V V V V V V V Cut germline DNA with restriction enzymes VV VVV VV VV C A range of fragment sizes is generated Blot with a V region probe Blot with a C region probe N.B. This example describes events on only ONE of the chromosomes

7. C V V V V V C V V V V V Size fractionate by gel electrophoresis VV VV C V Blot with a V region probe Blot with a C region probe Cut myeloma B cell DNA with restriction enzymes VVV Blot with a V region probe Blot with a C region probe C VV VV VV Size fractionate by gel electrophoresis - compare the pattern of bands with germline DNA V and C probes detect the same fragment Some V regions missing C fragment is larger cf germline VV VV C V Evidence for gene recombination

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

9. Further diversity in the Ig heavy chain VLVL JLJL CLCL L CHCH VHVH JHJH DHDH L Heavy chain: between 0 and 8 additional amino acids between J H and C H The D or DIVERSITY region Each light chain requires two recombination events: V L to J L and V L J L to C L Each heavy chain requires three recombination events: V H to J H, V H J H to D H and V H J H D H to C H

10. 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?

11. Diversity: Multiple Germline Genes 123 V H genes on chromosome 14 40 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 VH Locus: JH Locus: 9 J H genes 6 functional J H genes with products identified 3 pseudo J H genes DH Locus: 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 GGGACAGGGGGC GlyThrGlyGly GGGACAGGGGGC GlyGlnGly GGGACAGGGGGC AspArgGly 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 Frame 1 Frame 2 Frame 3

13. Diversity: Multiple germline genes 132 V  genes on the short arm of chromosome 2 29 functional V  genes with products identified 87 pseudo V  genes 15 functional V  genes - with no products identified 25 orphans V  genes on the long arm of chromosome 2 5 J  regions V  & J  Loci: 105 V genes on the short arm of chromosome 2 30 functional genes with products identified 56 pseudogenes 6 functional genes - with no products identified 13 relics (<200bp V of sequence) 25 orphans on the long arm of chromosome 2 4 J regions V & J Loci:

14. Estimates of combinatorial diversity Using functional V D and J genes: 40 VH x 27 DH x 6JH = 6,480 combinations D can be read in 3 frames: 6,480 x 3 = 19,440 combinations 29 V  x 5 J  = 145 combinations 30 V  x 4 J = 120 combinations = 265 different light chains If H and L chains pair randomly as H 2 L 2 i.e. 19,440 x 265 = 5,151,600 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

15. Problems? 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? 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?

16. Genomic organisation of Ig genes (No.s include pseudogenes etc.) D H 1-27J H 1-9 CC L H 1-123 V H 1-123 L  1-132 V  1-132 J  1-5CC L 1-105 V 1-105 C 1 J 1C 2 J 2C 3 J 3C 4 J 4

17. Ig light chain gene rearrangement by somatic recombination Germline VV JJ CC Spliced  mRNA Rearranged 1° transcript

18. 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 VV JJ CC Non-productive rearrangement Spliced mRNA transcript Light chain has a second chance to make a productive join using new V and J elements

19. Ig heavy chain gene rearrangement D H 1-27J H 1-9 CC V H 1-123 Somatic recombination occurs at the level of DNA which can now be transcribed BUT:

20. Problems? 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? 1.How is an infinite diversity of specificity generated from finite amounts of DNA? Combinatorial Diversity and genomic organisation can account for some diversity

21. Cell surface antigen receptor on B cells Allows B cells to sense their antigenic environment Connects extracellular space with intracellular signalling machinery Secreted antibody Neutralisation Arming/recruiting effector cells Complement fixation Remember Lecture 1, Slide 1? How does the model of recombination allow for two different forms of the protein?

22. Primary transcript RNAAAAAA CC Polyadenylation site (secreted) pAs Polyadenylation site (membrane) pAm The constant region has additional, optional exons C1C1C2C2C3C3C4C4 Each H chain domain (& the hinge) encoded by separate exons h Secretion coding sequence Membrane coding sequence

23. mRNA C1C1C2C2C3C3C4C4 AAAAA h Transcription Membrane IgM constant region C1C1C2C2C3C3C4C4 1° transcript pAm AAAAA h C1C1C2C2C3C3C4C4 DNA h Membrane coding sequence encodes transmembrane region that retains IgM in the cell membrane Fc Protein Cleavage & polyadenylation at pAm and RNA splicing

24. mRNA Secreted IgM constant region C1C1C2C2C3C3C4C4 AAAAA h C1C1C2C2C3C3C4C4 DNA h Cleavage polyadenylation at pAs and RNA splicing 1° transcript pAs C1C1C2C2C3C3C4C4 Transcription AAAAA h Secretion coding sequence encodes the C terminus of soluble, secreted IgM Fc Protein

25. Primary transcript RNAAAAAA CC Polyadenylation site (secreted) pAs Polyadenylation site (membrane) pAm The constant region has additional, optional exons C1C1C2C2C3C3C4C4 Each H chain domain (& the hinge) encoded by separate exons h Secretion coding sequence Membrane coding sequence

26. C1C1C2C2C3C3C4C4 pAs AAAAA h J8J8 J9J9 D V Primary transcript RNA C1C1C2C2C3C3C4C4 AAAAA h J8J8 D V mRNA The Heavy chain mRNA is completed by splicing the VDJ region to the C region RNA processing VLVL JLJL CLCL AAAAA CHCH h JHJH DHDH VHVH The H and L chain mRNA are now ready for translation

27. Problems? 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? 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

28. V, D, J flanking sequences V 723 9 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 VV 712 9 JJ 723 9 J 712 9 D 7 9 7 9 VHVH 723 9 JHJH 7 9

29. 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 VHVH 723 9 D 712 9 7 9 JHJH 723 9 HEPTAMER - Always contiguous with coding sequence NONAMER - Separated from the heptamer by a 12 or 23 nucleotide spacer VHVH 723 9 D 712 9 7 9 JHJH 723 9 √√

30. 23-mer = two turns 12-mer = one turn Molecular explanation of the 12-23 rule Intervening DNA of any length 23 V 97 12 DJ7 9

31. 23-mer 12-mer Loop of intervening DNA is excised Heptamers and nonamers align back-to-back The shape generated by the RSS’s acts as a target for recombinases 7 9 9 7 V1 V2 V3V4 V8 V7 V6 V5 V9 DJ V1 DJ V2 V3 V4 V8 V7 V6 V5 V9 An appropriate shape can not be formed if two 23-mer flanked elements attempted to join (i.e. the 12-23 rule) Molecular explanation of the 12-23 rule

32. V D J 7 12 9 7 23 9 712 9 723 9 V D J 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. Junctional diversity Mini-circle of DNA is permanently lost from the genome Signal joint Coding joint

33. V1 V2 V3V4 V9 DJ Looping out works if all V genes are in the same transcriptional orientation V1 V2 V3 V9 DJ Non-deletional recombination DJ 712 9 V4 723 9 V1 723 9 D 712 9 J How does recombination occur when a V gene is in opposite orientation to the DJ region? V4

34. DJ 712 9 V4 723 9 V4 and DJ in opposite transcriptional orientations D J 7 12 9 V4 723 9 1. D J 7 12 9 V4 723 9 3. D J 7 12 9 V4 723 9 2. DJ 712 9 V4 723 9 4. Non-deletional recombination

35. DJ 712 9 V4 723 9 1. DJ V4 712 9 723 9 3. V to DJ ligation - coding joint formation DJ 712 9 V4 723 9 2. Heptamer ligation - signal joint formation DJ V4 712 9 723 9 Fully recombined VDJ regions in same transcriptional orientation No DNA is deleted 4.

36. Problems? 3.How do V region find J regions and why don’t they join to C regions? The 12-23 rule 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 4.How does the DNA break and rejoin?

37. V 723 9 D 712 9 J V 723 9 7 9 712 9 D 7 9 J 723 9 712 9 V 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 Steps of Ig gene recombination

38. V D J 723 9 712 9 A number of other proteins, (Ku70:Ku80, XRCC4 and DNA dependent protein kinases) bind to the hairpins and the heptamer ends. 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 V D J 7 23 9 7 12 9 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. Steps of Ig gene recombination

39. 7 D 12 9 J Junctional diversity: P nucleotide additions 7 V 23 9 D 712 9 J V 723 9 TC CACAGTG AG GTGTCAC AT GTGACAC TA CACTGTG The recombinase complex makes single stranded nicks at random sites close to the ends of the V and D region DNA. 7 D 12 9 J 7 V 23 9 CACAGTG GTGTCAC GTGACAC CACTGTG TC AG AT TA DJ V TC AG AT TA U U The 2nd strand is cleaved and hairpins form between the complimentary bases at ends of the V and D region.

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

41. V TC AG U DJ AT TA U Endonuclease cleaves single strand at random sites in V and D segment V TC~GA AG DJ AT TA~TA The nucleotides that flip out, become part of the complementary DNA strand Generation of the palindromic sequence 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. V TC AG U DJ AT TA U Regions to be joined are juxtaposed The nicked strand ‘flips’ out (Palindrome - A Santa at NASA)

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

43. V DJ TCGACGTTATAT AGCTGCAATATA Junctional Diversity 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.

44. Problems? 3.How do V region find J regions and why don’t they join to C regions? The 12-23 rule 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 4.How does the DNA break and rejoin? Imprecisely to allow Junctional Diversity

45. Why do V regions not join to J or C regions? IF the elements of Ig did not assemble in the correct order, diversity of specificity would be severely compromised Full potential of the H chain for diversity needs V-D-J-C joining - in the correct order 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 DIVERSITY 2x DIVERSITY 1x VHVH DHDH JHJH C

46. Somatic hypermutation FR1FR2FR3FR4CDR2CDR3CDR1 Amino acid No. Variability 80 100 60 40 20 406080100120 Wu - Kabat analysis compares point mutations in Ig of different specificity. What about mutation throughout an immune response to a single epitope? How does this affect the specificity and affinity of the antibody?

47. Clone 1 Clone 2 Clone 3 Clone 4 Clone 5 Clone 6 Clone 7 Clone 8 Clone 9 Clone 10 CDR1CDR2CDR3 Day 6 CDR1CDR2CDR3CDR1CDR2CDR3CDR1CDR2CDR3 Day 8 Day 12 Day 18 Deleterious mutation Beneficial mutation Neutral mutation Lower affinity - Not clonally selected Higher affinity - Clonally selected Identical affinity - No influence on clonal selection Somatic hypermutation leads to affinity maturation 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. Cells with accumulated mutations in the CDR are selected for high antigen binding capacity – thus the affinity matures throughout the course of the response

48. 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. J regions C2C2CC C4C4C2C2C1C1C1C1C3C3CC CC Organisation of the functional human heavy chain C region genes

49. C2C2CC C4C4C2C2C1C1C1C1C3C3CC CC Switch regions The S  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. S3S3S1S1S1S1S2S2S4S4SS S2S2 SS Upstream of C regions are repetitive regions of DNA called switch regions. (The exception is the C  region that has no switch region).

50. C2C2CC C4C4C2C2C1C1C1C1C3C3CC CC CC CC C3C3 VDJ S3S3 CC CC C3C3 C1C1 S1S1 C1C1 C3C3 C1C1 C3C3 IgG3 produced. Switch from IgM VDJ C1C1 IgA1 produced. Switch from IgG3 VDJ C1C1 IgA1 produced. Switch from IgM Switch recombination At each recombination constant regions are deleted from the genome An IgE - secreting B cell will never be able to switch to IgM, IgD, IgG1-4 or IgA1