Chemistry/immunology Flashcards
From the following, identify a specific component of the adaptive immune system that is
formed in response to antigenic stimulation:
A. Lysozyme
B. Complement
C. Commensal organisms
D. Immunoglobulin (Ig)
D Ig is a specific part of the adaptive immune system and is formed only in response to a
specific antigenic stimulation. Complement, lysozyme, and commensal organisms all
act nonspecifically as a part of the adaptive immune system. These three components
do not require any type of specific antigenic stimulation
Which two organs are considered the primary lymphoid organs in which
immunocompetent cells originate and mature?
A. Thyroid and Peyer patches
B. Thymus and bone marrow
C. Spleen and mucosal-associated lymphoid tissue (MALT)
D. Lymph nodes and thoracic duct
B Bone marrow and the thymus are considered primary lymphoid organs because
immunocompetent cells either originate from them or mature in them. Some
immunocompetent cells mature or reside in bone marrow (the source of all
hematopoietic cells) until transported to the thymus, spleen, or Peyer patches, where
they process antigen or manufacture antibody. T lymphocytes, after originating in bone
marrow, travel to the thymus to mature and differentiate.
What type of B cells is formed after antigen stimulation?
A. Plasma cells and memory B cells
B. Mature B cells
C. Antigen-dependent B cells
D. Receptor-activated B cells
A Mature B cells exhibit surface Ig that may cross-link a foreign antigen, thus forming
the activated B cell and leading to capping and internalization of antigen. The activated
B cell gives rise to plasma cells that produce and secrete Igs and memory cells that
reside in lymphoid organs.
T cells travel from bone marrow to the thymus for maturation. What is the correct
order of the maturation sequence for T cells in the thymus?
A. Bone marrow to the cortex; after thymic education, released back to peripheral circulation
B. Maturation and selection occur in the cortex; migration to the medulla; release of mature T
cells to secondary lymphoid organs
C. Storage in either the cortex or medulla; release of T cells into the peripheral circulation
D. Activation and selection occur in the medulla; mature T cells are stored in the cortex until
activated by antigen
B Immature T cells travel from bone marrow to the thymus to mature into functional T
cells. Once in the thymus, T cells undergo a selection and maturation sequence that
begins in the cortex and moves to the medulla of the thymus. Thymic factors, such as
thymosin and thymopoietin, and cells within the thymus, such as macrophages and
dendritic cells, assist in this sequence. After completion of the maturation cycle, T cells
are released to secondary lymphoid organs to await antigen recognition and activation
Which cluster of differentiation (CD) marker is the most specific identifying marker for
mature T cells?
A. CD1
B. CD2
C. CD3
D. CD4 or CD8
C The CD3 marker appears during the early stages of T-cell development and can be
used to differentiate T cells from other lymphocytes
Which markers are found on mature, peripheral helper T cells?
A. CD1, CD2, CD4
B. CD2, CD3, CD8
C. CD1, CD3, CD4
D. CD2, CD3, CD4
D Mature, peripheral helper T cells have the CD2, CD3 (mature T cell), and CD4 (helper)
markers.
Which T cells express the CD8 marker and act specifically to kill tumors or virally
infected cells?
A. Helper T cells
B. Suppressor T cells
C. Cytotoxic T cells (TC cells)
D. Regulator T cells
C TC cells recognize antigen in association with major histocompatibility complex
(MHC) class I complexes and act against target cells that express foreign antigens.
These include viral antigens and the HLAs that are the target of graft rejection
How are TC cells and natural killer (NK) cells similar?
A. Require antibody to be present
B. Effective against virally infected cells
C. Recognize antigen in association with human leukocyte antigen (HLA) class II markers
D. Do not bind to infected cells
B Both TC and NK cells are effective against virally infected cells, and neither requires
antibody to be present to bind to infected cells. NK cells do not exhibit MHC class
restriction, whereas activation of TC cells requires the presence of MHC class I
molecules in association with the viral antigen.
What is the name of the process by which phagocytic cells are attracted to a substance,
such as a bacterial peptide?
A. Diapedesis
B. Degranulation
C. Chemotaxis
D. Phagotaxis
C Chemotaxis is the process by which phagocytic cells are attracted toward an area
where they detect a disturbance in the normal functions of body tissues. Products from
bacteria and viruses, complement components, coagulation proteins, and cytokines
from other immune cells may all act as chemotactic factors.
All of the following are immunologic functions of complement except:
A. Induction of an antiviral state
B. Opsonization
C. Chemotaxis
D. Anaphylatoxin formation
A Complement components are serum proteins that function in opsonization,
chemotaxis, and anaphylatoxin formation but do not induce an antiviral state in target
cells. This function is performed by interferons.
Which complement component is found in both the classic and alternative pathways?
A. C1
B. C4
C. Factor D
D. C3
D C3 is found in both the classic and alternative (alternate) pathways of the complement
system. In the classic pathway, C3b forms a complex on the cell with C4b2a that
enzymatically cleaves C5. In the alternative pathway, C3b binds to an activator on the
cell surface. It forms a complex with factor B called C3bBb, which, like C4b2a3b, can
split C5.
Which Ig(s) help(s) initiate the classic complement pathway?
A. IgA and IgD
B. IgM only
C. IgG and IgM
D. IgG only
C Both IgG and IgM are the Igs that help to initiate the activation of the classic
complement pathway. IgM is, however, a more potent complement activator.
What is the purpose of C3a, C4a, and C5a, the split products of the complement
cascade?
A. To bind with specific membrane receptors of lymphocytes and cause release of cytotoxic
substances
B. To cause increased vascular permeability, contraction of smooth muscle, and release of
histamine from basophils
C. To bind with membrane receptors of macrophages to facilitate phagocytosis and the
removal of debris and foreign substances
D. To regulate and degrade membrane cofactor protein after activation by C3 convertase
B C3a, C4a, and C5a are split products of the complement cascade that participate in
various biological functions, such as vasodilation and smooth muscle contraction.
These small peptides act as anaphylatoxins, for example, effector molecules that
participate in the inflammatory response to assist in the destruction and clearance of
foreign antigens.
How is complement activity destroyed in vitro?
A. Heating serum at 56°C for 30 minutes
B. Keeping serum at room temperature of 22°C for 1 hour
C. Heating serum at 37°C for 45 minutes
D. Freezing serum at 0°C for 24 hours
A Complement activity in serum, in vitro, is destroyed by heating serum at 56°C for 30
minutes. In test procedures where complement may interfere with the test system, it
may be necessary to destroy complement activity in the test sample by heat
inactivation
Which region of the Ig molecule can bind antigen?
A. Fragment antigen binding (Fab)
B. Fragment crystallizable (Fc)
C. Constant light (CL)
D. Constant heavy (CH)
A Fab is the region of the Ig molecule that can bind antigen. Two Fab fragments are
formed from hydrolysis of the Ig molecule by papain. Each consists of a light chain
and the VH and CH1 regions of the heavy chain. The variable regions of the light and
heavy chains interact, forming a specific antigen-combining site
Which Ig class(es) has (have) a J-chain?
A. IgM
B. IgE and IgD
C. IgM and surface IgA (sIgA)
D. IgG3 and IgA
C Both IgM and sIgA have a J-chain joining individual molecules together; the J-chain
in IgM joins five molecules and the J-chain in sIgA joins two molecules
Which region determines whether an Ig molecule can fix complement?
A. Variable heavy (VH)
B. Constant heavy (CH)
C. Variable light (VL)
D. Constant light (CL)
B The composition and structure of the constant region of the heavy chain determine
whether that Ig will fix complement. Fc is formed by partial Ig digestion with papain
and includes the CH2 and CH3 domains of both heavy chains. The complement
component C1q molecule will bind to the CH2 region of an IgG or IgM molecule
Which Ig appears first in the primary immune response?
A. IgG
B. IgM
C. IgA
D. IgE
B The first antibody to appear in the primary immune response to an antigen is IgM.
The titer of antiviral IgM (e.g., IgM antibody to cytomegalovirus [anti-CMV]) is more
specific for acute or active viral infection than IgG and may be measured to help
differentiate active infection from prior infection.
Which immunoglobulin appears in highest titer in the secondary response?
A. IgG
B. IgM
C. IgA
D. IgE
A A high titer of IgG characterizes the secondary immune response. Consequently, IgG
antibodies comprise about 80% of the total Ig concentration in normal serum
Which Ig can cross the placenta?
A. IgG
B. IgM
C. IgA
D. IgE
A IgG is the only Ig class that can cross the placenta. All subclasses of IgG can cross the
placenta, but IgG2 crosses more slowly. This process requires recognition of the Fc
region of the IgG by placental cells. These cells take up the IgG from maternal blood
and secrete it into fetal blood, providing humoral immunity to the neonate for the first
few months after delivery.
All of the following are functions of Igs except:
A. Neutralizing toxic substances
B. Facilitating phagocytosis through opsonization
C. Interacting with TC cells to lyse viruses
D. Combining with complement to destroy cellular antigens
C Tc cells lyse virally infected cells directly, without requirement for specific antibody.
The TC cell is activated by viral antigen that is associated with MHC class I molecules
on the surface of the infected cell. The activated TC cell secretes several toxins, such as
tumor necrosis factor, which destroy the infected cell and virions
Which Ig cross-links mast cells to release histamine?
A. IgG
B. IgM
C. IgA
D. IgE
D IgE is the Ig that cross-links with basophils and mast cells. IgE causes the release of
such immune response modifiers as histamine and mediates an allergic immune
response.
Which of the following cell surface molecules is classified as an MHC class II antigen?
A. HLA-A
B. HLA-B
C. HLA-C
D. HLA-DR
D The MHC region is located on the short arm of chromosome 6 and codes for antigens
expressed on the surface of leukocytes and tissues. The MHC region genes control
immune recognition; their products include the antigens that determine transplant
rejection. HLA-DR antigens are expressed on B cells. HLA-DR2, -DR3, -DR4, and -
DR5 antigens show linkage with a wide range of autoimmune diseases
What molecule on the surface of most T cells recognizes antigen?
A. IgT, a four-chain molecule that includes the tau heavy chain
B. MHC protein, a two-chain molecule encoded by the HLA region
C. CD3, consisting of six different chains
D. T-cell receptor (TcR), consisting of two chains: α-chain and β-chain
D T cells have a membrane bound receptor (TcR) that is antigen specific. This two-
chain molecule consists of a single α-chain, similar to an Ig light chain, and a single β-
chain, similar to an Ig heavy chain. Some T cells may express a γ-δ receptor instead of
the α-β molecule. There is no τ heavy chain. MHC and CD3 molecules are present on
T cells, but they are not the molecules that give antigen specificity to the cell.
Which of the following are products of HLA class III genes?
A. T-cell immune receptors
B. HLA-D antigens on immune cells
C. Complement proteins C2, C4, and factor B
D. Ig VL regions
C Complement components C2 and C4 of the classic pathway and factor B of the
alternative pathway are class III molecules. HLA-A, HLA-B, and HLA-C antigens are
classified as class I antigens, and HLA-D, HLA-DR, HLA-DQ, and HLA-DP antigens
are classified as class II antigens.
Which MHC class of molecule is necessary for antigen recognition by CD4-positive T
cells?
A. Class I
B. Class II
C. Class III
D. No MHC molecule is necessary for antigen recognition
B Helper T lymphocytes (CD4-positive T cells) recognize antigens only in the context
of a class II molecule. Because class II antigens are expressed on macrophages,
monocytes, and B cells, the helper-T-cell response is mediated by interaction with
processed antigen on the surface of these cells.
TcR is similar to Ig molecules in that it:
A. Remains bound to the cell surface and is never secreted
B. Contains V and C regions on each of its chains
C. Binds complement
D. Can cross the placenta and provide protection to a fetus
B The antigen binding regions of both the α- and β-chains of the TcR are encoded by V
genes that undergo rearrangement similar to that observed in Ig genes. The α-chain
gene consists of V and J segments, similar to an Ig light chain. The β-chain consists of
V, D, and J segments, similar to an Ig heavy chain. The α- and β-chains each have a
single C-region gene encoding the constant region of the molecule. Although answer A
is true for TcRs, it is not true for Igs that can be cell bound or secreted. Answers C and
D are true for certain Ig heavy-chain isotypes but are not true for the TcR
A superantigen, such as toxic shock syndrome toxin-1 (TSST-1), bypasses the normal
antigen-processing stage by binding to and cross-linking:
A. A portion of an Ig molecule and complement component C1
B. TLRs and an MHC class 1 molecule
C. A portion of an Ig and a portion of a TcR
D. A portion of a TcR and an MHC class II molecule
D A superantigen binds to the V β-portion of the TcR and an MHC class II molecule.
This binding can activate T cells without the involvement of an antigen-presenting cell.
In some individuals, a single V β-protein that recognizes TSST-1 is expressed on up to
10% to 20% of T cells. The simultaneous activation of this amount of T cells causes a
heavy cytokine release, resulting in the vascular collapse and pathology of toxic shock
syndrome.
Toll-like receptors (TLRs) are found on which cells?
A. T cells
B. Dendritic cells
C. B cells
D. Large granular lymphocytes
B TLRs are the primary antigen recognition protein of the innate immune system. They
are found on antigen-presenting cells, such as dendritic cells and macrophages. Eleven
TLRs have been described. TLRs recognize certain structural motifs common to
infecting organisms. TLR 4, for example, recognizes bacterial lipopolysaccharide
(LPS). The name TLR comes from its similarity to the Toll protein in Drosophila.
Macrophages produce which of the following proteins during antigen processing?
A. IL-1 and IL-6
B. γ-Interferon
C. IL-4, IL-5, and IL-10
D. Complement components C1 and C3
A Interleukin-1 (IL-1) and IL-6 are proinflammatory macrophage-produced cytokines.
In addition to their inflammatory properties, they activate T-helper cells during antigen
presentation. γ-Interferon, IL-4, IL-5, and IL-10 are all produced by T cells.
Complement components are produced by a variety of cells but are not part of the
macrophage antigen-presentation process.
T-regulator cells, responsible for controlling autoimmune antibody production, express
which of the following phenotypes?
A. CD3, CD4, CD8
B. CD3, CD8, CD25
C. CD3, CD4, CD25
D. CD8, CD25, CD56
C T-regulator cells are believed to be the primary immune suppressor cells and express
CD3, CD4, and CD25. CD25 is the IL-2 receptor. CD25 may be expressed by
activated T cells, but is constitutively expressed by the T-regulator cells. CD25
expression on T-regulator cells occurs in the thymus and is regulated by the FOXP3
protein.
The interaction between individual antigen and antibody molecules depends on several
types of bonds, such as ionic bonds, hydrogen bonds, hydrophobic bonds, and van der
Waals forces. How is the strength of this attraction characterized?
A. Avidity
B. Affinity
C. Reactivity
D. Valency
B Affinity refers to the strength of a single antibody–antigen interaction. Avidity is the
strength of interactions between many different antibodies in a serum against a
particular antigen (i.e., the sum of many affinities).
The detection of precipitation reactions depends on the presence of optimal proportions
of antigen and antibody. A patient’s sample contains a large amount of antibody, but
the reaction in a test system containing antigen is negative. What has happened?
A. Performance error
B. Low specificity
C. A shift in the zone of equivalence
D. Prozone phenomenon
D Although performance error and low specificity should be considered, if a test system
fails to yield the expected reaction, excessive antibody preventing a precipitation
reaction is usually the cause. Prozone occurs when antibody molecules saturate the
antigen sites, preventing cross-linking of the antigen–antibody complexes by other
antibody molecules. Because antigen and antibody do not react at equivalence, a
visible product is not formed, leading to a false-negative result
A laboratory is evaluating an enzyme-linked immunosorbent assay (ELISA) for
detecting an antibody to cyclic citrullinated peptide (CCP), which is a marker for
rheumatoid arthritis (RA). The laboratory includes serum from healthy volunteers and
from patients with other connective tissue diseases in the evaluation. These specimens
determine which factor of the assay?
A. Sensitivity
B. Precision
C. Bias
D. Specificity
D Specificity is defined as a negative result in the absence of the disease. The non–RA
specimens would be expected to test negative if the assay has high specificity.
Precision is the ability of the assay to repeatedly yield the same results on a single
specimen. Both bias and sensitivity calculations would include specimens from RA
persons. Although those specimens would be included in the evaluation, they are not
listed in the question.
The positive and negative control values for an ELISA procedure are below their
acceptable ranges. What is the most likely cause?
A. Decay of the positive and negative controls
B. Incomplete washing following specimen addition
C. Overly long incubation times
D. Decay of the antibody–enzyme conjugate
D The antibody–enzyme conjugate is very sensitive to storage conditions and is easy to
dissociate. Control specimens are unlikely to decay, and the other options would lead
to higher values.
What is the interpretation when an Ouchterlony plate shows crossed lines between wells
1 and 2 (antigen is placed in the center well and antisera in wells 1 and 2)?
A. No reaction between wells 1 and 2
B. Partial identity between wells 1 and 2
C. Nonidentity between wells 1 and 2
D. Identity between wells 1 and 2
C Crossed lines indicate nonidentity between wells 1 and 2. The antibody from well 1
recognizes a different antigenic determinant than the antibody from well 2.
A weight lifter taking many supplements is tested monthly for thyroid-stimulating
hormone (TSH) in a direct capture assay, which uses a streptavidin–biotin indicator
system. She has had normal TSH levels for the past 3 months on specimens collected in
the late evening. This month she comes in right after breakfast for her blood draw. The
TSH level is three times her previous level. What may be the cause of this difference?
A. Diurnal variation in TSH levels
B. Exogenous biotin in her system from a supplement taken that morning
C. Reduced thyroid function caused by an unidentified pathology
D. Pipetting error
B High levels of exogenous biotin may be present in blood up to 10 to 15 hours after
ingestion and will usually cause falsely elevated values in avidin–biotin direct ELISA
assays and competitive immunoassays but lower values in sandwich (immunometric)
assays. Monthly variation in TSH should not approach the level seen in this situation
What comprises the indicator system in an indirect ELISA for detecting antibody?
A. Enzyme-conjugated antibody + chromogenic substrate
B. Enzyme conjugated antigen + chromogenic substrate
C. Enzyme + antigen
D. Substrate + antigen
A ELISA measures antibody by using immobilized reagent antigen. The antigen is fixed
to the walls of a tube or bottom of a microtiter well. Serum is added (and incubated)
and the antibody binds, if present. After washing, the antigen–antibody complexes are
detected by adding an enzyme-labeled antiimmunoglobulin (anti-Ig). The unbound
enzyme label is removed by washing, and the bound enzyme label is detected by
adding chromogenic substrate. The enzyme catalyzes the conversion of substrate to a
colored product.
What outcome results from improper washing of a tube or well after adding the
enzyme–antibody conjugate in an ELISA system?
A. Result will be falsely decreased
B. Result will be falsely increased
C. Result will be unaffected
D. Result is impossible to determine
B If unbound enzyme-conjugated anti-Ig is not washed away, it will catalyze the
conversion of the substrate to a colored product, yielding a falsely elevated result.
What would happen if the color reaction phase is prolonged in one tube or well of an
ELISA test?
A. Result will be falsely decreased
B. Result will be falsely increased
C. Result will be unaffected
D. Impossible to determine
B If the color reaction is not stopped within the time limits specified by the procedure,
the enzyme will continue to act on the substrate, producing a falsely elevated test
result.
The absorbance of a sample measured by ELISA is greater than the highest standard.
What corrective action should be taken?
A. Extrapolate an estimated value from the highest reading
B. Repeat the test using a standard of higher concentration
C. Repeat the assay using one half the volume of the sample
D. Dilute the test sample
D Usually, when a test sample reads at a value above the highest standard in an ELISA,
the sample is diluted and measured again. In those instances where no additional
clinical value can be obtained by dilution, the result may be reported as greater than the
highest standard (citing the upper reportable limit of the assay).
A patient was suspected of having a lymphoproliferative disorder. After several
laboratory tests were completed, the patient was found to have an IgMκ paraprotein. In
what sequence should the laboratory tests leading to this diagnosis have been
performed?
A. Serum protein electrophoresis (SPE) followed by immunofixation electrophoresis (IFE)
B. Ig levels, followed by SPE
C. Total lymphocyte count, followed by Ig levels
D. Ig levels, followed by urine protein electrophoresis
A SPE should be performed initially to detect the presence of an abnormal Ig that
demonstrates restricted electrophoretic mobility. A patient producing only monoclonal
light chains may not show an abnormal serum finding because the light chains may be
excreted in urine. A positive finding for either serum or urine should be followed by
IFE on the positive specimen. This is required to confirm the presence of monoclonal
Ig and to identify the heavy and light chain types.
An IFE performed on a serum sample showed a narrow dark band in the lanes
containing anti-γ and anti-λ. How should this result be interpreted?
A. Abnormally decreased IgG concentration
B. Abnormal test result demonstrating monoclonal IgGλ
C. Normal test result
D. Impossible to determine without densitometric quantitation
B A narrow dark band formed in both the lane containing anti-γ and anti-λ indicates the
presence of a monoclonal IgG-λ. A diffuse dark band would indicate a polyclonal
increase in IgG that often accompanies chronic inflammatory disorders, such as
systemic lupus erythematosus (SLE).
Which type of nephelometry is used to measure immune complex formation almost
immediately after reagent has been added?
A. Rate
B. Endpoint
C. Continuous
D. One-dimensional
A Rate nephelometry is used to measure the formation of small immune complexes as
they are formed under conditions of antibody excess. The rate of increase in the
photodetector output is measured within seconds or minutes, and the rate increases
with increasing antigen concentration. Antigen concentration is determined by
comparing the rate for the sample with that for standards by using an algorithm that
compensates for nonlinearity. In endpoint nephelometry, reactions are read after
equivalence. Immune complexes are of maximal size but may have a tendency to settle
out of solution, thereby decreasing the amount of scatter
An antinuclear antibody (ANA) test was performed by using immunofluorescence
microscopy assay (IFA), and a clinically significant pattern and titer were reported.
Positive and negative controls performed as expected. However, the clinical evaluation
of the patient was not consistent with the reported pattern. What is the most likely
explanation for this situation?
A. The clinical condition of the patient changed since the sample was tested
B. The pattern of fluorescence was misinterpreted
C. The control results were misinterpreted
D. The wrong cell line was used for the test
B In an IFA for antinuclear antibodies, the fluorescence pattern must be correlated
correctly with the specificity of the antibodies. Both pathological and nonpathological
antibodies can occur, and antibodies may be detected at a significant titer in a patient
whose disease is inactive. Failure to correctly identify subcellular structures may result
in misinterpretation of the antibody specificity or a false-positive result caused by
nonspecific fluorescence
What corrective action should be taken when a specific pattern cannot be identified in a
specimen with a positive ANA IFA?
A. Repeat the test using a larger volume of sample
B. Call the physician
C. Have another medical laboratory scientist read the slide
D. Dilute the sample and retest
D An unexpected pattern may indicate the presence of more than one antibody. Diluting
the sample may help to clearly show the antibody specificities, if they are found in
different titers. If the pattern is still atypical, a new sample should be collected and the
test repeated or the specimen should be tested by an alternative method, such as ELISA
or multiplex.
Which statement best describes passive agglutination reactions used for serodiagnosis?
A. Such agglutination reactions are more rapid because they are a single-step process
B. Reactions require the addition of a second antibody
C. Passive agglutination reactions require biphasic incubation
D. Carrier particles for antigen, such as latex particles, are used
D Most agglutination tests used in serology employ passive or indirect agglutination,
where carrier particles are coated with the antigen. The carrier molecule is of sufficient
size so that the reaction of the antigen with antibody results in the formation of a
complex that is more easily visible.
What has happened in a titer, if tube Nos. 5 to 7 show a stronger reaction than tube
Nos.1 to 4?
A. Prozone reaction
B. Postzone reaction
C. Equivalence reaction
D. Poor technique
A In tubes Nos.1 to 4, insufficient antigen is present to give a visible reaction because
excess antibody has saturated all available antigen sites. After dilution of antibody,
tubes Nos.1 to 4 have the equivalent concentrations of antigen and antibody to allow formation of visible complexes.
What is the titer in tube No. 8 if tube No. 1 is undiluted and dilutions are doubled?
A. 64
B. 128
C. 256
D. 512
B The antibody titer is reciprocal of the highest dilution of serum giving a positive
reaction. For doubling dilutions, each tube has one half the amount of serum as the
previous tube. Because the first tube was undiluted (neat), the dilution in tube No. 8 is
(1/2)^7 and the titer equals 2^7 or 128.
The directions for a slide agglutination test instruct that after mixing the patient’s
serum and antigen-coated latex particles, the slide must be rotated for 2 minutes. What
would happen if the slide were rotated for 10 minutes?
A. Possible false-positive result
B. Possible false-negative result
C. No effect
D. Depends on the amount of antibody present in the sample
A Failure to follow directions, as in this case where the reaction was allowed to proceed
beyond the recommended time, may result in a false-positive reading. Drying on the
slide may lead to a possible erroneous positive reading.
Which outcome indicates a negative result in a complement fixation test?
A. Hemagglutination
B. Absence of hemagglutination
C. Hemolysis
D. Absence of hemolysis
C In complement fixation, hemolysis indicates a negative test result. The absence of
hemolysis indicates that complement was fixed in an antigen–antibody reaction and,
therefore, that the specific complement binding antibody was present in the patient’s
serum. Consequently, it was not available to react in the indicator system.
What effect does selecting the wrong gate have on the results when cells are counted by
flow cytometry?
A. No effect
B. Failure to count the desired cell population
C. Falsely elevated results
D. Impossible to determine
B Gating is the step performed to select the correct cells to be counted. Failure to
properly perform this procedure will result in problems in isolating and counting the
desired cells. It is impossible to determine if the final result would be falsely elevated
or falsely lowered by problems with gating.
Which statement best describes immunophenotyping?
A. Lineage determination by detecting antigens on the surface of the gated cells by using
fluorescent antibodies
B. Identification of cell maturity by using antibodies to detect antigens within the nucleus
C. Identification and sorting of cells by front and side scatter of light from a laser
D. Analysis of cells collected by flow cytometry by using traditional agglutination reactions
A Immunophenotyping refers to classification of cells (lineage and maturity assignment)
with use of a panel of fluorescent-labeled antibodies directed against specific surface
antigens on the cells. Antibodies are referred to by their CD number. Monoclonal
antibodies having a common CD number do not necessarily bind to the same epitope
but recognize the same antigen on the cell surface. Reactivity of the selected cells with
a panel of antibodies differentiates lymphoid cells from myeloid cells and identifies the
stage of cell maturation
A flow cytometry scattergram of a bone marrow sample shows a dense population of
cells located in-between normal lymphoid and normal myeloid cells. What is the most
likely explanation?
A. The sample was improperly collected
B. An abnormal cell population is present
C. The laser optics are out of alignment
D. The cells are most likely not leukocytes
B Lymphoid cells and myeloid cells display in predictable regions of the scatterplot
because of their characteristic size and density. Lymphoid cells cause less forward
scatter and side scatter from the laser compared with myeloid cells. A dense zone of
cells in-between those regions is caused by the presence of a large number of abnormal
cells, usually blasts. The lineage of the cells can be determined by immunophenotyping
with a panel of fluorescent-labeled antibodies.
Which serum antibody response usually characterizes the primary (early) stage of
syphilis?
A. Antibodies against syphilis are undetectable
B. Detected 1 to 3 weeks after appearance of the primary chancre
C. Detected in 50% of cases before the primary chancre disappears
D. Detected within 2 weeks after infection
B During the primary stage of syphilis, about 90% of patients develop antibodies between
1 and 3 weeks after the appearance of the primary chancre.
What substance is detected in the sample by the rapid plasma reagin (RPR) and
Venereal Disease Research Laboratory (VDRL) tests for syphilis?
A. Cardiolipin
B. Anticardiolipin antibody (ACA)
C. Anti–Treponema pallidum antibody
D. T. pallidum
B Reagin is the name for a nontreponemal antibody that appears in the serum of
individuals with syphilis and is detected by the RPR and VDRL assays. Reagin reacts
with cardiolipin, a lipid-rich extract of beef heart and other animal tissues
What type of antigen is used in the RPR card test?
A. Live treponemal organisms
B. Killed suspension of treponemal organisms
C. Cardiolipin
D. Tanned sheep cells
C Cardiolipin is extracted from animal tissues, such as beef hearts, and attached to carbon
particles. In the presence of reagin, the particles will agglutinate.
Which of the following is the most sensitive test to detect congenital syphilis?
A. VDRL
B. RPR
C. T. pallidum particle agglutination (TP-PA)
D. Polymerase chain reaction (PCR)
D PCR will amplify a very small amount of DNA from T. pallidum and allow for the
detection of the organism in the infant. Antibody tests, such as VDRL and RPR, may
detect maternal antibody only and do not indicate if the infant has been infected.
A biological false-positive reaction is least likely with which test for syphilis?
A. VDRL
B. TP-PA
C. RPR
D. All are equally likely to yield a false-positive result
B The TP-PA test is more specific for T. pallidum compared with nontreponemal tests,
such as the VDRL and RPR tests, and would be the least likely to yield a biological
false-positive result. Nontreponemal tests have a biological false-positive rate of 1% to
10%, depending on the patient population tested. False-positive findings are caused
commonly by infectious mononucleosis (IM), SLE, viral hepatitis, and human
immunodeficiency virus (HIV) infection
A 12-year old girl has symptoms of fatigue and localized lymphadenopathy. Laboratory
tests reveal peripheral blood lymphocytosis, positive RPR, and positive spot test for IM.
What test should be performed next?
A. HIV screen
B. VDRL
C. Epstein-Barr virus (EBV)–specific antigen test
D. TP-PA test
D The patient’s symptoms are nonspecific and could be attributed to many potential
causes. However, the patient’s age, lymphocytosis, and serological results point to IM.
The rapid spot test for antibodies seen in IM is highly specific. The EBV-specific
antigen test is more sensitive but is unnecessary when the spot test is positive. HIV
infection is uncommon at this age and is often associated with generalized
lymphadenopathy and a normal or reduced total lymphocyte count. IM antibodies are
commonly implicated as a cause of biological false-positive nontreponemal test results
for syphilis. Therefore, a treponemal test for syphilis should be performed to document
this phenomenon in this case
Which test is most likely to be positive in the tertiary stage of syphilis?
A. Treponemal-specific antibody
B. RPR
C. VDRL
D. Reagin screen test (RST)
A A treponemal-specific antibody test is more likely to be positive compared with a
nontreponemal test in the tertiary stage of syphilis. In some cases, systemic lesions
have subsided by the tertiary stage, and the nontreponemal tests become seronegative.
Although the treponemal-specific antibody test is the most sensitive test for tertiary
syphilis, it will be positive in both treated and untreated cases.
What is the most likely interpretation of the following syphilis serological results?
RPR: reactive; TP-PA: nonreactive
A. Neurosyphilis
B. Secondary syphilis
C. Syphilis that has been successfully treated
D. Biological false positive
D A positive reaction with nontreponemal antigen and a negative reaction with a
treponemal antigen is most likely caused by a biological false-positive nontreponemal
test result.
Which specimen is the sample of choice to evaluate latent or tertiary syphilis?
A. Serum sample
B. Chancre fluid
C. Cerebrospinal fluid (CSF)
D. Joint fluid
C Latent syphilis usually begins after the second year of untreated infection. In some
cases, the serological tests become negative. However, if neurosyphilis is present, CSF
serology will be positive and the CSF will display increased protein and pleocytosis
characteristic of central nervous system infection
Interpret the following quantitative RPR test results.
RPR titer: weakly reactive—1:4; reactive—1:8 to 1:64
A. Excess antibody, prozone effect
B. Excess antigen, postzone effect
C. Equivalence of antigen and antibody
D. Impossible to interpret; testing error
A This patient may be in the secondary stage of syphilis and is producing large amounts
of antibody to T. pallidum sufficient to cause a prozone reaction as a result of antibody
excess in the test. The test became strongly reactive only after the antibody was
diluted
Tests to identify infection with HIV fall into which three general classification types of
tests?
A. Tissue culture, antigen, and antibody tests
B. Tests for antigens, antibodies, and nucleic acid
C. DNA probe, DNA amplification, and Western blot tests
D. ELISA, Western blot, and Southern blot tests
B The fourth- and fifth-generation HIV assays detect both antibodies to HIV and the
HIV p24 antigen. Molecular assays can be used to resolve discrepant screening results
or to confirm results, as well as to quantitate the amount of virus present
Which tests are considered screening tests for HIV?
A. ELISA, chemiluminescent, and rapid antibody tests
B. IFA, Western blot, radioimmunoprecipitation assay
C. Culture, antigen capture assay, DNA amplification
D. Reverse transcriptase and messenger RNA (mRNA) assay
A The fourth- and fifth-generation HIV assays detect both antibody and the p24 antigen.
These assays are available in ELISA, automated chemiluminescent systems, and
mutilplex systems and in a rapid card format.
Which tests are the recommended confirmatory tests for HIV?
A. ELISA and rapid antibody tests
B. HIV-1,2 antibody differentiation assays, and qualitative PCR test
C. Culture, antigen capture assay, quantitative PCR
D. Reverse transcriptase and mRNA assay
B The current HIV testing algorithm begins with a screening assay, followed by an
HIV-1,2 antibody differentiation assay (HIV-1,2 supplemental assay) and, if those
results are discordant, an HIV qualitative PCR assay is performed. Western blot is not
included in this algorithm
How do fourth- and fifth-generation HIV tests reduce the time from infection to the test
becoming positive?
A. They are PCR tests detecting viral RNA
B. They detect p24 antigen in addition to HIV antibody
C. They detect proviral DNA
D. They detect antibodies to more antigens than earlier generations of HIV tests
B Including the p24 antigen in the fourth- and fifth-generation tests allows for the
detection of HIV infection approximately 1 week earlier compared with the third-
generation antibody-only assays and up to 3 weeks earlier compared with Western
blot.
A woman who has had five pregnancies subsequently tests positive for HIV on a fourth-
generation assay and is negative on an HIV-1,2 differentiation assay and a follow-up
molecular assay. The initial reactivity may be caused by:
A. Possible cross-reaction with herpes or EBV antibodies
B. Interference from medication
C. Cross-reacting antibodies elicited during pregnancy
D. Possible technical error; a repeat specimen should be requested
C Pregnancy is a common cause of false-positive HIV screening results.
Interpret the following results for HIV testing:
Fourth-generation ELISA: positive; repeat ELISA:
positive; HIV 1,2 antibody differentiation assay:
negative; qualitative HIV RNA rtPCR assay: positive
A. False-positive fourth-generation assay
B. False-negative antibody differentiation assay
C. Indeterminate; further testing indicated
D. HIV p24 antigen detected on fourth-generation ELISA
D The fourth-generation HIV assay detects antibody and the p24 antigen but does not
differentiate between those results. In this case, the antibody-confirming test is
negative, suggesting the initial reactive fourth-generation test result is either a false-
positive one or is caused by the presence of p24 in the specimen. The positive
molecular assay confirms the presence of the virus in the specimen. This usually
occurs in early infection, prior to antibody being produced.
What is the most likely explanation when antibody tests for HIV are negative but the
PCR test is positive?
A. Probably not HIV infection
B. Patient is in the “window phase” before antibody production
C. Tests were performed incorrectly
D. Clinical signs may be misinterpreted
B In early seroconversion, patients may not be making antibodies in sufficient amounts
to be detected by antibody tests. The period between infection with HIV and the
appearance of detectable antibodies is called the window phase. This period has been
reduced to a few weeks by antibody and antigen-detecting fourth- and fifth-generation
assays, and an algorithm that includes PCR testing.
What is the main difficulty associated with the development of an HIV vaccine?
A. The virus has been difficult to culture; antigen extraction and concentration are extremely
laborious
B. Human trials cannot be performed
C. Different strains of the virus are genetically diverse
D. Anti-idiotype antibodies cannot be developed
C Vaccine development has been difficult primarily because of the genetic diversity
among different strains of the virus, and new strains are constantly emerging. HIV-1
can be divided into two main subtypes designated M (for main) and O (for outlier).
The M group is further divided into nine subgroups, designated A through J (there is
no E subgroup), based on differences in the nucleotide sequence of the gag gene. Two
remaining subtypes are designated N (non-M and non-O) and P (a subtype related to
SIVgor). A vaccine that is effective for all of the subgroups of HIV-1 has yet to be
developed.
What criteria constitute the classification system for HIV infection?
A. CD4-positive T-cell count and clinical symptoms
B. Clinical symptoms, condition, duration, and strength of reactivity on a fourth-generation
HIV test
C. Presence or absence of lymphadenopathy
D. Strong fourth-generation HIV test reactivity and CD8-positive T-cell count
A The classification (not diagnostic) system for HIV infection is based on a
combination of CD4-positive T-cell count (helper T cells) and various categories of
clinical symptoms. Classification is important in determining treatment options and the
progression of the disease.
What is the advantage of fourth-generation rapid HIV tests over earlier rapid HIV
tests?
A. They use recombinant antigens
B. They detect multiple strains of HIV
C. They detect p24 antigen
D. They are quantitative
C Both third-generation and fourth-generation rapid tests for HIV use recombinant and
synthetic HIV antigens conjugated to a solid phase. The multivalent nature of these
tests allows for detection of less common subgroups of HIV-1 and simultaneous
detection of both HIV-1 and HIV-2. However, the fourth-generation assays also use
solid-phase antibodies to p24 antigen to detect its presence. Because p24 antigen
appears before antibodies to HIV, fourth-generation tests can detect infection 4 to 7
days earlier compared with tests based on antibody detection alone.
Which CD4:CD8 ratio is most likely in a patient with AIDS?
A. 2:1
B. 3:1
C. 2:3
D. 1:3
D An inverted CD4:CD8 ratio (less than 1.0) is a common finding in a patient with
AIDS. The Centers for Disease Control and Prevention (CDC) requires a CD4-positive
(helper T) cell count of less than 200/μL or 14% in the absence of an AIDS-defining
illness (e.g., Pneumocystis carinii pneumonia) in the case surveillance definition of
AIDS
What is the most likely cause when a fourth-generation HIV assay is positive for all
controls and samples?
A. Improper pipetting
B. Improper washing
C. Improper addition of sample
D. Improper reading
B Improper washing may not remove unbound, enzyme-conjugated antihuman Ig, and
every sample may appear positive.
Which method is used to test for HIV infection in infants who are born to HIV-positive
mothers?
A. ELISA
B. Western blot test
C. PCR test
D. Viral culture
C Fourth- and fifth-generation ELISA and chemiluminescent assays reflect the presence
of maternal antibody. The PCR test uses small amounts of blood and does not rely on
the antibody response. PCR amplifies small amounts of viral nucleic acid and can
detect less than 20 copies of viral RNA per milliliter of plasma. These qualities make
PCR ideal for the testing of infants. Nucleic acid methods for HIV RNA include both
qualitative (for diagnosis) and quantitative (for monitoring) reverse-transcriptase real-
time PCR (RT-PCR) assays.
What constitutes a diagnosis of viral hepatitis?
A. Abnormal test results for liver enzymes
B. Clinical signs and symptoms
C. Positive results for hepatitis markers
D. All of these options
D To diagnose a case of hepatitis, the physician must consider clinical signs as well as
the results of laboratory tests that measure liver enzymes and hepatitis markers
Which of the following statements regarding infection with hepatitis D virus (HDV) is
true?
A. Occurs in patients with HIV infection
B. Does not progress to chronic hepatitis
C. Occurs in patients with hepatitis B virus (HBV) infection
D. Is not spread through blood or sexual contact
C HDV is an RNA virus that requires the surface antigen or envelope of the HBV for
entry into the hepatocyte. Consequently, HDV can infect only patients who are
coinfected with hepatitis B.
All of the following hepatitis viruses are spread through blood or blood products except:
A. Hepatitis A virus (HAV)
B. HBV
C. HCV
D. HDV
A HAV is spread through the fecal–oral route and is the cause of infectious hepatitis.
HAV has a shorter incubation period (2–7 weeks) than HBV (1–6 months). Epidemics
of HAV can occur, especially when food and water become contaminated with raw
sewage. Hepatitis E virus is also spread via the oral–fecal route and, like HAV, has a
short incubation period
Which hepatitis B marker is the best indicator of early acute infection?
A. Hepatitis B surface antigen (HBsAg)
B. Hepatitis B e-antigen (HBeAg)
C. Hepatitis B core antibody (anti-HBc)
D. Hepatitis B surface antibody (anti-HBs)
A HBsAg is the first marker to appear in HBV infection. It is usually detected within 4
weeks of exposure (prior to the rise in transaminases) and persists for about 3 months
after serum enzyme levels return to normal.
Which is the first antibody detected in serum after infection with HBV?
A. Anti-HBs
B. Anti-HBc IgM
C. Anti-HBe
D. All are detectable at the same time
B Antibody to the hepatitis B core antigen (anti-HBc) is the first detectable hepatitis B
antibody. It persists in serum for years after infection and is found in the serum of
asymptomatic carriers of HBV. Because levels of total anti-HBc are high after
recovery, IgM anti-HBc is a more useful marker for acute infection. Both anti-HBc and
anti-HBs can persist for life, but only anti-HBs is considered protective.
Which antibody persists in low-level carriers of HBV?
A. IgM anti-HBc
B. IgG anti-HBc
C. IgM anti-HBe
D. IgG anti-HBs
B IgG anti-HBc can be detected in carriers who are HBsAg and anti-HBs negative.
These persons are hepatitis B DNA positive also and, thus, are presumed infective,
even though the level of HBsAg is too low to detect. No specific B core IgG test is
available, however. This patient would be positive in the anti-B core total antibody
assay and negative in the anti-HB core IgM test.
What is the most likely explanation when a patient has clinical signs of viral hepatitis
but tests negative for HAV IgM, HBsAg, and HCV antibody?
A. Tests were performed improperly
B. The patient does not have hepatitis
C. The patient may be in the “core window”
D. Clinical evaluation was performed improperly
C The patient may be in the “core window,” the period of HBV infection when both the
surface antigen and surface antibody are undetectable. The IgM anti-HBc and the anti-
HBc total antibody assays, along with the hepatitis B DNA PCR assay would be the
only detectable markers in the serum of a patient in the core window phase of HBV
infection.
Which hepatitis B markers should be performed on blood products?
A. HBsAg and anti-HBc
B. Anti-HBs and anti-HBc
C. HBeAg and HBcAg
D. Anti-HBs and HBeAg
A Blood products are tested for HBsAg, an early indicator of infection, and anti-HBc, a
marker that may persist for life. Following recovery from HBV infection, some
patients demonstrate negative serology for HBsAg and anti-HBs but are positive for
anti-HBc. Such patients are considered infective.
Which hepatitis antibody confers immunity against reinfection with HBV?
A. Anti-HBc IgM
B. Anti-HBc IgG
C. Anti-HBe
D. Anti-HBs
D Anti-HBs appears later in infection compared with anti-HBc and is used as a marker
for immunity after infection or vaccination, rather than for diagnosis of current
infection.
Interpret the following results for EBV infection: IgG and IgM antibodies to viral
capsid antigen (VCA) are positive.
A. Infection in the past
B. Infection with a mutual enhancer virus, such as HIV
C. Current infection
D. Impossible to interpret; need more information
C IgM and IgG antibodies to VCA are found in a current infection with EBV. The IgG
antibody may persist for life, but the IgM anti-VCA disappears within 4 months after
the infection resolves
Rapid mono tests use latex particles coated with which of the following?
A. Guinea pig antigen
B. Beef proteins
C. Horse proteins
D. Sheep proteins
B Rapid mono tests detect a heterophile antibody directed against beef proteins.
Although these antibodies may also react with horse or sheep red blood cells (RBCs),
those proteins are not used in these tests
Which test, other than serological markers, is most consistently elevated in viral
hepatitis?
A. Antinuclear antibodies
B. Alanine aminotransferase (ALT)
C. Absolute lymphocyte count
D. Lactate dehydrogenase
B ALT is a liver enzyme and may be increased in hepatic disease. Highest levels occur
in acute viral hepatitis, reaching 20 to 50 times the upper limit of normal.
If only anti-HBs is positive, which of the following can be ruled out?
A. HBV vaccination
B. Distant past infection with HBV
C. Hepatitis B immune globulin (HBIG) injection
D. Chronic HBV infection
D Persons with chronic HBV infection show a positive test result for anti-HBc (IgG or
total) and HBsAg but not anti-HBs. Patients with active chronic hepatitis have not
become immune to the virus
Blood products are tested for which virus before being transfused to newborns?
A. EBV
B. Human T-lymphotropic virus II (HTLV-II)
C. CMV
D. HDV
C CMV can be life threatening if transmitted to a newborn through a blood product.
HTLV-II is a rare virus, which like HIV, is a T-cell tropic RNA retrovirus. The virus
has been associated with hairy cell leukemia, but this is not a consistent finding.
What is the endpoint for the antistreptolysin O (ASO) latex agglutination assay?
A. Highest serum dilution that shows no agglutination
B. Highest serum dilution that shows agglutination
C. Lowest serum dilution that shows agglutination
D. Lowest serum dilution that shows no agglutination
B The latex test for ASO includes latex particles coated with streptolysin O. Serial
dilutions are prepared and the highest dilution showing agglutination is the endpoint.
A streptozyme test was performed, but the result was negative, even though the patient
showed clinical signs of a streptococcal throat infection. What should be done next?
A. Either ASO or anti-deoxyribonuclease B (anti-DNase B) test
B. Another streptozyme test using diluted serum
C. Antihyaluronidase test
D. Wait for 3 to 5 days and repeat the streptozyme test
A The streptozyme test is used for screening and contains several of the antigens
associated with streptococcal products. Because some patients produce an antibody
response to a limited number of streptococcal products, no single test is sufficiently
sensitive to rule out infection. Clinical sensitivity is increased by performing additional
tests when initial results are negative. The streptozyme test generally yields more false-
positive and false-negative results compared with the ASO and anti-DNase B tests. A
positive result occurs in a smaller number of patients with recent streptococcal
infections in the antihyaluronidase test compared with the ASO and anti-DNase B
tests
Rapid assays for influenza that utilize specimens obtained from nasopharyngeal swabs
detect:
A. IgM anti-influenza
B. IgA anti-influenza
C. IgA–influenza antigen immune complexes
D. Influenza nucleoprotein antigens
D The rapid influenza assays are antigen detection methods. They are designed to detect
early infection, before antibody is produced.
How can interfering cold agglutinins be removed from a test sample?
A. Centrifuge the serum and remove the top layer
B. Incubate the clot at 1°C to 4°C for several hours and then remove the serum
C. Incubate the serum at 56°C in a water bath for 30 minutes
D. Use an anticoagulated sample
B Cold agglutinins will attach to autologous RBCs if incubated at 1°C to 4°C. The
absorbed serum will be free of cold agglutinins
All tubes (dilutions) except the negative control are positive for cold agglutinins. This
indicates:
A. Contaminated RBCs
B. A rare antibody against RBC antigens
C. The sample was stored at 4°C prior to separating serum and cells
D. Further serial dilution is necessary
D Cold agglutinins may be measured in patients who have cold agglutinin disease, that
is, cold autoimmune hemolytic anemia. In such cases, titers can be as high as 106. If all
tubes (dilutions) for cold agglutinins are positive, except the negative control, then a
high titer of cold agglutinins is present in the sample. Further serial dilutions should be
performed.
All positive cold agglutinin tubes remain positive after 37°C incubation except the
positive control. What is the most likely explanation for this situation?
A. High-titer cold agglutinins
B. Contamination of the test system
C. Antibody other than cold agglutinins
D. Faulty water bath
C Cold agglutinins do not remain reactive above 30°C, and agglutination must disperse
after incubation at 37°C. The most likely explanation when agglutination remains after
37°C incubation is that a warm alloantibody or autoantibody is present.
Which increase in antibody titer (dilution) best indicates an acute infection?
A. From 1:2 to 1:8
B. From 1:4 to 1:16
C. From 1:16 to 1:256
D. From 1:64 to 1:128
C A fourfold (two-tube) or greater increase in antibody titer is usually indicative of an
acute infection. Although answers A and B show a fourfold rise in titer, answer C
shows a 16-fold rise in titer and is the most definitive. In most serological tests, a
single high titer is insufficient evidence of acute infection unless specific IgM
antibodies are measured because age, individual variation, immunologic status, and
history of previous exposure (or vaccination) cause a wide variation in normal serum
antibody titers.
Which of the following positive antibody tests may be an indication of recent
vaccination or early primary infection for rubella in a patient with no clinical
symptoms?
A. Only IgG antibodies positive
B. Only IgM antibodies positive
C. Both IgG and IgM antibodies positive
D. Fourfold rise in titer for IgG antibodies
B If only IgM antibodies are positive, this result indicates recent vaccination or early
primary infection.
Why is laboratory diagnosis difficult in cases of Lyme disease?
A. Clinical response may not be apparent upon initial infection; IgM antibody may not be
detected until 3 to 6 weeks after the infection
B. Laboratory tests may be designed to detect whole Borrelia burgdorferi, not flagellar
antigen found early in infection
C. Most laboratory tests are technically demanding and lack specificity
D. Antibodies formed initially to B. burgdorferi may cross react in antigen tests for
autoimmune diseases
A Lyme disease is caused by B. burgdorferi, a spirochete, and typical clinical
symptoms, such as rash or erythema chronicum migrans, may be absent in some
infected individuals. Additionally, IgM antibody is not detectable by laboratory tests
until 3 to 6 weeks after a tick bite, and IgG antibody develops later.
Serological tests for which disease may give a false-positive result if the patient has
Lyme disease?
A. HIV
B. Syphilis
C. EBV
D. Hepatitis C
B Lyme disease is caused by a spirochete, and positive results may occur with some
specific treponemal antibody tests for syphilis
In monitoring a patient with HIV infection, which parameter may be expected to be the
most sensitive indicator of the effectiveness of antiretroviral treatment?
A. HIV antibody titer
B. CD4:CD8 ratio
C. HIV viral load
D. Absolute total T-cell count
C The HIV viral load will rise or fall in response to treatment more quickly compared
with any of the other listed parameters. The absolute CD4 count is also an indicator of
treatment effectiveness and is used in resource-poor areas that might not have facilities
for molecular testing. Note, however, that the absolute CD4 count is not one of the
choices.
A renal transplant recipient is found to have a rising creatinine level and reduced urine
output. The physician orders a “urine PCR” assay. When you call to find out what
organism the physician wants to identify, you are told:
A. HCV
B. Legionella pneumophila
C. EBV
D. BK virus
D BK virus is a polyoma virus that can cause renal and urinary tract infections. The
virus is an opportunistic pathogen and has become a well-recognized cause of poor
renal function in kidney transplant recipients. Antibody testing is not practical or
useful for this infection. The principal diagnostic assays are urinary cytology, and
specific BK virus PCR testing in urine and serum. Although L. pneumophila can be
diagnosed through a urinary antigen assay, that organism is not a primary cause of
renal insufficiency in transplant recipients
A newborn is to be tested for vertically transmitted HIV infection. Which of the
following tests is most useful?
A. HIV PCR
B. CD4 count
C. Rapid HIV antibody test
D. HIV IgM antibody test
A Neonatal HIV diagnosis is performed by screening for the presence of the virus. The
current antibody tests are either an IgG-specific assay or an IgG/IgM combination
assay. Thus, an infant whose mother is HIV positive will also be positive in the HIV
antibody assay. Although the CD4 count may be a useful assay to determine disease
activity, there are many causes of reduced CD4 numbers, so this assay should not be
used to diagnose HIV infection
Which of the following fungal organisms is best diagnosed by an antigen detection test
as opposed to an antibody detection assay?
A. Histoplasma
B. Cryptococcus
C. Candida
D. Aspergillus
B The Cryptococcus antibody response is not a reliable indicator of a current infection;
thus, an antigen assay is normally used to monitor the disease. The antigen assay may
be used for serum or CSF and will decline in response to treatment much faster than a
traditional antibody test. A urinary antigen test is available for histoplasmosis, and a
serum galactomannan assay is available for Aspergillus. Those two assays perform
better than antibody detection. No antigen test is available for Candida, and thus,
antibody detection is the best serological procedure for this organism.
Your cytology laboratory refers a Papanicolaou smear specimen to you for an assay
designed to detect the presence of a virus associated with cervical cancer. You perform:
A. An ELISA for anti-human simplex virus 2 (anti-HSV-2) antibodies
B. A molecular assay for HSV-2
C. An ELISA for human papilloma virus (HPV) antibodies
D. A molecular assay for HPV
D Cervical cell atypia and cervical cancer are associated with specific high-risk
serotypes of HPV infections. Although HPV antibody assays are available, they are not
serotype specific, nor do they relate to disease activity. Thus, molecular probe assays
are the tests of choice to detect high-risk HPV infection. Although HSV-2 is associated
with genital herpesvirus, that virus has not been shown to cause cervical cancer
An immunosuppressed patient has unexplained anemia. The physician suspects a
parvovirus B19 infection. The parvovirus IgM test result is negative. The next course of
action is to tell the physician that:
A. The patient does not have parvovirus
B. A convalescent specimen is recommended in 4 weeks to determine if a fourfold rise in
titer has occurred
C. A parvovirus PCR is recommended
D. A recent transfusion for the patient’s anemia may have resulted in a false-negative result
and the patient should be retested in 4 weeks
C A negative IgM assay rarely rules out an infection. Although a convalescent specimen
may be useful in many cases, in an immunosuppressed patient, the convalescent
specimen may remain negative in the presence of an infection. Thus, a parvovirus PCR
test is the preferred choice in this case. A false-negative result could be caused by
multiple whole blood or plasma transfusions, but retesting for antibody a month later
would not be beneficial to the patient.
What is a general definition for autoimmunity?
A. Increase of tolerance to self-antigens
B. Loss of tolerance to self-antigens
C. Increase in clonal deletion of mutant cells
D. Manifestation of immunosuppression
B Autoimmunity is a loss of tolerance to self-antigens and the subsequent formation of
autoantibodies.
An ANA test is performed on a specimen from a 55-year-old woman who has
unexplained joint pain. The IFA result shows a titer of 40 and a homogeneous pattern.
The appropriate follow-up for this patient is:
A. Anti-DNA assay
B. Extractable nuclear antigen (ENA) testing
C. Retest ANA in 3 to 6 months
D. CH50 complement assay
C Approximately 25% of women in this age range may have low titer–positive ANA
assays with no demonstrable connective tissue disease. A patient with anti-DNA–
positive SLE would be expected to have a much higher titer (greater than 160) in an
IFA. A similar titer would be expected for an ENA positive specimen, although the
pattern would be speckled. Complement testing would not be indicated with this low
titer in a 55-year-old female.
Which disease is likely to show a rim (peripheral) pattern in an immunofluorescence
(IF) microscopy test for ANA?
A. Mixed connective tissue disease (MCTD)
B. RA
C. SLE
D. Scleroderma
C The rim or peripheral pattern seen in indirect immunofluorescence techniques is most
commonly found in cases of active SLE. The responsible autoantibody is highly
correlated to anti–double-stranded DNA (anti-dsDNA).
A patient’s specimen is strongly positive in an ANA ELISA. Which of the following
would not be an appropriate follow-up to this result?
A. IFA on human epithelial type 2 (HEp-2) cells
B. Specific ENA ELISA tests
C. Specific anti-DNA ELISA
D. Rheumatoid factor (RF) assay
D The ANA ELISA is a screening assay. A positive result may be followed up by more
specific antibody ELISA tests or an ANA IFA to determine pattern and titer. The ANA
ELISA does not screen for RF.
What type of antibodies is represented by the homogeneous pattern in the IFA for
ANAs?
A. Antihistone antibodies
B. Anticentromere antibodies
C. Anti-ENA (anti-Smith [anti-Sm] and anti-ribonucleoprotein [anti-RNP]) antibodies
D. Anti-RNA antibodies
A Antihistone antibodies (and also anti-DNA antibodies) cause the solid or homogeneous
pattern, which is commonly found in patients with SLE, RA, MCTD, and Sjögren
syndrome. Antibodies to the centromere of chromosomes is a marker for the CREST
(calcinosis, Raynaud phenomenon, esophageal dysfunction, sclerodactyly, and
telangiectasia) form of systemic sclerosis.
What disease is indicated by a high titer of anti-Sm antibody?
A. MCTD
B. RA
C. SLE
D. Scleroderma
C High titer anti-Sm is indicative of SLE. Anti-Sm is an antibody against saline ENAs.
and causes a speckled pattern of immunofluorescence.
Which disease is least likely when a nucleolar pattern occurs in an IFA for ANAs?
A. MCTD
B. Sjögren syndrome
C. SLE
D. Scleroderma
A All of the diseases except MCTD may cause a nucleolar pattern of
immunofluorescence. Nucleolar fluorescence is caused by anti-RNA antibodies and is
seen in about 50% of patients with scleroderma.
What antibodies are represented by the nucleolar pattern in the IFA for ANAs?
A. Antihistone antibodies
B. Anti-dsDNA antibodies
C. Anti-ENA (anti-Sm and anti-RNP) antibodies
D. Anti-RNA antibodies
D Anti-RNA antibodies are represented by the nucleolar pattern. This pattern may be
seen in most systemic autoimmune diseases and is especially common in patients with
scleroderma. Anti-RNA and anti-Sm are not usually found in patients with MCTD.
This is a syndrome involving aspects of SLE, RA, scleroderma, and polymyositis. The
immunofluorescence pattern most often seen in MCTD is the speckled pattern caused
by anti-RNP.
Which test would best distinguish between SLE and MCTD?
A. Multiplex or ELISA test for anti-Sm and anti-RNP
B. IFA using Crithidia as substrate
C. Slide agglutination testing
D. Laboratory tests cannot distinguish between these disorders
A Line blots, multiplex, and ELISA assays, using purified or recombinant antigens, are
available for differentiating anti-RNP from anti-Sm. Anti-Sm with or without anti-
RNP is found in approximately one third of patients with SLE. Anti-RNP in the
absence of anti-Sm is found in over 95% of patients with MCTD
An ANA test on HEp-2 cells shows nucleolar staining in interphase cells and dense
chromatin staining in mitotic cells. The most likely cause of this staining pattern is:
A. Antifibrillarin antibody
B. Anti–ribosomal P antibody
C. A serum with nucleolar and homogeneous patterns
D. Technical artifact
A Antifibrillarin antibody has this appearance. Ribosomal P antibody has nucleolar
staining and a background homogeneous and cytoplasmic stain. A combination
nucleolar/homogeneous specimen will also show homogeneous staining in the
interphase cells. This pattern is not seen in typical technical artifacts
Which immunofluorescence pattern indicates the need for ENA testing by multiplex,
line blots, or ELISA assays?
A. Homogeneous or solid
B. Peripheral or rim
C. Speckled
D. Nucleolar
C A speckled pattern is often caused by the presence of antibodies against the ENAs,
such as Sm, RNP, SSA, and SSB. Homogeneous and rim patterns suggest antibodies to
dsDNA. The homogeneous pattern may also be seen with antibodies to
deoxyribonuclear protein, which is not an ENA. Nucleolar patterns often indicate
antibodies to RNA or fibrillarin.
Which of the following is used in rapid slide tests for detection of RFs?
A. Whole IgM molecules
B. Fc portion of the IgG molecule
C. Fab portion of the IgG molecule
D. Fc portion of the IgM molecule
B RFs react with the Fc portion of the IgG molecule and are usually IgM. This is the
basis of rapid agglutination tests for RA. Particles of latex or cells are coated with IgG.
Addition of serum containing RF results in visible agglutination
Which of the following methods is least likely to give a definitive result for the diagnosis
of RA?
A. Nephelometric measurement of anti-IgG
B. Agglutination testing for RF
C. Anti-CCP
D. IFA for ANAs
D Patients with RA often show a homogeneous pattern of fluorescence in tests for
ANAs. However, this pattern is seen in a wide range of systemic autoimmune diseases
and in many normal persons at a titer below 10. The first two methods listed may be
used to identify anti-IgG, which is one laboratory criterion used to establish a diagnosis
of RA. Anti-CCP is an additional laboratory criterion used in the RA diagnostic
algorithm.
Which disease might be indicated by antibodies to smooth muscle?
A. Atrophic gastritis
B. Autoimmune hepatitis
C. Myasthenia gravis
D. Sjögren syndrome
B Antibodies to smooth muscle are found in the serum of up to 70% of patients with
active chronic hepatitis and up to 50% of patients with primary biliary cirrhosis
Antibodies to thyroid peroxidase may appear in which of the following diseases?
A. Graves disease and Hashimoto thyroiditis
B. Myasthenia gravis
C. Granulomatous thyroid disease
D. Addison disease
A Antibodies to thyroid peroxidase may be detected in both Graves disease
(hyperthyroidism) and Hashimoto thyroiditis (hypothyroidism). If a positive result is
found to thyroid peroxidase, thyroxine levels and clinical presentation can be used to
distinguish between the two diseases.
What is the main use of laboratory tests to detect antibodies to islet cells and insulin in
cases of insulin-dependent diabetes mellitus (IDDM)?
A. To regulate levels of injected insulin
B. To diagnose IDDM
C. To rule out the presence of other autoimmune diseases
D. To screen susceptible individuals prior to destruction of β-cells
D Fasting hyperglycemia and hemoglobin A1C levels are the primary findings used to
diagnose IDDM. For individuals with an inherited susceptibility to the development of
IDDM, laboratory tests for the detection of antibodies to islet cells and insulin may
help to initiate early treatment before complete destruction of β-cells
A patient presents with clinical symptoms of celiac disease. Tests for anti-tissue
transglutaminase and antigliadin antibodies are negative. Which of the following tests
should be ordered?
A. IgG level
B. HLA DQ2 and DQ8 typing
C. HLA DR3 and DR7 typing
D. IgM level
B Although antibodies to tissue transglutaminase and gliadin are often found in celiac
disease, their combined sensitivity is less than 100%. Celiac disease is almost
exclusively associated with the presence of HLA DQ2 and/or HLA DQ8. These HLA
genes are not diagnostic of celiac disease, but provide a testing alternative in antibody-
negative individuals who meet the clinical diagnostic criteria for celiac disease.
A specimen appears to have a perinuclear staining pattern in an antineutrophil
cytoplasmic antibody (ANCA) immunofluorescent assay using ethanol-fixed
neutrophils, suggesting the possibility of a perinuclear ANCA (pANCA). On which of
the following substrates would this specimen display cytoplasmic speckling?
A. Formalin-fixed neutrophils
B. Unfixed neutrophils
C. HEp-2 cells
D. Rabbit kidney tissue
A Antibodies to neutrophil cytoplasmic antigen demonstrating a perinuclear pattern of
fluorescence indicate a diagnosis of vasculitis. However, atypical ANCAs and ANAs
also demonstrate a perinuclear staining pattern on ethanol-fixed neutrophils. To
differentiate these from pANCA, specimens appearing as pANCAs on ethanol-fixed
cells are tested on formalin-fixed neutrophils. The myeloperoxidase-containing
granules that coalesce around the nuclear membrane during ethanol fixation will
remain in the cytoplasm during formalin fixation. Thus, pANCAs will have a
cytoplasmic (cANCA) pattern on a formalin-fixed slide, but ANAs will retain a
perinuclear pattern and the fluorescence will be diminished.
Which of the following is a description of a type I hypersensitivity reaction?
A. Ragweed antigen cross-links with IgE on the surface of mast cells, causing release of
preformed mediators and resulting in symptoms of an allergic reaction
B. Anti-Fya from a pregnant woman crosses the placenta and attaches to the Fya antigen-
positive RBCs of the fetus, destroying the RBCs
C. Immune complex deposition occurs on the glomerular basement membrane of the kidney,
leading to renal failure
D. Exposure to poison ivy causes sensitized T cells to release lymphokines that cause a
localized inflammatory reaction
A Type I immediate hypersensitivity (anaphylactic) responses are characterized by IgE
molecules binding to mast cells via the Fc receptor. Cross-linking of surface IgE
caused by binding of allergens causes the mast cell to degranulate, releasing histamine
and other chemical mediators of allergy. Answer B describes a type II reaction; C
describes a type III reaction; and D describes a type IV reaction
Which in vitro test measures IgE levels against a specific allergen?
A. Histamine release assay
B. Radioimmunosorbent test (RIST)
C. Fluorescent allergosorbent test (FAST)
D. Precipitin radioimmunosorbent test (PRIST)
C FAST is a fluorescent assay that measures specific IgE; RIST and PRIST tests are
radioimmunoassays that measure total IgE. The FAST procedure and a
chemiluminescent assay have replaced the radioallergosorbent test (RAST), for
measuring allergen specific IgE. The histamine release assay is a more general assay
Why is skin testing the most widely used method to test for a type I hypersensitivity
reaction?
A. It causes less trauma and is more cost effective than other methods
B. It has greater sensitivity than in vitro measurements
C. It is more likely to be positive for IgE-specific allergens compared with other methods
D. It may be used to predict the development of further allergen sensitivity
B Skin testing is considered much more sensitive than in vitro tests that measure either
total or antigen-specific IgE.
A patient who is blood group O is accidentally transfused with group A blood and
develops a reaction during the transfusion. What antibody is involved in this type II
reaction?
A. IgM
B. IgE
C. IgG and IgE
D. IgG
A IgG and IgM are the antibodies involved in a type II cytotoxic reaction. Naturally
occurring anti-A in the form of IgM is present in the blood of a group O individual and
would cause an immediate transfusion reaction. Cell destruction occurs when
antibodies bind to cells causing destruction via complement activation, thereby
triggering intravascular hemolysis.
Which test would measure the coating of RBCs by antibody as occurs in hemolytic
transfusion reactions?
A. Indirect antiglobulin test (IAT)
B. Direct antiglobulin test (DAT)
C. ELISA
D. Hemagglutination
B The DAT measures antibody that has already coated RBCs in vivo. DAT and direct
IFAs use anti-Ig to detect antibody-sensitized cells.
Which test detects antibodies that have attached to tissues, resulting in a type-II
cytotoxic reaction?
A. Indirect immunofluorescence
B. Direct immunofluorescence (DIF)
C. Immunofixation electrophoresis (IFE)
D. Hemagglutination
B The direct IFA detects the presence of antibody that may cause a type II cytotoxic
reaction. For example, renal biopsies from patients with Goodpasture syndrome exhibit
a smooth pattern of fluorescence along the basement membrane after reaction with
fluorescein isothiocyanate (FITC)–conjugated anti-Ig. The reaction detects antibodies
against the basement membrane of the glomeruli
Which of the following conditions will most likely result in a false-negative DAT test?
A. Insufficient washing of RBCs
B. Use of heavy chain–specific polyclonal anti-human Ig
C. Use of excessive centrifugal force
D. Use of a sample obtained by finger puncture
A Insufficient washing can cause incomplete removal of excess or unbound Igs and other
proteins, which may neutralize the antiglobulin reagent.
Which of the following tests will detect circulating immune complexes in the serum of
some patients with systemic autoimmune diseases, such as RA?
A. Direct IFA
B. Enzyme immunoassay (EIA)
C. Assay of cryoglobulins
D. IAT
C Most autoimmune diseases involve the formation of antigen–antibody complexes that
deposit in tissues, causing local inflammation and necrosis induced by complement
activation, phagocytosis, white blood cell (WBC) infiltration, and lysosomal damage.
Some patients make monoclonal or polyclonal antibodies with RF activity that bind to
serum Igs, forming aggregates that are insoluble at 4°C. These circulating immune
complexes are detected by allowing a blood sample to clot at 37°C, transferring the
serum to a sedimentation rate tube, and then incubating the serum at 4°C for 3 days.
What immune elements are involved in a positive skin test for tuberculosis?
A. IgE antibodies
B. T cells and macrophages
C. NK cells and IgG antibody
D. B cells and IgM antibody
B T cells and macrophages are the immune elements primarily responsible for the clinical
manifestations of a positive tuberculosis test. Reactions usually take 48 to 72 hours to
reach peak development and are characteristic of localized type IV cell-mediated
hypersensitivity. The skin reaction is characterized by a lesion containing a
mononuclear cell infiltrate.
A patient receives a transfusion of packed RBCs and fresh frozen plasma (FFP) and
develops an anaphylactic, nonhemolytic reaction. She reports receiving a transfusion 20
years earlier. She had no reaction to the previous transfusion, but she did feel “poorly”
a few weeks later. Which of the following transfused substances most likely elicited the
reaction?
A. IgA
B. Group A antigen
C. Rho (D) antigen
D. An antigen belonging to the Duffy system
A The fact that this is a nonhemolytic reaction suggests that a non-RBC antigen may be
involved. Selective IgA deficiency occurs in approximately one in 700 individuals and
is often asymptomatic. Individuals deficient in IgA may make an antibody against the
α-heavy chain if they are exposed to IgA via a transfusion. This antibody may lead to a
serum sickness reaction if the IgA is still present after antibody formation. This could
explain the “poor feeling” the patient had after the initial transfusion. A subsequent
transfusion may lead to an Arthus reaction if IgG anti-IgA is present or an anaphylactic
reaction if IgE anti-IgA is present.
A patient deficient in the C3 complement component would be expected to mount a
normal:
A. Type I and IV hypersensitivity response
B. Type II and IV hypersensitivity response
C. Type I and III hypersensitivity response
D. Type II and III hypersensitivity response
A Complement is involved in types II and III hypersensitivity; thus an individual
deficient in C3 will be deficient in those responses. The complement deficiency should
have no effect on IgE (type I) or cell-mediated (type IV) hypersensitivities
Which of the following symptoms in a young child may indicate an immunodeficiency
syndrome?
A. Anaphylactic reactions
B. Severe rashes and myalgia
C. Recurrent bacterial, fungal, and viral infections
D. Weight loss, rapid heartbeat, breathlessness
C An immunodeficiency syndrome should be considered in a young child who has a
history of recurrent bacterial, fungal, and viral infections manifested after the
disappearance of maternal IgG. Immunodeficiency disorders may involve deficiencies
in production and/or function of lymphocytes and phagocytic cells or a deficiency in
production of a complement factor. Choice of laboratory tests is based on the patient’s
clinical presentation, age, and history.
What screening test should be performed first in a young patient suspected of having an
immune dysfunction disorder?
A. Complete blood count (CBC) and WBC differential
B. Chemotaxis assay
C. Complement levels
D. Bone marrow biopsy
A The first screening tests performed in the initial evaluation of a young patient who is
suspected of having an immune dysfunction are the CBC and differential. WBCs that
are decreased in number or abnormal in appearance may indicate further testing
Which test should be performed when a patient has a reaction to transfused plasma
products?
A. Ig levels
B. T-cell count
C. Hemoglobin levels
D. RBC enzymes
A A reaction to plasma products may be found in an IgA-deficient person who has
formed anti-IgA antibodies. Ig levels would aid in this determination. Selective IgA
deficiency is the most common immunodeficiency disease and is characterized by
serum IgA levels below 5 mg/dL. IgA is usually absent from secretions, but the B-cell
count is usually normal.
What is the “M” component in monoclonal gammopathies?
A. IgM produced in excess
B. Heavy chain produced in excess
C. Malignant proliferation of B cells
D. Monoclonal antibody or cell line
D The “M” component refers to any monoclonal protein or cell line produced in a
monoclonal gammopathy, such as multiple myeloma
In testing for DiGeorge syndrome, what type of laboratory analysis would be most
helpful in determining the number of mature T cells?
A. CBC
B. Dihydrorhodamine reduction (DHR) test
C. T-cell mitogen assays
D. Flow cytometry
D DiGeorge syndrome is caused by a developmental failure or hypoplasia of the thymus
and results in deficiency of T lymphocytes and cell-mediated immune function. The T-
cell count is low, but the level of Igs is usually normal. Flow cytometry is most helpful
in determining numbers and subpopulations of T cells.
A child suspected of having an inherited humoral immunodeficiency disease is given
diphtheria/tetanus vaccine. Two weeks after the immunization, his level of antibody to
the specific antigens is measured. Which result is expected for this patient if he, indeed,
has a humoral deficiency?
A. Increased levels of specific antibody
B. No change in the level of specific antibody
C. An increase in IgG-specific antibody but not IgM-specific antibody
D. Increased levels of nonspecific antibody
B In a patient with immunodeficiency, the expected levels of specific antibody to the
antigens in the vaccine would be decreased or not present. This response provides
evidence of deficient antibody production.
Which disease may be expected to show an IgM spike on an electrophoretic pattern?
A. Hypogammaglobulinemia
B. Multicystic kidney disease
C. Waldenström macroglobulinemia
D. Wiskott-Aldrich syndrome
C Waldenström macroglobulinemia is a malignancy of plasmacytoid lymphocytes
involving both bone marrow and lymph nodes. The malignant cells secrete monoclonal
IgM and are in transition from B cells to plasma cells. In contrast to multiple myeloma,
osteolytic bone lesions are not found.
Interpret the following description of an IFE assay of urine. Dense wide bands in both
the κ- and λ-lanes. No bands present in the heavy-chain lanes.
A. Normal
B. Light-chain disease
C. Increased polyclonal Fab fragments
D. Multiple myeloma
C Heavy wide bands seen with both anti-κ and anti-λ antisera indicate excessive
polyclonal light-chain excretion. Light-chain disease would show a heavy restricted
band for one of the light-chain reactions, but not both. The finding of excess λ- and κ-
chains indicates a polyclonal gammopathy with increased Ig turnover and excretion of
the light chains as Fab fragments.
Free monoclonal light chains are often present in the serum of patients with multiple
myeloma and may be useful for disease monitoring. Which of the following assays would
be recommended to detect the presence of free light chains in serum?
A. SPE
B. Urine immunofixation
C. Nephelometry
D. ELISA
C Free light chains in serum are a sensitive indicator of a monoclonal gammopathy. They
are often not present in sufficient quantity to show a band on a protein electrophoresis
gel. Detecting light chains in urine is not an indicator of what the serum levels may be.
Serum Ig heavy and light chains are most commonly measured by using rate or
endpoint nephelometry. ELISA assays are most often used to measure specific
antibody levels, not to quantitate Ig heavy- or light-chain isotypes.
What type of disorders would show a decrease in C3, C4, and CH50?
A. Autoimmune disorders, such as SLE and RA
B. Immunodeficiency disorders, such as common variable immunodeficiency
C. Tumors
D. Bacterial, viral, fungal, or parasitic infections
A The pattern of decreased C3, C4, and CH50 indicates classic pathway activation. This
results in consumption of complement and is associated with SLE, serum sickness,
subacute bacterial endocarditis, and other immune complex diseases. The
inflammatory response seen in malignancy and acute infections gives rise to an
increase in complement components. Immunodeficiency caused by an inherited
deficiency in complement constitutes only about 1% of immunodeficiency diseases.
Such disorders reduce the CH50 but involve a deficient serum level of only one
complement factor.
What is measured in the CH50 assay?
A. RBC quantity needed to agglutinate 50% of antibody
B. Complement needed to lyse 50% of RBCs
C. Complement needed to lyse 50% of antibody-sensitized RBCs
D. Antibody and complement needed to sensitize 50% of RBCs
C CH50 is the amount of complement needed to lyse 50% of standardized antibody-
sensitized sheep RBCs. It is expressed as the reciprocal of the serum dilution resulting
in 50% hemolysis. Low levels are associated with deficiency of some complement
components and active systemic autoimmune diseases in which complement is being
consumed.
All of the following tests measure phagocyte function except:
A. Leukocyte adhesion molecule analysis
B. DHR assay
C. Nitro blue tetrazolium (NBT) test
D. IL-2 assay
D The DHR assay and the older NBT tests are used to diagnose chronic granulomatous
disease, an inherited disorder in which phagocytic cells fail to kill microorganisms
because of a defect in peroxide production (respiratory burst). Leukocyte adhesion
deficiency is associated with a defect in the production of integrin molecules on the
surface of WBCs and their granules. IL-2 is a cytokine produced by activated T-helper
and B cells. It causes B-cell proliferation and increased production of antibody,
interferon, and other cytokines. IL-2 can be measured by EIA and is used to detect
transplant rejection, which is associated with an increase in the serum and urine levels.
A patient had surgery for colorectal cancer, after which he received chemotherapy for 6
months. The test for carcinoembryonic antigen (CEA) was normal at this time. One year
later, the bimonthly level of CEA was elevated (above 10 ng/mL). An examination and
biopsy revealed recurrence of a small tumor. What was the value of the results provided
by the CEA test in this clinical situation?
A. Diagnostic information
B. Information for further treatment
C. Information on the immunologic response of the patient
D. No useful clinical information in this case
B CEA is a glycoprotein that is elevated in about 60% of patients with colorectal cancer
and one third or more patients with pulmonary, gastric, and pancreatic cancers. CEA
may be positive in smokers and in patients with cirrhosis, Crohn disease, and other
nonmalignant conditions. Because sensitivity for malignant disease is low, CEA is not
recommended for use as a diagnostic test. However, an elevated CEA level after
treatment is evidence of tumor recurrence and the need for second-look surgery.
A carbohydrate antigen 125 assay (CA-125) was performed on a woman with ovarian
cancer. After treatment, the levels fell significantly. An examination performed later
revealed recurrence of the tumor, but the CA-125 levels remained low. How can this
finding be explained?
A. Test error
B. CA-125 was the wrong laboratory test; α-fetoprotein (AFP) is a better test to monitor
ovarian cancer
C. CA-125 may not be sensitive enough when used alone to monitor tumor development
D. CA-125 is not specific enough to detect only one type of tumor
C CA-125 is a tumor associated carbohydrate antigen that is elevated in 70% to 80% of
patients with ovarian cancer and about 20% of patients with pancreatic cancer.
Although an increase in CA-125 may indicate recurrent or progressive disease, low
levels do not necessarily indicate the absence of tumor growth.
How is HLA typing used in the investigation of genetic diseases?
A. For prediction of the severity of the disease
B. For genetic linkage studies
C. For direct diagnosis of disease
D. Is not useful in this situation
B HLA typing is useful in predicting some genetic diseases and for genetic counseling
because certain HLA types show strong linkage to some diseases. HLA typing is not
specifically used to diagnose a disease or assess its severity. In linkage studies, a
disease gene can be predicted because it is located next to the locus of a normal gene
with which it segregates. For example, the relative risk of developing ankylosing
spondylitis is 87% in persons who are positive for HLA-B27. Analysis of family
pedigrees for the linkage marker and disease can be used to determine the probability
that a family member will inherit the disease gene.
Which of the following substances, sometimes used as a tumor marker, is increased two-
or threefold in a normal pregnancy?
A. Alkaline phosphatase (ALP)
B. Calcitonin
C. Adrenocortocotropic hormone (ACTH)
D. Neuron-specific enolase
A Isoenzymes of ALP are sometimes used as tumor markers, but they have a low
specificity because they are also increased in nonmalignant diseases. These include
placental-like (heat-stable) ALP isoenzymes, which are found (infrequently) in some
malignancies, such as cancer of the lung; bone-derived ALP, which is a marker for
metastatic bone cancer; and the fast-migrating liver isoenzyme, which is a marker for
metastatic liver cancer. ACTH is secreted as an ectopic hormone in some patients with
cancer of the lung. Calcitonin is a hormone produced by the medulla of the thyroid and
is increased in the serum of patients with medullary thyroid carcinoma. Neuron-
specific enolase is an enzyme that is used as a tumor marker primarily for
neuroblastoma.
What is the correct procedure upon receipt of a test request for human chorionic
gonadotropin (hCG) on the serum of a 60-year-old man?
A. Return the request; hCG is not performed on men
B. Perform a qualitative hCG test to see if hCG is present
C. Perform the test; hCG may be increased in testicular tumors
D. Perform the test but use different standards and controls
C hCG is normally tested for in pregnancy; it is increased in approximately 60% of
patients with testicular tumors and a lower percentage of those with ovarian,
gastrointestinal, breast, and pulmonary tumors. Malignant cells secreting hCG may
produce only the β-subunit; therefore, qualitative and quantitative tests that detect both
intact hCG and free β-subunits provide better sensitivity than either test by itself
Would an hCG test using a monoclonal antibody against the β-subunit of hCG likely be
affected by an increased level of follicle-stimulating hormone (FSH)?
A. Yes, the β-subunit of FSH is identical to that of hCG
B. No, the test would be specific for the β-subunit of hCG
C. Yes, a cross reaction would occur because of structural similarities
D. No, the structure of FSH and hCG are not at all similar
B Luteinizing hormone, FSH, and hCG share a common α-subunit but have different β-
subunits. A test for hCG using a monoclonal antibody would be specific for hCG,
provided that the antibody was directed against an antigenic determinant on the
carboxy terminal end of the β-subunit.
What is an advantage of performing a prostate-specific antigen (PSA) test for prostate
cancer?
A. PSA is stable in serum and not affected by digital rectal examination
B. PSA is increased only in prostatic malignancy
C. A normal serum level rules out malignant prostatic disease
D. The percentage of free PSA is elevated in persons with malignant disease
A PSA is a glycoprotein with protease activity that is specific for the prostate gland. High
levels may be caused by prostate malignancy, benign prostatic hypertrophy, or
prostatitis, but PSA is not increased by physical examination of the prostate. PSA has a
sensitivity of 80% and a specificity of about 75% for prostate cancer. The sensitivity is
sufficiently high to warrant its use in screening tests, but its sensitivity for stage A
cancer is below 60%. Most of the serum PSA is bound to protease inhibitors, such as
α1-antitrypsin and α1-antichymotrypsin. Patients with borderline PSA levels (4–10
ng/mL) and a low percentage of free PSA are more likely to have cancer of the prostate
compared with patients with a normal percentage of free PSA.
Which method is the most sensitive for quantitation of AFP?
A. Double immunodiffusion
B. Electrophoresis
C. Enzyme immunoassay
D. Particle agglutination
C AFP is a glycoprotein that is produced in about 80% to 90% of patients with hepatoma
and in a lower percentage of patients with other tumors, including retinoblastoma,
breast cancer, uterine cancer, and pancreatic cancer. The upper reference limit for
serum is only 10 ng/mL, which requires a sensitive method of assay, such as EIA. The
high analytical sensitivity of chemiluminescent immunoassays permits detection of
reduced AFP levels in maternal serum associated with Down syndrome, as well as
elevated levels associated with spina bifida
Select the best donor for a man, blood type AB, in need of kidney transplantation.
A. His brother, type AB, HLA matched for class II antigens
B. His mother, type B, HLA matched for class I antigens
C. His cousin, type O, HLA matched for major class II antigens
D. Deceased donor, type O, HLA matched for some class I and II antigens
A A twin or sibling donor of the same blood type and HLA matched for class II antigens
is the best donor in this situation. Class II antigens (HLA-D, -DR, -DQ, and -DP)
determine the ability of the transplant recipient to recognize the graft. The HLA genes
are located close together on chromosome 6, and crossover between HLA genes is
rare. Siblings with closely matched class II antigens most likely inherited the same
class I genes. The probability of siblings inheriting the same HLA haplotypes from
both parents is 1:4.
Interpret the following microcytotoxicity target cell results:
A9 and B12 cells = damaged; A1 and Aw19 cells = intact.
A. Positive for A1 and Aw19; negative for A9 and B12
B. Negative for A1 and Aw19; positive for A9 and B12
C. Error in test system; retest
D. Impossible to determine
B The microcytotoxicity test is based on the reaction of specific antisera and HLA
antigens on test cells. Cells damaged by the binding of antibody and complement are
detected with a supravital dye, such as eosin.
Which method, classically used for HLA-D typing, is often used to determine the
compatibility between a living organ donor and a recipient?
A. Flow cytometry
B. Mixed lymphocyte culture (MLC)
C. Primed lymphocyte test
D. Restriction fragment length polymorphism (RFLP)
B Flow cytometry can be used in transplantation to type serologically defined HLA
antigens. The one-way mixed lymphocyte reaction is used to identify HLA-D antigens
on the donor’s lymphocytes and is used for crossmatching living donors with
transplant recipients. The assay is time consuming and would not be used as part of a
workup for a transplant from a deceased donor. HLA-D incompatibility is associated
with the recognition phase of allograft rejection. The primed lymphocyte test is used to
identify HLA-DP antigens.
SITUATION: Cells type negative for all HLA antigens in a complement-dependent
cytotoxicity assay. What is the most likely cause?
A. Too much supravital dye was added
B. Rabbit complement is inactivated
C. All leukocytes are dead
D. Antisera is too concentrated
B Inactive rabbit complement may not become fixed to antibodies that have bound test
leukocytes; therefore, no lysis of cells will occur. When the supravital dye is added, all
cells will appear negative (exclude the dye) for all HLAs
What method may be used for tissue typing instead of serological HLA typing?
A. PCR
B. Southern blotting
C. RFLP
D. All of these options
D PCR, Southern blotting, and testing for RFLPs may all be used to identify HLA
genes. Many laboratories use PCR technology for the routine determination of HLA
type
Which of the following serial dilutions contains an incorrect factor?
A. 1:4, 1:8, 1:16
B. 1:1, 1:2, 1:4
C. 1:5, 1:15, 1:45
D. 1:2, 1:6, 1:12
D All the dilutions are multiplied by the same factor in a progression except the last one:
1:2 to 1:6 is × 3, whereas 1:6 to 1:12 is × 2. Threefold dilutions of a 1:2 dilution would
result in a 1:6 followed by a 1:18
A patient was tested for syphilis by the RPR method and was reactive. A TP-PA test was
performed and the result was negative. Subsequent testing showed the patient to have a
high titer of ACAs by the ELISA method. Which routine laboratory test is most likely to
be abnormal for this patient?
A. Activated partial thromboplastin time (APTT)
B. Anti–smooth muscle antibodies
C. Aspartate aminotransferase (AST)
D. C3 assay by immunonephelometry
A Approximately 50% to 70% of patients with ACAs also have the lupus anticoagulant
(LAC) in their serum. LAC is an Ig that interferes with in vitro coagulation tests:
prothrombin time (PT), APTT, and dilute Russell viper venom (DRVV) time. These
tests require phospholipid for the activation of factor X. About 30% of patients with
antibodies to cardiolipin or phospholipids have a biological false-positive RPR result.
Anti–smooth muscle is most commonly associated with chronic active hepatitis, and
increased AST with necrotic liver diseases. Although ACA and LAC may be
associated with SLE, the majority of patients with these antibodies do not have SLE
and would have a normal C3 level
Inflammation involves a variety of biochemical and cellular mediators. Which of the
following may be increased within 72 hours after an initial infection?
A. Neutrophils, macrophages, antibody, complement, α1-antitrypsin
B. Macrophages, T cells, antibody, haptoglobin, fibrinogen
C. Neutrophils, macrophages, complement, fibrinogen, C-reactive protein
D. Macrophages, T cells, B cells, ceruloplasmin, complement
C The correct list, in which all mediators are involved in an inflammatory response
within 72 hours after initial infection, is neutrophils, macrophages, complement,
fibrinogen, and C-reactive protein. Phagocytic cells, acute phase reactants, and
fibrinolytic factors enter the site of inflammation. Antibody and lymphocytes do not
enter until later.
An 18-month-old boy has recurrent sinopulmonary infections and septicemia. Bruton
thymidine kinase deficiency is suspected. Which test result would be markedly
decreased?
A. Serum IgG, IgA, and IgM
B. Total T-cell count
C. Both B- and T-cell counts
D. Lymphocyte proliferation with phytohemagglutinin stimulation
A The patient with Bruton thymidine kinase deficiency presents with clinical symptoms
related to recurrent infections, demonstrated in the laboratory by decreased or absent
Igs. Peripheral blood B cells are absent or markedly reduced, but T cells are normal in
number and function. Because phytohemagglutinin is a T-cell mitogen, the lymphocyte
proliferation test using PHA would be normal for this patient.
A patient received 5 units of FFP and developed a severe anaphylactic reaction. He has a
history of respiratory and gastrointestinal infections. Post-transfusion studies showed all
5 units to be ABO compatible. What immunologic test would help to determine the
cause of this transfusion reaction?
A. Complement levels, particularly C3 and C4
B. Flow cytometry for T-cell counts
C. Measurement of Igs
D. NBT test for phagocytic function
C The patient had an anaphylactic reaction to a plasma product. This, combined with the
history of respiratory and gastrointestinal infections, suggests a selective IgA
deficiency. Measurement of Igs would be helpful in this case. A low serum IgA and
normal IgG substantiate the diagnosis of selective IgA deficiency. Such patients
frequently produce anti-IgA, which is often responsible for a severe transfusion
reaction when ABO-compatible plasma is administered.
IFE revealed excessive amounts of polyclonal IgM and low concentrations of IgG and
IgA. What is the most likely explanation of these findings and the best course of action?
A. Proper amounts of antisera were not added; repeat both tests
B. Test specimen was not added properly; repeat both procedures
C. Patient has common variable immunodeficiency; perform B-cell count
D. Patient has immunodeficiency with hyper-M; perform CD40 ligand (CD154) analysis
D Low plasma concentrations of IgG and IgA and an abundance of IgM is consistent
with the CD40 ligand deficiency. Most cases are X-linked and result from a mutation
of the gene TNFSF5, which encodes a receptor needed for switching Ig production.
Patients with common variable immunodeficiency have low serum IgG, IgA, and IgM.
SITUATION: A 54-year-old man was admitted to the hospital after having a seizure.
Many laboratory tests were performed, including an RPR, but none of the results was
positive. The physician suspects a case of late (tertiary) syphilis. Which test should be
performed next?
A. Repeat RPR, followed by VDRL
B. Treponemal test, such as TP-PA on serum
C. VDRL on CSF
D. No laboratory test is positive for late (tertiary) syphilis
B Serum antibody tests, such as RPR and VDRL, are often negative in cases of late
syphilis. However, treponemal tests remain positive in greater than 95% of cases. The
VDRL test on CSF is the most specific test for diagnosis of neurosyphilis because
treponemal test results remain positive after treatment. It should be used as the
confirmatory test when the serum treponemal test result is positive. However, the CSF
VDRL is limited in sensitivity and would not be positive if the serum treponemal-
specific antibody test was negative.
A 19-year-old girl came to her physician complaining of a sore throat and fatigue. Upon
physical examination, lymphadenopathy was noted. Reactive lymphocytes were noted
on the differential, but a rapid test for antibodies to IM was negative. Liver enzymes
were only slightly elevated. What test(s) should be ordered next?
A. Hepatitis testing
B. EBV serological panel
C. HIV confirmatory testing
D. Bone marrow biopsy
B An EBV serological panel would give a more accurate assessment than a rapid slide
IM test. The time of appearance of the various antibodies to the viral antigens differs
according to the clinical course of the infection.
A patient came to his physician complaining of a rash, severe headaches, stiff neck, and
sleep problems. Laboratory tests of significance were an elevated sedimentation rate
(ESR) and slightly increased liver enzymes. Further questioning of the patient revealed
that he had returned from a hunting trip in upstate New York 4 weeks ago. His
physician ordered a serological test for Lyme disease, and the assay was negative. What
is the most likely explanation of these results?
A. The antibody response is not sufficient to be detected at this stage
B. The clinical symptoms and laboratory results are not characteristic of Lyme disease
C. The patient likely has early-stage HBV infection
D. Laboratory error has caused a false-negative result
A The antibody response to B. burgdorferi may not develop until several weeks after
initial infection. The antibody test should be repeated 2 to 4 weeks later. To confirm a
positive EIA, samples that initially test positive or equivocal (indeterminate) should be
retested using a second enzyme immunoassay or immunoblot method. Regardless of
the laboratory results, if the physician suspects Lyme disease, treatment should begin
immediately.
A patient received 2 units of RBCs following surgery. Two weeks after the surgery, the
patient was seen by his physician and was found to have mild jaundice and slightly
elevated liver enzymes. Hepatitis testing, however, was negative. What should be done
next?
A. Nothing until more severe or definitive clinical signs develop
B. Repeat hepatitis testing immediately
C. Repeat hepatitis testing in a few weeks
D. Check blood bank donor records and contact donor(s) of transfused units
C The level of HBsAg may not have reached detectable levels, and antibodies to HBc
and HCV would not have yet developed. Waiting 1 or 2 weeks and repeating the tests
may reveal evidence of hepatitis virus infection
A hospital employee received the final dose of the hepatitis B vaccine 3 weeks ago. She
wants to donate blood. Which of the following results are expected from the hepatitis
screen, and will she be allowed to donate blood?
A. HBsAg, positive; anti-HBc, negative—she may donate
B. HBsAg, negative; anti-HBc, positive—she may not donate
C. HBsAg, positive; anti-HBc, positive—she may not donate
D. HBsAg, negative; anti-HBc, negative—she may donate
D She may donate if she is symptom free. The response to hepatitis B vaccine would
include a positive result for anti-HBs, a test not normally a part of routine donor
testing. She will be negative for HBsAg and anti-HBc; however, transient antigen
positivity (less than 2 weeks) may be seen following vaccination
A pregnant woman came to her physician with a maculopapular rash on her face and
neck. Her temperature was 37.7°C. Rubella tests for both IgG and IgM antibody were
positive. What positive test(s) would reveal a diagnosis of congenital rubella syndrome
in her baby after birth?
A. Positive rubella tests for both IgG and IgM antibody
B. Positive rubella test for IgM
C. Positive rubella test for IgG
D. No positive test is revealed in congenital rubella syndrome
B A finding of IgG is not definitive for congenital rubella syndrome because IgG
crosses the placenta from the mother; however, demonstration of IgM, even in a single
neonatal sample, is diagnostic.
SITUATION: A patient with RA has acute pneumonia but a negative result on throat
culture. The physician suspects an infection with Mycoplasma pneumoniae and requests
an IgM-specific antibody test. The test is performed directly on serial dilutions of serum
less than 4 hours old. The result is positive, giving a titer of 1:32. However, the test is
repeated 3 weeks later, and the titer remains at 1:32. What test should be performed to
determine if the patient is truly infected with M. pneumoniae?
A. IgG anti-M. pneumoniae
B. Cold agglutinins
C. M. pneumoniae PCR or other molecular assay
D. Respiratory culture
C The IgM-specific antibody test for M. pneumoniae detects antibodies to mycoplasmal
membrane antigens and, unlike cold agglutinins, is specific for M. pneumoniae. A
positive result (titer of 1:32 or higher) occurs during the acute phase in about 87% of
M. pneumoniae infections and does not need to be confirmed by assay of convalescent
serum. However, Mycoplasma IgM may last a year or more, thus its presence does not
always indicate a current infection. PCR performed on a respiratory specimen is the
definitive test and should be performed if there is a question about an IgM result.
A patient with ovarian cancer who has been treated with chemotherapy is being
monitored for recurrence by using serum CA-125, CA-50, and CA 15–3. Six months
after treatment the CA 15–3 is elevated, but the CA-125 and CA-50 remain low. What
is the most likely explanation of these findings?
A. Ovarian malignancy has recurred
B. CA 15–3 is specific for breast cancer and indicates metastatic breast cancer
C. Testing error occurred in the measurement of CA 15–3 caused by poor analytical
specificity
D. The CA 15–3 elevation is spurious and probably benign
A Although CA-125 is the most commonly used tumor marker for ovarian cancer, not
all ovarian tumors produce CA-125. Greatest sensitivity in monitoring for recurrence is
achieved when several markers known to be increased in the malignant tissue type are
measured simultaneously and when the markers are elevated (by malignancy) prior to
treatment. In addition to limited sensitivity, no single tumor marker is entirely specific.
Carbohydrate and other oncofetal antigens are produced by several malignant and
benign conditions. Although testing errors may occur in any situation, measurements
of carbohydrate antigens use purified monoclonal antibodies with very low cross
reactivities.
A patient had a PSA level of 60 ng/mL the day before surgery to remove a localized
prostate tumor. One week after surgery, serum PSA was determined to be 8 ng/mL by
the same method. What is the most likely cause of these results?
A. Incomplete removal of the malignancy
B. Cross reactivity of the antibody with another tumor antigen
C. Testing too soon after surgery
D. Hook effect with the PSA assay
C When monitoring the level of a tumor marker for treatment efficacy or recurrence, the
half-life of the protein must be considered when determining the testing interval. PSA
has a half-life of almost 4 days and would not reach normal levels after surgery for
approximately 3 to 4 weeks. The hook effect is the result of very high antigen levels
giving a lower than expected result in a double antibody sandwich assay when both
antibodies and sample are added at the same time
What is the main advantage of the recovery and reinfusion of autologous stem cells?
A. It slows the rate of rejection of transplanted cells
B. It prevents graft-versus-host disease
C. No HLA testing is required
D. Engraftment occurs in a more efficient sequence
B The main advantage to the patient from the reinfusion of autologous stem cells is that
the procedure prevents graft-versus-host disease, especially in the
immunocompromised patient. Although HLA testing is not required, this is not the
primary advantage for patient care.
A patient with symptoms associated with SLE and scleroderma was evaluated by
immunofluorescence microscopy for ANAs by using the HEp-2 cell line as substrate.
The cell line displayed a mixed pattern of fluorescence that could not be separated by
serial dilutions of the serum. Which procedure would be most helpful in determining
the antibody profile of this patient?
A. Use of a different tissue substrate
B. Absorption of the serum using the appropriate tissue extract
C. Requesting a new specimen
D. ELISA tests for specific antibodies
D Many patients with multiorgan autoimmune disease display symptoms that overlap
two or more diseases and have complex mixtures of serum autoantibodies. The HEp-2
substrate is the most sensitive cell line for immunofluorescent microscopy because it
contains cells in various mitotic stages, which exposes the serum to more antigens. Use
of a nonhuman substrate, such as Crithidia, may help identify dsDNA antibodies but
would not aid in differentiating all of the antibodies in a complex mixture. The best
methods are ELISA, line blots, and multiplex tests because they are more specific than
immunofluorescence microscopy for identifying antibodies to specific antigens. These
assays are often used to measure antibodies to ENAs, which may be partially or
completely lost during fixation of cells used for immunofluorescent microscopy. These
antibodies cause a speckled pattern and are seen in a wide range of autoimmune
diseases. Identification of the anti-ENA specificities is helpful in differentiating these
diseases.
A patient with joint swelling and pain tested negative for serum RF by both latex
agglutination and ELISA methods. What other test would help establish a diagnosis of
RA in this patient?
A. Anti-CCP
B. ANA testing
C. Flow cytometry
D. Complement levels
A Antibodies to CCP are often found in RF-negative patients with RA. The absence of
RFs from serum does not rule out a diagnosis of RA, and more than half the patients
who are diagnosed with RA present initially with a negative serum RF result. Both RF
and anti-CCP are criteria assays for diagnosing RA, and at least one of them must be
positive for a confirmed diagnosis.
A transplant recipient began to show signs of rejection 8 days after transplantation, and
the organ was removed. What immune elements might be found in the rejected organ?
A. Antibody and complement
B. Primarily antibody
C. Macrophages
D. T cells
D Acute rejection occurs within 3 weeks of transplantation. The immune element most
likely to be involved in an acute rejection is the T cell in a type IV, delayed
hypersensitivity (cell-mediated) reaction. Preformed antibody, and possibly
complement, is usually involved in hyperacute (immediate) rejection and chronic
rejection.
An initial and repeat fourth-generation HIV screening test are both positive. The
antibody differentiation assay is negative, as is the qualitative RNA PCR test. The
patient shows no clinical signs of HIV infection, and the patient’s CD4 T-cell count is
normal. Based on these results, which conclusion is correct?
A. Patient is diagnosed as HIV-1 positive
B. Patient is diagnosed as HIV-2 positive
C. Results are inconclusive
D. Patient is diagnosed as HIV-1 negative
D The fourth-generation algorithm detects infection approximately 2 weeks after
exposure. This patient has negative follow-up test results for both antibody and nucleic
acid, so she would be considered HIV negative. The lack of symptoms would not be
consistent with a very recent infection; the patient would likely have an acute retroviral
syndrome. However, a repeat test may be performed if clinical suspicion remains high.
A woman who has been pregnant for 12 weeks is tested for toxoplasmosis. Her IgM
ELISA titer is 2.6 (reference range less than 1.6), and her IgG ELISA value is 66
(reference range less than 8). The physician asks you if these results indicated an
infection during the past 12 weeks. Which of the following tests would you recommend
to determine if the woman was infected during her pregnancy?
A. Toxoplasmosis PCR on amniotic fluid
B. Toxoplasmosis IgM on amniotic fluid
C. Toxoplasmosis IgG avidity
D. Amniotic fluid culture
C Although IgM is positive, in toxoplasmosis, specific IgM may remain detectable for a
year or more following infection. IgG avidity, or the strength of binding of a serum to
the antigen of interest, is a useful method to determine if an infection is recent or in the
distant past. IgG avidity will increase with time following an infection. Amniotic fluid
testing is not useful for determining when the mother might have been infected.
On January 4, an SPE on a specimen obtained at your hospital in North Dakota from a
58-year-old patient shows a band at the β—γ junction. The specimen was also positive
for RF. You recommend that an immunofixation test be performed to determine if the
band represents a monoclonal Ig. Another specimen is obtained 2 weeks later by the
physician in his office 30 miles away, and whole blood is submitted to you for IFE. The
courier placed the whole blood specimen in an ice chest for transport. In this specimen,
no β-γ band is seen in the serum protein lane, and the IgM lane is very faint. The RF on
this specimen was negative. The physician wants to know what went wrong in your
laboratory. Your response is:
A. Nothing went wrong in our laboratory; the patient had an infection 2 weeks ago, and it
had cleared up
B. Something went wrong in our laboratory—we likely mislabeled one of the specimens;
please resubmit a new specimen, and we will test it at no charge
C. We will run a second specimen after 2-mercaptoethanol treatment, which will eliminate
IgM aggregates and allow for more sensitive monoclonal IgM detection
D. Please redraw another specimen from the patient, and this time, separate the serum from
the clot in your office before placing the specimen on ice and sending it to us by courier
D The most likely cause of the discrepant results is the presence of a type II
cryoglobulin. This is a monoclonal RF. The protein likely precipitated during the
courier ride and was, thus, in the clot when the laboratory separated the serum.
A patient undergoing dialysis is positive for both HBsAg and anti-HBs. The physician
suspects a laboratory error. Do you agree?
A. Yes; the patient should not test positive for both HBsAg and anti-HBs
B. No; incomplete dialysis of a patient in the core window phase of HBV infection will yield
this result
C. No; it is likely the patient has recently received a hepatitis B booster vaccination within
the past week, and this could have caused these results
D. Perhaps; a new specimen should be submitted to clear up the confusion
C HBsAg will remain detectable at low levels following a vaccination for up to 1 to 2
weeks. Thus, patients who have received a second injection of hepatitis B vaccine may
have anti-HBs and detectable antigen for a brief period. This has been reported more
frequently in patients undergoing dialysis and in pediatric populations.
You are evaluating an ELISA assay as a replacement for your IFA ANA test. You test
50 specimens in duplicate on each assay. The ELISA assay uses a HEp-2 extract as its
antigen source. The correlation between the ELISA and IFA tests is only 60% (30 of 50
specimens agree). Which of the following is the next best course of action?
A. Test another 50 specimens
B. Perform a competency check on the medical laboratory scientists who performed the tests
C. Order a new lot of both kits and then retest on the new lots
D. Refer the discrepant specimens for testing by another method
D In this situation, you have already tested the specimens in duplicate. Testing an
additional 50 specimens will not change the fact that you have 20 discrepant
specimens. The best course of action is to determine what antibodies are actually
present in these specimens. Then, you can determine whether ELISA or IFA is a better
procedure for detecting the most clinically relevant antibodies. You could perform
clinical chart reviews as an alternative, but obtaining that data would be difficult and
much of it may be subjective
A solution that has a transmittance of 1.0%T would have an absorbance of:
A. 1.0
B. 2.0
C. 1.0%
D. 99.0%
B
A = 2.0 – log %T
A = 2.0 – log 1.0
The log of 1.0 = 0
A = 2.0
Which formula correctly describes the relationship between absorbance and percent
transmittance (%T)?
A. A = 2 – log %T
B. A = log T
C. A = –log %T
D. A = 2 – %T
A Absorbance is proportional to the inverse log of transmittance:
A = –log T = log 1/T
Multiplying the numerator and denominator by 100 gives:
A = log (100/100 × T)
100 × T = %T, substituting %T
for 100 × T gives:
A = log 100/%T
A = log 100 – log %T
A = 2.0 – log %T
For example, if %T = 10.0, then:
A = 2.0 – log 10.0
log 10.0 = 1.0
A = 2.0 – 1.0 = 1.0
A green-colored solution would show highest transmittance at:
A. 475 nm
B. 525 nm
C. 585 nm
D. 620 nm
B Green light consists of wavelengths from 500 to 550 nm. A green-colored solution with
a transmittance maximum of 525 nm and a 50-nm bandpass transmits light of 525 nm
and absorbs light below 475 nm and above 575 nm. A solution that is green would be
quantitated using a wavelength that it absorbs strongly, such as 450 nm.
In absorption spectrophotometry:
A. Absorbance is directly proportional to transmittance
B. Percent transmittance is directly proportional to concentration
C. Percent transmittance is directly proportional to the light path length
D. Absorbance is directly proportional to concentration
D Beer’s law states that A = a × b × c, where a is the absorptivity coefficient (a constant),
b is the path length, and c is concentration. Absorbance is directly proportional to both
b and c. Doubling the path length results in incident light contacting twice the number
of molecules in solution. This causes absorbance to double, the same effect as doubling
the concentration of molecules.
Which wavelength would be absorbed strongly by a red-colored solution?
A. 450 nm
B. 585 nm
C. 600 nm
D. 650 nm
A A solution transmits light corresponding in wavelength to its color, and usually absorbs
light of wavelengths complementary to its color. A red solution transmits light of 600
to 650 nm and strongly absorbs 400 to 500 nm light.
SITUATION: A technologist is performing an enzyme assay at 340 nm using a visible-
range spectrophotometer with a tungsten light source. After setting the wavelength and
adjusting the readout to zero %T with the light path blocked, a cuvette with deionized
water is inserted. With the light path fully open and the 100%T control at maximum,
the instrument readout will not rise above 90%T. What is the most appropriate first
course of action?
A. Replace the source lamp
B. Insert a wider cuvette into the light path
C. Measure the voltage across the lamp terminals
D. Change the wavelength to 335 nm
A Visible spectrophotometers are usually supplied with a tungsten or quartz halogen
source lamp. Tungsten lamps produce a continuous range of wavelengths from greater
than 320 to 2,000 nm. Output increases as wavelength becomes longer peaking at
around 1,000 nm, and is poor below 400 nm. As the lamp envelope darkens with age,
the amount of light reaching the photodetector at 340 nm becomes insufficient to set
the blank reading to 100%T. Changing the wavelength to 335 nm will decrease the
assay sensitivity as well as result in less light reaching the photodetector
Which type of monochromator produces the purest monochromatic light in the
ultraviolet (UV) range?
A. A diffraction grating and a fixed exit slit
B. A sharp cutoff filter and a variable exit slit
C. Interference filters and a variable exit slit
D. A prism and a variable exit slit
D Diffraction gratings and prisms both produce a continuous range of wavelengths. A
diffraction grating produces a uniform separation of wavelengths. A prism produces
much better separation of high-frequency light because refraction is greater for higher-
energy wavelengths. Instruments using a prism and a variable exit slit can produce UV
light of high purity (very narrow bandpass). The adjustable slit is required to allow
sufficient light to reach the detector to set 100%T.
Which monochromator specification is required to measure the true absorbance of a
compound having a natural absorption bandwidth of 30 nm?
A. 50-nm bandpass
B. 25-nm bandpass
C. 15-nm bandpass
D. 5-nm bandpass
D Bandpass refers to the range of wavelengths passing through the sample. The narrower
the bandpass, the greater is the photometric resolution. Bandpass can be made smaller
by reducing the width of the exit slit. Accurate absorbance measurements require a
bandpass less than one-fifth the natural bandpass of the chromophore
Which photodetector is most sensitive to low levels of light?
A. Barrier layer cell
B. Photodiode
C. Diode array
D. Photomultiplier tube
D The photomultiplier tube uses dynodes of increasing voltage to amplify the current
produced by the photosensitive cathode. It is 10,000 times as sensitive as a barrier
layer cell, which has no amplification. A photomultiplier tube requires a DC-regulated
lamp because it responds to light fluctuations caused by the alternating current (AC)
cycle.
Which condition is a common cause of stray light?
A. Unstable source lamp voltage
B. Improper wavelength calibration
C. Dispersion from second-order spectra
D. Misaligned source lamp
C Stray light is caused by the presence of any light other than the wavelength of
measurement reaching the detector. It is most often caused by second-order spectra,
deteriorated optics, light dispersed by a darkened lamp envelope, and extraneous room
light.
A linearity study is performed on a visible spectrophotometer at 650 nm and the
following absorbance readings are obtained:
Concentration of Standard Absorbance
10.0 mg/dL 0.20
20.0 mg/dL 0.41
30.0 mg/dL 0.62
40.0 mg/dL 0.79
50.0 mg/dL 0.92
The study was repeated using freshly prepared standards and reagents, but results
were identical to those shown. What is the most likely cause of these results?
A. Wrong wavelength used
B. Insufficient chromophore concentration
C. Matrix interference
D. Stray light
D Stray light is the most common cause of loss of linearity at high-analyte
concentrations. Light transmitted through the cuvette is lowest when absorption is
highest. Therefore, stray light is a greater percentage of the detector response when
sample concentration is high. Stray light is usually most significant when
measurements are made at the extremes of the visible spectrum because lamp output
and/or detector response are low.
Which type of filter is best for measuring stray light?
A. Wratten
B. Didymium
C. Sharp cutoff
D. Neutral density
C Sharp cutoff filters transmit almost all incident light until the cutoff wavelength is
reached. At that point, they cease to transmit light. Because they give an “all or none
effect,” only stray light reaches the detector when the selected wavelength is beyond
the cutoff.
Which of the following materials is best suited for verifying the wavelength calibration
of a spectrophotometer?
A. Neutral density filters
B. Potassium dichromate solutions traceable to the National Bureau of Standards reference
C. Wratten filters
D. Holmium oxide glass
D Wavelength accuracy is verified by determining the wavelength reading that gives the
highest absorbance (or transmittance) when a substance with a narrow natural
bandwidth (sharp absorbance or transmittance peak) is scanned. For example,
didymium glass has a sharp absorbance peak at 585 nm. Therefore, an instrument
should give its highest absorbance reading when the wavelength dial is set at 585 nm.
Holmium oxide produces a very narrow absorbance peak at 361 nm; likewise, the
hydrogen lamp of a UV spectrophotometer produces a 656-nm emission line that can
be used to verify wavelength. Neutral density filters and dichromate solutions are used
to verify absorbance accuracy or linearity. A Wratten filter is a wide-bandpass filter
made by placing a thin layer of colored gelatin between two glass plates and is
unsuitable for spectrophotometric calibration.
Why do many optical systems in chemistry analyzers utilize a reference light path?
A. To increase the sensitivity of the measurement
B. To minimize error caused by source lamp fluctuation
C. To obviate the need for wavelength adjustment
D. To reduce stray light effects
B A reference beam is used to produce an electrical signal at the detector with which the
measurement of light absorption by the sample is compared. This safeguards against
measurement errors caused by power fluctuations that change the source lamp
intensity. Although reference beams increase the accuracy of measurements, they do so
at the expense of optical sensitivity if some of the incident light is used to produce the
reference beam.
Which component is required in a spectrophotometer to produce a spectral absorbance
curve?
A. Multiple monochromators
B. A reference optical beam
C. Photodiode array
D. Laser light source
C There are two ways to perform spectral scanning for compound identification. One is
to use a stepping motor that continuously turns the monochromator so that the
wavelength aligned with the exit slit changes at a constant rate. A more practical
method is to use a diode array detector. This consists of a chip embedded with as many
as several hundred photodiodes. Each photodiode is aligned with a narrow part of the
spectrum produced by a diffraction grating and produces current proportional to the
intensity of the band of light striking it (usually 1–2 nm in range). The diode signals
are processed by a computer to create a spectral absorbance or transmittance curve.
The half bandwidth of a monochromator is defined by:
A. The range of wavelengths passed at 50% maximum transmittance
B. One half the lowest wavelength of optical purity
C. The wavelength of peak transmittance
D. One half the wavelength of peak absorbance
A Half bandwidth is a measure of bandpass made using a solution or filter having a
narrow natural bandwidth (transmittance peak). The wavelength giving maximum
transmittance is set to 100%T (or 0 A). Then, the wavelength dial is adjusted
downward, until a readout of 50%T (0.301 A) is obtained. Next, the wavelength is
adjusted upward until 50%T is obtained. The wavelength difference is the half
bandwidth. The narrower the half bandwidth, the better is the photometric resolution of
the instrument
The reagent blank corrects for absorbance caused by:
A. The color of reagents
B. Sample turbidity
C. Bilirubin and hemolysis
D. The intrinsic absorbance of both the reagents and sample matrix
A When a spectrophotometer is set to 100%T with the reagent blank instead of water,
the absorbance of reagents is automatically subtracted from each unknown reading.
The reagent blank does not correct for absorbance caused by interfering molecules in
the sample, such as bilirubin, hemoglobin, or lipids.
A plasma sample is hemolyzed and turbid. What is required to perform a sample blank
to correct the measurement for the intrinsic absorbance of the sample when performing
a spectrophotometric assay?
A. Substitute deionized water for the sample
B. Dilute the sample 1:2 with a standard of known concentration
C. Substitute saline for the reagent
D. Use a larger volume of the sample
C A sample blank is used to subtract the intrinsic absorbance of the sample usually
caused by hemolysis, icterus, turbidity, or drug interference. On automated analyzers,
this is accomplished by measuring the absorbance after the addition of sample and a
first reagent, usually a diluent. For tests using a single reagent, sample blanking can be
done prior to the incubation phase before any color develops. Substituting deionized
water for sample is done to subtract the absorbance of the reagent (reagent blanking).
Diluting the sample with a standard (standard addition) may be done when the
absorbance is below the minimum detection limit for the assay. Using a larger volume
of sample will make the interference worse
Which instrument requires a highly regulated direct current (DC) power supply?
A. A spectrophotometer with a barrier layer cell
B. A colorimeter with multilayer interference filters
C. A spectrophotometer with a photomultiplier tube
D. A densitometer with a photodiode detector
C When AC voltage regulators are used to isolate source lamp power, light output
fluctuates as the voltage changes. Because this occurs at 60 hertz (Hz), it is not
detected by eyesight or slow-responding detectors. Photomultiplier tubes are sensitive
enough to respond to the AC frequency and require a DC-regulated power supply
Which statement regarding reflectometry is true?
A. The relation between reflectance density and concentration is linear
B. Single-point calibration can be used to determine concentration
C. 100% reflectance is set with an opaque film called a white reference
D. The diode array is the photodetector of choice
C Reflectometry does not follow Beer’s law, but the relationship between concentration
and reflectance can be described by a logistic formula or algorithm that can be solved
for concentration. For example, K/S = (1 – R)2/2R, where K = Kubelka–Munk
absorptivity constant, S = scattering coefficient, R = reflectance density. K/S is
proportional to concentration. The white reference is analogous to the 100%T setting
in spectrophotometry and serves as a reference signal. Dr = log R0/R1, where Dr is the
reflectance density, R0 is the white reference signal, and R1 is the photodetector signal
for the test sample
Bichromatic measurement of absorbance can correct for interfering substances if:
A. The contribution of the interferent to absorbance is the same at both wavelengths
B. Both wavelengths pass through the sample simultaneously
C. The side band is a harmonic of the primary wavelength
D. The chromogen has the same absorbance at both wavelengths
A In bichromatic photometry, the absorbance of sample is measured at two different
wavelengths. The primary wavelength is at or near the absorbance maximum. An
interfering substance having the same absorbance at both primary and secondary (side
band) wavelengths does not affect the absorbance difference (Ad)
Which instrument requires a primary and secondary monochromator?
A. Spectrophotometer
B. Atomic absorption spectrophotometer
C. Fluorometer
D. Nephelometer
C A fluorometer uses a primary monochromator to isolate the wavelength for excitation
and a secondary monochromator to isolate the wavelength emitted by the
fluorochrome. Fluorescence is directly proportional to analyte concentration.
Which of the following statements about fluorometry is accurate?
A. Fluorometry is less sensitive than spectrophotometry
B. Fluorometry is less specific than spectrophotometry
C. Unsaturated cyclic molecules are often fluorescent
D. Fluorescence is directly proportional to temperature
C Increasing temperature results in more random collision between molecules by
increasing their motion. This causes energy to be dissipated as heat instead of
fluorescence. Temperature is inversely proportional to fluorescence. Fluorescence is
more sensitive than spectrophotometry because the detector signal can be amplified
when dilute solutions are measured. It is also more specific than spectrophotometry
because both the excitation and emission wavelengths are characteristics of the
compound being measured.
Which of the following components is not needed in a chemiluminescent immunoassay
analyzer?
A. Source lamp
B. Monochromator
C. Photodetector
D. Wash station
A Chemiluminescence is the production of light following a chemical reaction.
Immunoassays based on chemiluminescence generate light when the chemiluminescent
molecule becomes excited; therefore, a light source is not used. In immunoassay
platforms, chemiluminescent molecules, such as acridinium, can be used to label
antigens or antibodies. Alternatively, chemiluminescent substrates, such as luminol or
dioxetane phosphate, may be used. Light will be emitted when the enzyme-labeled
molecule reacts with the substrate. In such assays, free and bound antigen separation is
required and is usually accomplished by using paramagnetic particles bound to either
antibody or reagent antigen.
Which substance is used to generate the light signal in electrochemiluminescence?
A. Acridinium
B. Luminol
C. Dioxetane phosphate
D. Ruthenium
D All of these substances are chemiluminescent. Dioxetane phosphate is excited by
alkaline phosphatase (ALP). Acridinium and luminol are excited by hydrogen peroxide
(H2O2). In electrochemiluminesence, ruthenium is used to label the antibody. Antigen–
antibody complexes containing the ruthenium label are bound to paramagnetic
particles via a streptavidin–biotin reaction. The paramagnetic particles are attracted to
an electrode surface. The flow cell is washed with a solution containing tripropylamine
(TPA) to remove unbound ruthenium label. At the electrode surface, the TPA is
oxidized and the electrons excite the ruthenium, causing production of 620-nm light.
Light scattering when the wavelength is greater than 10 times the particle diameter is
described by:
A. Rayleigh’s law
B. The Beer–Lambert law
C. Mie’s law
D. The Rayleigh–Debye law
A Rayleigh’s law states that when the incident wavelength is much longer than the
particle diameter, there is maximum backscatter and minimum right-angle scatter.
Rayleigh–Debye’s law predicts maximum right-angle scatter when wavelength and
particle diameter approach equality. Mie’s law predicts maximum forward scatter
when particle diameter is much longer than wavelength. In nephelometry, the
relationship between wavelength and diameter determines the angle at which the
detector is located.
Which statement regarding nephelometry is true?
A. Nephelometry is less sensitive than absorption spectrophotometry
B. Nephelometry follows Beer’s law
C. The optical design is identical to a turbidimeter except that a helium–neon (HeNe) laser
light source is used
D. The detector response is directly proportional to concentration
D In nephelometry, the detector output is proportional to concentration (as opposed to
turbidimetry, where the detector is behind the cuvette). The detector(s) is (are) usually
placed at an angle between 25° and 90° to the incident light, depending on the
application. Nephelometers, like fluorometers, are calibrated to zero with the light path
blocked, and sensitivity can be increased up to 1,000 times by amplification of the
detector output, usually done by increasing the photomultiplier tube dynode voltage.
The purpose of the nebulizer in an atomic absorption spectrophotometer that uses a
flame is to:
A. Uniformly distribute the sample in the flame
B. Cause ejection of an outer shell electron
C. Reduce evaporation of the sample
D. Burn off organic impurities
A The atomizer of the atomic absorption spectrophotometer consists of either a
nebulizer and flame or a graphite furnace. The nebulizer disperses the sample into a
fine aerosol, distributing it evenly into the flame. Heat from the flame is used to
evaporate water and break the ionic bonds of salts, forming ground state atoms. The
flame also excites a small percentage of the atoms, which release a characteristic
emission line.
Interference in atomic absorption spectrophotometry caused by differences in viscosity
is called:
A. Absorption interference
B. Matrix effect
C. Ionization interference
D. Quenching
B Significant differences in aspiration and atomization result when the matrix of the
sample and unknowns differ. Differences in viscosity and protein content are major
causes of matrix error. Matrix effects can be reduced by using protein-based calibrators
and diluting both standards and samples prior to assay
A flameless atomic absorption spectrophotometer dehydrates and atomizes a sample
using:
A. A graphite capillary furnace
B. An electron gun
C. A thermoelectric semiconductor
D. A thermospray platform
A Flameless atomic absorption uses a hollow tube of graphite with quartz ends. The
tube is heated in stages by an electric current to successively dry, ash, and atomize the
sample. During the ash and atomization steps, argon is injected into the tube to
distribute the atoms. The furnace is more sensitive than a flame atomizer and more
efficient in atomizing thermostable salts. However, it is prone to greater matrix
interference and is slower than the flame atomizer because it must cool down before
introduction of the next sample.
When measuring lead (Pb) in whole blood using atomic absorption spectrophotometry,
what reagent is required to obtain the needed sensitivity and precision?
A. Lanthanum
B. Lithium
C. Triton X-100
D. Chloride
C A graphite furnace is preferred over a flame for measuring lead because it is
sufficiently sensitive to detect levels below 5 μg/dL, the cutoff needed for lead
screening of children. The matrix modifier consists of Triton X-100, ammonium
phosphate, and nitric acid. This allows for release of Pb from the red blood cells
(RBCs), and solubilization of cell stroma. The matrix modifier also prevents loss of Pb
caused by formation of lead halides and promotes interaction between Pb and the tube
wall, preventing its loss during the ashing cycle.
The ion-selective membrane used to measure potassium is made of:
A. High-borosilicate glass membrane
B. Polyvinyl chloride dioctylphenyl phosphonate ion exchanger
C. Valinomycin gel
D. Calomel
C Valinomycin is an antibiotic with a highly selective reversible-binding affinity for
potassium ions. Sodium electrodes are usually composed of a glass membrane with a
high content of aluminum silicate. Calcium and lithium ISEs are made from organic
liquid ion exchangers called neutral carrier ionophores. Calomel is made of mercury
covered with a paste of mercurous chloride (Hg°/Hg2Cl2) and is used as a reference
electrode for pH.
The reference potential of a silver–silver chloride electrode is determined by the:
A. Concentration of the potassium chloride filling solution
B. Surface area of the electrode
C. Activity of total anion in the paste covering the electrode
D. The concentration of silver in the paste covering the electrode
A The activity of any solid or ion in a saturated solution is unity. For a silver electrode
covered with silver chloride paste, the Nernst equation is E = E° – RT/nF × 2.3 log10
[Ag° × Cl–]/[AgCl]. Because silver and silver chloride have an activity of 1.0, and all
components except chloride are constants, the potential of the reference electrode is
determined by the chloride concentration of the filling solution.
E = Eo – RT/nF × 2.3 log10[Cl–] = E° – 59.2 mV × log[Cl–] (at room temperature)
Select the equation describing the potential that develops at the surface of an ion-
selective electrode (ISE).
A. van Deemter equation
B. van Slyke equation
C. Nernst equation
D. Henderson–Hasselbalch equation
C The van Deemter equation describes the relation between the velocity of mobile
phase to column efficiency in gas chromatography (GC). The Henderson–Hasselbalch
equation is used to determine the pH of a solution containing a weak acid and its salt.
van Slyke developed an apparatus to measure carbon dioxide (CO2) and oxygen (O2)
content by using a manometer.
Which of the following is required when measuring magnesium by atomic absorption
spectrophotometry?
A. A mercury vapor source lamp
B. A chopper to prevent optical interference from magnesium emission
C. A neutral density filter
D. A 285-nm reference beam to correct for background absorption
B Atomic absorption requires a lamp with a cathode made from the metal to be assayed.
The lamp emits the line spectrum of the metal, providing the wavelength that the atoms
can absorb. The chopper pulses the source light, allowing it to be discriminated from
light emitted by excited atoms. A diffraction grating eliminates light emitted by the
ideal gas in the lamp. Deuterium (wide bandpass light) or Zeeman correction (splitting
the incident light into side bands by a magnetic field) may be used to correct for
background absorption
Ion selective analyzers using undiluted samples have what advantage over analyzers
that use a diluted sample?
A. Can measure over a wider range of concentration
B. Are not subject to pseudohyponatremia caused by high lipids
C. Do not require temperature equilibration
D. Require less maintenance
B Ion-selective analyzers measure the electrolyte dissolved in the fluid phase of the
sample in millimoles per liter of plasma water. When undiluted blood is assayed, the
measurement is independent of colloids, such as protein and lipid. Hyperlipemic
samples cause falsely low sodium measurements when assayed by ion-selective
analyzers requiring dilution because lipids displace plasma water containing the
electrolytes. One drawback to undiluted or direct-measuring systems is that the
electrodes require more frequent deproteinization and usually have a shorter duty
cycle.
When measuring calcium by atomic absorption spectrophotometry, which is required?
A. An organic extraction reagent to deconjugate calcium from protein
B. An internal standard
C. A magnesium chelator
D. Lanthanum oxide to chelate phosphates
D An acidic diluent, such as hydrochloric acid (HCl), will displace calcium bound to
albumin. However, calcium forms a thermostable bond with phosphate that causes
chemical interference in atomic absorption. Lanthanum displaces calcium, forming
lanthanum phosphate, and eliminates interference from phosphates. Unlike in some
colorimetric methods for calcium (e.g., o-cresolphthalein complexone), magnesium
does not interfere because it does not absorb the 422.7-nm emission line from the
calcium–hollow cathode lamp.
The term RT/nF in the Nernst equation defines the:
A. Potential at the ion-selective membrane
B. Slope of the electrode
C. Decomposition potential
D. Isopotential point of the electrode
B In the term RT/nF, R = the molar gas constant, T = temperature in degrees Kelvin, F =
Faraday’s constant, and n = the number of electrons donated per atom of reductant.
The slope depends on the temperature of the solution and the valence of the reductant.
At room temperature, the slope is 59.2 mV for a univalent ion and 29.6 mV for a
divalent ion.
The response of a sodium electrode to a 10-fold increase in sodium concentration should
be:
A. A 10-fold drop in potential
B. An increase in potential of approximately 60 mV
C. An increase in potential of approximately 10 mV
D. A decrease in potential of approximately 10 mV
B The Nernst equation predicts an increase of approximately 60 mV per 10-fold
increase in sodium activity. For sodium:
E = E° + RT/nF × 2.3 log10[Na+]
RT/nF × 2.3 = 60 mV at 37°C.
Therefore:
E = E° + 60 mV × log10[Na+].
If sodium concentration is 10 mmol/L, then:
E = E° + 60 mV × log10[10] = E° + 60 mV.
If sodium concentration increases from 10 mmol/L to 100 mmol/L, then:
E = E° + 60 mV × log10[100] =
E° + 60 mV × 2 = E° + 120 mV.
Which of the electrodes below is a current-producing (amperometric) rather than a
voltage-producing (potentiometric) electrode?
A. Clark electrode
B. Severinghaus electrode
C. pH electrode
D. Ionized calcium electrode
A The Clark electrode is composed of two half cells that generate current, not voltage. It
is used to measure partial pressure of oxygen (PO2), and is based on an amperometric
method called polarography. When –0.8 V is applied to the cathode, O2 is reduced,
causing current to flow. Current is proportional to the PO2 of the sample.
Persistent noise from an ISE is most often caused by:
A. Contamination of the sample
B. Blocked junction at the salt bridge
C. Over-range from high concentration
D. Improper calibration
B Electrode noise most often results from an unstable junction potential. Most reference
electrodes contain a high concentration of the potassium chloride (KCl) internal
solution used to produce the reference potential. This forms a salt bridge with the
measuring half cell by contacting sample but is kept from equilibrating via a barrier,
called a junction. When this junction becomes blocked by salt crystals, the reference
potential will be unstable, resulting in fluctuation in the analyzer readout.
Which of the following statements accurately characterizes the coulometric titration of
chloride?
A. The indicator electrodes generate voltage
B. Constant current must be present across the generator electrodes
C. Silver ions are formed at the generator cathode
D. Chloride concentration is inversely proportional to titration time
B The Cotlove chloridometer is based on the principle of coulometric titration with
amperometric detection. Charge in the form of silver ions is generated by oxidation of
silver wire at the generator anode. Silver ions react with chloride ions, forming
insoluble silver chloride (AgCl). When all of the chloride is titrated, free silver ions are
detected by reduction back to elemental silver, which causes an increase in current
across the indicator electrodes (a pair of silver electrodes with a voltage difference of
about 1.0 V DC). Charge or titration time is directly proportional to chloride
concentration as long as the rate of oxidation remains constant at the generator anode.
Which element is reduced at the cathode of a Clark polarographic electrode?
A. Silver
B. Oxygen
C. Chloride
D. Potassium
B The Clark electrode is designed to measure O2. O2 diffuses through a gas-permeable
membrane covering the electrode. It is reduced at the cathode, which is made of
platinum or other inert metal. Electrons are supplied by the anode, which is made of
silver. The net reaction is:
4 KCl + 2 H2O + O2 + 4 Ag° → 4 AgCl + 4 KOH
Which of the following would cause a “response” error from an ISE for sodium when
measuring serum but not the calibrator?
A. Interference from other electrolytes
B. Protein coating the ion-selective membrane
C. An over-range in sodium concentration
D. Protein binding to sodium ions
B Response is the time required for an electrode to reach steady-state potential. Ion-
selective analyzers use a microprocessor to monitor electrode response, slope, drift,
and noise. When an electrode gives an acceptable response time when measuring an
aqueous calibrator, but not when measuring serum, the cause is often protein buildup
on the membrane
In polarography, the voltage needed to cause depolarization of the cathode is called the:
A. Half-wave potential
B. Isopotential point
C. Decomposition potential
D. Polarization potential
C In polarography, a minimum negative voltage must be applied to the cathode to cause
reduction of metal ions (or O2) in solution. This is called the decomposition potential.
It is concentration dependent (dilute solutions require greater negative voltage) and can
be determined by using the Nernst equation.
In the coulometric chloride titration:
A. Acetic acid in the titrating solution furnishes the counter ion for reduction
B. The endpoint is detected by amperometry
C. The titrating reagent contains a phosphate buffer to keep pH constant
D. Nitric acid (HNO3) is used to lower the solubility of AgCl
B Reduction of Ag+ back to Ag° generates the current, which signals the endpoint. The
titrating reagent contains HNO3, acetic acid, H2O, and either gelatin or polyvinyl
alcohol. The HNO3 furnishes nitrate, which is reduced at the generator cathode,
forming ammonium ions. The ammonium becomes oxidized back to nitrate at the
indicator anode. Gelatin or polyvinyl alcohol is needed to prevent pitting of the
generator anode. Acetic acid lowers the solubility of AgCl, preventing dissociation
back to Ag+
One mole per kilogram water (H2O) of any solute will cause which of the following:
A. Lower the freezing point by 1.86°C
B. Raise vapor pressure by 0.3 mm Hg
C. Lower the boiling point by 0.52°C
D. Lower osmotic pressure by 22.4 atm
A Both freezing point and vapor pressure are lowered by increasing solute
concentration. Boiling point and osmotic pressure are raised. Increasing solute
concentration of a solution opposes a change in its physical state and lowers the
concentration of H2O molecules.
Which of the following compounds can interfere with the coulometric chloride assay?
A. Bromide
B. Ascorbate
C. Acetoacetate
D. Nitrate
A Chloride assays based on either coulometric or chemical titration are subject to
positive interference from other anions and electronegative radicals that may be titrated
instead of chloride ions. These include other halogens, such as bromide, cyanide, and
cysteine.
Which of the following compounds contributes to the osmolality of plasma?
A. Cholesterol
B. Protein
C. Drug metabolites
D. Triglyceride
C Osmolality is the concentration (in moles [mol]) of dissolved solute per kilogram
solvent. Proteins and lipids are not in solution, and do not contribute to osmolality. The
nonionized solutes, such as glucose and urea, contribute 1 osmole per mole per
kilogram (Osmol/mol/kg) water, whereas dissociated salts contribute 1 Osmol/mol of
each dissociated ion or radical.
The freezing point osmometer differs from the vapor pressure osmometer in that only
the freezing point osmometer:
A. Cools the sample
B. Is sensitive to ethanol
C. Requires a thermoelectric module
D. Requires calibration with aqueous standards
B Alcohol enters the vapor phase so rapidly that it evaporates before the dew point of
the sample is reached. Therefore, ethanol does not contribute to osmolality as
measured by using the vapor pressure osmometer. Freezing-point osmometers measure
alcohol and can be used in emergency department (ED) settings to estimate ethanol
toxicity.
What component of a freezing point osmometer measures the sample temperature?
A. Thermistor
B. Thermocouple
C. Capacitor
D. Electrode
A A thermistor is a temperature-sensitive resistor. The resistance to current flow
increases as temperature falls. The temperature at which a solution freezes can be
determined by measuring the resistance of the thermistor. Resistance is directly
proportional to the osmolality of the sample.
What type of measuring circuit is used in a freezing point osmometer?
A. Electrometer
B. Potentiometer
C. Wheatstone bridge
D. Thermal conductivity bridge
C The resistance of the thermistor is measured using a network of resistors called a
Wheatstone bridge. When the sample is frozen, the bridge is balanced by using a
calibrated variable resistor so that no current flows to the readout. The resistance
required to balance the meter is equal to the resistance of the thermistor.
Which measurement principle is employed in a vapor pressure osmometer?
A. Seebeck
B. Peltier
C. Hayden
D. Darlington
A The term Seebeck effect refers to the increase in voltage across the two junctions of a
thermocouple caused by a difference in the temperature at the junctions. Increasing
osmolality lowers the dew point of a sample. When sample is cooled to its dew point,
the voltage change across the thermocouple is directly proportional to osmolality
The method for measuring iron or lead by plating the metal and then oxidizing it is
called:
A. Polarography
B. Coulometry
C. Anodic stripping voltometry
D. Amperometry
C Anodic stripping voltometry is used to measure lead. The cation of the metal is plated
onto a mercury cathode by applying a negative charge. The voltage of this electrode is
reversed until the plated metal is oxidized back to a cation. Current produced by
oxidation of the metal is proportional to concentration.
The term reverse phase is used in HPLC to indicate that the mobile phase is:
A. More polar than the stationary phase
B. Liquid and the stationary phase is solid
C. Organic and the stationary phase is aqueous
D. A stronger solvent than the stationary phase
A In reverse-phase HPLC, the separation takes place using a nonpolar sorbent
(stationary phase), such as octadecylsilane (C18). Solutes that are nonpolar are retained
longer compared with polar solutes. Most clinical separations of drugs, hormones, and
metabolites use reverse phase because aqueous mobile phases are far less toxic and
flammable
The term isocratic is used in high-performance liquid chromatography (HPLC) to mean
that the:
A. Mobile phase is at constant temperature
B. Stationary phase is equilibrated with the mobile phase
C. Mobile phase consists of a constant solvent composition
D. Flow rate of the mobile phase is regulated
C An isocratic separation uses a single mobile phase of constant composition, pH, and
polarity, and requires a single pump. Some HPLC separations use a gradient mobile
phase to increase distance between peaks. Gradients are made by mixing two or more
solvents by using a controller to change the proportions of solvent components
What is the primary means of solute separation in HPLC using a C18 column?
A. Anion exchange
B. Size exclusion
C. Partitioning
D. Cation exchange
C Stationary phases (column packings) used in HPLC separate solutes by multiple
means, but in reverse-phase HPLC using C18, the relative solubility between the
mobile phase and stationary phase is most important and depends on solvent polarity,
pH, and ionic strength. Stationary phases with particles less than 2.5 microns in
diameter are designated as ultraperformance liquid chromatography (UPLC) because
they increase column resolution over standard size sorbents
The most commonly used detector for clinical gas–liquid chromatography (GLC) is
based on:
A. UV light absorbance at 254 nm
B. Flame ionization
C. Refractive index
D. Thermal conductance
B Volatile solutes can be detected in GLC using flame ionization, thermal conductivity,
electron capture, and mass spectroscopy. In flame ionization, energy from a flame is
used to excite the analytes as they elute from the column. The flame is made by
igniting a mixture of hydrogen, carrier gas, and air. Current is produced when an outer
shell electron is ejected from the excited analyte.
In GC, the elution order of volatiles is usually based on the:
A. Boiling point
B. Molecular size
C. Carbon content
D. Polarity
A The order of elution is dependent on the velocity of the analyte. Usually, the lower the
boiling point of the compound, the greater is its velocity or solubility in carrier gas.
What type of detector is used in high-performance liquid chromatography with
electrochemical detection (HPLC–ECD)?
A. Calomel electrode
B. Conductivity electrode
C. Glassy carbon electrode
D. Polarographic electrode
C HPLC–ECD uses a glassy carbon-measuring electrode and a silver–silver chloride
reference. The analyte is oxidized or reduced by holding the glassy carbon electrode at
a positive voltage (oxidization) or negative voltage (reduction). The resulting current
flow is directly proportional to concentration. Phenolic groups, such as
catecholamines, can be measured by using HPLC–ECD.
Select the chemical that is used in most HPLC procedures to decrease solvent polarity.
A. Hexane
B. Nonane
C. Chloroform
D. Acetonitrile
D All of the compounds mentioned have nonpolar properties. Because most HPLC is
reverse phase (a polar solvent is used), hexane and nonane are too nonpolar.
Acetonitrile is more polar and less toxic than chloroform and along with methanol is a
common polarity modifier for HPLC
In thin-layer chromatography (TLC), the distance the solute migrates divided by the
distance the solvent migrates is the:
A. tR
B. Kd
C. Rf
D. pK
C Rf is the distance migrated by the solute divided by the distance migrated by the
solvent. The tR refers to the retention time of the solute in HPLC or GC. The Kd is the
partition coefficient and is a measure of the relative affinity of solutes for the stationary
phase. The solute with the greater Kd will be retained longer. The pK is the negative
logarithm of K, the ionization constant and is a measure of ionization.
Which reagent is used to extract alkaline drugs, such as cocaine metabolites, from
urine?
A. Acid and sodium chloride
B. Alkali and organic solvent
C. Chloroform and sodium acetate
D. Neutral solution of ethyl acetate
B Alkaline drugs, such as cocaine, amphetamine, and morphine, are extracted at alkaline
pH. Ideally, the pH for extracting alkaline drugs into an organic solvent should be 2 pH
units greater than the negative log of dissociation constant (pKa) of the drug. More than
90% of the drug will be nonionized and will extract in ethyl acetate or another organic
solvent.
What is the purpose of an internal standard in HPLC and GC methods?
A. To compensate for variation in extraction and injection
B. To correct for background absorbance
C. To compensate for changes in flow rate
D. To correct for coelution of solutes
A Internal standards should have the same affinity as the analyte for the extraction
reagents. Dividing peak height (or area) of all samples (standards and unknowns) by
the peak height (or area) of the internal standard reduces error caused by variation in
extraction recovery and injection volume
What is the confirmatory method for measuring drugs of abuse?
A. HPLC
B. Enzyme-multiplied immunoassay technique (EMIT)
C. Gas chromatography with mass spectroscopy (GC-MS)
D. TLC
C GC-MS determines the mass spectrum of the compounds eluting from the analytic column. Each substance has a unique and characteristic spectrum of mass fragments.
This spectrum is compared with spectra in a library of standards to determine the
percent match. A match of greater than 95% is considered confirmatory
The fragments typically produced and analyzed in methods employing mass
spectroscopy are typically:
A. Of low molecular size ranging from 10 to 100 daltons
B. Cations caused by electron loss or proton attachment
C. Anions caused by bombarding the molecule with an electron source
D. Neutral species formed after excited molecules form a stable resonance structure
B In most MS applications, cations of the molecule are measured. Cations can be
formed by various methods, the most common of which is electron bombardment
(electron ionization). The energy transferred to the molecule causes ejection of an outer
shell electron. MS can analyze sizes from trace metals through macromolecules.
Proteins are measured after conversion to cations by ionization procedures, such as
matrix-assisted laser desorption/ionization (MALDI), in which energy from a nitrogen
laser causes transfer of a proton from the matrix (an acid) to the protein.
What component is used in a GC-MS but not used in an LC-MS?
A. Electron source
B. Mass filter
C. Detector
D. Vacuum
A The mass spectrometer requires a sample that is suspended in a gas phase, and
therefore, the separated analytes from a GC directly transfer into the mass
spectrometer. Although chemical ionization of the sample is possible, most GC-MS
instruments utilize electron ionization. Electrons are produced by applying 70 electron
volts to a filament of tungsten or rhenium under vacuum. The electrons collide with the
neutral molecules coming from the GC, splitting them into fragments. The array of
fragments is a unique identifier of each molecule. A vacuum is needed in MS
instruments to control the path of ions as they move through the mass filter.
What process is most often used in LC-MS to introduce the sample into the mass filter?
A. Electrospray ionization (ESI)
B. Chemical ionization
C. Electron impact ionization
D. Fast atom bombardment
A HPLC instruments use solvent rather than gas to separate molecules. The sample is
converted into a gaseous state by ESI before it enters the mass filter. ESI uses a small-
bore tube that forms a 1- to 4-micron nozzle at the mass filter inlet and which is
charged by several kilovolts. The sample enters the tube along with inert drying gas.
The tube is heated to evaporate the solvent, but unlike the electron impact used in GC-
MS, the ionizer is not under vacuum. When a droplet of the sample reaches the nozzle,
it becomes highly charged. The size of the droplet is decreased as a result of
evaporation. This causes the charge density to become excessive, and the droplets
break apart. The tiny charged droplets repel each other and break apart again, forming
a plume. These particles are drawn into the mass filter by “ion optics” (a system of
repeller plates, counter-electrode, and magnets). ESI does not result in extensive
fragmentation, producing mostly the parent or “molecular” ion, a process called soft
ionization.
In mass spectroscopy, the term base peak typically refers to:
A. The peak with the lowest mass
B. The peak with the most abundance
C. A natural isotope of the molecular ion
D. The first peak to reach the mass detector
B The base peak is typically the “molecular ion” or parent ion, meaning that it is the
initial fragment made by releasing an electron. The cation thus formed has a charge of
+1, and therefore, its mass:charge (m/z) ratio is equal to its mass. The base peak is
used for selective ion monitoring (SIM). It is the most abundant and most stable ion
and gives the best sensitivity for quantitative analysis.
Which of the following is the most useful method when screening for errors of amino
acid and organic acid metabolism?
A. Two-dimensional TLC
B. GC
C. Electrospray ionization tandem-mass spectroscopy (MS/MS)
D. Inductively coupled-mass spectroscopy (ICP-MS)
C While two-dimensional TLC can separate both amino acids and organic acids, it is
not sufficiently sensitive for newborn screening. ESI allows a small alcohol-extracted
whole blood sample to be analyzed by two mass spectrometers without prior separation
by LC or GC. Disorders of both organic and fatty acid metabolism are identified by the
specific pattern of acylcarnitine ions produced. Amino acids are detected as amino
species that have lost a carboxyl group during ionization, a process called neutral loss.
In MS/MS, the first mass filter performs the same function as:
A. The ion source
B. The chromatography column
C. Extraction
D. The vacuum system
B A tandem mass spectrometer uses two or more mass filters in sequence. The first filter
functions as an ion trap. Once the sample is ionized, the filter selects molecular or
parent ions of interest by excluding ions outside a specified size range. Therefore, it
effectively separates the analyte(s) of interest from unwanted compounds. MS/MS uses
ESI to introduce the sample into the first mass filter, usually a quadrapole. The
radiofrequency (RF) and DC voltages of the quadrapole are set to optimize the
trajectory of the parent ions of interest and cause ejection of unwanted ions. The parent
ions are drawn into a second mass filter where they are bombarded by argon atoms.
The collisions result in the formation of mass fragments called daughter ions. This
process is called collision-induced dissociation, and the second filter is called a
collision chamber. The process can be repeated in a third mass filter that generates
granddaughter ions. A total-ion chromatogram is produced from these, enabling the
compound of interest to be identified and quantified. MS/MS is used to screen for
inborn errors of fatty acid, amino acid, and organic acid metabolism
SITUATION: A GC-MS analysis using nitrogen as the carrier gas shows an extensively
noisy baseline. A sample of the solvent (ethyl acetate) used for the extraction procedure
was injected and showed the same noise. Results of an Autotune test showed the
appearance of a base peak at 16 with two smaller peaks at 17 and 18. These results
indicate that:
A. The solvent is contaminated
B. The carrier gas is contaminated
C. There is electrical noise in the detector
D. The ion source is dirty
B All of these situations are sources of baseline noise in GC-MS. However, the peak at
16 indicates the presence of O2 in the carrier gas. O2 in the atmosphere also contains
small quantities of two isotopes with molecular weights of 17 and 18 because of the
presence of one and two extra neutrons, respectively
Why is vacuum necessary in the mass filter of a GC-MS?
A. Ionization does not occur at atmospheric pressure
B. It prevents collision between fragments
C. It removes electrons from the ion source
D. It prevents contamination
B Vacuum is needed in the mass filter of GC-MS instruments to prevent random
collisions between ions that would alter their trajectory or time of flight. The vacuum
prevents collision between the carrier gas molecules and the ions. However, in
spectrometers that use ESI, chemical ionization, and matrix-assisted laser
desorption/ionization-time of flight (MALDI-TOF) and surface-enhanced laser
desorption/ionization-time of flight (SELDI-TOF), the ion source is not under vacuum
What method is used to introduce the sample into a mass spectrometer for analysis of a
trace element?
A. ESI
B. Laser desorption
C. ICP ionization
D. Direct injection
C Mass spectrometers can be used to measure trace metals, but the atoms need to be
vaporized and ionized like molecules before they enter the mass filter. This is done by
introducing the sample into a very hot plasma (6,000–10,000°K) called a torch. The
torch is made by circulating argon through inner and outer quartz tubes. The tubes are
wrapped with a coil of wire that receives a radio frequency. This creates current flow
through the wire and a magnetic field at the torch end. Argon atoms are excited by the
current and magnetic field and ionize. When the argon is ignited by a spark, it forms a
plasma. The sample is mixed with argon at the other end to create an aerosol. When it
reaches the torch, the solvent is evaporated and the energy from the torch and
collisions with argon ions cause ejection of outer-shell electrons, forming cations of
the element. ICP-MS is used to measure any trace element that readily forms cations.
In real-time polymerase chain reaction (PCR), what value is needed to determine the
threshold?
A. Background signal
B. Melt temperature
C. Maximum fluorescence
D. Threshold cycle
A In real-time PCR, the fluorescence of the reporter probe is proportional to the
concentration of PCR products. For quantitation of PCR products, a well factor and
background fluorescence must be determined. Well-factor values are analogous to
cuvette blanks. They are used to correct the measurements from each well so that the
same concentration of fluorescent dye gives the same signal intensity regardless of the
well. The threshold is the lowest signal that indicates the presence of product. It can be
calculated manually from a real-time amplification curve by finding the average
standard deviation of the fluorescent signal (relative fluorescence units [RFU]) from
cycles 2 to 10. This is multiplied by 10 to obtain the threshold value in RFUs.
Which component is needed for a thermal cycler to amplify DNA?
A. Programmable heating and cooling unit
B. Vacuum chamber with zero head space
C. Sealed airtight constant-temperature chamber
D. Temperature-controlled ionization chamber
A The PCR for DNA amplification consists of three phases. Denaturation requires a
temperature of 90°C to 94°C and separates the double-stranded DNA. Annealing
requires a temperature of 40°C to 65°C and allows the primers to bind to the target
base sequence. Extension requires a temperature of 72°C and allows the heat-stable
polymerase to add complementary bases to the primer in the 5’ to 3’ direction. A cycle
consists of each temperature stage for a specific number of minutes and most
procedures require 20 or more cycles to generate a detectable quantity of target DNA.
Rapid heating and cooling is usually achieved using a thermoelectric block that is
cooled by forced air flow.
Given the following real-time PCR amplification curve, what is the threshold cycle
(Ct)?
A. 15
B. 20
C. 25
D. 30
C The maximum curvature of the plot approximates the Ct. A line is drawn from the
threshold value on the y-axis through the curve, and a perpendicular dropped to the x-
axis. The Ct is determined by the intersection point on the x-axis. The threshold is
usually determined by an algorithm but can be calculated manually as 10 times the
average standard deviation of the RFUs for cycles 2 to 10.
In addition to velocity, what variable is also needed to calculate the relative centrifugal
force (g force) of a centrifuge?
A. Head radius
B. Angular velocity coefficient
C. Diameter of the centrifuge tube
D. Ambient temperature in degrees Centigrade
A The relative centrifugal force (number times the force of gravity) is proportional to
the square of the rotor speed in revolutions per minute and the radius in centimeters of
the head (distance from the shaft to the end of the tube).
RCF = s2 × r × 1.118 × 10–5
where s is the speed in revolutions per minute (RPM), r is the radius in centimeters and
1.118 × 10–5 is a conversion constant.
When calibrating a semiautomatic pipet that has a fixed delivery of 10.0 μL using a
gravimetric method, what should be the average weight of deionized water transferred?
A. 10.0 μg
B. 100.0 μg
C. 1.0 mg
D. 10.0 mg
D Gravimetric and spectrophotometric analyses are the two methods used to verify pipet
volume accuracy and precision. Since spectrophotometric analysis involves dilution,
gravimetric analysis is associated with greater certainty. At 20°C, the density of pure
water is 0.99821 g/mL. Therefore, each microliter weighs almost exactly 1.0 mg.
Which of the following situations is likely to cause an error when weighing with an
electronic analytical balance?
A. Failure to keep the knife edge clean
B. Failure to close the doors of the balance before reading the weight
C. Oxidation on the surface of the substitution weights
D. Using the balance without allowing it to warm up for at least 10 minutes
B Electronic balances do not use substitution weights or knife edges to balance the
weight on the pan. Instead, they measure the displacement of the pan by the weight on
it by using electromagnetic force to return it to its reference position. Regardless of the
type of balance used, all need to be located on a firm weighing table free of vibration.
Doors must be closed to prevent air currents from influencing the weighing, and the
pan and platform must be clean and free of dust and chemical residue.
Which of the following represents the Henderson-Hasselbalch equation as applied to
blood pH?
A. pH = 6.1 + log HCO3–/PCO2
B. pH = 6.1 + log HCO3–/(0.03 × PCO2)
C. pH = 6.1 + log DCO2/HCO3–
D. pH = 6.1 + log (0.03 × PCO2)/HCO3–
B The Henderson–Hasselbalch equation describes the pH of a buffer comprising a weak
acid and its salt. pH = pKa + log salt/acid, where pKa is the negative logarithm of the
dissociation constant of the acid. In this case, the salt is sodium bicarbonate (NaHCO3)
and the acid is the dissolved CO2, which is equal to 0.03 (mmol/L per mm Hg) × PCO2.
The pKa includes both the hydration and dissociation constant for dissolved CO2 in
blood, 6.1.
What is the PO2 of calibration gas containing 20.0% O2, when the barometric pressure is
30 in?
A. 60 mm Hg
B. 86 mm Hg
C. 143 mm Hg
D. 152 mm Hg
C Convert barometric pressure in inches to millimeters of mercury by multiplying by
25.4 (mm/in). Next, subtract the vapor pressure of H2O at 37°C, 47 mm Hg, to obtain
dry gas pressure. Multiply dry gas pressure by the %O2:
25.4 mm/in × 30 in = 762 mm Hg
762 mm Hg – 47 mm Hg (vapor pressure) =
715 mm Hg (dry gas pressure)
0.20 × 715 mm Hg = 143 mm Hg PO2
What is the blood pH when the partial pressure of carbon dioxide (PCO2) is 60 mm Hg
and the bicarbonate concentration is 18 millimoles per liter (mmol/L)?
A. 6.89
B. 7.00
C. 7.10
D. 7.30
C Solve using the Henderson-Hasselbalch equation. pH = pK’ + log HCO3–/(0.03 ×
PCO2), where pK’, the negative logarithm of the combined hydration and dissociation
constants for dissolved CO2 and carbonic acid, is 6.1, and the solubility coefficient for
CO2 gas is 0.03.
pH = 6.1 + log 18/(0.03 × 60) = 6.1 + log 18/1.8
pH = 6.1 + log 10. Because log 10 = 1, pH = 7.10
Which of the following best represents the reference (normal) range for arterial pH?
A. 7.35 to 7.45
B. 7.42 to 7.52
C. 7.38 to 7.68
D. 6.85 to 7.56
A The reference range for arterial blood pH is 7.35 to 7.45 and is only 0.03 pH units
lower for venous blood owing to the buffering effects of hemoglobin (Hgb), known as
the chloride isohydric shift. Most laboratories consider less than 7.20 and greater than
7.60 the critical values for pH
Which of the following contributes the most to serum total CO2 (TCO2)?
A. PCO2
B. DCO2
C. HCO3–
D. Carbonium ion
C The total CO2 is the sum of the DCO2, H2CO3 (carbonic acid or hydrated CO2), and
bicarbonate (as mainly NaHCO3). When serum is used to measure total CO2, the DCO2
is insignificant because all the CO2 gas has escaped into the air. Therefore, serum total
CO2 is equivalent to the bicarbonate concentration. Total CO2 is measured by
potentiometry. An organic acid is used to release CO2 gas from bicarbonate and PCO2
is measured with a Severinghaus electrode. Alternatively, bicarbonate can be measured
by an enzymatic reaction using phosphoenol pyruvate carboxylase. The enzyme forms
oxaloacetate and phosphate from phosphoenol pyruvate and bicarbonate. The
oxaloacetate is reduced to malate by malate dehydrogenase and nicotinamide adenine
dinucleotide (NADH) is oxidized to NAD+. The negative reaction rate is proportional
to plasma bicarbonate concentration.
What is the PCO2 if the DCO2 is 1.8 mmol/L?
A. 24 mm Hg
B. 35 mm Hg
C. 60 mm Hg
D. 72 mm Hg
C Dissolved CO2 is calculated from the measured PCO2 × 0.0306, the solubility
coefficient for CO2 gas in blood at 37°C.
DCO2 = PCO2 × 0.03
Therefore, PCO2 = DCO2/0.03
PCO2 = 1.8 mmol/L ÷ 0.03 mmol/
L per mm Hg = 60 mm Hg
- What is the normal ratio of bicarbonate to dissolved carbon dioxide (HCO3–:DCO2) in
arterial blood?
A. 1:10
B. 10:1
C. 20:1
D. 30:1
C When the ratio of HCO3–:DCO2 is 20:1, the log of salt/acid becomes 1.3. Substituting
this in the Henderson-Hasselbalch equation and solving for pH gives pH = 6.1 + log
20; pH = 6.1 + 1.3 = 7.4. Acidosis results when this ratio is decreased and alkalosis
when it is increased
n the Henderson-Hasselbalch expression pH = 6.1 + log HCO3–/DCO2, the 6.1
represents:
A. The combined hydration and dissociation constant for CO2 in blood at 37°C
B. The solubility constant for CO2 gas
C. The dissociation constant of H2O
D. The ionization constant of NaHCO3
A The equilibrium constant, Kh, for the hydration of CO2 (DCO2 + H2O → H2CO3) is
only about 2.3 × 10–3M, making DCO2 far more prevalent than carbonic acid. The
dissociation constant, Kd, for the reaction H2CO3→H+ + HCO3– is about 2 × 10–4 M.
The product of these constants is the combined equilibrium constant, K‘. The negative
logarithm of K’ is the pK’, which is 6.103 in blood at 37°C.
In addition to NaHCO3, what other substance contributes the most to the amount of
base in blood?
A. Hgb concentration
B. Dissolved O2 concentration
C. Inorganic phosphorus
D. Organic phosphate
A The primary blood buffer bases preventing acidosis, in order of concentration, are
bicarbonate, deoxyhemoglobin, albumin, and monohydrogen phosphate. At
physiological pH, there is significantly more H2PO4–1 than HPO4–2, and phosphate is a
buffer system that is more efficient at preventing alkalosis than acidosis. Because all of
the blood buffer systems are in equilibrium, pH can be calculated accurately from the
concentration of bicarbonate and dissolved CO2 by using the Henderson-Hasselbalch
equation.
Which of the following effects results from exposure of a normal arterial blood sample
to room air?
A. PO2 increased PCO2 decreased pH increased
B. PO2 decreased PCO2 increased pH decreased
C. PO2 increased PCO2 decreased pH decreased
D. PO2 decreased PCO2 decreased pH decreased
A The PO2 of air at sea level (21% O2) is about 150 mm Hg. The PCO2 of air is only
about 0.3 mm Hg. Consequently, blood releases CO2 gas and gains O2 when exposed
to air. Loss of CO2 shifts the equilibrium of the bicarbonate buffer system to the right,
decreasing hydrogen ion (H+) concentration and blood becomes more alkaline.
Which of the following formulas for O2 content is correct?
A. O2 content = %O2 saturation/100 × Hgb g/dL × 1.39 mL/g + (0.0031 × PO2)
B. O2 content = PO2 × 0.0306 mmol/L/mm
C. O2 content = O2 saturation × Hgb g/dL × 0.003 mL/g
D. O2 content = O2 capacity × 0.003 mL/g
A Oxygen content is the sum of O2 bound to Hgb and O2 dissolved in the plasma. It is
dependent on the Hgb concentration and the percentage of Hgb bound to O2 (O2
saturation). Each gram of Hgb binds 1.39 mL of O2. The dissolved O2 is determined
from the solubility coefficient of O2 (0.0031 mL per dL/mm Hg) and the PO2. O2
content = % Sat/100 × Hgb in g/dL × 1.39 mL/g + (0.0031 × PO2)
The normal difference between alveolar and arterial PO2 (PAO2–PaO2 difference) is:
A. 3 mm Hg
B. 10 mm Hg
C. 40 mm Hg
D. 50 mm Hg
B The PAO2–PaO2 difference results from the low ratio of ventilation to perfusion in the
base of the lungs. The Hgb in the blood coming from the base of the lung has a lower
O2 saturation. This blood will take up O2 from the plasma of blood leaving well-
ventilated areas of the lung, thus lowering the mixed arterial PO2
A decreased PAO2–PaO2 difference is found in:
A. A/V (arteriovenous) shunting
B. V/Q (ventilation/perfusion) inequality
C. Ventilation defects
D. All of these options
C Patients with A/V shunts, V/Q inequalities, and cardiac failure will have an increased
PAO2–PaO2 difference. However, patients with ventilation problems have low alveolar
PO2 as a result of retention of CO2 in the airway. This reduces the PAO2–PaO2
difference
The determination of the O2 saturation of Hgb is best accomplished by:
A. Polychromatic absorbance measurements of a whole blood hemolysate
B. Near infrared transcutaneous absorbance measurement
C. Treatment of whole blood with alkaline dithionite prior to measuring absorbance
D. Calculation using PO2 and total Hgb by direct spectrophotometry
A Measurement of oxyhemoglobin, deoxyhemoglobin (reduced Hgb),
carboxyhemoglobin, methemoglobin, and sulfhemoglobin can be accomplished by
using direct spectrophotometry at multiple wavelengths and the absorptivity
coefficients of each pigment at those wavelengths. The O2 saturation is determined by
dividing the fraction of oxyhemoglobin by the sum of all pigments. This eliminates
much of the errors that occur in the other methods when the quantity of an abnormal
Hgb pigment is increased.
Correction of pH for a patient with a body temperature of 38°C would require:
A. Subtraction of 0.015
B. Subtraction of 0.01%
C. Addition of 0.020
D. Subtraction of 0.020
A The pH decreases by 0.015 for each degree Celsius above the 37°C. Because the
blood gas analyzer measures pH at 37°C, the in vivo pH would be 0.015 pH units
below the measured pH.
Select the anticoagulant of choice for blood gas studies.
A. Sodium citrate 3.2%
B. Lithium heparin 100 units/mL blood
C. Sodium citrate 3.8%
D. Ammonium oxalate 5.0%
B Heparin is the only anticoagulant that does not alter the pH of blood; heparin salts
must be used for pH and blood gases. Solutions of heparin are air equilibrated and
must be used sparingly to prevent contamination of the sample by gas in the solution.
What is the maximum recommended storage time and temperature for an arterial
blood gas sample drawn in a plastic syringe?
Storage Time Temperature
A. 10 min 2°C-8°C
B. 20 min 2°C-8°C
C. 30 min 2°C-8°C
D. 30 min 22°C
D Arterial blood gas samples collected in plastic syringes should be stored at room
temperature because cooling the sample allows O2 to enter the syringe. Storage time
should be no more than 30 minutes because longer storage results in a significant drop
in pH and PO2 and increased PCO2.
A patient’s blood gas results are as follows:
pH = 7.26 DCO2 = 2.0 mmol/L HCO3– = 29 mmol/L
These results would be classified as:
A. Metabolic acidosis
B. Metabolic alkalosis
C. Respiratory acidosis
D. Respiratory alkalosis
C Imbalances are classified as respiratory when the primary disturbance is with PCO2
because PCO2 is regulated by ventilation. PCO2 = DCO2/0.03 or 60 mm Hg (normal 35–
45 mm Hg). Increased DCO2 will increase H+ concentration, causing acidosis.
Bicarbonate is moderately increased, but a primary increase in NaHCO3 causes
alkalosis. Thus, the cause of this acidosis is CO2 retention (respiratory acidosis), and it
is partially compensated for by renal retention of bicarbonate.
A patient’s blood gas results are:
pH = 7.50 PCO2 = 55 mm Hg HCO3– = 40 mmol/L
These results indicate:
A. Respiratory acidosis
B. Metabolic alkalosis
C. Respiratory alkalosis
D. Metabolic acidosis
B A pH above 7.45 corresponds with alkalosis. Both bicarbonate and PCO2 are elevated.
Bicarbonate is the conjugate base and is under metabolic (renal) control, whereas PCO2
is an acid and is under respiratory control. Increased bicarbonate (but not increased
CO2) results in alkalosis; therefore, the classification is metabolic alkalosis, partially
compensated for by increased PCO2.
Which of the following will shift the O2 dissociation curve to the left?
A. Anemia
B. Hyperthermia
C. Hypercapnia
D. Alkalosis
D A left shift in the oxyhemoglobin dissociation curve signifies an increase in the
affinity of Hgb for O2. This occurs in alkalosis, hypothermia, and in those
hemoglobinopathies, such as Hgb Chesapeake, which increase the binding of O2 to
heme. A right shift in the oxyhemoglobin dissociation curve lowers the affinity of Hgb
for O2. This occurs in anemia which increases 2,3-diphosphoglycerate (2,3-DPG); with
increased body temperature, increased H+ concentration, hypercapnia (increased PCO2);
and in some hemoglobinopathies, such as Hgb Kansas.
Which set of results is consistent with uncompensated respiratory alkalosis?
A. pH 7.70 HCO3 30 mmol/L PCO2 25 mm Hg
B. pH 7.66 HCO3 22 mmol/L PCO2 20 mm Hg
C. pH 7.46 HCO3 38 mmol/L PCO2 55 mm Hg
D. pH 7.36 HCO3 22 mmol/L PCO2 38 mm Hg
B Respiratory alkalosis is caused by hyperventilation, inducing low PCO2. Very often, in
the early phase of an acute respiratory disturbance, the kidneys have not had time to
compensate, and the bicarbonate is within normal limits. In option A, the bicarbonate
is high and PCO2 low; thus, both are contributing to alkalosis and this would be
classified as a combined acid–base disturbance. In answer C, the pH is almost normal,
and both bicarbonate and PCO2 are increased. This can occur in the early stage of a
metabolic acid–base disturbance when full respiratory compensation occurs or in a
combined acid–base disorder. In answer D, both bicarbonate and PCO2 are within
normal limits (22–26 mmol/L, 35–45 mm Hg, respectively) as is the pH.
Which would be consistent with partially compensated respiratory acidosis?
A. pH PCO2 Bicarbonate
increased increased increased
B. pH PCO2 Bicarbonate
increased decreased decreased
C. pH PCO2 Bicarbonate
decreased decreased decreased
D. pH PCO2 Bicarbonate
decreased increased increased
D Acidosis = low pH; respiratory = disturbance of PCO2; a low pH is caused by
increased PCO2. In partially compensated respiratory acidosis, the metabolic
component of the buffer system, bicarbonate, is retained. This helps to compensate for
retention of PCO2 by titrating hydrogen ions. The compensatory component always
moves in the same direction as the cause of the acid–base disturbance.
In which circumstance will the reporting of calculated O2 saturation of Hgb based on
PO2, PCO2, pH, temperature, and Hgb be in error?
A. Carbon monoxide (CO) poisoning
B. Diabetic ketoacidosis
C. Oxygen therapy
D. Assisted ventilation for respiratory failure
A CO has about 200 times the affinity as O2 for Hgb and will displace O2 from Hgb at
concentrations that have no significant effect on the PAO2. Consequently, calculated O2
saturation will be erroneously high. Other cases in which the calculated O2Sat should
not be used include any hemoglobinopathy that affects O2 affinity and
methemoglobinemia. The other situations above affect the O2 saturation of Hgb in a
manner that can be predicted by the effect of pH, PO2, and PCO2 on the oxyhemoglobin
dissociation curve.
Which condition results in metabolic acidosis with severe hypokalemia and chronic
alkaline urine?
A. Diabetic ketoacidosis
B. Phenformin-induced acidosis
C. Renal tubular acidosis
D. Acidosis caused by starvation
C Metabolic acidosis can be caused by any condition that lowers bicarbonate. In
nonrenal causes, the kidneys will attempt to compensate by increased acid excretion.
However, in renal tubular acidosis (RTA), an intrinsic defect in the tubules prevents
bicarbonate reabsorption. This causes alkaline instead of acidic urine. Excretion of
bicarbonate as potassium bicarbonate (KHCO3) results in severe hypokalemia.
Which of the following mechanisms is responsible for metabolic acidosis?
A. Bicarbonate deficiency
B. Excessive retention of dissolved CO2
C. Accumulation of volatile acids
D. Hyperaldosteronism
A Metabolic acidosis is caused by bicarbonate deficiency and metabolic alkalosis by
bicarbonate excess. Respiratory acidosis is caused by PCO2 retention (defective
ventilation), and respiratory alkalosis is caused by PCO2 loss (hyperventilation).
Important causes of metabolic acidosis include renal failure, diabetic ketoacidosis,
lactate acidosis, and diarrhea.
Which of the following disorders is associated with lactate acidosis?
A. Diarrhea
B. Renal tubular acidosis
C. Hypoaldosteronism
D. Alcoholism
D Lactate acidosis often results from hypoxia, which causes a deficit of NAD+. This
promotes the reduction of pyruvate to lactate, regenerating NAD+ needed for
glycolysis. In alcoholic acidosis, oxidation of ethanol to acetaldehyde consumes the
NAD+. In diabetes, lactate acidosis can result from depletion of Krebs cycle
intermediates. Diarrhea and renal tubular acidosis result in metabolic acidosis via
bicarbonate loss. Hypoaldosteronism causes metabolic acidosis via H+ and potassium
ion (K+) retention.
Which of the following is the primary mechanism of compensation for metabolic
acidosis?
A. Hyperventilation
B. Release of epinephrine
C. Aldosterone release
D. Bicarbonate excretion
A In metabolic acidosis, the respiratory center is stimulated by chemoreceptors in the
carotid sinus, causing hyperventilation. This results in increased release of CO2.
Respiratory compensation begins almost immediately unless blocked by pulmonary
disease or respiratory therapy. Hyperventilation can bring the PCO2 down to
approximately 10 to 15 mm Hg.
The following conditions are all causes of alkalosis. Which condition is associated with
respiratory (rather than metabolic) alkalosis?
A. Anxiety
B. Hypovolemia
C. Hyperaldosteronism
D. Hypoparathyroidism
A Respiratory alkalosis is caused by hyperventilation, which leads to decreased PCO2.
Anxiety and drugs that stimulate the respiratory center, such as epinephrine, are
common causes of respiratory alkalosis. Excess aldosterone increases net acid
excretion by the kidneys. Low levels of parathyroid hormone (PTH) cause increased
bicarbonate reabsorption, resulting in alkalosis. Hypovolemia increases the relative
concentration of bicarbonate. This is common and is called dehydrational alkalosis,
chloride responsive alkalosis, or alkalosis of sodium deficit.
Which of the following conditions is associated with both metabolic and respiratory
alkalosis?
A. Hyperchloremia
B. Hypernatremia
C. Hyperphosphatemia
D. Hypokalemia
D Hypokalemia is both a cause and result of alkalosis. In alkalosis, hydrogen ions may
move from the cells into the extracellular fluid and potassium into the cells. In
hypokalemia caused by overproduction of aldosterone, hydrogen ions are secreted by
the renal tubules. This increase in net acid excretion results in metabolic alkalosis.
In uncompensated metabolic acidosis, which of the following will be normal?
A. Plasma bicarbonate
B. PCO2
C. p50
D. Total CO2
B The normal compensatory mechanism for metabolic acidosis is respiratory
hyperventilation. In uncompensated cases, PCO2 is not reduced, indicating a
concomitant problem in respiratory control.
Which of the following conditions is classified as normochloremic acidosis?
A. Diabetic ketoacidosis
B. Chronic pulmonary obstruction
C. Uremic acidosis
D. Diarrhea
A Bicarbonate deficit will lead to hyperchloremia unless the bicarbonate is replaced by
an unmeasured anion. In diabetic ketoacidosis, acetoacetate and other ketoacids replace
bicarbonate. The chloride remains normal or low, and there is an increased anion gap.
Which PCO2 value would be seen in maximally compensated metabolic acidosis?
A. 15 mm Hg
B. 30 mm Hg
C. 40 mm Hg
D. 60 mm Hg
A In metabolic acidosis, hyperventilation increases the ratio of bicarbonate to dissolved
CO2. The extent of compensation is limited by the rate of both gas diffusion and
diaphragm contraction. The lower limit is between 10 and 15 mm Hg PCO2, which is
the maximum compensatory effect.
A patient has the following arterial blood gas results:
pH = 7.56 PCO2 = 25 mm Hg
PO2 = 100 mm Hg HCO3– = 22 mmol/L
These results are most likely the result of which condition?
A. Improper specimen collection
B. Prolonged storage
C. Hyperventilation
D. Hypokalemia
C The pH is alkaline (reference range 7.35–7.45) and this can be caused by either low
PCO2 or increased bicarbonate. This patient has a normal bicarbonate (reference range
22–26 mmol/L) and a low PCO2 (reference range 35–45 mm Hg). Low PCO2 is always
caused by hyperventilation, and therefore, this is a case of uncompensated respiratory
alkalosis. The acute stages of respiratory disorders are often uncompensated.
Prolonged storage would cause the pH and PO2 to fall, and the PCO2 to rise.
Hypokalemia causes alkalosis, but usually is associated with the retention of CO2 as
compensation.
Why are three levels used for quality control of pH and blood gases?
A. Systematic errors can be detected earlier than with two controls
B. Analytical accuracy needs to be greater than for other analytes
C. High, normal, and low ranges must always be evaluated
D. A different level is needed for pH, PCO2, and PO2
A Error detection occurs sooner when more controls are used. Some errors, such as
those resulting from temperature error and protein coating of electrodes, are not as
pronounced near the calibration point as in the acidosis and alkalosis range. The
minimum requirement for blood gas quality control (QC) is one sample every 8 hours
and at three levels (acidosis, normal, alkalosis) every 24 hours. Three levels of control
are also used commonly for therapeutic drug monitoring and hormone assays because
precision differs significantly in the high and low ranges.
A single-point calibration is performed between each blood gas sample to:
A. Correct the electrode slope
B. Correct electrode and instrument drift
C. Compensate for temperature variance
D. Prevent contamination by the previous sample
B Calibration using a single standard corrects the instrument for error at the labeled
value of the calibrator but does not correct for analytic errors away from the set point.
A two-point calibration adjusts the slope response of the electrode, eliminating
proportional error caused by poor electrode performance.
In which condition would hypochloremia be expected?
A. Respiratory alkalosis
B. Metabolic acidosis
C. Metabolic alkalosis
D. All of these options
C Chloride is the major extracellular anion and is retained or lost to preserve
electroneutrality. Low chloride will occur in metabolic alkalosis because excess
bicarbonate is present. Low chloride also will occur in partially compensated
respiratory acidosis because the kidneys compensate by increased retention of
bicarbonate.
Given the following serum electrolyte data, determine the anion gap:
Na = 132 mmol/L K = 4.0 mmol/L
Cl = 90 mmol/L HCO3– = 22 mmol/L
A. 12 mmol/L
B. 24 mmol/L
C. 64 mmol/L
D. Cannot be determined from the information provided
B The anion gap is defined as unmeasured anions minus unmeasured cations in the
plasma or serum. It is calculated by subtracting the measured anions (bicarbonate and
chloride) from the measured cations (sodium plus potassium although some
laboratories ignore the potassium). A normal anion gap is approximately 12 to 20
mmol/L (8–16 mmol/L when potassium is not included).
Anion gap = (Na + K) – (HCO3 + Cl)
Anion gap = (132 + 4) – (90 + 22) = 24 mmol/L
