IMMUNOLOGY
Bios 328
a textbook-based study of immunology
Spring 2003
http://www.lehigh.edu/~sk08/Courses/Bios328/mainpage.htm
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?
Ig proteins are specified by genetic
“cassettes”
•Light chains are specified by “variable” (V),
“joining (J), and “constant” (C) gene
segments (aka “cassettes”).
DNA rearrangement
and
alternative RNA spicing
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”).
DNA rearrangement
and
alternative RNA splicing
Another view….
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?
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…
Cassettes rearrange…
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)
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:
A mouse has:
134 V
H, 13 D
H, 4 J
H segments
85 V
, 4 J
segments
and 2 V
, 3 J
segments
Thus, a mouse has:
134 13 4 = 6968 heavy chains
85
4 = 340 kappa chain
and 2
3 = 6 lambda chains
6968
(340 + 6) = 2,410,928 antibodies
PART III (the HARD part…)
2.4
10
6
< 10
10
So, there must be additional mechanisms of diversity
other than “fusing” “cassettes”
How does a RAG enzyme work?
Junctional flexibility
The “hairpin loop”
Junctional flexibility, “P” nucleotides,
and “N” nucleotides are added to
CDR3
Somatic hypermutation
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
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?