Antibody diversity is generated by multiple mechanisms, each of which contributes to the generation of antibodies with up to 1011 different amino acid sequences in their antigen-binding sites. Several of these mechanisms involve changes in the DNA sequences encoding the antibody heavy and light chain proteins. One mechanism that does not rely on changes to the DNA within the immunoglobulin heavy and light chain gene loci is, instead, dependent on: The contributions of amino acids from both the heavy chain and the light chain to form the antigen-binding site The random usage of V, D, and J gene segments to form the heavy chain V region sequence The random usage of k light chains versus l light chains to pair with the heavy chain The activity of TdT to add random nucleotides at the junctions between the V, J, and D region sequences The fact that heavy chain V regions contain an extra gene segment encoded by the D region compared to light chain V regions

M/C Which statement is the best with respect to how the mouse activates the interferon (INF) pathway with this virus based on the graph below? Infection of mice that are wild type or have the indicated protein or cell deletions 120 100 80 % Survival 40 60 60 20 20 0 0 2 4 T 6 Days after infection wild type RIGI minus Toll receptor minus INF a/ẞ minus CD8 T cell minus INF a/ẞ + CD8 T cell minus 8 10 12 Toll receptors are required for activation of the INF pathway. The INF pathway is activated through recognition of the virus by RIGI. Both Toll receptors and RIGI are required for activation of the INF pathway. Neither Toll receptors or RIGI are likely involved in INF pathway activation.

An immunodeficient mouse strain is identified, that has a single gene defect causing its disease. Mice with this defect have greatly impaired responses to protein antigens following subcutaneous immunization and also exhibit severely delays in the kinetics of their antibody responses. Analysis of their lymph nodes revealed profound alterations in the normal architecture, with a lack of organization of distinct T-cell and B-cell zones. A likely candidate for the defect in these mice is: The chemokine receptor CCR7, which recruits B cells, T cells, and dendritic cells to lymph nodes. The TNF family member LT-b, which is made by Lti cells. The chemokine receptor CXCR5 that recruits B cells to the lymph node follicles. Prox1, the transcription factor required for the development of lymphatic vessels. TNF-a, which is required for the development of FDCs.

Virus infections induce production of interferons that act on infected cells to enhance their recognition by CD8 cytotoxic T cells. To counter these mechanisms, viruses often encode proteins that interfere with antigen processing and presentation. In an experiment, cells infected with Virus X are treated with interferon and compared with uninfected cells treated with interferon. Proteasomes are isolated from the two cell populations and their enzymatic activities are compared. The data in figure below show the amino acid preferences for cleavage of peptides by the two samples of proteasomes. Based on these data, Virus X most likely encodes a protein that interferes with: The expression of MHC class I on the surface of the infected cell The rate at which peptides are produced from intact proteins in the infected cell The transport of peptides from the cytosol to the endoplasmic reticulum in the infected cell The replacement of constitutive proteasome subunits with immunoproteasome…

Infants and young children with deficiencies in specific complement components present with recurrent respiratory infections caused by extracellular bacteria. The peak age of susceptibility is between 6 and 12 months after birth. At this time, as shown in figure below, maternal antibodies acquired by the child during fetal gestation are nearly gone, but the child is not yet generating robust antibody responses to new infections, as indicated by the low circulating levels of IgG and IgA. As children with this immunodeficiency get older, they outgrow this disease and show no further evidence of these recurrent infections. Based on this information, name one likely gene deficiency (in the complement system) that could cause this primary immunodeficiency, and the specific complement pathway likely to be affected. Explain your answer.

Explain why transplanting skin graft from the black mouse to white mouse results in rejection (A) Why (A) is it called a primary response?

Knockout mice are mice in which certain genes are rendered irreversibly nonfunctional through the technique of homologous recombination. Which knockout mice would you use to study the two pathways of antigen processing and presentation?

Addition of immunoglobulin G (IgG) specific for hemoglobin to a solution of hemoglobin results in the formation of a red precipitate. In contrast, addition of the Fab fragments from this antibody to hemoglobin results in no such precipitate. What could explain this difference in results? Treatment with papain produces Fab fragments with different antigen specificity than the original IgG molecule. IgG can simultaneously bind two different antigens, whereas an Fab fragment can only bind one antigen at a time. The Fab fragments preferentially bind to other Fab fragments rather than to hemoglobin. The hemoglobin molecule antibody-binding sites can bind IgG molecules, but cannot bind Fab fragments.

¹C2²2S + E Sk3,0 where S is a T-cell of a particular variety, E is the corresponding (free, i.e. unbound) antigen, and C is a complex consisting of a T-cell bound to an antigen. Ø indicates the loss of T-cells due to decay as described above. 1. Let s, c and e be the concentration of S, C and E, respectively, in the body at time t. Assume that the background exposure to the disease results in a constant concentration eo of the antigen (free or bound) in the body. (b) (c) Consider the T-cell system. Assuming mass action kinetics, write down a system of ODEs for Show that ds dt S + E and k-1 where KM dc dt Show (include your working) that the 2D system of ODEs has two equilibria: (s*,c*) = (0,0) == : -k₁(eo — c)s + k_1c +2k₂c - k3s k-1+k2 k₁ (s*, c*) = ( k₂e0 k3 and KM, eo k2 dt' dt and KM = k₁(eo - c)s - k_1c - k₂c.