Immunology Test II

We take a Myeloma cell and an embryonic cell. From both, we isolate the DNA. Form the myeloma cell we isolate the mRNA for κ light chain mRNA and radiolabel it. This mRNA will later be used as probes for blotting. Take the DNA that we derived and digest it with restriction endonucleases. Take the restriction fragments and electrophoreses them. Probe with mRNA for κ light chain. The Embryonic DNA shows two bands, while the Myeloma DNA shows only one band.
This tells us that the embryonic cell DNA for κ light chain is longer and has two splice sites for the endonuclease while the Myeloma DNA is shorter and one of the restriction sites for the endonuclease has been spliced out of the DNA.

Therefore: There is rearrangement of genes at the DNA level.

The λ light chain DNA is located on human chromosome 22 and mouse chromosome 16.

The κ light chain DNA is located on human chromosone 2 and on mouse chromosome 6. ??

The 5` end and the 3` end of the nucleotide are conserved because of the splicing mechanism.

Variable genes: two of them, not super variable
Conserved genes
J genes: Joining regions, part of the variable end.

These regions are not clustered together.

All the variable alleles are at the 5` end. There are 85 of them.
There are 5 J alleles and one pseudogene. They are all together.
Any one of the 85 V alleles can join any one of the 4 J alleles.

There are 134 Variable alleles
There are 4 Joining alleles
There are 12 D (diversity) alleles

The 3` end contains exons that code for isotypes

1. V-J joining: Randomly choose one v and one J chain and splice out everything in the middle. If splicing goes wrong on one chromosome, you have another chromosome to play with. If it goes wrong on both chromosome the B cell will die.
2. Make Primary RNA transcript: In the primary transcript the 5` of the V chain is missing. After Splicing of the DNA enhancers are closer to the promoter but only to the closest Variable region. At the 3' end of the Ck region there is a transcription termination signal. So we get a primary transcript with only one leader sequence, one variable region (the one that was closest to the splice) and one or more Js as well as A Ck region.
3. Splicing of Primary RNA to get mRNA: The splicesome only recognizes the one J that is connected to the V region. It splices out all other Js, treating them as introns. The intron between the Leader sequence and V is spliced out. The mRNA now has only one L, one V, one J, and one Ck.
4. Poly A tail added to the mRNA
5. Translation
6. L sequence is spliced off before the polypeptide goes to the Golgi for processing.
7. We now have a light chain that is able to be secreted or membrane bound, depending on the treatment of the leader sequence.

1. There is a D to J splice at the DNA level.
2. There is a V to DJ splice at the DNA level.
3. Primary Transcript: Only V region most 3` is transcribed. There is one L, one V, one D, one or more J's, and two C's, Cμ and Cδ The C regions give the antibody their isotype. Naive B cells have only the IgM and IgD. Transcription is stopped after IgD. IgM does not have a transcription stop.
4. mRNA undergoes alternative splicing to produce either a transcript with IgM (with 3` Cμ) or IgD (with 3` Cδ). Both can coexist within a cell.

RSS are signal sequences that bring together V and J allels in light chains or D and J alleles in Heavy chains.
The RSS can be either one turn in length or two turn in length.
one turn has:5' conserved nonamer
3' conserved heptamer
12 bp in the middle
two turn has 5' conserved heptamer
3' conserved nonamer
23 bp in the middle
The nonamers can base pair together. The only way to splice is to recognize a two turn and one turn RSS.
RAG genes: recombination activation genes code for two enzymes RAG 1 and Rag 2 that do the DNA splicing.

RSS's (one one turn and one two turn)
RAG1 and/or RAG2

RAG genes: recombination activation genes code for two enzymes RAG 1 and Rag 2 that do the DNA splicing.
Rags recognize RSS's

Knock out one of the RAGs you till get splicing. So both do the same work.

1. RSS is recognized by Rag 1 and Rag 2
2. A single strand cleavage by rag 1 and rag 2 is made. This is made possible because the DNA twists around so that the two RSS's are above one another.
3. A double stranded cut is made and the chromosome is cut. There exist now three pieces of dna. One with the J allele, one with the V allele, and one with the RSS's and the undeeded portion of the dna. This is a dangerous step.
4. The chromosome is rejoined together.
i. A hairpin loop at the end of each of the two pieces was formed. The loop is made with a regular phosphodiester bond.
ii. The hairpin is nicked and straightened, then filled in.
1) This is when P nucleotide addition can take place. The nicking and filling in can make a palendromic sequence that was not present before.
2) N nucleotide addition can also occur ih H chains, when after the hairpin is cleave the enzyme TdT adds more nucleotides (other than the palendromic sequences). It can add up to 7 nucleotides.
3) Junctional Flexibility adds to variation. Different 3 if nucleotides can be cut at the RSS splice site. Also, some P's and N's can be spliced off after they are added.

4. The chromosome is rejoined together.
i. A hairpin loop at the end of each of the two pieces was formed. The loop is made with a regular phosphodiester bond.
ii. The hairpin is nicked and straightened, then filled in.
1) This is when P nucleotide addition can take place. The nicking and filling in can make a palendromic sequence that was not present before.

2) N nucleotide addition can also occur ih H chains, when after the hairpin is cleave the enzyme TdT adds more nucleotides (other than the palendromic sequences). It can add up to 7 nucleotides.

3) Junctional Flexibility adds to variation. Different 3 if nucleotides can be cut at the RSS splice site. Also, some P's and N's can be spliced off after they are added.

Additional variation can be introduced by AID (activation induced cytidine deaminase) that converts a C to a U. Over time mutations occur in B cells. AID is cis acting. When there is splicing AID becomes more active. AID adds to diversity and also increases the affinity of an Ab by directional mutation towards higher affinity.
in antibody formation not in TCR

A large number of genes that undergo "random" splicing at the DNA level
Alternative mRNA splicing
P nucleotide addition
N nucleotide addition
Junctional flexibility
Mutations via AID.

CDR 3 is the most variable region because it is made of VL-JL junction and VH-DH-JH junction while CDR1 and CDR2 are made just from V segments. CDR3 also has junctional flexibility, p-nucleotide addition, and n-nucleotide addition (only in H chain). All three CDRs are liable to somatic hypermutation.

Isotype switching:
We are not sure of the mechanism. It is most likely due to AID because gene Ko's of AID show no isotype switching.
In isotype switching a switch site (S) is used to loop out isotypes that are unneeded and are then spliced out. This is done at the DNA level.
Transcription stops after the isotype region
Certain cytokines influence switching:
IL4 induces IgE switch
Just a thought --> inhibit IL4, sown regute IgE, and reduce allergic response? A possible treatment.

MHC class I present in almost all nucleated cells, presents viral peptides to cytotoxic T cells

MHC class II are on B cells and antigen presenting cells. Present to T helper cells that induce B cells to produce antibody or make cytokines like gamma interferon.

Take two strains of inbred mice. Take a piece of skin from each mouse.
A transplant to the other strain results in rejection.
A transplant to the F1 hybrid of both mice is accepted.
A transplant from the F1 to either of the parents results in rejection.

IH-2K, IIAβ, IIAα, IIEβ, IIEα, IIIS, ID

K, IA, IE, S, D

class I MHC

class II MHC in mice

Class II MHC in mice

class III MHC in mice

class I MHC in mice

In the mouse the MHC class I gene is split in two.
Most rapid rejection of skin transplant occurs when the MCH class I H-2K.
The alpha-beta gene products are a heterodimer.
MHC class III is responsible for inducing complement and some cytokines

MHC class III is responsible for inducing complement and some cytokines

IIDPαβ, IIDQαβ, IIDRαβ (most variable), IIIC4C2BF, IB, IC, IA

LMP/TAP genes are responsible for antigen presentation.
DO Beta is under negative regulation
The Dr region can have multiple beta genes
In the human population there are hundreds of alleles for each site (we are polyallelic)
Humans are also polygenic, more than one gene codes for the MHC

There are 6 loci on the H2 (K, Abeta, Aalpha, E beta, E alpha, D)
An outbred mouse has 8 class II molecules (humans can have 12 or more).
An inbred mouse can have 2 different class I alleles
An outbred mouse can have four different class I

MHC molecules are promiscuous- they can bind to many but not all foreign proteins. They can be promiscuous because they can bind aa that the biochemically similar at the amino and carboxyl ends (the anchor amino acids)

Peptide binding domain is α1/β1
Can hold longer/bigger antigens (13-18 amino acids)
Peptides are usually bigger than the binding cleft and look like a hotdog in a bun.
The peptide binding cleft is open on both ends.
The binding cleft is made from Beta 1 and Alpha 1.
The biding cleft walls are alpha helices and the floor is a beta pleated sheet.
It is an alpha beta dimer
β1 most variable region (and α1?)
What are the loci of the alpha chain?
L, α1, α2, T(m+c) C
What are the loci of the Beta chain?
L, β1, β2, T(m+c) C

Peptide binding domain is α1/α2
It binds smaller peptides (8-10 amino acids) Preferred length is 9
The peptide binding cleft is closed at both ends and the peptide arches over the cleft (because of the close proximity of the 7 antiparallel β pleated sheets?)
Only the anchor residues at both ends of the peptide interact with the MHC molecule.
There is generally hydrophobic carboxyl-terminal anchor.
Region 1 and 2 are most variable
CD8 is an invariable adhesion molecule that binds to α3 and β2μ.
What are the loci ?
L, Α1, α2, α3, Tm,C,C

LCMV is a virus that usually doesn't infect humans. It is a mouse virus. You can infect a mouse with LCMV and harvest its spleen cells. The spleen cells are cytotoxic, meaning they are CD8, MHC class I restricted. They are specific for LCMV, are from a certain background mouse strain, and are restricted to a certain haplotype of MHC I. Then, take epithelial cells that express class I MHC. Chromium stain them. Chromium stain will not be released by cells unless they are killed. Percentage chromium released can be a quantitative measure of cell death. We find that:
1. Recognition of MHC I is independent of strain.
2. Recognition of MHC I is dependant on the Haplotype
i. Since MHC class I is coded for by H2K and HD which are the first and sixth region of the H2 complex, haplotype must match at least one of these two positions (1 and 6)
3. If haplotypes of the Tc cell match the target chromium stained MHC 1 haplotype k epithelial cell at the 1 or 6 region, then cell killing will occur. Otherwise it will not.

*If this was done in human cells the haplotye would have to match at the last (three?) positions.
*If this was done for class II MHC with cd4 TH cells and and antigen presenting cells then in the mouse the positions would have to match at 2-5 and in the human at the first 3 (?)

Syngenic - identical alleles at each loci

Conjenic- same alleles except at one loci

Conjenic strain - all background genes are the same except the MHC haplotye is different

Allogenic- foreign, different allel dissimilar at MHC loci

To make the antigen more antigenic mix the antigen with Freud's. Freud's can be complete or incomplete:
complete Freud's is mineral oil with dead microbacteria
Incomplete Freud's is mineral oil.
Freud's increases immune response because oil is hydrophobic and provides sustained release of whatever is mixed in with it.

To get a DTH response:
Mix antigen (lets say Fowl Gamma Globulin FGG) with complete Freud's
Inject into animal
Two weeks later inject again
There will be a large area of swelling, redness, and induration
The size of the area can be a quantitative measure of the DTH response
This 2nd DTH response is more rapid, with more redness, swelling, and induration

Adaptive transfer experiment:
1. Take a donor spleen of a mouse with a certain MHC haplotye. Take out lymphocyte CD4 cells that are primed. (meaning that they have been introduced to the antigen for a primary response)
2. The recipient has not had a primary response and does not have primed T cells, but it can process antigen and present it in Class II dependant antigen presenting cells.
3. Activated T cells respond within hours while an allotypic response takes weeks. (T his is why we can do the experiment)
4. Inject primed CD4 class II restricted, FGG specific cells from donor into recipient.
5. Inject FGG for the first time into the recipient.
6. A DTH response will only be seen when:
i. Recipient haplotype match the donor haplotype at the 2, 3, 4, or 5 positions
ii. The recipient need not match at the one and 6 positions that code for K and D
iii. The recipient need not match background strain type

To get a DTH response:
Mix antigen (lets say Fowl Gamma Globulin FGG) with complete Freud's
Inject into animal
Two weeks later inject again
There will be a large area of swelling, redness, and induration
The size of the area can be a quantitative measure of the DTH response
This 2nd DTH response is more rapid, with more redness, swelling, and induration

To make the antigen more antigenic mix the antigen with Freud's. Freud's can be complete or incomplete:
complete Freud's is mineral oil with dead microbacteria
Incomplete Freud's is mineral oil.
Freud's increases immune response because oil is hydrophobic and provides sustained release of whatever is mixed in with it.

1. Take a donor spleen of a mouse with a certain MHC haplotye. Take out lymphocyte CD4 cells that are primed. (meaning that they have been introduced to the antigen for a primary response)
2. The recipient has not had a primary response and does not have primed T cells, but it can process antigen and present it in Class II dependant antigen presenting cells.
3. Activated T cells respond within hours while an allotypic response takes weeks. (T his is why we can do the experiment)
4. Inject primed CD4 class II restricted, FGG specific cells from donor into recipient.
5. Inject FGG for the first time into the recipient.
6. A DTH response will only be seen when:
i. Recipient haplotype match the donor haplotype at the 2, 3, 4, or 5 positions
ii. The recipient need not match at the one and 6 positions that code for K and D
iii. The recipient need not match background strain type

To determine in vitro: Use equilibrium dialysis
Put certain MHC restricted molecules in one side and a certain antigen on other. The higher the affinity for the peptide, by the MHC the more peptide will leave its side and be taken out of solution.

Competition assay:
take MHC molecule (example IAd)
Put this MHC molecule on the column
Mix it with an antigen that we know binds to it well (example OVA) that is hot
So all the IAd on the column is now bound with OVA
Take other peptides that are not labeled and run them on the column.
Measure competition by measuring desorbance of labeled, OVA
Results:
MHC molecules bind better to some peptides than others because of the haplotype

T cell activation assay:
Take T OVA specific, IAd restricted cells and add antigen presenting cell which expresses Iad and have processed OVA. Measure activation by measuring proliferation or IL2 expression via ELISA
If you add a peptide that bind well to the MHC, then the T cell activation will be inhibited because the peptide will displace OVA but because the T cell is not specific to the other peptide no activation will take place
If you add a peptide that does not bind well to the MHC, then the OVA will not be displaced and inhibition of the T cell will not be profound.

take MHC molecule (example IAd)
Put this MHC molecule on the column
Mix it with an antigen that we know binds to it well (example OVA) that is hot
So all the IAd on the column is now bound with OVA
Take other peptides that are not labeled and run them on the column.
Measure competition by measuring desorbance of labeled, OVA
Results:
MHC molecules bind better to some peptides than others because of the haplotype

Take T OVA specific, IAd restricted cells and add antigen presenting cell which expresses Iad and have processed OVA. Measure activation by measuring proliferation or IL2 expression via ELISA
If you add a peptide that bind well to the MHC, then the T cell activation will be inhibited because the peptide will displace OVA but because the T cell is not specific to the other peptide no activation will take place
If you add a peptide that does not bind well to the MHC, then the OVA will not be displaced and inhibition of the T cell will not be profound.

Certain changes to the amino acid composition of a peptide will still allow for MHC binding, especially if they are similar in nature to the normal peptide. They will not however (usually) stimulate T cells. They might inhibit T cell activation in a competition T cell activation assay, but only if they bind well to the specific MHC

A Tc cell that is restricted to one MHC haplotype and specific for a certain antigen will cross react against a different MHC haplotype, (without the antigen). It may react against the MHC itself or against a processed peptide that has been processed differently by the allotypic MHC.
The Tc cell will then kill the uninfected foreign MHC presenting cell.
In addition, foreign cells shed their MHCs and your cells process the MHC as antigen and present them in your own MHCs to T cells that then can go and react against the allotypic MHC cells.

1. Take cells from mouse strain A and use mitomycin-C to inhibit mitosis of the cells. These cells have a certain MHC
2. Take T cells from mouse strain B (responder cells) and mix them with the strain A cells.
3. After 4 to 5 days, ass tridiated Thymidine
4. Measure radioactivity 18 hours later
Results:
The more discordance between the class II loci between the two strains, the more the responder T cells will proliferate

1. Take a guinea pig and inject it with antigen
2. Wait 7 days and remove lymph node cells
3. Separate out T cells (by column)
4. Take 2nd guinea pig of a different strain and separate out macrophages (class II APC)
5. Expose the macrophages to antigen
6. Mix macrophages with antigen from strain A with primed T cells from strain B
7. Measure T cell proliferation by thyamadine incorperation
Results:
If strains match --> T cells will respond against antigen in MHC and proliferate
If strains mismatch --> T cells will not respond
If you make f1 hybrid --> T cells will respond
Therefore you get T cell proliferation if and only if the antigen is presented in the correct MHC context.

Target cells express both class I and class II MHC
In this experiment there are rare cytotoxic T cells that are class II restricted
1. If the cell is transfected with normal virus both types of T cells respond
2. If the cell is transfected with noninfectious virus only class II restricted Tc cells respond
3. If the cell is transfected with normal virus and treated with emetine (That stops replication of virus) class I restricted T cells do not respond
4. If the cell is transfected with normal virus but treated with chloroquine (that prevents the formation of endocytotic vacuoles) class II T cells do not respond
5. Class I MHC do not present Hemagglutinin protein because it wasn’t processed
6. Class I MHC do present it when the target cell is transfected with the gene
7. Synthetic peptide activates both because of molar displacement outside of cell (doesn't have to do with processing)

Results:
In order for MHC class I to present antigen, the antigen must by produced in the cytoplasm by the cell machinery.
In order for MHC class II to present antigen formation of endocytotoic vacuoles are needed.

Class I presentation follows the endogenous pathway

exogenous pathway

In cross presentation the normal exogneous pathway becomes a class I pathway

Ubiquitin is a molecule that signal proteosomic destruction
1. A virus infected cell makes virus
2. Ubiquitin is added to the amino end of a lysine on the protein
3. Ubiquitin signals the protein to go to the proteosome
4. In the proteosome the anitgen gets choped up
5. The pieces move to the ER where MHC is being folded
6. Folding of MHC:
i. Calhexins are chaperone proteins that help the MHC fold.
ii. At this point the alpha 1 subunit is bound to Beta one but beta 2 and alpha 2 are still unbound
iii. Topasin links the class I MHC to TAP
iv. Tap is a transporter of antigenic peptides. It is a heterodimer that pumps proteosomed digested enzymes into the leumen of the ER and onto the MHC
v. Once the MHC is in the binding cleft, the MHC folding is finished
7. The MHC + antigen go to Golgi for processing and then are expressed on the cell surface

Ubiquitin is a molecule that signal proteosomic destruction

Ubiquitin is added to the amino end of a lysine on the protein

i. Calhexins are chaperone proteins that help the MHC fold.

Topasin links the class I MHC to TAP

iv. Tap is a transporter of antigenic peptides. It is a heterodimer that pumps proteosomed digested enzymes into the leumen of the ER and onto the MHC

1. Antigen is endocytosed by an APC
2. Endocytotic vacuoles lower their pH
3. The endocytotic vacuole fuses with a lysosome to make an endolysosome with even lover pH
4. Low pH causes antigenic molecule to be digested
5. In the ER a class II molecule is being formed
i. An invariant chain protein helps it fold
ii. The invariant chain trimerizes and sticks with three MHcs so you have a nonamer
iii. The invariant chain seems to prevent linkage of class II with topasin so that class II MHC cant be in close proximity to TAP and present molecules meant for class I.
6. The invariant chain takes class II molecule out of the lumen and leads it to an endolysosome
i. When fused this is called the MIIC
ii. Here we have invarient chain breakdown
iii. One part of the invariant chain remains in the MHC binding cleft. It is called CLP.
iv. HLA-DM somehow plays a role in displacing CLP
7. One CLP is removed from the binding cleft the digested foreign peptide to binds to the MHC
8. MHC + antigen is presented on cell surface

i. An invariant chain protein helps MHC II fold
ii. The invariant chain trimerizes and sticks with three MHcs so you have a nonamer

The invariant chain seems to prevent linkage of class II with topasin so that class II MHC cant be in close proximity to TAP and present molecules meant for class I.

iii. One part of the invariant chain remains in the MHC binding cleft in the MIIC. It is called CLP.

6. The invariant chain takes class II molecule out of the lumen and leads it to an endolysosome
i. When fused this is called the MIIC

HLA-DM somehow plays a role in displacing CLP

Melenoma cancer cells are highly invasive and proliferate into the blood and organs

There is a type of sugar α1,3 galactosil that is not present in primates because all primates are missing the enzyme α1,3 galactosil transferase.

Humans have a high titer Ab agains α1,3 galactosil because bacteria express it. So humans are positive for α1,3 galactosil and mice are negative because they naturally express this sugar.

So…

1. Make a KO α1,3 galactosil transferase in a mouse and now it is positive for antibody against α1,3 galactosil.
2. Take a melenoma cell from a normal mouse that expresses α1,3 galactosil on its surface
3. Transfuse into the KO mouse
4. Antibodies against α1,3 galactosil react against the melanoma cell
5. If you have a KO mouse that has melanoma and you transfect α1,3 galactosil positive melanoma cells the melanoma tumor will shrink because of cross presentation.
i. Antibody reacts agains α1,3 galactosil on melanoma cell
ii. Compliment is activated by Ab
iii. Cell lysis occus
iv. There is a heat shock protein signal (HSP) plus the exsistance of tumor specific antigens that were present on the cell surface leads to cross presentation of TSA in Antigen presenting cells.
v. When a dendritic cell has peptides form the same protein in class I and class II it activates Tc cells and TH1 cells against TSA that are present on all melanoma cells.

tumor specific antigens

The β chain of the TCR is analagus to the Heavy chain of an Antibody

The α chain is analagous to the L chain

1. Take a Tcell that is specific for KLH and Iaf restricted and fuse it with a cell that is OVA specific and IAk restricted.
If there are two receptors on that recognize MHC and antigen separatly, then the fusion cell will be both OVA and KLH specific and both IAf and IAk restricted in any combonation (OVA with IAf etc…)
If there is one receptor than the fusion cell will only be OVA-IAk or KLH-IAf
2. The fusion cell is presented with the flowing APC's:
i. OVA-Iak
ii. OVA-IAf
iii. KLH-IAk
iv. KLH-Iaf

Results: There is only one receptor

1. Take a T cell clone
2. Lyse
3. Ultracentrifuge
4. Get membrane
5. Solubalize to remove proteins from membrane
6. These proteins are a combo of the TCR and random cell surface proteins
7. Take the cell membrane proteins and inject into a mouse
8. Mouse makes antibody against all of the proteins
i. Most of the Ab will be against random proteins and be able to interact with many differnet T cell clones
ii. The Ab against the TCR will only be able to bind with the particular T cell clone form which it came. It will be able to inhibit that particular clone from reacting with antigen.
9. A negative result (no binding to antigen) means that this Ab is the one against the TCR
10. Take Ab and put it on beads in a colums
11. Pour solublized cell surface proteins over column
i. ONLY TCR STICKS
12. Lower pH or increase salt concentration to release TCR
13. Use SDS and mercaptoethenol to get rid of tertiary structure
14. Gel electrophorese/western blot
15. Get two bands on the gel
16. Purify intact TCR and perform x ray defraction

dimer of aplha and beta
alpha chain has two immunoglobulin domains (a1 and a2)
beta chain has to IG domains (b1 nad B 2)
there is a short cytoplasmic tails mean that TCR needs other molecules to activate the cell
the transmembrane region has one positivly charged amino acid
There are three CDR (variable regions) on the alpha and 3 CDRs on the beta chain
Six CDRs need to bind to MHC and antigen

There are three CDR (variable regions) on the alpha and 3 CDRs on the beta chain
Six CDRs need to bind to MHC and antigen

One of the amino acids of the TCR in the transmembrane region is actually charged +. It interacts with a signaling protein via the transmembrane region.
These proteins include ITAM, a tyrosine kinase and CD3 (accessory molecules)
ITAM couldbe either next to or below the CDR

Work like Ab genes. There are RSSs, Rag1 and Rag2 that splice certain regions.

1. Take mRNA from T
i. Throw out free mRNA that is not bound to Rough ER and therefore couldn't code for a membrane protein
2. Reverse transcribe into cDNA
3. Label with 32P and make ss
4. Mix with mRNA from a B cell
5. Separate ss (non hybridized) DNA form ds hybridized DNA on hydroxyapatite
i. Throw out all ds DNA
6. Use the ss labled mRNA as probes for a southern blot
i. Take one label and use it as probe for separate southerns of different T cell clones
1) If all the southerns are the same, then this cDNA is NOT from the TCR gene
2) If one of the southerns is differnet, then this cDNA codes for the TCR for that clone
You have found the cDNA that undergoes splicing to bring you the TCR
7. Sequence the cDNA to get code for TCR

There is only one constant region in alpha and two identical ones in beta (but only one allele)
Humans have hundreds of V's D's and J's.
*Remember, every person have every allele

First there is a V to J splice
Then the LVJ ---D DNA is transcribed and the intron btw VJ and D is removed
*The alpha chain has no D

First there is a J to D splice
The there is a V to JD splice
Then the LVJD---C region is transcribed and the intron btw VJD and C is removed

Of the CDRs, CDR3 is the most variable

Because you habe two C regions in the beta chain you have 2 chances per chromosome for correct splicing.

For the alpha chain, because you have v to J splicing first, you have many opportunities to splice correctly.

When beta chains are expressed with surrogate RAG1 and RAG2 are turned off. (remember that beta would be transcribed and expressed first) Then mitosis occurs, After mitosis RAG I and RAG II are turned on and the alpha chain is then transcribed. This increases the number of cells with correct splicing and increases the chances of cell survival.

Gamma and delta TCRs may recognize CD1 or be part of innate immunity

In TCR transcription there is also junctional flexibility, n nucleotide addition etc…

However, somatic mutation is missing in T cells because you don't want to recognize the MHC too well. (remember that somatic mutation leads to higher binding affinity) or not recognize the MHC at all

Usually:
CDR1α
Recognizes amino end of the processed peptide
CDR1β
Recognizes the carboxyl end of the processed peptide
CDR3α + β
Recognize the middle of the peptide
CDR2β + α
Recognize the MHC

α recognizes the αII domain of class I

Β recognizes the ????

The TCR is responsible for specificity of binding to the MHC-antigen

CD4 and CD8