The Molecular Genetics Of Immunoglobulins

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  • 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 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
  • 5. Proof of the Dreyer - Bennett hypothesis
    • Tools:
    • cDNA probes to distinguish V from C regions
    • Germline (e.g. placenta) and rearranged B cell DNA (e.g. from a myeloma B cell)
    C V V V V V V V V V Germline DNA
    • DNA restriction enzymes to fragment DNA
    C V V V V V Rearranged DNA
  • 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
  • 10. 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?
  • 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 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
  • 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?
    • 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?
    • 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 1-27 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
  • 20. Problems?
    • 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?
    • 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. 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
  • 27. Problems?
    • How do V region find J regions and why don’t they join to C regions?
    • How does the DNA break and rejoin?
    • 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
  • 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
  • 31. 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)
    Molecular explanation of the 12-23 rule 23-mer 12-mer
    • 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 V3 V4 V8 V7 V6 V5 V9 D J V1 D J V2 V3 V4 V8 V7 V6 V5 V9
  • 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.
  • 36. Problems?
    • How do V region find J regions and why don’t they join to C regions? The 12-23 rule
    • 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 does the DNA break and rejoin?
  • 37. 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 V 7 23 9 D 7 12 9 J V 7 23 9 7 23 9 7 12 9 D 7 12 9 J 7 23 9 7 12 9 V D J
  • 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
  • 44. Problems?
    • How do V region find J regions and why don’t they join to C regions? The 12-23 rule
    • How is an infinite diversity of specificity generated from finite amounts of DNA? Combinatorial Diversity, genomic organisation and Junctional Diversity
    • How can the same specificity of antibody be on the cell surface and secreted? Use of alternative polyadenylation sites
    • 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 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
  • 49. 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.
    • Upstream of C regions are repetitive regions of DNA called switch regions. (The exception is the C  region that has no switch region).
    C  2 C  C  4 C  2 C  1 C  1 C  3 C  C  S  3 S  1 S  1 S  2 S  4 S  S  2 S 
  • 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