quiz 5 Flashcards
(27 cards)
In a secondary immune response, the affinity of the antibody is higher than in the primary response, which preceded it. Which process makes the largest contribution to this increased affinity?
A.Differentiation of B cells into plasma cells
B.Loss of IgD production
C.Replacement of IgM by IgG
D.Repression of RAG-1 and RAG-2 activity
E.Somatic mutation of immunoglobulin variable region DNA
Answer Key: E
Feedback:
A: Incorrect: plasma cell differentiation is not unique to secondary immune responses.
B: Incorrect: loss of IgD expression occurs as B cells undergo isotype switching, but this does not affect the antigen-binding specificity of the antibody produced.
C: Incorrect: the isotype of the antibody produced does not directly affect the antigen specificity of the antibody.
D: Incorrect: V(D)J recombinase activity does not affect the antigen specificity of antibodies made in secondary immune responses.
E: CORRECT: somatic hypermutation of immunoglobulin variable region DNA during primary responses allows for the refinement of antibody affinity, so that higher affinity antibodies are produced quickly upon induction of a secondary immune response.
An example of a top-down stress management technique is which of the following?
A.Imagery
B.Exercise
C.Taking anxiolytics
D.Communication
Feedback: Explanation: Top-down refers to using the mind to affect the body. Imagery is the only top-down technique in the list.
A 70 kg man who has been exhibiting symptoms of a stroke for 30 minutes is brought to the emergency room. You determine that administration of the thrombolytic agent alteplase is the appropriate course of treatment. The volume of distribution of alteplase is 0.1 L/kg, and its elimination half-life is 5 minutes. The product instructions state that an intravenous bolus dose of 70 μg/kg should be administered, followed by an intravenous infusion of 700 μg/kg over the next 60 minutes.
How long will it take to achieve a steady state concentration of alteplase?
A.less than 1 minute
B.5 minutes
C.25 minutes
D.45 minutes
Answer Key: C
Feedback: The correct answer is C, 25 min. The steady state concentration is always reached in 4 to 5 half-lives. Thus, 5 times 5 min = 25 min.
When the sequences of different MHC class I molecules are compared, the variation between molecules is concentrated within which ONE of the following?
A.Areas of the molecule that bind CD4 and CD8
B.Areas of the molecule that bind to the TCR and the antigenic peptide
C. 2-microglobulin
D.Cytoplasmic domains of the molecule
E.Transmembrane domains of the molecule
Feedback: A: incorrect; MHC class I molecules don’t bind CD4. The α3 domain of MHC class I molecule binds CD8 which is invariant among MHC molecules
B: correct; these areas vary among different Class I alleles and allow expression of different peptides within the groove of the Class I molecule
C: incorrect; β-2 microglobulin is not associated with the binding site on Class I molecules and is encoded on a different chromosome
D: incorrect; variability among Class I molecules is associated with expression of the α 1 and α2 domains of the α chain of the Class I molecule on the cell surface of the antigen presenting cell
E: incorrect; variability among Class I molecules is associated with expression of the α 1 and α2 domains of the α chain of the Class I molecule on the cell surface of the antigen presenting cell
Compare Transcription and Replication. Which phrase describes a feature that is DIFFERENT between the two processes?
A.primer dependent
B.template dependent
C.nucleotide triphosphate precursors
D.direction of chain growth
E.nucleotides are selected by Watson/Crick Base-pairing rules.
Answer Key: A
Feedback: a is correct, transcription is not primer dependent
b is incorrect, both processes are template dependent
c is incorrect, both processes use nucleotide triphosphate precursors
d is incorrect, both nascent chains grow by addition of nucleotides to the 3’-end
e is incorrect, in both cases, the nucleotide to be added to the growing chain is dictated by Watson/Crick base pairing
Which polypeptide could be encoded by the following DNA TEMPLATE?
5’- ACACCACCCGTC - 3’
A.NH2 – Thr - Pro - Pro - Val – CO2H
B.NH2 – Leu - Pro - Thr - Thr – CO2H
C.NH2 – Cys - Gly - Gly - Gln – CO2H
D.NH2 – Asp - Gly - Trp - Cys – CO2H
Answer Key: D
Feedback:
This template will give rise to an mRNA with the complementary sequence:
5’- GACGGGUGGUGU- 3’: using the codon table, this could encode the following peptides:
In frame 1 - GAC GGG UGG UGU
Asp - Gly - Trp - Cys
In frame 2 - ACG GGU GGU GU
Thr - Gly - Gly - Val
In frame 3 - CGG GUG GUG G
Arg - Val - Val ?
a. Incorrect, not any of the possible peptides
b. Incorrect, not any of the possible peptides
c. Incorrect, not any of the possible peptides
d. Correct, this is the peptide encoded in frame 1.
A renal transplant patient is started on a daily oral dose of 1 g of cyclosporine. Immediately after the first dose, the serum concentration is determined to be 100 ng/ml. The next day, just prior to the second dose, the serum concentration is determined to be 75 ng/ml. What will the concentration of cyclosporine be at steady state?
A.150 ng/ml
B.200 ng/ml
C.300 ng/ml
D.400 ng/ml
E.500 ng/ml
Answer Key: D
Feedback: The correct answer is D. To answer this question you need to use the accumulation factor, which is 1/fraction eliminated in one dosing interval. In this case the fraction eliminated in one dosing interval is 0.25. So the accumulation factor is 1/0.25 and the answer is therefore 4 x the concentration after the first dose, or 4 x 100 = 400 ng/ml.
In a study of patients admitted to Walter Reed after blast injuries in combat, researchers found that length of hospitalization (measured in days) followed a skewed distribution with most patients discharged within 60 days but a few extremely long hospital stays (outliers). Based on this information, what is the best way to summarize length of hospitalization?
A.mean (standard deviation)
B.median (interquartile range)
C.mode (range)
D.percent (number of patients)
Answer Key: B
Feedback: a. incorrect: the mean will be highly influenced by the outliers in the data and will not reflect a typical hospitalization
b. correct: the median is an appropriate summary for quantitative data, and is not likely to be influenced by a few outliers
c. incorrect: the mode is the most common value, and is generally useful for summarizing data that can only take on a small number of values.
d. incorrect: percents are useful for summarizing categorical data, but length of hospitalization is quantitative.
A 10-year-old boy presents with atypical pneumonia caused by the fungus Aspergillus fumigatus. A complete blood count reveals that all types of white blood cells are present in normal amounts. Serum immunoglobulin
concentrations are in the high normal range and the patient makes a normal delayed-type hypersensitivity (DTH) response to tetanus toxoid. However, all the patient’s neutrophils are unable to reduce nitro blue tetrazolium (NBT) dye. Which enzyme is MOST LIKELY defective in this
patient?
A.Cytochrome oxidase
B.Eosinophil peroxidase
C.Granzyme B
D.Nitric oxide synthase
E.Phagocyte NADPH oxidase
Answer Key: E
Feedback:
A: incorrect; An oxidizing enzyme containing iron and a porphyrin, found in mitochondria and important in cell respiration as an agent of electron transfer from certain cytochrome molecules to oxygen molecules
B: incorrect; In the presence of H2O2 produced by eosinophils, and either chloride or bromide ions, eosinophil peroxidase provides a potent mechanism by which eosinophils kill multicellular parasites
C: incorrect; Granzyme B is one of the serine proteases located in the granules of CTLs and NK cells
D: incorrect; Nitric oxide synthase (NOS) produces nitric oxide (NO) and citrulline from arginine, molecular oxygen, and NADPH
E: correct; Failure to reduce NBTindicates a defect in the neutrophil oxidative burst, especially in phagocyte NADH oxidase enzyme which catalyzes the reaction between NADPH and oxygen to form superoxide
Which diagram correctly depicts a ribosome after the first peptidyltransferase step?
A.A
B.B
C.C
D.D
E.E
B
Feedback: a. Incorrect, the peptidyltransferase step has not yet occurred since there is no peptide bond
b. Correct, a peptide bond has formed between the valine on the valyl-tRNA in the A-site and the methionine on the methionyl-tRNA in the P-site, thus the methionine has been transferred to the valine, which is still attached to its tRNA.
c. Incorrect, in this depiction the valine is attached to the methionyl-tRNA and would have to have been inserted between the methionine and the tRNA. - this is not correct.
d. Incorrect, the peptide grows in N to C-terminal direction, such that the methionine will be on the N-terminus; in this depiction the peptide is growing in the wrong direction.
e. Incorrect, as in d peptide is growing in the wrong direction, with the added problem that the growing peptide is on the methionyl-tRNA and will be ejected from the ribosome upon translocation.
Which of the following correctly describes Kahneman’s “thinking fast and slow?”
A.Dual Process Theory
B.Self-determination Theory
C.Observational Learning
D.Field Theory
E.Encoding and Forgetting Curve
Feedback:
Dual process theory is a relatively recent theory of cognitive development that conceptualizes thought along a dyad of two processes: system 1 and system 2. Daniel Kahneman further describes this theory in terms of the speed of cognitive processes. In his book Thinking, Fast and Slow Kahneman describes the two systems using a variety of situations- from immediate associations to decision making.
The correct response would not be Self Determination Theory (SDT) since SDT is a theory of motivation and not directly related to thinking. The correct response is not observational learning since this is a theory of learning with some elements of personality development. Field theory is not a correct response since this is a meta-theory proposed by Lewin that conceptualizes and predicts all human motivation and behavior in terms of a dynamic field. Finally, Encoding and Forgetting Curve are terms attributed to work by German psychologist Hermann Ebbinghouse’s work on memory capacity.
A patient presents with symptoms of an adult-onset hereditary peripheral neuropathy. The family history is incomplete; individuals indicated with a question mark (?) in the pedigree below either died or lost contact with the rest of the family before the age of 40, when symptoms typically appear. Assuming that the disorder is fully penetrant, which mode of inheritance is incompatible with the information in the pedigree?
A.Autosomal dominant
B.Autosomal recessive
C.X-linked dominant
D.X-linked recessive
Answer Key: C
Feedback:
a. It would be consistent with autosomal dominant inheritance if individuals I-1 and III-4 had the disease.
b. For the individual IV-1 to be affected with an autosomal recessive disease both of his parents have to be carriers. His mother (III-4) inherited the disease allele from her father (II-1). II-1 is affected, therefore I-1 and I-2 are obligate heterozygotes. II-4 could have inherited the disease allele from one of her parents and passed it to her son (III-5). The pedigree is consistent with autosomal recessive inheritance.
c. Inconsistent. Individual II-1 received his X chromosome from the mother and Y chromosome from his father. If the disease was X-linked dominant his mother (I-2) had to be affected.
d. Assuming the disease is X-linked recessive, I-2 and III-4 are obligate heterozygotes (males inherit their X chromosome from their mothers and transmit it to their daughters). III-4 then transmitted the X with the disease allele to her son (IV-1) who is affected. The pedigree is consistent with the assumption.
A girl with early developmental and speech delay is able to progress to third grade but is failing fourth grade. Evaluation shows an elongated body habitus with somewhat lax joints. Genetic evaluation is recommended, and chromosomal analysis reveals a 47, XXX karyotype with no evidence of fragile X. Which of the following descriptions best fits this abnormality?
A.Euploidy
B.Polyploidy
C.Aneuploidy
D.Y chromosome translocation
E.X monosomy
Answer Key: C
Feedback:
a. Incorrect. Euploidy refers to a situation when one or more complete sets of chromosomes are present. In human the complete set includes 22 autosomes and one sex chromosome. Somatic cells are diploid and have two complete sets: 46, XX (or 46, XY).
b. Incorrect. Polyploidy is the presence of more than two complete sets of chromosomes. E.g., 69, XXX individual is polyploid with three complete sets of chromosomes.
c. Correct. In aneuploidies an extra one or few chromosomes (but NOT complete sets of chromosomes) are present or one or few chromosomes are missing. In this question an extra X chromosome is present, and the total number of chromosomes is 47.
d. Incorrect. The karyotype does not indicate the presence of Y in this individual.
e. Incorrect. Monosomy refers to the cases when only one (instead of usual two) copies of a particular chromosome is present.
A 70 kg man who has been exhibiting symptoms of a stroke for 30 minutes is brought to the emergency room. You determine that administration of the thrombolytic agent alteplase is the appropriate course of treatment. The volume of distribution of alteplase is 0.1 L/kg, and its elimination half-life is 5 minutes. The product instructions state that an intravenous bolus dose of 70 μg/kg should be administered, followed by an intravenous infusion of 700 μg/kg over the next 60 minutes. What is the serum concentration immediately after the bolus dose?
A.70 μg/L
B.460 μg/L
C.700 μg/L
D.4.9 mg/L
E.7.0 mg/L
Answer Key: C
Feedback: The correct answer is C, 700 µg/L. Remember that volume of distribution (the apparent volume into which the drug will be distributed) is equal to the amount of drug in the body divided by the drug’s concentration in plasma at time zero, or Vd = amount of drug in the body /drug concentration in plasma. Therefore, the serum concentration = drug amount/Vd or 70 µg/kg divided by 0.1 L/kg or 700 µg/L.
A patient presents with symptoms characteristic of familial hypercholesterolemia, a haploinsufficient disorder, but studies of the patient’s DNA by Southern blotting and PCR analysis do not reveal any deletions in either copy of the LDL receptor gene. One explanation for this apparent inconsistency is that this disorder shows
A.heteroplasmy
B.premutations
C.incomplete penetrance
D.variable expressivity
E.allelic heterogeneity
Answer Key: E
Feedback:
a. Incorrect. Heteroplasmy refers to the situation when a cell has two types of mitochondria: with and without a mutation in a mitochondrial DNA. This leads to increased variability of symptoms in patients with mitochondrial diseases, depending on a fraction of mutant mitochondria in an individual, and a fractio of cells with “mostly mutant” and “mostly normal” mitochondria in critical organs and tissues.
b. Incorrect. Premutation is a state in the development of unstable repeat mutation when the number of repeats is not high enough to cause the disease, but too high to remain stable during DNA replication and repair. A premutation is likely to develop into disease-causing mutation in the next generation, particularly, if the carrier of a permutation is a female.
c. Incorrect. Incomplete penetrance assumes lack of the symptoms.
d. Incorrect. For certain genes the same mutation can cause different symptoms in different individuals. This is called variable expressivity and does not apply to this case, because the symptoms of only one person are described.
e. Correct. Genetic diseases are caused by inactivation of a particular gene or genes, and any gene can be inactivated by different types of mutations: deletions, insertions, nonsense mutations, frameshift mutations, point mutations affecting splicing, RNA stability or transcription, etc. Such mutations are called allelic mutations (different mutations in the same gene), and the presence of more than one type of mutations represents allelic heterogeneity. Sometimes different allelic mutations cause slightly different symptoms. In our case the patient does not have any deletions in the gene, but he was not tested for other types of mutations. Lack of the deletion mutation in this patient assumes the presence of other type of mutations in LDL receptor and therefore, allelic heterogeneity for this gene.
The therapeutic index of a drug is defined as
A.TD50 / ED50 .
B.ED50 / TD50
C.TD50 /LD50 .
D.KD /Vd
E.the fraction of unchanged drug reaching the systemic circulation after oral administration
Answer Key: A
Feedback: The correct answer is A. This is the ONLY answer that both relates effective to toxic does (A, B or C) AND which has toxicity in the numerator and effective dose in the denominator so that the greater the quotient the greater the safety.
Answer Key: A
Feedback: The correct answer is A. This is the ONLY answer that both relates effective to toxic does (A, B or C) AND which has toxicity in the numerator and effective dose in the denominator so that the greater the quotient the greater the safety.
A renal transplant patient is started on a daily oral dose of 1 g of cyclosporine. Immediately after the first dose, the serum concentration is determined to be 100 ng/ml. The next day, just prior to the second dose, the serum concentration is determined to be 75 ng/ml.
If the elimination half-life of cyclosporine is 69 hr, and the volume of distribution in this patient is 300 L, what is the clearance?
A.0.3 L/hr
B.0.69 L/hr
C.3 L/h
D.6.9 L/h
E.30 L/h
Answer Key: C
Feedback: The correct answer is C. The relationship between clearance (CL) and volume of distribution (Vd) is (CL) = keVd, where ke is the rate elimination constant. Since ke = 0.693/t1/2, if the elimination half-life is 69 hr, then ke = 0.693/69 or 0.01. Thus if Vd is 300 L, then CL = 0.01 x 300 = 3 L/h. This is the hardest question you are likely to see on an NBME exam.
A 4-year-old child who has had repeated infections with staphylococci and streptococci (both bacteria that colonize extracellular spaces), has normal phagocytic function and delayed-type hypersensitivity (DTH) responses. A lymph node biopsy would MOST LIKELY reveal an absence of
A.CD8+ T cells.
B.germinal centers.
C.macrophages.
D.paracortical areas.
E.postcapillary venules.
Answer Key: B
Feedback: A: incorrect; since the DTH responses are normal, there is no evidence of a T- cell defect
B: correct; germinal centers are the location of B celles and antibody production and staph and strep are extracellular pathogens (which favors a humoral response)
C: incorrect; phagocytic function in the child is normal and macrophages are a major phagocytic cell
D: incorrect; the paracortex is the area within the secondary lymphoid organs where T cells reside and since the DTH responses are normal, there is no evidence of a T- cell defect.
E: incorrect; venules located in lymphoid tissue; these are the site of recirculation of lymphocytes from the blood to the lymphoid tissue; not associated with absence or presence of cells
Not all mutations in coding sequences change the amino acid sequence of the encoded protein because some codons
A.specify more than one amino acid.
B.are recognized by tRNAs carrying different amino acids.
[Correct]
C.specify the same amino acids as do others.
D.consist of more or fewer than three nucleotides.
E.are skipped during translation.
E.E
Answer Key: C
Feedback: a is incorrect, the code is unambiguous; no codon specifies more than one amino acid.
b is incorrect, a given codon will only be recognized by a tRNA that is charged with the amino acid specified by that codon.
c is correct, with the exception of methionine and tryptophan, there are multiple (2-6) codons that specify each amino acid; thus, if a mutation converts a codon into a synonym (codon specifying the same amino acid) the amino acid sequence of the encoded protein will not change. d is incorrect, all codons consist of three nucleotides
e is incorrect, the ribosomes decodes each consecutive codon in the mRNA, no codons are skipped during translation
At a DNA replication fork, both parental DNA strands act simultaneously as templates for synthesis of complementary daughter strands, but only one daughter strand is synthesized continuously. The other daughter strand is synthesized discontinuously because
A.DNA Pol III lacks a 5′ → 3′ exonuclease activity.
B.DNA Pol III requires a preexisting primer.
C.DNA Pol III is unable to add a dNTP to the 5’ end of a growing strand.
D.single-stranded binding proteins block one template strand.
E.it is synthesized by DNA Pol I, which is slower than DNA Pol III.
Answer Key: C
Feedback: a is incorrect; although DNA Pol III does lack a 5′ → 3′ exonuclease activity, this is not the reason that one strand is synthesized discontinuously.
b is incorrect; DNA polymerases do require a preexisting primer, but this is not the reason that one strand is synthesized discontinuously.
c is correct – since DNA polymerases can only add nucleotides to the 3′ end of the growing chain it is not possible for both strands to be synthesized continuously as this would require that one of the strands were being synthesized by adding nucleotides to the 5′ end of the growing chain.
d is incorrect; single-stranded binding proteins bind both template strand, but do not prevent their replication
e is incorrect; DNA Pol I is involved in maturing the Okazaki fragments into a continuous chain, but DNA Pol III synthesizes both the leading and lagging strands
A 42 year old female presents with dry mouth and dry eyes since 6months. Further history reveals she has bilateral pain in her wrists and knees. Which of the following antibody laboratory tests would be most helpful in confirming your clinical suspicion?
A.Anti-double-stranded DNA
B.Anti-Sm
C.SS-A
D.Scl-70
E.Anticentromere
Answer Key: C
Feedback: The correct answer is c, SS-A. SS-A and SS-B are most specific for Sjoegren Syndrome. Anti-ds-DNA and Anti-Sm are for Systemic Lupus Erythematosus, Scl-70 for Systemic sclerosis and Anticentromere for CREST syndrome. These are in a table on pg 215 in Robbins.
An investigator injects an experimental animal with a newly discovered bacterial strain to evaluate T lymphocyte activation. Macrophages engulf these organisms and bacterial-derived peptides are presented to CD4+ T lymphocytes. Which ONE of the following cell-surface molecules on the macrophage is MOST DIRECTLY involved in the presentation of the processed peptides?
A.CD28
B.Class II MHC
C.Fc receptor
D.Interleukin-2 (IL-2) receptor
E.Membrane immunoglobulin
Answer Key: B
Feedback: A: Incorrect – CD28 is expressed by T cells.
B: CORRECT – bacterium-derived peptides are presented to T cells by their insertion into the peptide-binding grooves of class II MHC molecules, which are then expressed on the surface of the macrophage.
C: Incorrect – Fc receptors participate in the binding of immunoglobulin molecules.
D: Incorrect – IL-2 receptor is not expressed by macrophages.
E: Incorrect – macrophages do not express membrane immunoglobulin.
Females occasionally have symptoms of X-linked recessive diseases such as Duchenne muscular dystrophy, hemophilia, or color blindness. Which of the following is the most likely explanation?
A.locus heterogeneity
B.somatic mosaicism
C.incomplete penetrance
D.heteroplasmy
E.germline mosaicism
Answer Key: B
Feedback:
a. Locus heterogeneity refers to the cases when defects in different genes can cause a similar clinical presentation. Theoretically, locus heterogeneity can apply separately for each of the X-linked condition if one imagines that mutations in some other hypothetical gene, which is not located on the X chromosome, cause similar symptoms. However, this particular question seems to be about general property of X-linked disorders, therefore there should be a better answer.
b. This is correct. Females are mosaic with respect to X chromosome inactivation. In general, approximately half of somatic cells have inactive maternal X and the other half have inactive paternal X. Since inactivation is random, the 1:1 ratio is not precisely maintained within the body and between individuals. If maternal X has a mutation, than females with lower fraction of inactive maternal X would have higher chance of developing symptoms. Appearance of symptoms will also depend on whether the tissue critical for the disease have larger or smaller fraction of inactive normal X.
c. Incorrect. Incomplete penetrance applies to cases where a mutant allele has been inherited but did not present clinically (i.e., the lack of expected symptoms).
d. Incorrect. Heteroplasmy is only relevant to mitochondrial diseases.
e. Incorrect. Germline mosaicism (i.e. two or more subpopulations of germ cells in the individual) is suspected when more than one child with a dominant disease is born to unaffected parents with no family history of this disease.
In an antibody-producing B cell, one heavy chain Ig allele has undergone recombination, while the other has a germline configuration. This is evidence that
A.a similar pattern will occur in light chain genes.
B.during development the initial rearrangement was productive.
C.during development the pre-B receptor was not expressed.
D.more than one allele may be expressed.
E.the B cell may make a different antibody later.
Answer Key: B
Feedback: A: incorrect; light chain genes re-arrange after a productive re-arrangement of a heavy chain allele ; re-arrangement of one of the light chains may or may not be productive
B: correct; if the first heavy chain re-arrangement had not been productive then re-arrangement would have been initiated in the second heavy chain allele and neither of the Ig alleles would be in germline configuration
C: incorrect; expression of the pre-B receptor occurs before re-arrangement of the heavy chain alleles during development
D: incorrect; re-arrangement occurs in a tightly regulated stepwise manner; if re-arrangement on the first heavy chain was not productive, only then is re-arrangement on the second allele initiated
E: incorrect: ; antibody specificity is encoded in the VJ and VDJ segments of the Ig genes and specificity is determined before heavy chain re-arrangement begins
