Antibody function and genes

Question 1

What is antibody valence?

Answer 1

Valence refers to the number of antigenic determinants an antibody molecule can bind.

Question 2

What is an epitope/antigenetic determinant?

Answer 2

Not all of an antigen binds specifically to an antibody; the part that actually interacts is usually 10 to 20 amino acids long, and is called an epitope or antigenic determinant. Typical proteins have several epitopes which elicit and bind to different antibodies.

Question 3

Please describe precipitation and agglutination.

Answer 3

The large immune complexes that are formed at or near equivalence tend to become insoluble and fall out of solution. When the antigen is a molecule, the phenomenon is called precipitation; when it’s a cell or cell-sized particle, it is called agglutination.

Question 4

What is affinity?

Answer 4

Affinity: The strength with which an antibody molecule binds an epitope (antigenic determinant) is called its affinity.

Question 5

Discuss why a line of precipitate may form in agar gel when antigen and antibody diffuse towards each other.

Answer 5

If you take a layer of agar gel, cut two holes in it, and put antibody in one and antigen in the other, they will begin diffusing radially out of their wells: In the area between the two wells, the situation is similar to a quantitative precipitin test; there is antigen excess near the antigen well, and antibody excess near the antibody well, so somewhere in between, eventually, equivalence must be reached. But: what happens at equivalence? The complex precipitates. We go back next day and look at our agar gel, and see… …a line of precipitate between the wells! This technique is called immunodiffusion.

Question 6

Compare and contrast precipitation and agglutination in terms of the nature of the antigens involved, and sensitivity of the tests.

Answer 6

Agglutination is more readily detected than precipitation, and so an agglutination test is more sensitive. When the antigen is a molecule, the phenomenon is called precipitation; when it’s a cell or cell-sized particle, it is called agglutination.

Question 7

List the components of complement in the order in which they become activated in the classical pathway. Name those that are also activated in the alternative pathway.

Answer 7

Complement is a large number of proteins, similar to the blood clotting system in that each exists in an inactive form, and when the first is activated the rest follow in a sort of cascade.There are at least three ways to activate the C cascade; the one that is most familiar is the classical pathway. More recently, an alternative and a lectin pathway have been described. Each pathway gets started differently â–ºbut all come together by C5.The classical pathway is activated by complexes of IgG or IgM antibody with antigen. There is a change in the Fc portions of the antibodies after interaction with antigen, which allows the binding and activation of C1q. C1 activates C4 and then C2, which together activate C3, which can then activate C5. â–ºClassical C counts: 1-4-2-3-5-6-7-8-9.

Question 8

Discuss the different ways in which complement is activated by IgG and IgM. Identify the complement components which are: opsonizing; lytic; anaphylatoxic; and chemotactic.

Answer 8

â–ºThe C1q must interact with two Fcs simultaneously; it does so either by finding two IgGs close together, or a single IgM (this reinforces the point that IgM is a much more efficient a C activator than is IgG).It is LYTIC, as just described, if the membrane attack complex is activated. Neisseria (gonorrhea, meningitis) are by far the most susceptible family of bacteria to C lysis.It is OPSONIZING. One split product of activated C3, namely C3b, adheres to membranes. Phagocytic cells (PMN, macrophages) have C3b receptors, and so can get a firm grip on an antigen if it is opsonized with C3b. As we said before, IgG is also opsonizing, because phagocytes have receptors for its Fc end called FcR (there are several different FcR).It is CHEMOTACTIC. The C5 activation product, C5a, is chemotactic for phagocytes, especially neutrophils. This explains much of the inflammation in an antibody-mediated reaction, and why PMN are the hallmarks of such a reaction.It is ANAPHYLATOXIC. C3a, C4a and C5a are all capable of releasing histamine non-specifically from mast cells or basophils. This means that there will be increased blood flow in the area of antigen deposition, and a better chance for inflammatory cells to get out of the blood and into the tissues. Sometime, a person with a lot of complement activation will break out in hives, and you can confuse what’s going on with an allergic reaction.

Question 9

Discuss how complement is important in immunity to bacteria even if the bacteria are resistant to lysis by C9. Identify the family of bacteria for which lysis is necessary for their clearance.

Answer 9

Remember - complement is both chemotactic and opsonizing. If it can’t lyse it’s subject it sure as hell can call in the guns (PMN’s and such).Regardless, here is how lysing works: The formation of the membrane attack (lytic) complex: C5, activated by any of the three pathways described, activates C6, C7, C8, and C9. C8 and C9 form a lesion on the target cell membrane which, on electron microscopy, looks like a hole; the cell loses its ability to regulate its osmotic pressure and lyses or pops.Nyseria: gonorrhea and meningitis

Question 10

Discuss how complement plays roles in both innate and adaptive immunity.

Answer 10

The alternative pathway seems to be a more primitive, early, less-specific sort of defense, since it can work even without waiting for antibody to be made.There is also the lectin pathway of complement activation, truly part of innate immunity. The lectin pathway is mediated by mannose-binding protein, MBP or MBL, a lectin. Lectins are proteins that bind (usually foreign) carbohydrates. MBP binds certain mannose–containing structures found in carbohydrates of bacteria but not humans.Adaptive: B-cells, antibodies, and the activation of the classical pathway.

Question 11

Describe how the lectin pathway of complement activation is triggered.

Answer 11

MBP is functionally similar to C1q in the classical complement pathway, so the lectin pathway goes MBP-4-2-3-5-6-7-8-9.

Question 12

Please define toxoid.

Answer 12

A toxoid is a bacterial toxin whose toxicity has been weakened or suppressed either by chemical (formalin) or heat treatment, while other properties, typically immunogenicity, are maintained.

Question 13

Please describe DNA recombination.

Answer 13

DNA recombination refers to the process that a DNA segment moves from one DNA molecule to another DNA molecule. Changing the relative positions of two pieces of DNA is called recombination.

Question 14

Please define primary rna transcript.

Answer 14

A primary transcript is an RNA molecule that has not yet undergone any modification after its synthesis.

Question 15

Please describe RNA splicing.

Answer 15

Splicing is a modification of an RNA after transcription, in which introns are removed and exons are joined.

Question 16

What is mRNA?

Answer 16

Seriously?mRNA is a molecule of RNA encoding a chemical blueprint” for a protein product.”

Question 17

What is a somatic mutation?

Answer 17

Alterations in DNA that occur after conception. Somatic mutations can occur in any of the cells of the body except the germ cells (sperm and egg) and therefore are not passed on to children.

Question 18

Explain, in terms of lymphocyte activation, how a self antigen might not itself elicit antibody, but might react with antibody elicited by a cross-reacting antigen.

Answer 18

If a non-self antigen X is very similar to a self antigen Y, X and Y can be cross-reacting antigens. A lymphocyte with a receptor complementary to non-self antigen X will bind self antigen Y with low affinity but the low affinity binding is not enough to activate production of antibody. When non-self antigen X binds its complementary receptor on the lymphocyte, it binds with high affinity and this tight binding activates antibody production. The antibodies produced can then go on to bind self antigen Y with low affinity and sometimes this binding is good enough to be inconvenient.eg. Group A streptococci antigens cross react with an antigen, laminin found in heart valves. Because of this cross reaction, streptococci antigens can activate production of antibodies that bind laminin and, in some people, this interaction can lead to a destructive inflammatory heart disease called rheumatic heart disease.

Question 19

Discuss the Clonal Selection Theory in term of: the number of different receptor specificities it postulates per cell; the role antigen plays in the initial expression of receptors; the role of antigen in clonal selection; an experiment which provides strong evidence for the theory; how it differs from an instructional theory; whether it is Darwinian or Lamarckian.

Answer 19

1 receptor specificity per cellEach cell has receptors for one specific antigen When antigen binds a cell, it activates it, resulting of the expansion of that clone and production of that antibody.Experiment: An antigen, X, was coupled with an intensely radioactive label, such that any cell it bound to would quickly die from radiation. If instruction was the mechanism, enough Ag given would bind all the animal’s B cells, because they would be “non-specific” until instructed; so no antibody could be made. If selection was correct, only the pre-existing B cells with randomly expressed receptors for X would bind and die; all others would survive. A week after getting radioactive X, the animal was immunized with nonradioactive X and with Y, and unrelated antigen. It responded normally to Y, but not to X. Selection of pre-existing clones was proven.Darwinian

Question 20

Define allotypic exclusion. Demonstrate your knowledge of the concept by first stating the number of chromosomes in a cell which bear H or L genes, and then the number that actually contribute to a single B cell’s antibody product.

Answer 20

Recall, allotype = the allele of the antibody chains found in the individual.Allotypic exclusion: So although any one cell is theoretically capable of making 2 H chains (by rearranging both maternal and paternal loci,) and 4 light chains (maternal and paternal, lambda and κ,) that never happens;â–º it makes only one of each—all other alleles are excluded.2 chromosomes in cells have H and L genes because we are diploidOnly one contributes to the antibody product, maternal or paternal- much like X chromosome inactivation

Question 21

Describe the somatic recombination model which explains how antibodies of the same specificity (that is, with the same CDRs and idiotype) can be found in two or more different classes (“class switching”).

Answer 21

A single mature B cell starts by making both IgM and IgD, which it puts into its membrane as receptors. Later it may switch to making IgG, IgE, or IgA. â–ºIn all cases, the V domain stays the same but the C region of the H chain changes. As may be anticipated by now, what happens is that the cell which has put its particular H-chain VDJ combination together with its mu and delta genes (as shown in the diagrams) goes back to its DNA, does a loop-out of mu and delta, and puts VDJ next to the C-region gene of gamma or epsilon or alpha, while excising and discarding the intervening DNA.The new mRNA, then, may be VDJα or VDJγ or VDJε. â–ºThus a cell which is making IgM can go on to make IgG, but a cell making IgG cannot go back to making IgM; the mu information is physically gone. “M to G” or “M to A” or “M to E” switching is common in antibody responses, and requires T cell help; without it, only IgM responses are possible.â–ºRemember that the cell switches heavy chain class, but doesn’t switch light chains; they remain the same throughout the B cell’s life.

Question 22

Show how breaking the variable region gene up into V, D and J subregions requires fewer genes.

Answer 22

Breaking the gene region up allows for more combinations so that the DNA can get folded and cut in more combinations.

Question 23

Calculate the minimal number of genes required to code for a million different antibody molecules, based on the (outdated) concept of “one gene-one H or L chain”.

Answer 23

1000 H* 1000L = 1,000,000 antibody molecules. Yay for multiplication!

Question 24

Explain why we commonly write V(D)J instead of VDJ.

Answer 24

The variable domain region of HEAVY chain genes is composed of multiple V, multiple D, and multiple J gene segments.The V region of LIGHT chains into only V and J segments, so generically we say ‘V(D)J.’The cell will choose one of its V’s, one D, and one J to make a VH domain gene region. L chain gene rearrangement is similar, â–ºbut they have only V and J segments, no D; and only one C domain gene, kappa or lambda, depending on the cell’s choice.

Question 25

Describe the essential difference between the somatic mutation and germ line hypotheses of immunological diversity.

Answer 25

There used to be two schools of thought about antibody diversity: one said that all the V genes were in the germ line; if you looked at a fertilized ovum you could predict all potential antibodies that potential individual could potentially make. The other said that only a few were there. It postulated that during embryonic lymphoid development these genes underwent repeated (somatic) mutation until a full complement of antibodies was produced.Both theories, it turns out, were right. As we’ve just discussed, a lot of our diversity is in the germ line (that is, in the individual V, D, and J segments you’re born with).Even more diversity is also generated by variable (“sloppy”) V/J and V/D joining.

Question 26

Define somatic hypermutation and distinguish it from the somatic mutation mechanism that produces N-region diversity.

Answer 26

The recombined V(D)J unit is “hypermutable”; each time a B cell divides after antigenic stimulation there is a good chance that one of the daughters will make a slightly different antibody. Selection of the best-fitting mutants after antigenic stimulation allows a gradual increase of affinity during an immune response—an exceptionally nice design feature called affinity maturation.For T cells, somatic mutation after contact with antigen does not seem to take place.N- region diversity: The V-D and D-J joins are “sloppy.” The cell uses randomizing mechanisms: First, exonucleases for chewing away a few nucleotides after the DNA is cut but before two gene segments (D to J, V to DJ) are joined. Second, an enzyme called terminal deoxynucleotidyl transferase, TdT, which doesn’t use a template addsa few nucleotides randomly as well.â–ºThus you can’t predict the sequence at the joining area (â–ºwhich is called an “N” region); it might be obvious that V7 has joined to D2, let’s say, but in this cell there’s an extra alanine and tyrosine there, and in that one there is a leucine missing. This produces a lot more diversity. There is a price for it: two times out of three the N region, being of random length, will create a frame-shift mutation, that is, a nonsense codon which terminates transcription.