1. The Molecular Genetics of Immunoglobulins

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

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

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

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? C L V L ~ 95  ~ 100  L C L V L ~ 95  ~ 100  J L Extra amino acids provided by one of a small set of J or JOINING regions L C L V L ~ 208  L

9. Further diversity in the Ig heavy chain 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 V L J L C L L C H V H J H D H L

12. Reading D segment in 3 frames GGGACAGGGGGC GlyThrGlyGly G GGACAGGGG GC GlyGlnGly GG GACAGGGGG C 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

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

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

17. Ig light chain gene rearrangement by somatic recombination Germline V  J  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  J  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 Somatic recombination occurs at the level of DNA which can now be transcribed BUT: D H 1-27 J H 1-9 C  V H 1-123

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

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

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

26. The Heavy chain mRNA is completed by splicing the VDJ region to the C region RNA processing C  1 C  2 C  3 C  4 pAs AAAAA h J 8 J 9 D V Primary transcript RNA C  1 C  2 C  3 C  4 AAAAA h J 8 D V mRNA V L J L C L AAAAA C H AAAAA h J H D H V H The H and L chain mRNA are now ready for translation

28. 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 V  7 23 9 V  7 12 9 J  7 23 9 J  7 12 9 D 7 12 9 7 12 9 V H 7 23 9 J H 7 23 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 V H 7 23 9 D 7 12 9 7 12 9 J H 7 23 9 HEPTAMER - Always contiguous with coding sequence NONAMER - Separated from the heptamer by a 12 or 23 nucleotide spacer V H 7 23 9 D 7 12 9 7 12 9 J H 7 23 9 √ √

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

32. 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 V D J 7 12 9 7 23 9 7 12 9 7 23 9 V D J Signal joint Coding joint

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

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

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

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

39. 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. 7 D 12 9 J 7 V 23 9 D 7 12 9 J V 7 23 9 TC CACAGTG AG GTGTCAC AT GTGACAC TA CACTGTG 7 D 12 9 J 7 V 23 9 CACAGTG GTGTCAC GTGACAC CACTGTG TC AG AT TA D J V TC AG AT TA U U

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

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

42. 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 V TC ~ GA AG D J AT TA ~ TA V TC ~ GA AG D J AT TA ~ TA CACACCTTA TTCT T GCAA Complementary bases anneal V D J DNA polymerases fill in the gaps with complementary nucleotides and DNA ligase IV joins the strands TC ~ GA AG AT TA ~TA CACACCTTA TTCT T GCAA D J TA ~ TA Exonucleases nibble back free ends V TC ~ GA CACACCTTA TTCT T GCAA V TC D TA GTT AT AT AG C

43. Junctional Diversity 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. V D J TC GA CGTT AT AT AG CT GCAA TA TA

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 V H D H J H C

46. Somatic hypermutation What about mutation throughout an immune response to a single epitope? How does this affect the specificity and affinity of the antibody? FR1 FR2 FR3 FR4 CDR2 CDR3 CDR1 Amino acid No. Variability 80 100 60 40 20 20 40 60 80 100 120 Wu - Kabat analysis compares point mutations in Ig of different specificity.

47. 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 Clone 1 Clone 2 Clone 3 Clone 4 Clone 5 Clone 6 Clone 7 Clone 8 Clone 9 Clone 10 CDR1 CDR2 CDR3 Day 6 CDR1 CDR2 CDR3 CDR1 CDR2 CDR3 CDR1 CDR2 CDR3 Day 8 Day 12 Day 18 Deleterious mutation Beneficial mutation Neutral mutation

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

50. 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 C  2 C  C  4 C  2 C  1 C  1 C  3 C  C  C  C  C  3 VDJ S  3 C  C  C  3 VDJ C  1 S  1 C  1 C  3 VDJ C  1 C  3 VDJ IgG3 produced. Switch from IgM VDJ C  1 IgA1 produced. Switch from IgG3 VDJ C  1 IgA1 produced. Switch from IgM