ICL 2.0: TCR & Ig Diversity

Question 1

what does RAG stand for?

Answer 1

recombination activating gene

assembly of immunoglobin genes is tightly controlled and requires the recombination activating gene (RAG) gene products

Question 2

what does AID stand for?

Answer 2

activation-induced cytidine deaminase enzyme

Question 3

what does TdT stand for?

Answer 3

terminal deoxynucleotidyl transferase enzyme

Question 4

how is the adaptive immune system able to have such a specific response to an antigen?

Answer 4

the highly specialized immune response is produced because each B lymphocyte and each T lymphocyte has a different receptor able to recognize a distinct unique structure (called epitope) on a pathogen (antigen)

consequently, during an adaptive immune response only B and T lymphocytes bearing receptors that recognize the infecting pathogen are selected for the adaptive immune response

these selected B and T lymphocytes proliferate and differentiate into effector lymphocytes through the process called clonal selection and clonal expansion

Question 5

what is clonal selection?

Answer 5

the processes that select lymphocytes for proliferation and differentiation into effector lymphocytes

this is needed because during an adaptive immune response only lymphocytes bearing receptors that recognize the infecting pathogen are selected for the response

When clonal selection gets out of control, this is how you get lymphomas!

Question 6

how are lymphocytes formed?

Answer 6

a common progenitor called pluripotent hematopietic stem cells derive multidifferential stem cells, which derive lymphoid stem cells and subsequently lymphocytes

multidifferential stem cells derive also myloid progenitor cells which next give rise to red blood cells (erythrocytes), and platelets (megakarocytes)

the same myloid progenitor cells derive also neutrophil, eosinophil, and basophil lines

Question 7

how do B and T cells get receptor diversity?

Answer 7

it stems from the fact that the diversity of receptors for B and T cells are produced by mixing of gene fragments encode receptors

the assembly of these fragments may encode endless possibilities for the extensive diversity of immunoglobin (Ig) or B cell receptor (BCR) and T cell receptor (TCR)

Question 8

what is allelic exclusion?

Answer 8

allelic exclusion control ensures that each B cell expresses only one type of binding site because an antibody binding site is determined by the combination of a particular heavy and light chain

it protects a single receptor specific for each B cell clone

there’s two copies of each Ig gene cluster but clonal selection requires just one H and one L chain to be produces so that there’s only one receptor type per B cell

if there’s no allelic exclusion then there would be 4 light chains and 2 heavy

Question 9

what is the numbers paradox?

Answer 9

we think there’s about 10ˆ9 different antibodies, each with their own specific receptors

but the number of genes in the human genome is no more than 2x10ˆ4 genes

the calculation of required specificities and available number of genes suggests that there’s 50,000 more Ab proteins than genes in the cell….

so how is it possible to achieve such high diversity numbers with minimal number of available genes???

this can be accomplished through the process of gene fragments rearrangements in a fashion producing practically an infinite numbers of possible specificities

Question 10

what’s the structure of an antibody?

Answer 10

Y shaped - consists of two heavy chains and two light chains

the AA sequence of each H and L chain is known as a variable region or V region and a constant C region

the variability portion in V region is the reason for the great diversity of antigen-binding specificities because paired V regions of H and L chains form the antigen binding site

remaining parts of H and L chains have much more limited variations in amino acid sequence and are called the constant region

slide 11

Question 11

what forms the antigen binding site?

Answer 11

the variability portion in V region is the reason for the great diversity of antigen-binding specificities because paired V regions of H and L chains form the antigen binding site

Each B cell and each T cell recognizes only one epitope!

Question 12

what is the combinatorial joining of Ig gene segments theory?

Answer 12

one gene even when assembled from different fragments is producing one protein

extensive diversification of germline sequences and combinatorial joining in the evolution of immunoglobulin heavy chain and light chain produces multiple specificities

there are 3 fragments in H chain = V, D, and J and there are 2 fragments in L chain = V and J for rearrangement of the variable region

after rearrangement, mRNA is produced, spliced and transcribed into a protein

if there are 5 fragments of 2 regions and one fragment from each region would make a variable fragment; then you do the math and there should be 25 possibilities of recombination rearrangements

when one of these possibilities is selected B cell does not express other possibilities

Question 13

what is the combinatorial joining of Ig gene segments in light chains of Ig?

Answer 13

there are 2 fragments in the V region of L chain = V and J for rearrangement of the variable region

there are 35 Vϰ fragments with 5 J fragments = 175 possibilities

there are 30Vλ fragments with 4 J fragments = 120 possibilities

175+120 possibilities = 300 possibilities for the L chain

at the moment that one of these 300 possibilities is successfully selected, other options are not used

Question 14

what is the combinatorial joining of Ig gene segments in heavy chains of Ig?

Answer 14

there are 3 fragments in the variable region of H chain = V, D, and J

there are 40 V fragments, 23 D fragments, and 6 J fragments = 5520 possibilities

again, when one of these 5520 possibilities is successfully selected, other options are not used

Question 15

how does a B cell decide which recombination group to pick to make up its epitope?

Answer 15

a single B cell selects one of 5520 combinations of the H chain and one of 300 combinations for the L chain to encode these two fragments as its antigen (epitope) recognition site

Question 16

what does RSS stand for?

Answer 16

recombination signal sequences

Question 17

what does the RSS do?

Answer 17

RSS = recombination signal sequences

each V, D, or J gene segment in H chain and each V and J gene segment in L chain are flanked by the RSS

Question 18

what are the types of RSS?

Answer 18

1. a nanomer (9-bp) and heptamer (7-bp) separated by a 12 bp spacer
2. a nanomer (9-bp) and heptamer (7-bp) separated by a 23-bp spacer

the heptamer (7-bp) is located at the nicking site of the gene fragment and nanomer (9-bp) as a positioning site for enzyme involved in cutting

Question 19

which RSS type flanks H chain fragments?

Answer 19

V and J fragments are flanked by RSS with a 23-bp spacer

D fragments are flanked by RSS with a 12-bp spacer

Question 20

which RSS type flanks L chain fragments?

Answer 20

V fragments are flanked by RSS with a 23-bp spacer

J fragments are flanked by RSS with a 12-bp spacer

Question 21

what is the 12/23 rule?

Answer 21

only gene segments flanked by an RSS of different length can rearrange efficiently

the signaling system with 12-bp and 23-bp spacers create very effective mechanism for proper rearrangements as only a 23-bp spacer will connect to a 12-bp spacer. These two RSS signals create a 12/23 rule

only a pair of dissimilar spacer RSS signals will be efficiently recombined = this mechanism guarantees efficient rearrangements of L and H chains

this insures that there’s not V-V joining, for example

Question 22

what do RAG1 and RAG2 do?

Answer 22

the process of rearrangements must be controlled and executed by enzymes which are able to separate, shuffle and rejoin the V, D, and J fragments for H chain as well as V and J fragments for L chain –> RAG1 and RAG2 are at the ends of VDJ genes that separate, shuffle, and rejoin the VDJ genes

RAG1 and RAG2 encode enzymes that play an important role in the rearrangement and recombination of gene fragments of immunoglobulin and T cell receptor molecules during the process of recombination

the diversity of antibodies (or BCRs) and TCRs are achieved by VDJ recombination

they work as multi-subunit complexes to cleave a single double stranded DNA between the antigen receptor coding segment and RSS signal

Question 23

when are RAG1 and RAG2 active?

Answer 23

both RAG1 and RAG2 enzymes are active exclusively during the development process of B and T cells

Question 24

what happens when there’s a RAG1 or RAG2 mutation?

Answer 24

RAG mutations lead to a Severe combined ImmunoDeficiency = SCID

there is partial V(D)J recombinase activity in an inherited disorder called Omenn syndrome

this is an absence of circulating B cells and infiltration of skin with activated oligoclonal T cells

Question 25

what is Omenn syndrome?

Answer 25

an autosomal recessive form of severe combined immunodeficiency (SCID) characterized by erythroderma, desquamation, alopecia, chronic diarrhea, failure to thrive, lymphadenopathy, eosinophilia, hepatosplenomegaly

caused by amino acid substitution mutations in the RAG genes that reduce their activity to 1–25% of normal

Question 26

give an example of how RAG works

Answer 26

RAG enzymes flank a 23-bp spacer and a 12-bp spacer

one RAG complex binds to the 23-bp spacer and second RAG complex to the 12-bp spacer to attach RSS signal site containing a 12-bp spacer to RSS signal site containing a 23-bp spacer

this fulfills the 12/23 rule

the sequences are broken at the heptamer sequences

the fragment between joined ends is excised as a circle

Question 27

what is the combinatorial joining of heavy and light chains theory?

Answer 27

to calculate the possibilities for V regions rearrangements: 3 L variable chains with 4 H variable chains would produce 12 antibodies, each with a unique antigen specificity

but don’t forget there’s 5520 possible H-chain combination and 300 possible L chain combinations

so that gives you 1.7x10^6 possible Ab molecules!

also, there can be imprecise joining of V and J fragments in L chains as well as V,D, and J fragments for H chains which would produce additional diversity! aka changing a couple nucleotides at the joining site could give you extra combinations because AAs would change

Question 28

how does addition of untemplated nucleotides increase Ab diversity?

Answer 28

another way to increase diversity is addition of N-nucleotides or non-templated nucleotides producing an additional N-region diversification

this process is under the control of TdT

TdT adds N-nucleotides to the V, D and J exons during antibody gene recombination enabling the phenomenon of additional functional diversity

this process occurs more commonly in H-chains; and it occurs at amino (N) end of V, D or J segments during joining

N-nucleotides or non-templated nucleotides are believed to exist only to create diversity at VJ and VDJ recombination

Question 29

what does TdT do?

Answer 29

it’s an enzyme that’s a specialized DNA polymerase

it’s expressed in immature, pre-B and pre-T cells

Question 30

where is the process of non-templated nucleotides additions more likely to occur?

Answer 30

H chains

it occurs at amino (N) end of D or J segments during joining

Question 31

what are the three ways to increase Ab diversity?

Answer 31

1. combinatorial joining of H and L chains
2. imprecise VJ or VDJ joining = junctional diversification
3. N-region diversification
4. somatic hypermutation (B cells only**)

ALL THE SAME WAYS TO INCREASE DIVERSITY IS PRESENT IN B AND T CELLS DURING THEIR MATURATION!

Question 32

what is somatic hypermutation?

Answer 32

a process that allows B cells to mutate the genes that they use to produce antibodies specific for certain pathogens; it involves a programmed process of mutation affecting the variable regions in immunoglobulin genes

somatic hypermutation improves specificity of antibody = affinity

it is associated with affinity maturation of BCR and Ig produced by this B cell

somatic hypermutation requires the activity of AID**

unlike germline mutation somatic hypermutation affects only individual B cell and such mutations are NOT transmitted to the offspring!!!

SOMATIC HYPERMUTATION IS EXCLUSIVELY FOR B CELLS! NEVER FOR T CELLS!

Question 33

what does AID do in somatic hypermutation?

Answer 33

somatic hypermutation requires the activity of activation induced cytidine deaminase (AID)

AID is an enzyme that creates deliberate mutations in DNA by resining the amino group from a cytidine base which turns it into a uridine; which is recognized as a thymidine

aka it changes a CG base pair into a UG base pair and since the cell’s DNA repair machinery recognizes it as a TG base pair, it finally converts up as a TA base pair

so this causes mutations that produce antibody diversity but that same mutation can lead to B-cell lymphoma!!

Question 34

what does AID deficiency cause?

Answer 34

autosomal recessive form of the hyper-IgM syndrome (HIGM2)

characterized by normal or elevated serum IgM levels with absence of IgG, IgA, and IgE, resulting in profound susceptibility to bacterial infections

AID deficiency causes hyper IgM syndrome because you can’t delete anything and our bodies are just gonna produce the first coding region they come across so you can’t do class switching without AID

Question 35

what are CDRs?

Answer 35

CDR = complementarity determining regions

the CDRs within antibody or T cell receptor proteins are complementary to the antigen’s shape (they’re in the variable region of Abs)

CDR regions determine the protein’s affinity and specificity for a unique antigen

CDR regions are the most variable part of the antibody and TCR and they contribute to the diversity of these molecules. Such adjustments during the immune response allow the antibody to mount very specific response

Question 36

which genes does a B cell have?

Answer 36

each B cell expresses only one ϰ OR λ light chain gene

it also expresses only one heavy chain gene

ϰ = chromosome 2

λ = chromosome 22

H = chromosome 14

Question 37

what is class switching?

Answer 37

class switching (CSR) = isotope switching

switching puts the same antigen specificity
onto different constant regions with differing functions

it’s a biological mechanism that changes a B cell’s production of immunoglobulin from one type to another, such as from the isotype IgM to the isotype IgG

on chromosome 14 where the heavy chain is, there are 5 different regions that determine which class the antibody will be (IgG, IgM, IgD, IgA, or IgE)

we have different classes of antibodies because they function differently in order to provide us with the best protection against invading pathogens so we need to be able to make all of these types of antibodies

AID enzyme helps to delete the portion of DNA in between the switch sequences so that different classes of antibody can be made

IgM is always the first to be made because it’s coding region is first in the linear organization of DNA so in order for transcription factors to reach the regions that code for other types of antibodies, we need to be able to skip over the IgM portion or any antibody region that comes before whichever antibody we’re trying to make

AID deletes the preceding gene regions so we’re able to produce different kinds of antibodies

AID deficiency causes hyper IgM syndrome because you can’t delete anything and our bodies are just gonna produce the first coding region they come across so you can’t do class switching without AID

Question 38

what are the 5 Ig classes?

Answer 38

IgM

IgD

IgG

IgE

IgA

Question 39

what is the function of AID?

Answer 39

it’s a B cell-specific enzyme that induces DNA damage by cytidine deamination and accounts for the B cell predisposition to cancer-associated chromosome translocations

AID is the master regulator of secondary antibody diversification – involved in somatic hypermutation, class switch recombination and gene conversion

Question 40

how are the constant regions of the H chain diversified?

Answer 40

germ line arrangement of H chain constant region exons

there are secreted and membrane-bound forms

Question 41

how are T cell receptors different than B cell receptors?

Answer 41

TCRs are:

– monovalent

– have no secreted version

– have no equivalent of class switching or hypermutation

Question 42

how are T cell receptors the same as B cell receptors?

Answer 42

Like B cell receptors, TCRs:

– undergo V(D)J joining via RAG

– are subject to N-region diversification

– use variable regions on two chains to define specificity

– rearrange one chain first () and express with surrogate

Question 43

what is the structure of the TCR antigen recognition site?

Answer 43

TCR antigen recognition site is composed of TCRα and TCRβ chains

Question 44

which chromosome are the TCRα and TCRβ chain genes located?

Answer 44

TCRα = chromosome 14

TCRβ = chromosome 7

Question 45

what is the TCRα chain structure?

Answer 45

it’s similar to the L chain of antibody

it has two fragments: V and J

70-80 V fragments

61 J fragments =

4880 rearrangement possibilities

Question 46

what is the TCRβ chain structure?

Answer 46

it’s similar to the H chain of antibody

3 fragments = V, J, D

Question 47

how is the TCRαβ receptor produced?

Answer 47

rearrangement of TCRα and TCRβ chains

the TCRα- and TCRb-chain genes are composed of segments that are joined by somatic recombination during development of the T cell

for TCRα, a V gene segment rearranges to a J gene segment to create a functional V-region exon. Next, the transcription and splicing of the VJ exon to C exon generates the mRNA that is translated to yield the TCRα chain protein

for TCRβ, there are three fragments V, D, and J which are rearranged to generate a functional VDJ exon that is transcribed and spliced to join to a C fragment

the rearranged TCRalpha and TCRbeta mRNAs are translated to produce a single TCRalpha/beta receptor for a T cell