Generation of lymphocyte antigen receptors !!! Flashcards

1
Q

The wide range of antigen specificities in the antigen-receptor repertoire is due to the _______ in the amino acid sequence at the _______-_____ site which is composed of the variable regions of the receptor.

A

variation; antigen-binding

*variation lies within the variable region of antigen receptors

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2
Q

(T/F) Each B cell and T cell produces an antigen receptor with a unique antigen-binding site.

A

True!

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3
Q

There are millions of lymphocytes in an individual with a unique antigen receptor, but there are only ~20,000-25,000 protein coding genes in the human genome?

How is this diverse antigen-receptor repertoire generated?

A

This diverse antigen-receptor repertoire is generated (in part) by the GENETIC MANIPULATION of the DNA sequence that encodes the VARIABLE REGION of the antigen receptor.

The variable region of the antigen receptors are encoded in several pieces called GENE SEGMENTS.

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4
Q

Different gene segments are assembled in developing lymphocytes by DNA ________ to form a complete V region.

The selection of gene segments occurs at ________.

A

recombination

RANDOM

*the large number of possible combinations accounts for much of the diversity of the antigen receptor repertoire.

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5
Q

(T/F) The germline DNA encoding variable region is also affected by DNA recombination.

A

False!

Germline DNA encoding variable region remains intact.

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6
Q

Generation of antigen receptors occurs during _______ ________.

A

lymphocyte development

*the gene rearrangement occurs here!

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7
Q

(T/F) In an immunoglobulin, there are two identical heavy chains and two identical light chains, generating two identical antigen binding sites.

A

True!

This confers the ability of an Ig to bind SIMULTANEOUSlY to two identical antigenic structures.

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8
Q

What are the two types of light chains found in an Ig?

Can a given Ig have either of the two or a mix of both?

A

Kappa and Lambda

A given Ig has EITHER kappa or lambda - NEVER ONE OF EACH.

*no functional difference has been found between an Ig having either.

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9
Q

Differentiate the variable region of an Ig from the constant region.

A

Variable region: The N-terminal sequences (top) of both the heavy and light chains vary greatly between different antibodies.

Constant region: The remainder of the sequence is CONSTANT between immunoglobulins of the same isotype. It confers effector functions.

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10
Q

Antigen binding site (variable region of Ig) dictates antigen _______ of the receptor.

A

specificity

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11
Q

What are hypervariable regions or complementarity determining regions?

A

HVs aka CDRs display high levels of sequence variation between different receptors and form LOOPS at the OUTER edge of the protein surface for antigen binding.

Both the heavy chain variable region - V(H) and light chain variable region - V(L) contain three CDRs. The pairing of heavy and light chain brings together these loops to form a single hyper variable surface (antigen-binding site).

There are a total of 12 CDRs on an Ig.

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12
Q

The variable regions of the Ig receptor are generated by 4 main mechanisms.

What are they?

A

1) Combinatorial diversity

2) Junctional diversity

3) Joining of heavy and light chains

4) Somatic hypermutation

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13
Q

Match the mechanisms to their definitions:

1) Combinatorial diversity

2) Junctional diversity

3) Joining of heavy and light chains

4) Somatic hypermutation

A) diversity created due to insertion of non-template NTs at the joints between gene segments that encode for the variable region of both light and heavy chains

B) diversity created due to high level of mutations introduced in the DNA sequence of the rearranged variable region in mature activated B cells

C) diversity created by the combinatorial joining of gene segments that encode for the variable region of both light and heavy chain

D) diversity created by combinatorial joining of different heavy and light chain variable regions

A

Combinatorial diversity: diversity created by the combinatorial joining of gene segments that encode for the variable region of both light and heavy chains.

Junctional diversity: diversity created due to insertion of non-template NTs at the joints between gene segments that encode for the variable region of both light and heavy chains

Joining of heavy and light chains: diversity created by combinatorial joining of different heavy and light chain variable regions

Somatic hypermutation: diversity created due to high level of mutations introduced in the DNA sequence of the rearranged variable region in mature activated B cells

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14
Q

While combinatorial diversity, junctional diversity, and joining of heavy and light chains occurs in the bone marrow during ___ _________ and _____ require antigen, somatic hypermutation occurs in the peripheral lymphoid organs once development is __________ and ____ require antigen binding.

A

B cell development; does not

Complete; does

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15
Q

The DNA sequence that encodes for the variable region consists of multiple gene segments that are eventually joined together.

1) How many gene segments are joined together to encode the light chain variable region? What are they?

2) How many gene segments are joined together to encode the heavy chain variable region? What are they?

A

1) 2 genes; variable gene segment (VL) and joining gene segment (JL)

2) 3 genes; variable gene segment (Vh), joining gene segment (Jh), and diversity gene segment (Dh)

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16
Q

What is somatic RECOMBINATION?

A

The JOINING TOGETHER of separate gene segments that encode for the variable region.

It is also known as V(D)J recombination or GENE REARRANGEMENT.

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17
Q

(T/F) Somatic and mietoic recombination are the same process.

A

False!

Somatic is used here to distinguish it from meiotic recombination that occurs during gamete production.

Somatic recombination occurs in lymphocytes and in non lymphoid cells, the variable region gene segments REMAIN in their germline (original) configuration.

18
Q

Which one of these statements is true?

1) The generation of the variable region occurs in a specific temporal sequence.

2) This sequence is independent to B cell development in the bone marrow.

3) The constant region of the light and heavy chain are encoded in the same exon.

4) The variable region is joined to the rearranged constant region by RNA splicing to remove the intervening L-V and J-C introns.

A

1!

2) This sequence CORRELATES with distinct stages of B cell development in the bone marrow.

3) The constant region of the light and heavy chain are encoded in a SEPARATE exon.

4) The CONSTANT region is joined to the rearranged VARIABLE region by RNA splicing to remove the intervening L-V and J-C introns.

19
Q

Are there only one copy of each gene segment of the V gene region?

Why is this important?

A

No! There are MULTIPLE COPIES of the V, D, J gene segments within the germline DNA.

It is the RANDOM SELECTION of one gene segment of each type that contributes to the large DIVERSITY of V regions among immunoglobulins.

Selected genes are joined together by somatic recombination, creating a complete variable region.

20
Q

In the germline, each locus (light and heavy chains) encodes for a _______ variable and constant region.

Each locus is located on a ________ chromosome and organized _______.

A

COMPLETE (encodes the appropriate gene segments)

different; differently

*lamba light chain and kappa light chain locus also differ from each other.

21
Q

Somatic recombination is guided by specialized sequences called _________ _______ _________.

These are ________ sequences that are located directly _____ to the V, D, and J gene segments to facilitate recombination.

A

Recombination Signal Sequences (RSS)

NON-coding; adjacent

22
Q

Describe the components of a RSS.

A

RSS consists of a conserved HEPTAMER, which is followed by a non-conserved SPACER (12 or 23 bp), which is, in turn, followed by a conserved NONAMER.

23
Q

What is the 12/23 rule?

A

The space between the heptamer and nonamer of a RSS is always 12 or 23 base pairs.

A gene segment flanked by an RSS with a 12 bp spacer usually joins with another gene segment flanked by an RSS with a 23 bp spacer.

24
Q

Why is 12 + 23 the ideal conformation for the recombination process of the different gene segments?

A

12 bp corresponds to approximately one turn of the DNA helix and 23 bp corresponds to two turns of the DNA helix.

Either of these lengths may bring heptamer and nonamer to the same side of the DNA helix to allow interactions with proteins that facilitate the recombination process.

It also helps to ensure that V gene segments are joined to D or J gene segments and not to other V gene segments.

25
Q

Describe the process that generates the two DNA ends during somatic recombination.

A

RAG1 and RAG2 endonuclease bind RSSs flanking the coding sequences to be joined. The RAG complex recruits the other RSS, creating a loop.

The RAG endonuclease creates breaks in the DNA BETWEEN each coding segment and its RSS.

Each cut creates:
1. A 3’ OH group which reacts with a phosphodiester bond on the opposite DNA strand to generate a hairpin.

  1. Blunt double stranded break at the ends of the RSSs.

These two types of DNA ends are resolved in different ways: generation of coding joints (for genes) and signal joints (for RSSs).

26
Q

(T/F) RAG1 and RAG2 are expressed in all cells all the time.

A

False!

RAG1 and RAG2 expression is restricted to lymphocytes during specific developmental stages. This prevents somatic recombination occurring in other cells.

27
Q

Match the steps of generating a coding joint during somatic recombination:

1) Step 1
2) Step 2
3) Step 3
4) Step 4

A) DNA-pk associates with Artemis which has endonuclease activity. Artemis opens the DNA hairpin at a random site.

B) The DNA ends are joined together by DNA ligase IV which forms a complex with a DNA repair protein, XRCC4.

C) The Ku70:80 heterodimer forms a ring around her DNA and associates a protein kinase, DNA-dependent protein kinase (DNA-pk)

D) The cut ends are modified by Tdt and exonuclease (junctional diversity)

A

Step 1: The Ku70:80 heterodimer forms a ring around the DNA and associates a protein kinase, DNA-dependent protein kinase (DNA-pk)

Step 2: DNA-pk associates with Artemis which has endonuclease activity. Artemis opens the DNA hairpin at a random site.

Step 3: The cut ends are modified by Tdt and exonuclease (junctional diversity)

Step 4: The DNA ends are joined together by DNA ligase IV which forms a complex with a DNA repair protein, XRCC4.

28
Q

Match the steps of generating a signal joint during somatic recombination:

1) Step 1
2) Step 2

A) DNA ligase IV:XRCC4 join the two ends to form a signal joint, which ultimately get removed from cells during development.

B) The ku70:80 heterodimer binds to the RSS but the ends are not further modified.

A

Step 1: The ku70:80 heterodimer binds to the RSS but the ends are not further modified.

Step 2: DNA ligase IV:XRCC4 join the two ends to form a signal joint, which ultimately get removed from cells during development.

29
Q

1) When does junctional diversity occur?

2) What are the two types of nucleotides added?

A

1) Junctional diversity occurs during generation of a coding joint in somatic recombination. It occurs when the cut ends are modified by Tdt and exonuclease.

2)
P- nucleotides: palindromic sequences created by ARTEMIS

N-nucleotides: non-template sequences added to the ends of the P-nucleotides by the enzyme Terminal Deoxynucleotidyl Transferases (Tdt). impt for junctional diversity

30
Q

Briefly explain how P nucleotides are added by artemis.

A

Artemis cleaves the DNA hairpin at a RANDOM site. A ss DNA end results from this cleavage.

DNA synthesis enzymes fill in complementary NTs on the ss tails which generate PALINDROMIC sequences.

31
Q

Briefly explain how N nucleotides are added by Tdt.

A

After opening of the palindromic sequences added by Artemis, N nucleotides are added by Tdt.

These are NON-TEMPLATED. After addition of random ~20 NTs, single stranded stretches may have generated some complementary base pairs by chance.

Any unpaired NTs are removed by exonuclease and the gaps are filled by DNA repair enzymes.

32
Q

An advantage of imprecise recombination of Tdt is a productive rearrangement, while the disadvantage is non-productive rearrangement.

Differentiate productive rearrangement from non-productive rearrangement.

A

Productive rearrangement: a successful in-frame DNA rearrangement that leads to a functional Ig molecule or TCR, creating a large diversity.

Non-productive rearrangement: production of OUT-OF-FRAME DNA sequences that lead to non-functional TCR (Ig) chains.

33
Q

(T/F) Combinatorial and junctional diversity happen both at heavy and light chains for all gene segments.

A

True!

34
Q

The alpha and beta glycoprotein chains of the TCR consists of a variable and constant domain.

Like Ig molecules, the antigen specificity of the TCR is defined by the ______ region.

A

Variable

*structure is similar between Ig and TCR, thus similar mechanisms and enzymes as B cells are used to created antigen receptor diversity.

35
Q

What are three main mechanisms used to generate the variable regions of the TCR?

A

1) Combinatorial diversity
2) Junctional diversity
3) Joining of alpha and beta chains

All of these occur in the thymus during T cell development and do not require antigen.

The sequence of generating the variable sequence correlates with distinct stages of T cell development in the thymus.

36
Q

Which one of the statements is true:

1) You can not uncouple B/T cell receptors development from B/T cell development.

2) Somatic hypermutation does not occur in T cells (only in mature activated B cells).

A

Both are true!

37
Q

List each statement as True or False:

1) The TCR alpha and beta chains consist of a variable amino terminal region and a downstream constant region.

2) The organization of the gene segments is different than the Ig gene segments.

3) The organization of the alpha and beta chain loci differ slightly from each other.

A
  1. True
  2. False. Organization of the gene segments is homologous to that of the Ig gene segments.
  3. True.
38
Q

Which gene segments does the TCR alpha locus contain?

Which gene segments does the TCR beta locus contain?

A

TCR alpha locus: V and J gene segments (like Ig light chain)

TCR beta locus: V, D, and J gene segments (like Ig heavy chains)

39
Q

(T/F) Functional alpha and beta chain variable regions are generated in the same way that complete Ig variable regions are created.

A

True!

40
Q

Recombination signal sequences (RSS) in TCR genes have the ____ sequences and use the _____ enzymes/processes for somatic recombination as those used for Ig somatic recombination.

A

Same; same