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The Molecular Genetics of Immunoglobulins

The Molecular Genetics of Immunoglobulins. Recall Structure . Numerous V region genes are preceded by Leader or signal sequences (60-90 bp ) exons interspersed with introns .

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The Molecular Genetics of Immunoglobulins

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  1. The Molecular Genetics of Immunoglobulins

  2. Recall Structure • Numerous V region genes are preceded by Leader or signal sequences (60-90 bp) exons interspersed with introns. • Heavy chain contains V (Variable), D (Diversity), J (Joining) and C (Constant) region gene segments. V - D - J – C • Light chain contains V, J, and C region gene segments. V - J - C • Constant region genes are sub-divided into exons encoding domains (CH1,CH2, CH3, CH4)

  3. CHARACTERISTICS OF IMMUNOGLOBULIN GENERE-ARRANGEMENT 1. Involves Allelic Exclusion. – Only one of two parental alleles of Ig is expressed in a B cell. – Either kappa or lambda light chain is expressed by a B cell (light chain isotype exclusion). 2. Ig rearrangement occurs prior to antigen exposure. A. Heavy chain re-arrangement – Re-arrangement occurs in a precise order: – Heavy chain re-arranges before Light chain. – D-J joining occurs first to form DJ and is followed by V-DJ joining to form VDJ. – Just as in light chain the production of μ heavy chain by re-arrangement of one allele inhibits re-arrangement on other allele. If re-arrangement on first allele is non-productive (due to mutations, deletions or frame shifts that generate stop codons), then re-arrangement on the second allele is stimulated.

  4. Allelic exclusion: only one chromosome is active in any one lymphocyte

  5. Light chain re-arrangement • Kappa chain (κ) rearranges before lambda (λ) chain V-joining occurs. • Productive arrangement on one allele blocks re-arrangement on other allele. • If kappa protein is produced, re-arrangement of lambda chain is blocked. • Otherwise lambda chain undergoes re-arrangement.

  6. Questions? • 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?

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

  8. V V Germline DNA V V V V V V V C V V V C Rearranged DNA V V • DNA restriction enzymes to fragment DNA 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)

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

  10. V C V C V V V C V V Size fractionate by gel electrophoresis Blot with a V region probe Blot with a C region probe Size fractionate by gel electrophoresis Blot with a V region probe Blot with a C region probe V V V V V V V V V C V V C V V V V V V V V V - compare the pattern of bands with germline DNA V V Evidence for gene recombination 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

  11. 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?

  12. L VL JL CL L VH JH DH CH Further diversity in the Ig heavy chain Heavy chain: between 0 and 8 additional amino acids between JH and CH The D or DIVERSITY region Each heavy chain requires three recombination events: VH to JH, VHJH to DH and VHJHDH to CH Each light chain requires two recombination events: VL to JL and VLJL to CL

  13. 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

  14. Vk & Jk Loci: • 132 Vk genes on the short arm of chromosome 2 • 29 functional Vk genes with products identified • 87 pseudo Vk genes • 15 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 2 • 30 functional genes with products identified • 56 pseudogenes • 6 functional genes - with no products identified • 13 relics (<200bp Vl of sequence) • 25 orphans on the long arm of chromosome 2 • 4 Jl regions Diversity: Multiple germline genes

  15. 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 Genomic organisation of Ig genes (No.s include pseudogenes etc.)

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

  17. Questions? • 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?

  18. Remember • 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 How does the model of recombination allow for two different forms of the protein?

  19. 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

  20. 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

  21. 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

  22. VH DH JH C 2x 1x DIVERSITY DIVERSITY 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

  23. Rearrangement of V, D, and J gene segments is guided by flanking DNA sequences

  24. 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

  25. 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

  26. 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

  27. 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)

  28. 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

  29. 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.

  30. 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.

  31. 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

  32. 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.

  33. 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

  34. V D J TCGACGTTATAT AGCTGCAATATA Junctional Diversity Germline-encoded nucleotides Palindromic (P) nucleotides - 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.

  35. Problems? • 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 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?Imprecisely to allow Junctional Diversity

  36. Variable addition and subtraction of nucleotides at the junctions between gene segments contributes to diversity in the third hypervariable region • Of the three hypervariable loops in the protein chains of immunoglobulins, two are encoded within the V gene segment DNA. The third (HV3 or CDR3) falls at the joint between the V gene segment and the J gene segment, and in the heavy chain is partially encoded by the D gene segment. • In both heavy and light chains, the diversity of CDR3 is significantly increased by the addition and deletion of nucleotides at two steps in the formation of the junctions between gene segments. The added nucleotides are known as P-nucleotidesandN-nucleotides

  37. As the total number of nucleotides added by these processes is random, the added nucleotides often disrupt the reading frame of the coding sequence beyond the joint. • Such frameshifts will lead to a nonfunctional protein, and DNA rearrangements leading to such disruptions are known as nonproductive rearrangements. • As roughly two in every three rearrangements will be nonproductive, many B cells never succeed in producing functional immunoglobulin molecules, and junctional diversity is therefore achieved only at the expense of considerable wastage.

  38. Imprecise joining generates diversity

  39. Some rearrangements are productive, others are non-productive: frame shift alterations are non-productive 1 in 3 in phase VL to VJ 1 in 3 in phase VH to DHJH Only 11% of cells mature and leave

  40. 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.

  41. 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

  42. 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

  43. 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

  44. Rearrangement of V, D, and J gene segments is guided by flanking DNA sequences • A system is required to ensure that DNA rearrangements take place at the correct locations relative to the V, D, or J gene segment coding regions. • V gene segment joins to a D or J and not to another V. • DNArearrangements are in fact guided by conserved noncoding DNA sequences that are found adjacent to the points at which recombination takes place. • These sequences consist of a conserved block of seven nucleotides—the heptamer 5′CACAGTG3′—which is always contiguous with the coding sequence, followed by a nonconserved region known as the spacer, which is either 12 or 23 nucleotides long. • This is followed by a second conserved block of nine nucleotides—the nonamer 5′ACAAAAACC3′ . • The spacer varies in sequence but its conserved length corresponds to one or two turns of the DNA double helix. • This brings the heptamer and nonamer sequences to the same side of the DNA helix, where they can be bound by the complex of proteins that catalyzes recombination. The heptamer-spacer-nonamer is called a recombination signal sequence (RSS).

  45. 12/23 Rule • Recombination only occurs between gene segments located on the same chromosome. • It generally follows the rule that only a gene segment flanked by a RSS with a 12-base pair (bp) spacer can be joined to one flanked by a 23 bp spacer RSS. This is known as the 12/23 rule. • For the heavy chain, a DH gene segment can be joined to a JH gene segment and a VH gene segment to a DH gene segment, but VHgene segments cannot be joined to JHgene segments directly, as both VH and JHgene segments are flanked by 23 bp spacers and the DHgene segments have 12 bp spacers on both sides

  46. The diversity of the immunoglobulin repertoire is generated by four main processes • Antibody diversity is generated in four main ways. • The gene rearrangement that combines two or three gene segments to form a complete V-region exon generates diversity in two ways. • First, there are multiple different copies of each type of gene segment, and different combinations of gene segments can be used in different rearrangement events. This combinatorial diversity is responsible for a substantial part of the diversity of the heavy- and light-chain V regions. • Second, junctional diversity is introduced at the joints between the different gene segments as a result of addition and subtraction of nucleotides by the recombination process. • A third source of diversity is also combinatorial, arising from the many possible different combinations of heavy- and light-chain V regions that pair to form the antigen-binding site in the immunoglobulin molecule. • Somatic mutation

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