Transcript Document

The Molecular Genetics of Immunoglobulins
©Dr. Colin R.A. Hewitt
The molecular genetics of immunoglobulins
How can the bifunctional nature of antibodies be explained genetically?
Dreyer & Bennett (1965)
For a single isotype of antibody there may be:
• A single C region gene encoded in the GERMLINE and separate
from the V region genes
• Multiple choices of V region genes available
• A mechanism to rearrange V and C genes in the genome so that
they can fuse to form a complete Immunoglobulin gene.
This was genetic heresy as it violated the then accepted
notion that DNA was identical in every cell of an individual
Genetic models of the 1960’s were also
unable to explain:
•
How B cells shut down the Ig genes on just one of their chromosomes.
All other genes known at the time were expressed co-dominantly. B cells
expressed a light chain from one parent only and a heavy chain from one
parent only (evidence from allotypes).
•
A genetic mechanism to account for increased antibody affinity in an
immune response
•
How a single specificity of antibody sequentially switched isotype.
•
How the same specificity of antibody was secreted and simultaneously
expressed on the cell surface of a B cell.
Proof of the Dreyer - Bennett hypothesis
V
V
V
V
V
V
V
V
V
V
V
V C
V
Single germline C gene
separate from multiple V genes
C
Rearranging V and C genes
V
Aim: to show multiple V genes and rearrangement to the C gene
Proof of the Dreyer - Bennett hypothesis
V
V
V
V
V
V
V
V
V
V
V
Germline DNA
V
V C
C
Rearranged DNA
V
Tools:
• cDNA probes to distinguish V from C regions
• DNA restriction enzymes to fragment DNA
• Germline (e.g. placenta) and rearranged B cell DNA (e.g. from a myeloma B cell)
N.B. This example
describes events on
only ONE of the
chromosomes
V
V
V
V
V
Cut germline DNA with
restriction enzymes
V
C
V
V
V
V
C
V
V
V
V
V
Size fractionate
by gel
electrophoresis
V
V
V
V
V
C
A range of fragment
sizes is generated
V
C
V
V
V
V
V
V
Blot with a V
region probe
Blot with a C
region probe
Evidence for gene recombination
V
Cut myeloma B cell
V
DNA with restriction
enzymes
V
Size fractionate
by gel
electrophoresis
Blot with a V
region probe
V
VC
V
V
Blot with a C
region probe
V C
V
V
V C
Size fractionate
by gel
electrophoresis
V
V
Blot with a V
region probe
Blot with a C
region probe
V
C
V
V
V
V
V
V
V
V and C probes detect the same fragment
Some V regions missing
C fragment is larger cf germline
V
V
V
- compare the pattern of bands
with germline DNA
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:
VL
CL
~ 95aa
~ 100aa
L
LV
L
CL
~ 208aa
Where do the extra
13 amino acids
come from?
LV
L
~ 95aa
JL
CL
~ 100aa
Extra amino acids
provided by one of a
small set of J or
JOINING regions
Further diversity in the Ig heavy chain
L VH JH DH
CH
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
L VL
JL
CL
Each light chain requires two recombination events:
VL to JL and VLJL to CL
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
VH Locus: • 123 VH genes on chromosome 14
•
•
•
•
JH Locus:
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
• 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
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
GGGACAGGGGGC
GlyThrGlyGly
Frame 1
GGGACAGGGGGC
GlyGlnGly
Frame 2
GGGACAGGGGGC
AspArgGly
Frame 3
Diversity: Multiple germline genes
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
Estimates of combinatorial diversity
Using functional V D and J genes:
40 VH x 27 DH x 6JH = 6,480 combinations
D can be read in 3 frames: 6,480 x 3 = 19,440 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.
19,440 x 265 = 5,151,600 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?
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?
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?
Genomic organisation of Ig genes
(No.s include pseudogenes etc.)
LH1-123
VH 1-123
DH1-27
Lk1-132
Vk1-132
Ll1-105
Vl1-105
JH 1-9
Jk 1-5
Cl1 Jl1
Cl2 Jl2
Cm
Ck
Cl3 Jl3
Cl4 Jl4
Ig light chain gene rearrangement by somatic
recombination
Vk
Germline
Rearranged
1° transcript
Spliced mRNA
Jk
Ck
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
Non-productive rearrangement
Light chain has a second chance to make
a productive join using new V and J elements
Spliced mRNA transcript
Ck
Ig heavy chain gene rearrangement
VH 1-123
DH1-27
JH 1-9
Cm
Somatic recombination occurs at the level of DNA which can
now be transcribed
BUT:
Problems?
1. How is an infinite diversity of specificity generated from
finite amounts of DNA?
Combinatorial Diversity and genomic organisation can
account for some diversity
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?
Remember Lecture 1, Slide 1?
• 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?
The constant region has additional, optional exons
Cm
Primary transcript RNA
Each H chain domain (&
the hinge) encoded by
separate exons
Cm1
h
Cm2
AAAAA
Secretion
coding
sequence
Cm3
Polyadenylation
site (secreted)
pAs
Polyadenylation
site (membrane)
pAm
Cm4
Membrane
coding
sequence
Membrane IgM constant region
DNA
Cm1
h
Cm2
Cm3
Cm4
Transcription
1° transcript
pAm
Cm1
Cleavage &
polyadenylation at pAm
and RNA splicing
mRNA
h
Cm2
Cm3
Cm1 h Cm2 Cm3 Cm4
Cm4
AAAAA
AAAAA
Membrane coding
sequence encodes
transmembrane region
that retains IgM in the
cell membrane
Protein
Fc
Secreted IgM constant region
DNA
Cm1
h
Cm2
Cm3
Cm4
Transcription
1° transcript
pAs
Cm1
h
Cm2
Cm3
Cm4
AAAAA
Cleavage polyadenylation
at pAs and RNA splicing
mRNA
Cm1 h Cm2 Cm3 Cm4
Protein
AAAAA
Secretion coding
sequence encodes the
C terminus of soluble,
secreted IgM
Fc
The constant region has additional, optional exons
Cm
Primary transcript RNA
Each H chain domain (&
the hinge) encoded by
separate exons
Cm1
h
Cm2
AAAAA
Secretion
coding
sequence
Cm3
Polyadenylation
site (secreted)
pAs
Polyadenylation
site (membrane)
pAm
Cm4
Membrane
coding
sequence
RNA processing
Primary transcript RNA
pAs
V
D J8
Cm1
J9
Cm2
Cm3
D J8 Cm1 h Cm2 Cm3 Cm4
V
mRNA
h
Cm4
AAAAA
The Heavy chain mRNA is completed by splicing the VDJ
region to the C region
The H and L chain mRNA are now ready for translation
VL
JL
VH
DH JH
AAAAA
CL
h
CH
AAAAA
AAAAA
Problems?
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
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?
V, D, J flanking sequences
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
Vl
7
Vk
7
23
12
7
23
9
12
9
7
12
9
9
9
VH
9
D
23
7
12
9
7
Jl
7
Jk
9
23
7
JH
Recombination signal sequences (RSS)
HEPTAMER - Always contiguous with
coding sequence
9
VH 7
23
√
VH
9
7
12
23
7
D 7
9
9
12
9
9
12
7
D
NONAMER - Separated from
the heptamer by a 12 or 23
nucleotide spacer
7 JH
23
7
12
9
√
9
23
7
JH
12-23 RULE – A gene segment flanked by a 23mer RSS can only be linked to a segment
flanked by a 12mer RSS
Molecular explanation of the 12-23 rule
12-mer = one turn
23-mer = two turns
23
V7
Intervening DNA
of any length
9
12
9
7D J
Molecular explanation of the 12-23 rule
V4
V1
V8
V9
V3
V2
V7
V6
V3
V4
V2
V5
9
9
23-mer
• Heptamers and nonamers
align back-to-back
7
7
V7
V8
V9
12-mer
V1
V6
Loop of
intervening
DNA is
excised
DJ
• The shape generated by the
RSS’s acts as a target for
recombinases
V5
DJ
• An appropriate shape can not be formed if two 23-mer flanked elements
attempted to join (i.e. the 12-23 rule)
Junctional diversity
Mini-circle of DNA is
permanently lost from the
genome
9
7
V
7
12
23
9
9
23
Coding joint
7
7
12
9
Signal joint
DJ
VDJ
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.
Non-deletional recombination
V1
V1
V1
V3
V2
7
V2
V3
9
23
V4
23
V4
7
V9
DJ
9
V9
V4
Looping out works if all V
genes are in the same
transcriptional orientation
9
DJ
12
7
D J
How does recombination occur
when a V gene is in opposite
orientation to the DJ region?
9
12
7
DJ
Non-deletional recombination
9
23
7 V4
9 12 7 D J
1.
2.
9
7 V4
23
9
3.
V4 and DJ in opposite
transcriptional orientations
23
9
23
7 V4
7 V4
4.
9
23
7 V4
9 12 7 D J
2.
1.
9
9
23
7
12
9
23
9
V4
12
7
7
V4
D J
Heptamer ligation - signal
joint formation
7 D J
3.
V4
9
23
9
4.
9
12
23
7
7
D J
7
7
12
9
V to DJ ligation coding joint
formation
V4 D J
Fully recombined VDJ regions in same transcriptional orientation
No DNA is deleted
Problems?
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
3. How do V region find J regions and why don’t they join to C
regions?
The 12-23 rule
4. How does the DNA break and rejoin?
Steps of Ig gene recombination
V
7 23
V
7 23
D J
9 12 7
7 23
9 12 7
9
7 23
D J
The two RAG1/RAG 2 complexes
bind to each other and bring the V
region adjacent to the DJ region
9
9 12 7 9 12 7
V
9
Recombination activating
gene products, (RAG1 & RAG
2) and ‘high mobility group
proteins’ bind to the RSS
9
• 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
D J
Steps of Ig gene recombination
V
7
23
9
D J
9 12 7
V D J
D
J
9 12 7 7 23 9
V
A number of other proteins, (Ku70:Ku80,
XRCC4 and DNA dependent protein
kinases) bind to the hairpins and the
heptamer ends.
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.
Junctional diversity: P nucleotide additions
7 23
V
AT GTGACAC
J D TA CACTGTG
9
9 12 7
V
TC CACAGTG
AG GTGTCAC
7
7
9
23
12
9
The recombinase complex makes single
stranded nicks at random sites close to the
ends of the V and D region DNA.
TC
AG
TC CACAGTG
AG GTGTCAC
7
GTGACAC
CACTGTG
7
V V
AT
AT
J JDTA DTA
9
23
12
9
The 2nd strand is cleaved and hairpins form between
the complimentary bases at ends of the V and D
region.
D J
V3
V2
V4
CACAGTG
GTGTCAC
7
GTGACAC
CACTGTG
7
9
23
12
V5
9
V9
Heptamers are ligated by
DNA ligase IV
TC
AG
V
AT
J DTA
V
V8
V7
TC
AG
V and D regions juxtaposed
AT
TA
V6
D J
Generation of the palindromic sequence
V
V
V
TC
AG
TC
AG
TC~GA
AG
AT
TA
D J
AT
TA
D J
Regions to be joined are juxtaposed
Endonuclease cleaves single strand at
random sites in V and D segment
The nicked strand ‘flips’ out
AT
TA~TA
D J
The nucleotides that flip out, become
part of the complementary DNA strand
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.
(Palindrome - A Santa at NASA)
Junctional Diversity – N nucleotide additions
V
TC~GA CACTCCTTA
AT
AG
TTCTTGCAA
TA~TA
D J
Terminal deoxynucleotidyl transferase
(TdT) adds nucleotides randomly to
the P nucleotide ends of the singlestranded V and D segment DNA
V
TC~GA CACACCTTA
AT
AG
TTCTTGCAA TA~TA
D J
Complementary bases anneal
V
TC~GACACACCTTA
D J
Exonucleases nibble back free ends
V
TC
CACACCTTA
TC~GA
GTT ATAT
AT
AGC
TTCTTGCAA
TA
TA~TA
AG
D J
DNA polymerases fill in the gaps
with complementary nucleotides
and DNA ligase IV joins the strands
TTCTTGCAA TA~TA
Junctional Diversity
V
TCGACGTTATAT
AGCTGCAATATA D
J
TTTTT Germline-encoded nucleotides
TTTTT Palindromic (P) nucleotides - not in the germline
TTTTT 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?
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
3. How do V region find J regions and why don’t they join to C
regions?
The 12-23 rule
4. How does the DNA break and rejoin?
Imprecisely to allow Junctional Diversity
Why do V regions not join to J or C regions?
VH
DH
JH
C
IF the elements of Ig did not assemble in the correct order, diversity of
specificity would be severely compromised
2x
DIVERSITY
Full potential of the H
chain for diversity needs
V-D-J-C joining - in the
correct order
1x
DIVERSITY
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
Somatic hypermutation
FR1
CDR1 FR2 CDR2
FR3
CDR3
FR4
100
Variability
80
60
40
20
20
40
60
80
100
120
Amino acid No.
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?
Somatic hypermutation leads to affinity maturation
Day 6
Day 8
Day 12
Day 18
CDR3
CDR1
CDR2
CDR3
CDR1
CDR2
CDR3
CDR1
CDR2
CDR3
CDR1
CDR2
Clone 1
Clone 2
Clone 3
Clone 4
Clone 5
Clone 6
Clone 7
Clone 8
Clone 9
Clone 10
Cells with
accumulated
mutations in
the CDR are
selected for
high antigen
binding
capacity –
thus the
affinity
matures
throughout
the course of
the response
Deleterious mutation Lower affinity - Not clonally selected
Beneficial mutation Higher affinity - Clonally selected
Neutral mutation
Identical affinity - No influence on clonal selection
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.
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.
Organisation of the functional human heavy chain C region genes
J regions
Cm
Cd
Cg3
Cg1 Ca1 Cg2
Cg4
Ce
Ca2
Switch regions
Cm
Sm
Cd
Cg3
Sg3
Cg1
Sg1
Ca1
Sa1
Cg2
Sg2
Cg4
Sg4
Ce
Se
Ca2
Sa2
• Upstream of C regions are repetitive regions of DNA called
switch regions. (The exception is the Cd region that has no
switch region).
• 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.
Switch recombination
Cm
Cd
Cg3
Cg1
Ca1
Cg2
Cg4
Cd
Ce
Cd
Ca2
Sg3
Cg3
Cg3
Cm
Sg1
Cm
Cg1
VDJ
Cg3
VDJ
Ca1
VDJ
Ca1
VDJ
Cg3
VDJ
Ca1
VDJ
Ca1
IgG3 produced.
Switch from IgM
IgA1 produced.
Switch from IgG3
IgA1 produced.
Switch from IgM
At each recombination constant regions are deleted from the genome
An IgE - secreting B cell will never be able to switch to IgM, IgD, IgG1-4 or IgA1