Brock Biology of Microorganisms (15th Edition)
15th Edition
ISBN: 9780134261928
Author: Michael T. Madigan, Kelly S. Bender, Daniel H. Buckley, W. Matthew Sattley, David A. Stahl
Publisher: PEARSON
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Chapter 27.6, Problem 1CR
Q Polymorphism implies that each different MHC protein binds a different peptide motif. For the MHC class I polymorphisms, how many different MHC proteins are expressed in an individual? How many by the entire human population?
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The extensive polymorphism of MHC genes in the population is thought to represent an evolutionary response to outflank the evasive strategies of pathogens. This polymorphism makes it difficult for pathogens to eliminate all potential MHC binding epitopes from their proteins. Based on this reasoning, it would seem advantageous for each individual to encode more than three different MHC class I and three different MHC class II genes per chromosome copy. If some individuals in the population had MHC loci that encoded 10 different MHC class I and 10 different MHC class II genes, the T cell repertoire in those individuals would likely be:
Much more diverse than in the rest of the individuals of that population
Much better at recognizing rare pathogens not encountered by most individuals in that population
Much less diverse than the rest of the individuals in that population
Much more alloreactive than the T cells found in the other individuals of that population
Very reactive to bacterial…
A cell line carrying a mutation in a single gene is found to express very low levels of MHC class I on its surface. When infected with influenza virus, these cells are not recognized nor are they killed by a CD8 T cell line specific for an influenza peptide bound to the MHC class I protein expressed by these cells. Incubation of the mutant cell line with a large excess of this peptide in the cell culture medium overnight leads to the results shown below. What is the most likely candidate for the gene that is defective in the mutant cell line?
In rare instances, B cells can be found that have two immunoglobulin light chain alleles, both of which are rearranged in frame, and can encode functional light chain proteins. Yet, on the surface of the B cell, only one of the two light chain proteins is detected in the membrane-bound immunoglobulin receptor. The reason these rare cells have two functional light chain rearrangements but only express one of the two light chains as part of the B-cell receptor is:
One of the two light chains is formed from rearrangement of a V gene segment that is a pseudogene.
One of the two light chain proteins doesn’t form a stable complex with the heavy chain expressed in this cell.
One of the two light chain alleles is not transcribed efficiently, and produces only low levels of protein.
One of the two light chain alleles uses a V gene segment that is not targeted very often by the RAG recombinase.
One of the two light chains is rapidly degraded after synthesis due to improper folding.
Chapter 27 Solutions
Brock Biology of Microorganisms (15th Edition)
Ch. 27.1 - Prob. 1MQCh. 27.1 - Prob. 2MQCh. 27.1 - Distinguish between clonal deletion and clonal...Ch. 27.1 - QWhy is it necessary that all three defining...Ch. 27.2 - Identify the intrinsic and extrinsic properties of...Ch. 27.2 - Describe an epitope recognized by an antibody, and...Ch. 27.2 - Give an example for each: natural and artificial...Ch. 27.2 - QWhat properties are required for a vaccine to...Ch. 27.3 - Summarize antibody production starting with...Ch. 27.3 - Differentiate among antibody classes using...
Ch. 27.3 - Prob. 3MQCh. 27.3 - QDescribe the structural and functional...Ch. 27.4 - Draw a complete Ig molecule and identify...Ch. 27.4 - Describe antigen binding to the CDR1, 2, and 3...Ch. 27.4 - Describe the recombination events that produce a...Ch. 27.4 - QWhich Ig chains are used to construct a complete...Ch. 27.5 - Identify the cells that display MHC class I and...Ch. 27.5 - Compare the MHC I and MHC II protein structures...Ch. 27.5 - Define the sequence of events for processing and...Ch. 27.5 - QDescribe the basic structure of class I and class...Ch. 27.6 - Define polymorphism and polygeny as they apply to...Ch. 27.6 - How does a single MHC protein present many...Ch. 27.6 - QPolymorphism implies that each different MHC...Ch. 27.7 - Prob. 1MQCh. 27.7 - Identify diversity-generating mechanisms unique to...Ch. 27.7 - Describe and compare the structural features of Ig...Ch. 27.7 - QWhat diversity-generating mechanisms function to...Ch. 27.8 - Describe the mechanism used by Tc cells to...Ch. 27.8 - Describe the effector system (the cell-killing...Ch. 27.8 - Compare and contrast the roles and activities of...Ch. 27.8 - QWhat mechanism do Tc cells use to identify and...Ch. 27.9 - Discriminate between immediate hypersensitivity...Ch. 27.9 - Provide examples and mechanisms for an...Ch. 27.9 - QHow do immediate and delayed-type...Ch. 27.10 - Describe the binding site for superantigens on T...Ch. 27.10 - Compare and contrast the immunodeficiency observed...Ch. 27.10 - Prob. 3MQCh. 27.10 - Prob. 1CRCh. 27 - Antibodies of the IgA class are probably more...Ch. 27 - Prob. 2AQCh. 27 - Polymorphism implies that each different MHC...Ch. 27 - What problems would arise if a person had a...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Most of the enzymes involved in immunoglobulin gene rearrangement are ubiquitously expressed in all cells of the body. However, the specific recombination events between V, J, and D gene segments that generate antibody diversity occur only in developing B cells. How do RAG-1 and RAG-2 ensure that recombination takes place at antibody gene segments?arrow_forwardVariable addition and subtraction of nucleotides at the junctions between gene segments contributes to the diversity of [Qi] the third hypervariable region. The addition and subtraction of nucleotides at the junctions between V, D, and J gene segments creates antibody proteins with wide variations in the numbers of amino acids in their CDR3 regions. This variability in CDR3 length is important as: Overall variability in CDR3 sequence is needed to create a sufficiently diverse antibody repertoire. The CDR3 region is more important in binding antigen than the CDR1 and CDR2 regions are. Some light chains bind better to heavy chains with longer CDR3 region sequences. Longer CDR3 sequences generally create antibodies with higher affinity for the antigen. Some antibodies bind relatively flat surfaces and others bind deep clefts in the antigen.arrow_forwardThe immunoglobulin molecules of a particular mammalian species have kappa and lambda light chains and heavy chains. The kappa gene consists of 250 V and 8 J segments. The lambda gene contains 200 V and 4 J segments. The gene for the heavy chain consists of 300 V, 8 J, and 4 D segments. If just somatic recombination and random combinations of light and heavy chains are taken into consideration, how many different types of antibodies can be produced by this species?arrow_forward
- Each immunoglobulin (Ig) domain is composed of a structure known as a ‘b-sandwich,’ which consists of two b sheets covalently linked by a disulfide bond. Only a subset of the ~110 amino acids in each domain are required to establish this overall structure, and it is these amino acids that are highly conserved when comparing Ig domains to each other. What might be the advantage of this structure for use as antibody variable domains?arrow_forwardWhy is it important that cell surface markers be similar when transplanting donor tissue into a host? How do the alleles of the glycosyltransferase (GT) gene control the presence of A and B antigens on the surface of red blood cells?arrow_forwardSuppose that a particular species of vertebrate animal has immunoglobulin heavy-chain genes that include 55 V segments, 21 D segments, and 8 J segments, and has immunoglobulin light-chain genes that include 41 V segments and 6 J segments. How many different immunoglobulin structures could be produced by mature B-lymphocytes? (Assume that segments are spliced without any junctional variation or imprecision).arrow_forward
- gamma:delta T-cell receptors are also generated by gene rearrangement. Some T cells express gamma:delta T-cell receptors rather than alpha:beta T-cell receptors. The organization of the a locus and the d locus helps to ensure that each T cell cannot express both types of T-cell receptors. The mechanism involved is that: The rearrangement of a T-cell receptor a gene deletes the d locus on that allele. The rearrangement of a T-cell receptor d gene deletes the a locus on that allele. The RAG recombinase enzymes are down-regulated immediately after the first T-cell receptor genes rearrange. The a:b T-cell receptor signals the T cell to delete the d locus. The g:d T-cell receptor signals the T cell to delete the a locus.arrow_forwardMice and humans with inactivating mutations in the gene encoding activation-induced cytidine deaminase (AID) have an immunodeficiency disease known as 'hyper IgM type 2'. Since AID is the enzyme that catalyzes the conversion of cytosines in the DNA to uracils, thereby initiating the process of somatic hypermutation, why do individuals with this deficiency only produce IgM antibodies?arrow_forwarda. What are the number of total chromosomes that carries the genes encoding heavy and light chain polypeptides in a single B cell (in human)? b. From the total number of chromosomes carrying genes for immunoglobulin peptides from question a, how many encode the heavy and light chains actually expressed by a single B cell?arrow_forward
- Alpha-1 antitrypsin has codominant inheritance. M genes express normal levels. S and Z genes have low expression. Which of the following is most likely to develop emphyema? A person with: 1) two M genes who does smoke 2) two S genes who does smoke 3) one M and one S gene who does not smoke 4) one M gene and one S gene who does smoke 5) two M genes who does not smoke 6) two S genes who does not smokearrow_forwardIn a set of experiments, T cells from wild-type (WT) or bm12 mice were mixed in vitro with antigen-presenting cells (APCs), in the presence or absence of the superantigen staphylococcal enterotoxin B (SEB), and T cell proliferation was measured. The data from these experiments are shown in the figure below. What is the explanation for the results in Rows 1–4 of the table?. Why does the T cell response to SEB (Rows 5–8) show a different pattern than the response to bovine insulin? Note: Epitope mapping studies identified amino acid residues 1–14 of the bovine insulin A chain as the peptide recognized by CD4 T cells from wild-type mice.arrow_forwardWhen a mixture of different IgG antibody proteins are treated with the enzyme papain, each antibody is cleaved into three roughly equal size fragments. From each original antibody, two of the three fragments are identical to each other, and represent the ‘arms’ of the antibody ‘Y’. These fragments are known as Fab fragments. The third fragment is known as the Fc region, because this fragment will crystallize when purified. The reason a mixture of Fc fragments will crystallize is because: It is the only part of the antibody protein that can easily be purified at the high concentrations needed for crystallization. It has no disulfide bonds holding the domains together, as disulfide bonds will inhibit crystallization. It is the only fragment of the antibody that still has disulfide bonds, so it remains intact during the crystallization process. The Fc fragments of IgG are much more water soluble than the Fab fragments. All Fc fragments generated from a mixture of IgG molecules have the…arrow_forward
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