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Molecular Genetics > Immunogenetics 1 and 2 > Flashcards

Immunogenetics 1 and 2 Flashcards

1
Q

Antigen vs Antibody

A

Antigen: molecules that elicit an immune reaction

Antibody: proteins that binds to antigens and mark them for destruction by phagocytic cells

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2
Q

The Organisation of the Immune System:

Humoral Immunity vs Cellular Immunity vs Clonal Selection.

A

Humoral immunity: the production of antibodies by B cells

Cellular immunity: depends on T cells

Clonal Selection: primary response, memory cells,
secondary response

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3
Q

Immune responses are classified as

A

humoral and cellular immunity

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4
Q

Immune responses are classified as
humoral and cellular immunity STEPS = 5

A
  1. Lymphocytes originate from STEM CELLS in the BONE MARROW
  2. B cells MATURE in the BONE MARROW
  3. When B cells encounter ANTIGENS, they MATURE into PLASMA CELLS, which secrete ANTIBODIES that bind to the antigen.
    HUMORAL IMMUNITY
  4. T cells mature in the THYMUS and ENTER CIRCULATION..
  5. They ATTACK by BINDING HOST CELLS AND LYSING THEM/
    CELLULAR IMMUNITY
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5
Q

B Cell Development - complex co-ordinated
process….

A
  1. SOMATIC RECOMBINATION - IN B CELLS:

—LARGE SCALE GENOMIC REARRANGEMENTS

PRODUCE FUNCTIONAL RECEPTORS

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6
Q

What are B cells or B lymphocytes?????

A

play a critical role in the immune response as
the producers of immunoglobulin (Ig) or
antibodies

IMMUNOGLOBULIN (Ig) or ANTIBODIES

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7
Q

Understanding Recombination… In Meiosis

A

During meiosis, the
chromosomes DUPLICATE, then CROSSING OVER (‘recombine’) to produce a HAPLOID GAMETE (sperm/egg)

The gamete derives GENETIC VARIANTS from both parents

Meiosis is the BASIS for HEREDITY…

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8
Q

ANTIBODY IN DETAIL

A
  1. Antibody = protein that binds specifically to a
    particular substance (antigen). Each antibody
    molecule has a unique structure that enables it to
    bind specifically to its antigen
  2. All antibodies have the SAME OVERALL STRUCTURE
  3. Antibodies are produced by plasma cells
    (differentiated B cells) in RESPONSE TO INFECTION OR IMMUNISATION.
  4. Antibodies bind & NEUTRALISE PATHOGENS (organisms that cause disease when they infect
    their host) or PREPARE them for UPTAKE &
    DESTRUCTION by OTHER CELLS OF THE IMMUNE RESPONSE.
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9
Q

An immune response to antigen produced through
CLONAL EXPANSION

10***

A
  1. In large POOL of B lymphocytes each is SPECIFIC for ONE ANTIGEN.
  2. When an ANTIGEN BINDS to a B cell, the B cell divides…
  3. and gives RISE TO CLONES of B cells, all specific for the same antigen.
  4. this proliferation of lymphocytes is thePRIMARY IMMUNE RESPONSE.
  5. some cells differentiate into ANTIBODY-SECRETING PLASMA CELL.
  6. Antibodies are for the SPECIFIC antigen.
  7. MEMORY CELLS REMAIN IN CIRCULATION.
  8. if a second exposure of the same antigen occurs…

9…. the ANTIGEN binds to the MEMORY CELL,…

  1. …which rapdily give rise to aVSECONDARY IMMUNE RESPONSE.
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10
Q

If B cells can produce antibodies that can bind to
specific antigens how can you generate enough
diversity to cope with all the micro-organisms
capable of causing disease???

A
  1. Virtually any substance can bring about an antibody response - is complex involving many different antibody molecules
  2. The complete collection of antibodies (antibody
    repertoire) in an individual consists of as many as

10^ 11
or perhaps even more = 10 0000000000!!!!!!!!!!!!!!!!!

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11
Q

What is the origin of this diversity??

Two possible explanations

A
  1. The GERMLINE THEORY: there is a separate gene for each different antibody chain & the antibody
    repertoire is inherited
  2. The SOMATIC DIVERSIFICATION theory: limited number of
    variable region sequences that undergo alteration
    within B cells during the lifetime of an individual to
    generate the observed repertoire.
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12
Q

WHICH IS MOST ACCURATE THEORY FOR DIVERSITY?

A

Cloning the genes that encode immunoglobulins lead to understanding - Igs are in fact generated during B cell development by DNA rearrangements

  1. Combine & assemble different gene segments
    from relatively small group of INHERITED sequences at each locus.
  2. Therefore BOTH theories are partially correct
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13
Q

SUMMARISE THE OVERALL VARIABILITY OF ANTIBODIES.

A

The enormous variability of antibodies derives from a SMALL number of genes.

Two groups of genes become linked together by somatic recombination.

Provides the basis for generating a vast array of different gene products

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14
Q

An Ig protein molecule consists of:

A

4 polypeptide chains:

2 identical heavy chains

2 identical light chains

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15
Q

explain Light chain vs Heavy Chain?

A

Light chain
- two classes - KAPPA AND LAMBDA
functionally equivalent, an - antibody can have
EITHER lambda or kappa chains

  • N-termini have variable (V) regions which need to
    bind foreign antigen;
  • the remaining C-terminal
    segments are constant (C) regions.
  1. HEAVY CHAIN - have different alternatives for the constant region which specify the tissues in which Ig will be expressed & the Ig class (A,D,E,G,M).
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16
Q

Antibody Structure

A
  1. two identical subunits.
  2. Each subunit contains a LIGHT chain & a HEAVY chain bound
    together by DISULFIDE BRIDGES
  3. Amino acid sequencing has revealed large areas that are similar in all antibodies - CONSTANT REGIONS
  4. There are also
    VARIABLE REGIONS
    where amino ACID
    sequence differs
    from antibody to
    antibody.
    - Variable region (V), Joining region (J), Constant (C)
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17
Q

The genes which encode the different types of chainsin Igs are located on

A

DIFFERENT chromosomes & are organised as clusters of numerous gene
segments.

Each cluster is unusual in that the coding sequences
for specific segments of each chain are often present in numerous copies that are Sequentially repeated.

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18
Q

what does V,D,J,C Mean?

A

V= variable, D = diversity, J = joining, C = constant

Combinatorial V-D-J joining (possibel outcomes)
Multiply each column respectively.

Possible combinatorila associations of heavy and light chains
ADD LIGHT CHAINS TOGTHER, THEN MULTIPLY HEAVY CHAIN TO THAT NUMBER

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19
Q

To make a human Kappa light chain…. = 6

A
  1. there are 30-35 different V gene segments,…
  2. 5’ J gene segments…
  3. and 1 C gene segment in GERMLINE DNA
  4. V2 can be moved next to J3 through somatic recombination, producing the DNA fond in mature B cell.
  5. the V-J-C pre-mRNA is processed so the mature mRNA contains sequences for only one V,J and C gene segment.
  6. This mRNA is translated into a functional light chain.
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20
Q

Understanding B cells and Clusters of it…

A
  1. As each Ig gene cluster in an INDIVIDUAL B cell
    only ever gives rise to one Ig polypeptide, an
    entire cluster can functionally be regarded as a SINGLE , albeit UNUSUAL type of gene.
  2. However, the individual gene segments in
    these clusters cannot be seen as classical
    exons.
    - Individual gene segments in these
    clusters are sometimes composed of coding &
    noncoding DNA.
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21
Q

L sequence?

A

L sequence is a leader sequence - specifies secretion - needed in light & heavy chains

22
Q

slide 20

A

Example 1: Constant region of human/mouse k LIGHT CHAIN Ig is encoded by single Ck sequence, the variable regions are encoded by a combination of a Vk region & a Jk region

23
Q

SLIDE 21

A

Example 2: The heavy chain locus is different (more complex -
more constant regions, also contains diversity region lacking in
light chain).

The variable region is encoded by a combination
of VH gene segment, a JH gene segment & also a DH gene segment.

Additionally there is also a variety of constant region
gene segments.

24
Q

Express functional antibody molecule???

A

Programmed DNA rearrangements at the Ig locus occurs during the maturation of the B cell

25
Q

EXPLAIN Programmed DNA rearrangements at the Ig locus occurs during the maturation of the B cell - 3

SOMATIC RECOMBINATION, ALTERNATIVE DNA SPLICING

A
  1. The unique arrangement of gene segments in the Ig
    gene clusters reflects the unusual way in which somatic recombinations are required in B cells before a functional Ig gene can be assembled & expressed.
  2. Occurs by process of SOMATIC RECOMBINATION
    Such events bring together different combinations of the different gene segments in different individual B cells.
  3. Regarded as cell specific ALTERNATIVE DNA SPLICING
    events
26
Q

original germline gene organisation = 3

A
  1. This is opposed to alternative RNA splicing which brings about different combinations of exons at the RNA level.
  2. As a result, theORIGINAL GERMLINE GENE ORGANISATION
    is altered:
    gene segments that are distant in the germline are spliced together at the DNA level.
  3. Choice of which of the many repeated gene segments are recombined to give a functional Ig unit is cell specific, individual B cells produce different Igs.
27
Q

TO MAKE LIGHT CHAIN - a SOMATIC RECOMBINATION …2

A

To make a light chain - a somatic recombination
event brings together one of the V segments &
one of the J segments (V-J joining).

Thereafter, SPLICING to the single C sequence occurs at the RNA level.

28
Q

To make a heavy chain - two successive somatic
recombinations 3

A
  1. To make a heavy chain - two (2) successive somatic
    recombinations are required.
  2. First - a D-J joining
    & then a V-D-J joining.
  3. Subsequently, the V-D-J
    sequence is spliced at the RNA level to the nearest C sequence… but as B cell matures, gets VDJ joined to different C gene = heavy
    chain switch
29
Q

Light chain:

Recall: L is the leader
sequence for secretion
from B cell! = 10

A

DNA ——-
1. Germline DNA

  1. Somatic recombination
  2. D-J joined rearranged DNA
  3. V-J or V-DJ joined rearranged DNA
  4. TRANSCRIPTION

RNA …………..

  1. Primary transcript RNA
  2. SPLICING
  3. mRNA
  4. TRANSLATION

——-PROTEIN

  1. Polypeptide chain
30
Q

Mechanism of Variable-region DNA rearrangements =

NUCLEOTIDE SEQUENCE OF RSSs = 4

A

1 * Recombination signal sequences (RSSs) direct recombination.

2 * RSS flank germ-line V,D, J gene segment. One RSS is located 3’ to each V gene segment, 5’ to each J gene segment and on both sides of each D gene segment.

3 * Contains conserved palindromic heptamer & AT-rich nonamer separated by intervening sequence

4 * 12-23bp correspond to 1 or 2 turns of DNA helix – one turn or two turn
RSS

31
Q

LOCATION of RSSs in germ-line immunoglobulin DNA

A

1 * Vk signal – one turn, Jk – two turn.

2 * In lamdba light chain, order reversed. In heavy chain, VH and JH have
two turns. DH – one turn.

3 * Signal sequences having one turn can only join with sequences having
two turns. Ensures that Vl only joins Jl and not to another Vl.

4 * Likewise, VH, DH and JH join in proper order.

32
Q

Gene segments are joined by recombinases explain 3

A
  1. V-(D)-J recombination which takes place at the junctions between RSSs and coding sequences, catalysed by V(D)J
    recombinase.
  2. 1990, first report of two recombination-activating genes; RAG-1 and RAG-2, whose protein acts synergistically and required to mediate V-(D)-J joining.
  3. RAG-1 and RAG-2 are only lymphoid specific gene products that have so far been shown to be involved in V-(D)-J rearrangement.
33
Q

Gene segments are joined by recombinases what is includes

A

Binding of RAG1/2, HMG PROTEINS, SYNAPSIS, RAG 1/2 HMG proteins, CLEAVAGE AND PROCESSING OF SIGNAL AND CODING JOINTS, Artemis, TdT (Terminal deoxynucleotidyl transferase), DNA ligae Iv, NHEJ proteins, GENERATION OF FUNCTIONAL Ig VARIABLE REGION GENE., CODING JOINT, SIGNAL JOINT.

34
Q
A

STEP 1: RAG1/2 and HMG proteins binds to the RSS and catalyse synapse formation between a V and a J gene segment.

STEP 2: RAG1/2 performs a single stranded nick at the exact 5’ border pf the heptameric RSSs bordering both the V and the J segments.

STEP 3: the hydroxyl group attacks the phosphate group on the non-coding strands of the V segment to yield a covalently-sealed hairpin coding end and a blunt signal end.

STEP 4: Ligation of the signal ends

STEP 5: Opening of the hairpin can result in a 5’ overhang, a 3’ over-hang, or a blunt end.

STEP 6: Cleavage of the Hairpin generates sites for P nucleotide addition.

STEP 7: Ligation of LIGHT CHAIN V and J regions

STEP 8: IN HEAVY CHAIN VD and DJ joints only; Exonuclease cleavage results in loss of coding nucleotides at joint - can occur on either or both sides of joints.

STEP 9: Non-templated nucleotides are added to the coding joint by TDT.

STEP 10: Ligation of Heavy Chain by DNA ligase IV and NHEJ proteins

35
Q

There are three types of functional Ig gene loci in
human cells -

A
  • heavy (H)

& 2 light (K, L) chains.

36
Q

explain There are three types of functional Ig gene loci in human cells - heavy (H) & 2 light (K, L) chains.

4

A
  1. As these occur on both maternal and paternal
    homologues there are six chromosomal segments in
    which DNA rearrangements can result in production of Ig chain.
  2. However, an individual B cell is monospecific - it
    produces only one type of Ig molecule….HOW??
  3. allelic exclusion…a light chain or heavy chain can be synthesised from maternal or paternal chromosomes in any one B cell but not from both….monoallelic expression.
  4. choice appears random – or is it?
37
Q

explain There are three types of functional Ig gene loci in human cells - heavy (H) & 2 light (K, L) chains.

appears random – or is it?

A
  1. Because of allelic exclusion, the Ig heavy and
    light chains of only one
    parental chromosome are
    expressed per cell
  2. Most likely, in each B cell, productive DNA
    rearrangements are attempted at all six Ig alleles, but the chances of productive arrangements in more than one cluster is not high.
  3. Also appears to be a NEGATIVE FEEDBACK REGULATION: a
    functional rearrangement at one heavy or light chain
    allele suppresses arrangements occurring at the other alleles.
38
Q

EXPLAIN: Somatic hypermutation adds diversity in already rearranged gene segments

A
  1. Additional antibody diversity is generated in rearranged variable
    regions by a process called somatic hypermutation
  2. Individual nucleotides in VJ or VDJ units are replaced with alternatives, altering specificity of encoded Igs
  3. SH targeted V regions located within a DNA sequence containing
    1500 nucleotides (whole VJ, VDJ)
  4. Frequency of 10-3 per bp per generation (100,000 fold higher than
    spontaneous mutation rate). V region = 600bp = 1 mutation per
    every 2 cell divisions
  5. Most are nucleotide substitutions (mostly random, clustered).
    Increase in antigen affinity = affinity maturation
39
Q

Most are nucleotide substitutions (mostly random, clustered).

Increase in antigen affinity

A

= affinity maturation

40
Q

CDR =

A

Complementarity
determining regions

(regions where Abs
complement an antigen’s
shape)

41
Q

Although a B cell produces only one type of Ig molecule, the heavy chain class can

A

change during cell lineage

42
Q

Explain CLASS SWITCHING…

A
  1. Initial synthesis of IgM only in immature B cells
  2. VDJ splices at RNA level to Cµ
  3. Later synthesis of both IgM and IgD by immature cells
  4. VDJ unit spliced at RNA level to Cd via alternative
    RNA splicing
  5. Synthesis of IgG, IgE or IgA by mature B cells
  6. VDJ spliced to a Cg, Ce, Ca at the DNA level = Somatic
    recombination event. Deletion of intervening sequence.
43
Q

Immunoglobulin Classes - in case of heavy chains

A

different alternatives for constant regions

44
Q

mmunoglobulin Classes - in case of heavy chains,
different alternatives for constant regions.

Specify tissue specific expression and Ig class = 5

A

Class Type of heavy
chain Location

  1. IgA = a Predominant Ig in seromucous
    secretions (saliva, milk),
  2. IgD = d =
    Low in serum but present in large quantities on surface of circulating B cells
  3. IgE e = On surface = membranes of
    basophils and mast cells
  4. IgG = g = Major serum Ig
  5. IgM= µ = Predominant “early” antibody
45
Q

Further recombination to G, A, or E constant regions gives

A

secretory antibodies with specificity to same antigen but with different immune functions.

  1. IgG - binds complement & binds
  2. Fc receptors on macrophages and neutrophils
  3. IgA - constant region recognised by Fc receptor on secretory
    epithelial cells for secretion to saliva, tears, milk, intestinal…
  4. IgE - Bind Fc receptors on mast cells and basophils causing
    secretion of cytokines
46
Q

We have seen that
development of a B cell
proceeds through

A

several stages marked
by the rearrangement
& expression of the
immunoglobulin genes.

47
Q

We have seen that
development of a B cell
proceeds through
several stages marked
by the rearrangement
& expression of the
immunoglobulin genes.

How is this process
controlled?

A

Transcription factors
and other proteins

48
Q

B CELL DEVELOPMENT AND Ig expressed

A

——BONE MARROW——–

  1. Hematopoietic stem cell = none
  2. Lymphoid cell = none

—partial heavy-chain gene rearrangement

  1. Pro-B Cell = none

–complete heavy chain gene rearrangement

  1. Pre B-cell = micro heavy chain + surrogate light chain

—light chain gene rearrangement

  1. IMMATURE B CELL = mlgM

— Change in RNA processing

——-PERIPHERAL LYMPHOID ORGANS ———

  1. MATURE B CELL = mlgM +mlgD

—-Antigen stimulation
—-activated B Cell
—- differentiation
7. IgM-secreting plasma cells = IgM

  1. Memory B Cells of various isotypes

Class Switching = IgG, IgA, IgE = plasma cells secreting various isotypes

49
Q

Development of committed B lineage cells from

A

multipotent progenitor requires the coordinated
activity of many transcription factors.

50
Q

Analysis of regulatory regions of B lineage associated genes & analysis of differential gene expression have
led to the = 3

A
  1. identification of transcription factors
    functioning in early B lineage cells.
  2. More recently, creation of mice lacking expression (KO mice) of transcription factors provide wealth of
    information for their role in lineage commitment.
  3. Factors have been shown to function in a hierarchy &
    also in a combinatorial manner to establish the
    expression of genes in the developing B cell.
51
Q

T-cell receptors are structurally similar to
Immunoglobulins = 4

A
  1. Two polypeptide chains
    – alpha and beta
  2. Each has variable and
    constant region
  3. Genes encode alpha
    and beta chain
    organised like Ig and
    undergo somatic
    recombination
  4. Eg. Human alpha chain
    = 44-46 V, 50 J and
    single C segments

SLIDE 49 IMAGE

Molecular Genetics (21 decks)

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