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Immunology > Antibodies > Flashcards

Antibodies Flashcards

1
Q

What produces antibodies?

A

B cell lymphocytes, and are part of the humoral component of the adaptive immune respose.

B cells originate in bone marrow as resting lymphocyte and become activated by an antigen to differentiate and divide into four daughter cells of either Memory B cells or plasma cells which secrete antibodies.

The specificity of antibody made by plasma cell is determined by the type of antigen receptor on the B cell.

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2
Q

How are Fab arms and Fc portion of antibody defined?

A

On the basis of enzyme digestion by papain which cleaves antibody ino 2 arms and Fc.

Fab- Fragment antigen binding

Fc- fragment crystallisable which has the effector function.

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3
Q

Describe the binding flexiblity

A

Particualrly in IgG, the flexible hinge of the antibody allows it to bind to 2 antigens spaced at different distances adapting to the many arrangements on a pathogen surface.

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4
Q

Describe the binding flexiblity

A

Particualrly in IgG, the flexible hinge of the antibody allows it to bind to 2 antigens spaced at different distances adapting to the many arrangements on a pathogen surface.

This allows for tight binding and longer occupancy time and therefore increases efficient effector function to kill pathogen.

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5
Q

What are the two different types of LIGHT chain on all Fab arms?

A

1) Lambda
2) Kappa

present in all antibody classes; in an individual antibody however, there will only be on type.

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6
Q

List the heavy chain types in each class of antibody

A 
Heavy chain determines class of antibody: 
IgG - GAMMA
IgA -ALPHA
IgM - MICRO
IgE- E
IgD- delta
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7
Q

What is the immuno-globulin fold?

A

The folded structure of the gobular domains; each domain has 2 beta sheets linked by a disulphide bridge to form a B barrel; each sheet made up of several beta strands with loops at the end; sheets lie over eachother.

There are more strands on the variable light domain compared to the constant light domain

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8
Q

What is shown if the amino acids of variable domains on heavy and light chains are compared

A

In both heavy and light varible domains there are three areas of hypervariable loops surrounded by less varible framework regions.

The three loops are joining the strands of sheet; in 3D the loops all lie at one end of the chain.

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9
Q

What forms the antigen binding site?

A

The hypervariable loops; they are known as complementarity determining regions (CDRs)
When Vheavy and Vlight are paired together there are 6 hypervarible loops (three from each) located close together

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10
Q

What forms the antigen binding site?

A

The hypervariable loops; they are known as complementarity determining regions (CDRs)
When Vheavy and Vlight are paired together there are 6 hypervarible loops (three from each) located close together which make a 3D specific antigen binding site structure.

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11
Q

What is a continuous or linear epitope?

A

Epitopes in protein antigens are a single stretch of polypeptide chain

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12
Q

What is a discontinuous epitope?

A

epitope may come from different areas depending on the folded antigen protein chain; the epitope is not all together.

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13
Q

What do epitope bind to?

A

Antigen epitopes fit into pockets, grooves, extended surfaces, projections on antibody antigen binding site.

The epitope are all accessible; a polivirus shows all its epitopes on the surface of its protein for binding.

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14
Q

How many different specificities have been calculated for a functional immune system?

A

10 xe11!

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15
Q

How are so many antigen specificities achieved?

A

Ig genes are organised in a special way. Ig genes are in a fragmented form that cannot be expressed

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16
Q

How are so many antigen specificities achieved?

A

Ig genes are organised in a special way. Ig genes are in a fragmented form that cannot be expressed. This is because for heavy and light chain genes there are families of gene segments arranged sequentially along the chromosome.

When B cells develop these gene segments are REARRANGED in order to make functional light and heavy chain genes and express antibodies.

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17
Q

What does germline configuration refer to?

A

The segments of gene fragments are inherited from egg and sperm

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18
Q

What gene segments encode Variable HEAVY chains?

A

On chromosome 14; where there are:
-40 V gene segment

downstream -23 D

and further downstream-6 J segments.

and finally constant gene segments downstream from there.

Where V encodes most of V domain, J encodes the joining segment and D encodes diversity segment

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19
Q

What gene segments encode Variable LIGHT chains?

A

V and J segments

Where V encodes V domain, J encodes the joing segment

On Kappa light chain on chromosome 2:
40 V segments
5 J segments
and constant gene segments.

On lambda chain: segments are arrangements differently

20
Q

What gene segments encode Variable HEAVY chains?

A

On chromosome 14; where there are:
-46 V gene segment

downstream -23 D

and further downstream-6 J segments.

and finally 9 constant gene segments downstream from there.

Where V encodes most of V domain, J encodes the joining segment and D encodes diversity segment

21
Q

What gene segments encode Variable LIGHT chains?

A

V and J segments

Where V encodes V domain, J encodes the joing segment

On Kappa light chain on chromosome 2:
38 V segments
5 J segments
and constant gene segments.

On lambda chain on chromosome 22: segments are arrangements differently with 33 V, 5 J and 5 constant

22
Q

When does the rearrangement of gene segments occur?

A

H chains rearrange D and J segments during EARLY pro B cell development in Bone marrow. They rearrange V-DJ during LATE pro B cell development.

Light chains rearrange V-J during pre B cell development

23
Q

Describe the rearrangement process in Heavy chains.

A

In pro B cells. First one of D gene segments will become joined to a J gene segment, eg D3 joins J4, where intervening DNA is spliced out irreversibly.

Step 2, where V3 segments joins DJ segment and intervening DNA is spliced out.

This produces a V-D-J gene. Downstream of this is constant region.

Step 3: Transcription of this gene produces primary RNA transcript containing required constant region (determining the class of antibody- u =IgM)

Step 4: There is RNA splicing to get rid of intervening DNA to produce messenger mRNA

Step 5: Translation creates u chain polypeptide

24
Q

What does somatic recombination refer to?

A

recombination in cells other than germ cells, here the developing B cells.

25
Q

Describe the rearrangement process in KAPPA Light chains.

A

In pre B cells, later than heavy chain rearragment.

Step 1: V segments join the J segment (eg, V4 joins J2) and DNA in between V and J is spliced out.

Step 2: Transcription of this gene makes primary RNA

Step 3: Splicing of RNA to remove DNA between J segment and constant segment to make mRNA.

Step 4: Translation of mRNA makes a kappa chain polypeptide.

26
Q

What are repeated rearrangements in Light chains?

A

If V segment joins J segment but results in an non-productive joint, another V-J combination with remaining J segments on RNA is tried so a functional antibody is produce.

27
Q

How does B cell know which parts of DNA to splice out of gene during segment rearranging?

A

Specific sequences called Recombination Signal Sequences (RSS) These are conserved and non coding.

28
Q

How does B cell know which parts of DNA to splice out of gene during segment rearranging?

A

Specific sequences called Recombination Signal Sequences (RSS) These are conserved and non coding.

It consists of a heptamer sequence, followed by non conserved 23 bp, and then a nonamer sequence. Or nonamer, 12bp, heptamer segment. These are are at the inside of each of ligating segments.

There are a set of enzymes which recombine V-D-J; these are V(D)J RECOMBINASES.

29
Q

What is the The 12, 23 rule?

A

RSS sequences is always 12bp recombine with 23bp spacer; never 12 and 12 for example.

30
Q

What components are only made in lymphocytes?

A

RAG1 and RAG2: recombination activating gene which allow recombination to occur in B cells These associate with each other and other proteins to form a RAG complex.

31
Q

Give examples of other enzymes in nucleated cells other than RAG complex. What are their functions?

A
  • DNA ligase IV
  • DNA-dependant protein kinase
  • Artemis nuclease
  • Ku protein

Funtions include:

  • repair of ds DNA
  • bending of DNA
  • modifying broken DNA ends
32
Q

Describe the process of DNA splicing involving the RAG complex.

A

1) RAG complex binds to 23 spacer RSS and another RAG complex binds to 12 spacer RSS
2) Binding aligns two RSS and cleaves DNA at the ends of gene segments.
3) DNA hairpin is formed (starting and ending at the ends of each segment which when cleaved is held in place by the RAG complexes.
4) broken ends of hairpin and segments are rejoined by DNA repair enzymes.
5) A coding joint is formed in chromosome containing ligated segments. A single joint is formed between ends on hairpin.

33
Q

Describe additional sequence diversity in the hypervariable loops (eg, CDR3) that is found when coding joints are formed

A

If D and J ligated; where DNA is cleaved and an OPEN DNA hairpin is made.

The complex open hair pins at different positions; when it opens hairpins it makes the double strand palindrome DNA into single strand DNA of P nucleotides.

The ends of the P nucleotide sequence is modified by exonucleases that remove nucleotides or TdT which adds nucleotides.

34
Q

What is a P nucleotide and What is an N nucleotide?

A

P nucleotide, is the single stranded end from which complementary bases orginally on adjaent double strandsare on same stran.

N nucleotide are non-templated

35
Q

Describe additional sequence diversity in the hypervariable loops (eg, CDR3) that is found when coding joints are formed; What is junctional diversity?

A

If D and J ligated; where DNA is cleaved and an OPEN DNA hairpin is made.

The complex open hair pins at different positions; when it opens hairpins it makes the double strand palindrome DNA into single strand DNA of P nucleotides.

The ends of the P nucleotide sequence are modified by exonucleases that remove nucleotides or TdT which adds nucleotides.

These ends anneal; removing nucleotides that don’t pair with exonucleases and nucleotides are filled in where needed giving a combination of original P nucleotides and recruited N nucleotides of completely new piece of sequence at the joining point.

36
Q

What are the four main processes of diversity in antibodies?

A

In naive B cells:
1) Combinatorial diversity where many copies of gene segmenrs combine in different ways.

2) Junctional diversity where nucleotides are added or subtracted at the joints between segments.
3) Many combination of light and heavy chains varible regions pair up to make different antigen binding sites.

In activated B cells:
4) Somatic hypermutation produces POINT MUTATION into the rearranged variable region once B cell is activated.

37
Q

What does comparison with human and mouse constant regions reveal about our Ig evolution?

A

We have had a duplication of IgG , and IgA regions.

38
Q

Describe the heavy chain locus

A

Constant heavy chain are 3’ to J segments with various exons within each C gene which correspond to a class of Ig.

39
Q

What is isotype switching?

A

Retains same V-D-J segment rearranged but switch the constant region so a different antibody isotope.

40
Q

What antibodies are expressed in developing naive B cells?

A

IgM and IgG expressed on naive cell surface.

This is done through alternative splicing where the same transcript is spliced in two different ways.

41
Q

What is allelic exclusion?

A

Each B cell will only produce antibody with a SINGLE antigen specificity.

Therefore only 1 copy of heavy chain locus and only 1 copy of light chain locus are used to undergo rearrangement.

42
Q

What is the BCR?

A

The B cell recepor is made of IgM in the membrane via hydrophobic regions at the C terminus. IgM ACTS AS ANTIGEN BINDING SITE. It is associated with two invariant chains called immunoglobulin alpha and beta (Iga, Ig-beta) which is important for signalling into B cell after antigen binds.

Therefore the first antigen produce will be IgM; plasma cells will make antibody the same as membrane bound receptor on presursor.

43
Q

How are transmembrane Ig and secreted Ig distinguished? How are the made to be different?

A

Transmembrane has hydrophobic tail to anchor Ig into membrane but secreted Ig has a hydrophilic secretory tail.

Alternative RNA splicing is used to include or exclude the appropriate exons for each.

44
Q

What two ways do helper T cells help B cells

A

1) Co-stimulatory signal through binding of CD40 ligand on B cell surface and CD40L on T cell surface
2) Secretion of cytokines with aid induction of isotype switching.

Helper T cells activate B cell to produce antibody to tackle viral coat protein.

45
Q

Which antibody classes also have subclasess?

A

IgA and IgG

46
Q

How are the different classes of antibodies produced?

A 
Through isotope switching or class switching 
Where Varible heavy exons with different Constant heavy genes the specificity is the same across different classes.

This recombination depends on enzyme AID to excise previous Constant gene via the switch regions upstream from constant genes; the excised switch region and old constant gene is done via a hairpin loop formation.

Immunology (19 decks)

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