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 germline and 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)

H

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 : chromosome 2 : chromosome 22

VH1 VH65VH2 DH1-------27 JH1-----6 C

Ig gene sequencing complicated the model

Structures of germline VL genes were similar for V, and V,However there was an anomaly between germline and

rearranged DNA:

Where do the extra 13 amino acids

come from?

CLVL

~ 95 ~ 100

L CLVL

~ 95 ~ 100

JL

Extra amino acids provided by one of a

small set of J or JOINING regions

L

CLVL

~ 208

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 V genes on the short arm of chromosome 2• 29 functional V genes with products identified• 87 pseudo V genes• 16 functional V genes - with no products

identified• 25 orphans V genes on the long arm of

chromosome 2• 5 J regions

V & J Loci:

• 105 V 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 V sequence)• 25 orphans on the long arm of chromosome 22• 4 J regions

V & J 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

GGGACAGGGGGC GlyGlnGly

GGGACAGGGGGC AspArgGly

Analysis of D regions from different antibodiesOne 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 combinationsD can be read in 3 frames: 5,520 x 3 = 16,560 combinations

29 Vx 5 J = 145 combinations30 Vx 4 J = 120 combinations

= 265 different light chainsIf 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 genesJUNCTIONAL 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 CLH1-123VH 1-123

L1-132V1-132 J 1-5 C

L1-105V1-105 C1 J1 C2 J2 C3 J3 C4 J4

Ig light chain gene rearrangement by somatic recombination

Germline

V J C

SplicedmRNA

Rearranged1° 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

V J C

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 CVH 1-123

Somatic recombination occurs at the level of DNA which can now be transcribed

C1 C2 C3 C4pAs

AAAAAhJ8 J9DV

Primary transcript RNA

C1 C2 C3 C4 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 fortwo different forms of the same protein?

Primary transcript RNA AAAAA

C

Polyadenylation site (secreted)

pAs

Polyadenylation site (membrane)

pAm

The constant region has additional, optional exons

C1 C2 C3 C4

Each H chain domain (& the hinge) encoded by

separate exons

h

Secretioncoding

sequence

Membranecoding

sequence

mRNAC1 C2 C3 C4 AAAAAh

Transcription

Membrane IgM constant region

C1 C2 C3 C41° transcriptpAm

AAAAAh

C1 C2 C3 C4DNA h

Membrane coding sequence encodes

transmembrane regionthat retains IgM in the

cell membrane

Fc

Protein

Cleavage & polyadenylation at pAm and RNA splicing

mRNA

Secreted IgM constant region

C1 C2 C3 C4 AAAAAh

C1 C2 C3 C4DNA h

Cleavage polyadenylation at pAs and RNA splicing

1° transcriptpAs

C1 C2 C3 C4

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

V 7 23 9

Sequencing up and down stream of V, D and J elementsConserved sequences of 7, 23, 9 and 12 nucleotides in an

arrangement that depended upon the locus

V 7 12 9 J7239

J7129

D7129 7 12 9

VH 7 23 9 JH7239

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 sequence

NONAMER - Separated fromthe heptamer by a 12 or 23

nucleotide spacer

VH 7 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 DNAof 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

DJ

712

9V47239

1.

DJ

712

9

V47239

3.

DJ7

129

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 J7129

V47239

2.

Heptamer ligation - signal joint formation

D JV47 12 97239

Fully recombined VDJ regions in same transcriptional orientationNo 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 9D7129 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 9TC CACAGTGAG GTGTCAC

AT GTGACACTA 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 9CACAGTGGTGTCAC

GTGACACCACTGTG

TCAG

ATTADJ

V TCAG

ATTA

UU

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 9CACAGTGGTGTCAC

7 12 9GTGACACCACTGTG

V TCAG U

DJ ATTA U

Heptamers are ligated by DNA ligase IV

V and D regions juxtaposed

V TCAG U D JAT

TA

U

V TCAG U D JAT

TA

U

Endonuclease cleaves single strand at random sites in V and D segment

V TC~GAAG 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.

V TCAG U D JAT

TA

U Regions to be joined are juxtaposed

The nicked strand ‘flips’ out

(Palindrome - A Santa at NASA)

Junctional Diversity – N nucleotide additions

V TC~GAAG 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

V TC~GAAG D JAT

TA~TA

CACACCTTA

TTCTTGCAA Complementary bases anneal

V D JDNA polymerases fill in the gaps with complementary nucleotides and DNA ligase IV joins the strands

TC~GAAG

ATTA~TA

CACACCTTA

TTCTTGCAA

D JTA~TAExonucleases nibble back free endsV TC~GACACACCTTA

TTCTTGCAA

V TCDTA

GTT AT AT

AG C

P-nucleotide and N-nucleotide addition during joining.

Generation of Antibody DiversityGeneration 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

TTTTTTTTTTTTTTT

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 hypermutationFR1 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 1Clone 2Clone 3Clone 4Clone 5Clone 6Clone 7Clone 8Clone 9Clone 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 selectedHigher affinity - Clonally selected

Identical affinity - No influence on clonal selection

Somatic hypermutation leads to affinity maturation

Hypermutation is T cell dependentMutations 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 C2CC4C2C1C1C3CC

Organisation of the functional human heavy chain C region genes

C2CC4C2C1C1C3CC

Switch regions

• The S 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.

S3 S1 S1 S2 S4 S S2S

• Upstream of C regions are repetitive regions of DNA called switch regions. (The exception is the C 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)