Immunoglobulins
Generation of Diversity
Introduction• Immunologist estimate that each person has
the ability to produce a range of individual
antibodies capable of binding to a total of well
over 1010 epitopes
• According to the germline theory, a unique
gene encodes each antibody
• Unfortunately, for this theory to be true the
number of antibody genes would need to be
100-1000-fold greater than the entire human
genome
Theories
• An alternative theory, the somatic mutation theory, holds that a single germline immunoglobulin gene undergoes multiple mutations that generate immunoglobulin diversity. This scheme, however, requires an unimaginable mutation rate
• The immune system has developed a much more elegant solution- the chromosomal rearrangement of separate gene segments, which employs some elements of the germlineand somatic mutation theories
Gene Rearrangement
• Each light and heavy chain is encoded by a series of
genes occurring in clusters along the chromosome
• In humans, the series of genes encoding the k light
chain, λ light chain, and the heavy chain are located
on chromosomes 2, 22, and 14 respectively
• When a cell becomes committed to the B lymphocyte
lineage, it rearranges the DNA, encoding its light and
heavy chains by cutting and splicing together some of
the DNA sequences, thus modifying the sequence of
the variable region gene
Tonegawa’s demonstration• 1976—used restriction enzymes and DNA probes to show that
germ cell DNA contained several smaller DNA segments compared to DNA taken from developed lymphocytes (myeloma cells)
Hk l
1 gene 1 transcript 1 protein
Antibody specificities more than 1,000,000,000,000
Human genome about 30,000 genes
Human Antibody genesH: chromosome 14 k: chromosome 2 l: chromosome 22
VH1 VH65VH2 DH1-------27 JH1-----6 Cg
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?
CLVL
~ 95aa ~ 100aa
L CLVL
~ 95aa ~ 100aa
JL
Extra amino acids
provided by one of a
small set of J or
JOINING regions
L
CLVL
~ 208aa
L
Further diversity in the Ig heavy chain
VL JL CLL
CHVH JH DHL
Heavy chain: between up to 8 additional amino acids between JH
and CH
The D or DIVERSITY region
Each light chain requires two recombination events:
VL to JL and VLJL to CL
Each heavy chain requires three recombination events:
JH to DH, JHDH to VH and JHDH VHto CH
Problems?
1. How is an infinite diversity of specificity generated
from finite amounts of DNA?
2. How can the same specificity of antibody be on the
cell surface and secreted?
3. How do V region find J regions and why don’t they
join to C regions?
4. How does the DNA break and rejoin?
Diversity: Multiple germline genes
• 132 Vk genes on the short arm of chromosome 2
• 29 functional Vk genes with products identified
• 87 pseudo Vk genes
• 16 functional Vk genes - with no products
identified
• 25 orphans Vk genes on the long arm of
chromosome 2
• 5 Jk regions
Vk & Jk Loci:
• 105 Vl genes on the short arm of chromosome
22
• 30 functional genes with products identified
• 56 pseudogenes
• 6 functional genes - with no products identified
• 13 relics (<200bp of Vl sequence)
• 25 orphans on the long arm of chromosome 22
• 4 Jl regions
Vl & Jl Loci:
Diversity: Multiple Germline Genes
• 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
VH Locus:
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
Reading D segment in 3 frames
GGGACAGGGGGCGlyThrGlyGly
GGGACAGGGGGCGlyGlnGly
GGGACAGGGGGCAspArgGly
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
Estimates of combinatorial diversity
Using functional V, D and J genes:
40 VH x 27 DH x 6JH = 5,520 combinations
D can be read in 3 frames: 5,520 x 3 = 16,560 combinations
29 Vk x 5 Jk = 145 combinations
30 Vl x 4 Jl = 120 combinations
= 265 different light chains
If H and L chains pair randomly as H2L2 i.e.
16,560 x 265 = 4,388,400 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
Problems?
2. How can the same specificity of antibody be on
the cell surface and secreted?
3. How do V region find J regions and why don’t
they join to C regions?
4. How does the DNA break and rejoin?
1. 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?
Genomic organisation of Ig genes(Numbers include pseudogenes etc.)
DH1-27 JH 1-9 Cm
LH1-123
VH 1-123
Lk1-132
Vk1-132 Jk 1-5 Ck
Ll1-105
Vl1-105 Cl1 Jl1 Cl2 Jl2 Cl3 Jl3 Cl4 Jl4
Ig light chain gene rearrangement by somatic
recombination
Germline
Vk Jk Ck
Spliced mRNA
Rearranged
1° transcript
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
Vk Jk Ck
Non-productive
rearrangement
Spliced mRNA transcript
Light chain has a second chance
to make a productive join using
new V and J elements
Ig heavy chain gene rearrangement
DH1-27 JH 1-9 CmVH 1-123
Somatic recombination occurs at the level of DNA
which can now be transcribed
Cm1 Cm2 Cm3 Cm4
pAs
AAAAAhJ8 J9DV
Primary transcript RNA
Cm1 Cm2 Cm3 Cm4 AAAAAhJ8DVmRNA
The Heavy chain mRNA is completed by splicing
the VDJ region to the C region
RNA processing
VL JL CL AAAAA
CH AAAAAhJHDHVH
The H and L chain mRNA are now ready for translation
Problems?
2. How can the same specificity of antibody be on
the cell surface and secreted?
3. How do V region find J regions and why don’t
they join to C regions?
4. How does the DNA break and rejoin?
1. How is an infinite diversity of specificity generated
from finite amounts of DNA?
Combinatorial Diversity and genomic organisation
can account for some diversity
• Cell surface antigen receptor on B cells
Allows B cells to sense their antigenic environment
Connects extracellular space with intracellular signalling
machinery
• Secreted antibody functions
Neutralisation
Arming/recruiting effector cells
Complement fixation
Remember These Facts?
How does the model of recombination allow for
two different forms of the same protein?
Primary transcript RNA AAAAA
Cm
Polyadenylation
site (secreted)
pAs
Polyadenylation
site (membrane)
pAm
The constant region has additional, optional exons
Cm1 Cm2 Cm3 Cm4
Each H chain domain (&
the hinge) encoded by
separate exons
h
Secretion
coding
sequence
Membrane
coding
sequence
mRNACm1 Cm2 Cm3 Cm4 AAAAAh
Transcription
Membrane IgM constant region
Cm1 Cm2 Cm3 Cm41° transcript
pAm
AAAAAh
Cm1 Cm2 Cm3 Cm4DNA h
Membrane coding
sequence encodes
transmembrane region
that retains IgM in the
cell membrane
Fc
Protein
Cleavage &
polyadenylation at pAm
and RNA splicing
mRNA
Secreted IgM constant region
Cm1 Cm2 Cm3 Cm4 AAAAAh
Cm1 Cm2 Cm3 Cm4DNA h
Cleavage polyadenylation
at pAs and RNA splicing
1° transcript
pAs
Cm1 Cm2 Cm3 Cm4
Transcription
AAAAAh
Secretion coding
sequence encodes the
C terminus of soluble,
secreted IgM
Fc
Protein
Alternative RNA processing generates transmembrane or secreted Ig
Secreted & membrane forms of the heavy chain by alternative ( differential ) RNA processing of primary transcript.
(a)
Synthesis, assembly, and secretion of the immunoglobulin molecule.
Problems?
3. How do V region find J regions and why don’t
they join to C regions?
4. How does the DNA break and rejoin?
1. How is an infinite diversity of specificity generated
from finite amounts of DNA?
Combinatorial Diversity and genomic organisation
accounts for some diversity
2. How can the same specificity of antibody be on
the cell surface and secreted?
Use of alternate polyadenylation sites
V, D, J flanking sequences
Vl 7 23 9
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
Vk 7 12 9 Jk7239
Jl7129
D7129 7 12 9
VH7 23 9 JH
7239
Gene rearrangements are made at recombination signal
sequences (RSS). RSSs are heptamer-nonamer sequences
Each RSS contains a conserved heptamer, a conserved nonamer and a
spacer of either 12 or 23 base pairs.
Generic light chain locus
Generic heavy chain locus
There is a RSS downstream of every V gene segment,
upstream of every J gene segment and flanking every D gene
segment
V JD
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
VH 7 23 9
D7129 7 12 9
JH7239
HEPTAMER - Always contiguous with
coding sequenceNONAMER - Separated from
the heptamer by a 12 or 23
nucleotide spacer
VH7 23 9
D7129 7 12 9
JH7239
√ √
1. Rearrangements only occur between segments on the same
chromosome.
2. A heptamer must pair with a complementary heptamer; a nonamer
must pair with a complementary nonamer.
3. One of the RSSs must have a spacer with 12 base pairs and the
other must be 23 base pairs (the 12/23 rule).
- RSS having a one-turn spacer can join only with RSS having a two-turn spacer
: one-turn / two-turn joining rule
- This ensures that V,D,J segments join in proper order & that segments of the
same type do not join each other.
- The enzymes recognizing RSS : recombination-activating genes.
( RAG-1, -2), lymphoid-specific gene products
23-mer = two turns 12-mer = one turn
Molecular explanation of the 12-23 rule
Intervening DNA
of any length23
V 97
12
D J79
23-mer
12-mer
Loop of
intervening
DNA is
excised
• Heptamers and nonamers
align back-to-back
• The shape generated by the
RSS’s acts as a target for
recombinases
7
9
97
V1 V2 V3 V4
V8V7
V6V5
V9 D J
V1 D J
V2
V3
V4
V8
V7
V6
V5
V9
• 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
V D J712
9
723
9
7 12 97239
V D J
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
Signal jointCoding joint
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
Non-deletional recombination
D J7129V47239
V1 7 23 9 D7129 J
How does recombination occur
when a V gene is in opposite
orientation to the DJ region?V4
D J7129V47239V4 and DJ in opposite
transcriptional orientations
V47239
1.
V47239
3.
V47239
2.
D J7129
V472394.
Non-deletional recombination
D J7129
V47239
1.
D J
V4
7129
7239
3.
V to DJ ligation -
coding joint
formation
D J
7129
V47239
2.
Heptamer ligation - signal
joint formation
D JV47 12 97239
Fully recombined VDJ regions in same transcriptional orientation
No DNA is deleted
4.
Problems?
3. How do V region find J regions and why don’t
they join to C regions?
The 12-23 rule
1. How is an infinite diversity of specificity generated
from finite amounts of DNA?
Combinatorial Diversity and genomic organisation
accounts for some diversity
2. How can the same specificity of antibody be on
the cell surface and secreted?Use of alternative polyadenylation sites
4. How does the DNA break and rejoin?
V 7 23 9
D7 12 9J
V 7 23 9
7 23 9
7 12 9
D7129 J
7 23 9
7 12 9
V
DJRecombination 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
DJ
7 23 9
7 12 9
A number of other proteins,
(Ku70:Ku80, XRCC4 and DNA
dependent protein kinases) bind
to the hairpins and the heptamer
ends.
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
72
39
71
29
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.
Steps of Ig gene recombination
7D 12 9J
Junctional diversity: P nucleotide additions
7V 23 9
D7 12 9J
V 7 23 9
TC CACAGTG
AG GTGTCAC
AT GTGACAC
TA CACTGTG
The recombinase complex makes single
stranded nicks at random sites close to the
ends of the V and D region DNA.
7D 12 9J
7V 23 9CACAGTG
GTGTCAC
GTGACAC
CACTGTG
TC
AG
AT
TADJ
VTC
AG
AT
TA
The 2nd strand is cleaved and hairpins form
between the complimentary bases at ends of the
V and D region.
V2V3
V4
V8
V7V6
V5
V9
7 23 9CACAGTG
GTGTCAC
7 12 9GTGACAC
CACTGTG
VTC
AG
DJAT
TA
Heptamers are ligated by
DNA ligase IV
V and D regions juxtaposed
VTC
AGD J
AT
TA
VTC
AG D JAT
TA
Endonuclease cleaves single
strand at random sites in V and D
segment
VTC~GA
AG D JAT
TA~TAThe nucleotides that flip out,
become part of the complementary
DNA strand
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.
VTC
AGD J
AT
TA Regions to be joined are juxtaposed
The nicked strand ‘flips’ out
(Palindrome - A Santa at NASA)
Junctional Diversity – N nucleotide additions
VTC~GA
AG D JAT
TA~TA
Terminal deoxynucleotidyl
transferase (TdT) adds
nucleotides randomly to the P
nucleotide ends of the single-
stranded V and D segment DNA
CACTCCTTA
TTCTTGCAA
VTC~GA
AG D JAT
TA~TA
CACACCTTA
TTCTTGCAAComplementary bases anneal
V D JDNA polymerases fill in the
gaps with complementary
nucleotides and DNA ligase IV
joins the strands
TC~GA
AGAT
TA~TA
CACACCTTA
TTCTTGCAA
D JTA~TAExonucleases nibble back free endsV
TC~GACACACCTTA
TTCTTGCAA
VTC
DTA
GTT AT AT
AG C
P-nucleotide and N-nucleotide addition during joining.
Generation of Antibody Diversity
P and N region nucleotide alteration
adds to diversity of V region
• During recombination some nucleotide bases are cut
from or add to the coding regions (p nucleotides)
• Up to 15 or so randomly inserted nucleotide bases are
added at the cut sites of the V, D and J regions (n
nucleotides_
• TdT (terminal deoxynucleotidyl transferase) a unique
enzyme found only in lymphocytes
• Since these bases are random, the amino acid
sequence generated by these bases will also be
random
V D JTCGACGTTATATAGCTGCAATATA
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.
Problems?
3. How do V region find J regions and why don’t
they join to C regions?
The 12-23 rule
1. How is an infinite diversity of specificity generated
from finite amounts of DNA?
Combinatorial Diversity, genomic organisation
and Junctional Diversity
2. How can the same specificity of antibody be on
the cell surface and secreted?
Use of alternative polyadenylation sites
4. How does the DNA break and rejoin?
Imprecisely to allow Junctional 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
DIVERSITY
2x
DIVERSITY
1x
VH DH JH C
Additional Degrees of Variation
• Somatic hypermutation: Stimulated memory B
cells accumulate small mutations on the VL or
VH leading to affinity maturation to antigens that
are frequently or chronically present
• Isotype switching
Somatic hypermutation
FR1 FR2 FR3 FR4CDR2 CDR3CDR1
Amino acid No.
Variability80
100
60
40
20
20 40 60 80 100 120
Wu - Kabat analysis compares
point mutations in Ig of different
specificity.
What about mutation throughout an immune response to a single
epitope?
How does this affect the specificity and affinity of the antibody?
Clone 1
Clone 2
Clone 3
Clone 4
Clone 5
Clone 6
Clone 7
Clone 8
Clone 9
Clone 10
CD
R1
CD
R2
CD
R3
Day 6
CD
R1
CD
R2
CD
R3
CD
R1
CD
R2
CD
R3
CD
R1
CD
R2
CD
R3
Day 8 Day 12 Day 18
Deleterious mutationBeneficial mutationNeutral mutation
Lower affinity - Not clonally selected
Higher affinity - Clonally selectedIdentical 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
Allelic Exclusion• A single B cell can express only one VL and one VH
allele to the exclusion of all others
• Both must be on the same member of the chromosome pair-either maternal or paternal
• The restriction of VL and VH expression to a single member of the chromosome pair is termed allelic exclusion
• The presence of both maternal and paternal allotypes in the serum reflects the expression of different alleles by different population of B cells
Allelic exclusion: only one chromosome is
active in any one lymphocyte
Model to account for allelic exclusion:If one allele arranges nonproductively, a B cell still can rearrange the other allele productively; once a productive rearrangement( 33%) have occurred, the recombination machinery is turned off. ( the protein product acts as a signal to prevent further gene rearrangement)
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 Ca2CeCg4Cg2Ca1Cg1Cg3CdCm
Organisation of the functional human heavy chain C region
genes
Ca2CeCg4Cg2Ca1Cg1Cg3CdCm
Switch regions
• The Sm 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.
Sg3 Sg1 Sa1 Sg2 Sg4 Se Sa2Sm
• Upstream of C regions are repetitive regions of DNA called
switch regions. (The exception is the Cd region that has no
switch region).
7 means of generating antibody
diversity
Generation of Antibody Diversity
• Germ line diversity.
• Combinatorial diversity.
• Junctional diversity.
• Somatic hypermutation ( affinity maturation)