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P ART I

P ART I. Immnoglobulins are proteins Proteins are specified by genes There are too few genes to specify all the antibodies. i. e. , ~32,000 genes < 10,000,000,000 Ab’s How is Ig diversity specified genetically?. Ig proteins are specified by genetic “cassettes”.

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P ART I

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  1. PART I • Immnoglobulins are proteins • Proteins are specified by genes • There are too few genes to specify all the antibodies. • i. e., ~32,000 genes < 10,000,000,000 Ab’s • How is Ig diversity specified genetically?

  2. Ig proteins are specified by genetic “cassettes” • Light chains are specified by “variable” (V), “joining (J), and “constant” (C) gene segments (aka “cassettes”).

  3. DNA rearrangementandalternative RNA spicing

  4. Ig proteins are specified by genetic “cassettes” • Heavy chains are specified by “variable” (V), “diversity” (D), “joining (J), and “constant” (C) gene segments (aka “cassettes”).

  5. DNA rearrangementandalternative RNA splicing

  6. Another view….

  7. PART II • Cassettes rearrange… • How does this happen? • How do you get one “V” fusing to one “J” (in a light chain)? • In a heavy chain, a “D” fuses with a “J”; then the fused DJ cassette fuses with a “V” cassette… • The orderliness of this process implies that there are genetic instructions. What are they?

  8. Cassettes rearrange… • The heptamer is a palindrome • (i.e., it exhibits two-fold rotational symmetry.) • The nonamer is AT-rich • “Turns” refer to the DNA helix…

  9. Cassettes rearrange…

  10. One turn – two turn rule… • one turn and two turn are “recombination signal sequences” • one turn only reacts with a two turn • Recombination signal sequences are the substrates of enzymes RAG-1 and RAG-2 (“RAG” = recombination-activating gene)

  11. So… cassettes are marked by RSS(i.e., they are substrates for recombination.) Thus, cassettes can be fused. What is the consequence? Look at mouse:

  12. A mouse has: 134 VH, 13 DH, 4 JH segments 85 V6, 4 J6 segments and2 V8, 3 J8 segments Thus, a mouse has: 134 C 13 C 4 = 6968 heavy chains 85 C 4 = 340 kappa chain and2 C 3 = 6 lambda chains 6968 C (340 + 6) = 2,410,928 antibodies

  13. PART III (the HARD part…) 2.4 C 106 < 1010 So, there must be additional mechanisms of diversity other than “fusing” “cassettes” How does a RAG enzyme work?

  14. Junctional flexibility

  15. The “hairpin loop”

  16. Junctional flexibility, “P” nucleotides, and “N” nucleotides are added to CDR3

  17. Somatic hypermutation

  18. One turn – two turn rule… • one turn only reacts with a two turn • crossover between direct repeats (same transcriptional orientation) leads to deletion • crossover between indirect repeats leads to inversion

  19. PART IV • How do immunoglobulins assemble? • Some immunoglobulins are in the surface membrane of immature B-cells while other immunoglobulins of the same idiotype are secreted by mature B-cells. What’s the difference? • Similarly, identical variable regions can be shared among different isotypes. How? • B-cells are diploids with two sets of genetic instructions. How does just one set get expressed?

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