immuno 12 gene rearrangement Flashcards
What does the germline DNA of heavy and light chains contain?
Light chain: Variable region genes, Joining region genes, and a constant region gene (kappa or lambda).
Heavy chain: Variable region gene, Diversity region gene, Joining region gene, and a constant region gene (Mu μ).
How is each of the regions of the germline dna distributed?
Leading segment then V and D are far from each other, while J and C are close.
In the light chain, V is far away and J and C are close.
T/F the heavy chain gene is rearranged before the light chain.
True
What heavy chain class do progenitor B cells, precursor B cells , and immature B cells form?
IgM
What is pre-BCR?
pre- B-cell receptor which is the antigen receptor with a genetically rearranged heavy chain.
How is heavy chain rearrangement done?
The first process is J and D recombination, this will cause J and D regions to bind. This will bring the C in close proximity.
Next we have VDJ recombination with the bound DJ segment, it will bind to the variable region bringing the Constant mu segment close to everything else.
The constant segment consists of 4 regions, each encoding for a part of the constant segments on the heavy chain (4 parts in IgM) .
RNA splicing will then happen, giving VDJC RNA.
How many segments are there in the VDJ regions of the heavy chain?
1-40 V
1-23 D
1-6 J
What is a precursor B-cell?
A B-cell that has undergone heavy chain but not light chain recombination.
How many V,J, and C segments are in the light chain region?
V 1-38
J 1-5
C 1 (kappa or lambda)
How is kappa light chain rearranged?
V is far from J and C.
J and V recombine and bind together, pulling C closer.
It is then transcribed into RNA then spliced.
How can diversity and specificity be increased in a way other than genetic rearrangement?
By adding new nucleotides during the VJ recombination process.
How many hypervariable regions are found on each light and heavy chain?
3 on each so 12 in 1 antibody.
What is a recombination signal sequence?
Each segment of V and J have a sequence (RSS) that consists of a heptamer and a nonamer with 12 or 23 base pairs in between them.
This sequence is required for VDJ recombination.
What is the 23/12 rule?
Segments bound to RSS motifs with 12 bp spacers can’t bind to other segments with 12 bp spacers, the same thing applies for 23 bp spacers.
Describe the 2 ways of recombination initiation.
1- Deletion:
The J and V sections have the nonamer part of the RSS close to each other, the heptamer ends are cut by RAG1 and RAG2 from both segments, giving rise to a deleted segment from the heptamer of V segment to heptamer of J segment, this segment is deleted from the DNA and forms a loop shape. The newly cut V and J ends form a hairpin loop via enzymes (K mu enzymes in kappa light chain), DNA protein kinase will break the hairpin loop again and terminal deoxynucleotidyl transferase adds nucleotides between V and J randomly. Finally, DNA ligase ligates the ends together.
2- Inversion:
If the nonamer in one of the RSSs is closer to the heptamer of the other RSS, then inversion happens (the DNA twists to let V and J get close to each other). The rest of the process is the same as in deletion, but the deleted segment looks different.
