Immunology 3

Question 1

Structure of immunoglobulins/antibodies

Answer 1

Y shaped molecule composed of
2x Light chains (25 kDa) (outsides of top of y)
2x Heavy chains (50 kDa)

2 disulphide bonds link the heavy chains
2 further disulphide bonds, one linking each light chain to its heavy chain partner

Question 2

Functions of immunoglobulin and where they occur:

Answer 2

• To recognise and bind antigen
• To then elicit effector functions
  These two functions are performed by different portions of the antibody molecule.
• N terminus of light and heavy chains = variable reason- antigen binding - 2 sites
• C terminus of light and heavy chains= Constant region - recruits additional immune molecules/ cells to destroy pathogens.

Question 3

Domains of light and heavy chains:

Answer 3

Each light chain is composed of 1 variable domain and 1 constant domain.
Each heavy chain is composed of 1 variable and 3 constant domain.
The constant domains are found at the C terminal ends of the polypeptides. The variable domains are found at the N terminal ends of the polypeptide chains.

Question 4

Papain digestion produces:

Answer 4

Papain digestion generates Fab fragments which have 1 antigen binding site and the Fc region. The Fc region binds to Fc receptors and elicits immune functions.

Question 5

Digestion with Pepsin:

Answer 5

Digestion with Pepsin generates small peptide fragments as it chops up a large proportion of the constant region. It also generates F(ab’)2 fragment. This can recognise antigen, but it has 2 antigen binding sites as the 2 antigen binding sites have not been separated.

Question 6

2 types of light chains:

Answer 6

κ and λ
Antibodies have one of either- but never both each.
2:1 ratio.

Question 7

immunoglobulin classes:

number and determination?

Answer 7

5 different classes (and additional subtypes) of Ig, class is determined heavy chains present,
γ α μ δ ε
IgG, IgA, IgM, IgD, IgE.

Question 8

Immunoglobulin fold

Answer 8

Heavy and light chains contain similar repeated domains. The light chain has 2 Ig folds and each heavy chain has 4 Ig folds.

Question 9

Secondary protein structure of Igs and bonds:

Answer 9

Ig domain (fold) composed of 2 antiparallel β pleated sheets held together by a disulphide bond.
The β strands are linked by flexible loops.

Question 10

Antigen binding forces:

Answer 10

Electrostatic forces
Hydrophobic forces
Hydrogen bonds
Van der Waals forces

Question 11

Antigen and antibody binding interrupted by:

Answer 11

1. High salt concentrations
2. Extreme pH
3. Detergents
4. High concentrations of purified epitope

Question 12

FC region- what is it?

Answer 12

FC = fragment crystallisation

Area of the antibody that binds to Fc receptors on immune cells to elicit immune functions

Question 13

What determines functions of Igs:

and what are the functions?

Answer 13

• Determined by the Fc regions
  1. Neutralisation. Antibodies bind to pathogen and toxins.
  2. Coating of pathogen with antibody = opsonization.
  3. Activation of the complement pathway to act as opsonin or lysis.

Question 14

Neutralisation function of Ig

Answer 14

1. Neutralisation. Pathogens or toxins gain entry to host cells by binding to certain cellular receptors.
   Antibodies binding to pathogens or toxins can block the interactions of pathogen with receptors. This prevents microbial species that use host cells as sites for replication, such as viruses or bacteria like mycobacteria species which hide from the immune system inside cells.

Question 15

Opsonization function of Igs

Answer 15

2. An antibody that is specific for a microbial molecule will bind to and coat the surface of the microbe.
   This antibody can then directly act as an opsonin and be recognised by Fc receptors on the surface of phagocytic cells, triggering engulfment.

Question 16

Complement pathway activation function of Igs

Answer 16

3. The activation of the complement pathway.
   The classical pathway of complement can be activated by binding the antibody that has already bound to a microbe.
   3 possible functional outcomes of complement: Opsonisation, anaphlylatoxin production and the MAC.

Question 17

Central (primary) lymphoid organs:

+significance?

Answer 17

Thymus
Bone marrow
Where the B and T cells develop from progenitors and differentiate into naive lymphocytes. B cells develop in the bone marrow and T cells in the thymus.

Question 18

Peripheral (secondary) lymphoid organs

Answer 18

• Appendix
• Lymph nodes
• Adenoids
• Tonsils
• Spleen
• Payers patches

Question 19

The total no. diversity of antibody specificities?

Answer 19

1 x 10 11.

Question 20

Why do we have need for a diverse repertoire of antibodies?

Answer 20

Required to recognise a vast array of more rapidly evolving pathogens.

Question 21

How do we achieve variation in Igs? issue with space- how is this solved?

Answer 21

Variation in immunoglobulin is derived from changes in the amino acid sequence and therefore DNA sequences encoding the CDRs.
Solution using somatic recombination has evolved to provide this basis of vertebrate adaptive immunity.
This mechanism is used in both B and T cell progenitors to generate the genes encoding the Ig and BCR of B cells and the TCR of T cells.

Question 22

Somatic recombination mechanism?

Answer 22

V regions of Ig genes are encoded in multiple pieces = gene segments
Multiple choices for each gene segment = GERMLINE DIVERSITY
Segments randomly selected, brought together in different combinations
This provides COMBINATORIAL DIVERSITY

Question 23

Controlling rearranagement?

Answer 23

This process is guided and controlled by conserved noncoding DNA sequences that are adjacent to the points at which segments joins.
They are called Recombination Signal Sequences, they have a structure that consists of a conserved heptamer (7 base pair sequence) followed by a spacer of either 12 or 23 base pairs followed by a conserved nonamer (a sequence of 9 base pairs).

Question 24

3 mechanisms that contribute to Ig and TCR diversity?

Answer 24

Germline diversity, with different gene segments being available.
Combinatorial diversity from segments being brought together in different combinations and also different combinations of heavy and light chains.
Finally junctional diversity comes from variety being added during the joining process.

Question 25

Junctional diversity

Answer 25

DNA is cut at the boundary of the RSS and temporary loops are formed, then reopened and rejoined to the other segments. To make these joints both strands of the DNA have to be the same length and sometimes this means there needs to be trimming or filling in of bases. This creates altered sequences at the joints between segments which is call junctional diversity.

Question 26

Types of junctional diversity and enzymes involved?

Answer 26

The position of the cut adds P nucleotides to the sequence, whilst the TDT adds N nucleotides.
The RAG proteins, encoded by the Recombination Activating Genes cleave the DNA making the ends that are then processed and joined. TdT (terminal deoxynucelotidyl transferase) will add the P nucleotides to the sequence to make equal length strands that can be rejoined.