2. 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?

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

4. DNA rearrangement and alternative RNA spicing

5. 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”).

6. DNA rearrangement and alternative RNA splicing

7. Another view….

8. P ART 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?

9. 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…

10. Cassettes rearrange…

11. 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)

12. 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:

13. A mouse has: 134 V H, 13 D H, 4 J H segments 85 V 6, 4 J 6 segments and 2 V 8, 3 J 8 segments Thus, a mouse has: 134 C 13 C 4 = 6968 heavy chains 85 C 4 = 340 kappa chain and 2 C 3 = 6 lambda chains 6968 C (340 + 6) = 2,410,928 antibodies

14. P ART III (the HARD part…) 2.4 C 10 6 < 10 10 So, there must be additional mechanisms of diversity other than “fusing” “cassettes” How does a RAG enzyme work?

16. Junctional flexibility

17. The “hairpin loop”

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

19. Somatic hypermutation

20. 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

21. P ART 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?