LabCE BOC 100Q Adaptive Test #2

The absorbance of a 40.0 mg/dL blood urea nitrogen (BUN) standard is 0.758. The absorbance of the patient's serum specimen is 0.220. What is the patient's serum BUN concentration (to the nearest tenths)? a) 11.0 b) 11.2 c) 11.6 d) 12.0

c) 11.6 Using Beer's law, the standard absorbance divided by the standard concentration will equal the unknown absorbance divided by the unknown concentration. The unknown could be a control or patient sample. By manipulating the formula, we have concentration of unknown = (concentration of standard X absorbance of unknown)/absorbance of standard. By plugging in our values, the concentration of the patient's BUN serum value = (40.0 mg/dL X 0.220)/ 0.758 = 11.6 mg/dL.

Which type of lipoprotein is the least dense? a) Chylomicrons b) VLDL c) LDL d) HDL

a) Chylomicrons The various lipoprotein particles were originally separated by ultracentrifugation into different density fractions. The least dense were the chylomicrons, followed by the VLDL (very-low-density lipoproteins) and LDL (low-density-lipoproteins) and the last was the HDL (high-density lipoproteins). This separation still is used as the basis for the most commonly used lipoprotein classification system.

Given the following information, calculate the results in mg/24 hrs for a 24-hour urine protein.Total volume for 24 hours = 2,400 mL Urine protein = 2.7 mg/dL a) 64.8 mg/24 hrs b) 10.87 mg/24 hrs c) 57.5 mg/24 hrs d) 5.89 mg/24 hrs

a) 64.8 mg/24 hrs 2.7 mg/dL X 2400 mL/24 hr X 1 dL/100 mL = 2.7 mg X 2400 100 = 64.8 mg/24 hr

From the IgG molecule illustration, which region is the heavy chain? a) A b) B c) C d) D

a) A A = Heavy chain B = Light chain C = Antigen binding site D = Variable region

What is the nuclear appearance of an Endolimax nana trophozoite? a) Blot-like karyosome, no peripheral chromatin b) Central karyosome, even peripheral chromatin c) Large karyosome, no peripheral chromatin d) Eccentric karyosome, uneven peripheral chromatin

a) Blot-like karyosome, no peripheral chromatin The nuclear appearance of Endolimax nana trophozoite has a blot-like karyosome but lacks peripheral chromatin. As examples, an Entamoeba hartmanii cyst has a central karyosome with even peripheral chromatin; an Endolimax nana cyst possesses a large karyosome with no peripheral chromatin and Entamoeba coli trophozoites, is a parasite with eccentric karyosome and uneven peripheral chromatin.

Which of the following Mycobacterium species is positive for nitrate reduction? a) Mycobacterium tuberculosis b) Mycobacterium bovis c) Mycobacterium simiae d) Mycobacterium ulcerans

a) Mycobacterium tuberculosis Out of the choices, Mycobacterium tuberculosisis the only one that will reduce nitrates. Mycobacterium bovis, <Mycobacterium simiae and Mycobacterium ulcerans will not reduce nitrates.

All of the terms below are a measure of central tendency, EXCEPT: a) Mean b) Molarity c) Median d) Mode

b) Molarity Molarity is the numbers of moles of a solute per liter of solution. This is not a measure of central tendency. Mean is the average of a group of numbers. Median is the central number in a set of numbers when the numbers are arranged in sequential order. Mode is the number that occurs most frequently in a group of numbers.

Which of the following tests would you employ to examine a CSF for syphilis: a) RPR b) VDRL c) Both d) Neither

b) VDRL The VDRL is the only non-treponemal test approved by the CDC for CSF. The CSF-VDRL test is done to diagnose syphilis in the brain or spinal cord, which is typically a sign of late-stage syphilis. Serum screening tests (VDRL and RPR) are better at detecting early/middle-stage syphilis.

Which type of electrophoresis is used to separate larger DNA fragments (>50 kb)? a) Polyacrylamide gel electrophoresis b) Capillary electrophoresis c) Isoelectric focusing d) Pulsed field electrophoresis

d) Pulsed field electrophoresis Fragments of DNA > 50 kb will not separate in most types of electrophoresis due to limited mobility from their large size. In pulsed-field electrophoresis, pulses of electricity are applied in alternating dimensions to better allow for migration. Polyacrylamide gel electrophoresis, or PAGE, is used for small DNA fragments or single-stranded DNA samples. Capillary electrophoresis typically serves as an alternative testing method to high-performance liquid chromatography (HPLC) and is used to separate organic chemicals and inorganic ions. Isoelectric focusing separates proteins based on their isoelectric point, or the pH at which the protein has no charge and does not move.

When a Specked Nuclear Antibody (ANA) pattern is observed, what follow-up test for antibodies related to Systemic Lupus Erythematosus (SLE) is not beneficial? a) Testing for antibodies to Smith (Sm) b) Testing for all antibodies to the extractable nuclear antigens (ENA) c) Testing for antibodies to SS-A/SS-B+ d) Testing for antibodies to U1RNP+ and dcSSc

d) Testing for antibodies to U1RNP+ and dcSSc U1RNP+ and Scl-70 antibodies are extractable nuclear antigens that yield a speckled ANA pattern, but they are not associated with Systemic Lupus Erythematosus (SLE). U1RNP+ is associated with mixed connective tissue disease/undifferentiated connective tissue disease; Scl-70+ is associated with Diffuse cutaneous scleroderma. If other criteria suggest that the patient is suffering from Systemic Lupus Erythematosus (SLE), the presence of Smith (Sm) and SS-A/SS-B+ are the most important antibodies to identify. Speckled ANA patterns can be followed up with testing for antibodies to the extractable nuclear antigens (ENA), which include: Sm (Smith), U1-RNP, SSA/ SSB, and Scl-70. If other criteria suggest that the patient is suffering from Systemic Lupus Erythematosus (SLE), the presence of Smith (Sm) and SS-A/SS-B+ are the most important antibodies to identify.

Which of the following reagents is often used to perform a direct antiglobulin test (DAT) to determine if either IgG or complement (or both) is attached to a patient's red cells? a) Low-ionic strength solution (LISS) b) Polyethylene glycol (PEG) c) Anti-C3d d) Polyspecific AHG

d) Polyspecific AHG Polyspecifc AHG can detect both IgG and complement molecules attached to patient red cells. LISS and PEG are potentiators used in indirect antiglobulin testing procedures. Anti-C3d is most often used after a DAT with polyspecific AHG is positive. A differential DAT is performed by testing patient cells against anti-IgG in one test tube and anti-C3d in another test tube to see which specific proteins are attached to the patient's red cells.

Regarding hemoglobin synthesis, which of the following constitutes the alpha globin chain coding? a) Two gene loci; one each on chromosome 11. b) Four gene loci; two each on chromosome 11. c) Two gene loci; one each on chromosome 16. d) Four gene loci; two each on chromosome 16.

d) Four gene loci; two each on chromosome 16. Humans need four gene loci that code for the alpha hemoglobin chain in order to make their full component of normal hemoglobins. Two alpha chain gene loci are found on each chromosome 16. Beta globin gene (and other beta-like genes) are coded on chromosome 11.

A hepatic function panel A consists of which of the following tests? a) AST, ALT, Alkaline Phosphatase, Total Protein, Albumin, Total Bilirubin, Direct Bilirubin b) Glucose, BUN, Creatinine, Sodium, Potassium, Chloride, CO2, Calcium c) Cholesterol, Lipoprotein, HDL, Triglycerides d) Potassium, Sodium, Chloride, CO2

a) AST, ALT, Alkaline Phosphatase, Total Protein, Albumin, Total Bilirubin, Direct Bilirubin The Hepatic Function Panel A consists of AST, ALT, Alkaline Phosphatase, Total Protein, Albumin, Total Bilirubin, Direct Bilirubin. The Basic Metabolic Panel consists of Glucose, BUN, Creatinine, Sodium, Potassium, Chloride, CO2, Calcium. The Lipid Panel consists of Cholesterol, Lipoprotein, HDL, Triglycerides. The Electrolyte Panel consists of potassium, sodium, chloride, carbon dioxide. These analytes are also typically ordered within other panels as well, including the basic metabolic and complete metabolic panels.

What is the cause of iron overload in hereditary hemochromatosis? a) Absorption of excessive amounts of iron in the small intestine b) Ingestion of excessive amounts of iron from diet or supplements c) Inability of the body to excrete normal amounts of dietary iron d) Failure of developing red blood cells to incorporate iron into protoporphyrin IX

a) Absorption of excessive amounts of iron in the small intestine Hereditary hemochromatosis is a genetic disorder typically involving a deficiency of hepcidin due to a mutation in the hepcidin gene or genes whose products regulate the expression of hepcidin. These mutations cause increased iron absorption in the small intestine, leading to iron overload. Ingestion of excessive amounts of dietary iron results in secondary hemochromatosis. Iron is recycled when cells die; the body does not have a mechanism for iron excretion. Failure of developing red blood cells to incorporate iron into protoporphyrin IX results in sideroblastic anemia.

Which of the following is the proper designation for the pluripotential stem cell that is a precursor for both myeloid and lymphoid cell lines? a) CFU-S b) CFU-GEMM c) G-CSF d) CFU-GM

a) CFU-S CFU-S stands for colony-forming unit spleen - it is the pluripotential stem cell that gives rise to all cell lines. CFU-GEMM is a multilineage precursor for granulocyte, erythrocyte, macrophage, and megakaryocyte. G-CSF is the precursor committed to granulocyte cell lines. CFU- GM is the precursor for granulocyte and monocyte cell lines.

Pheochromocytoma is a tumor of the adrenal medulla that results in elevated urinary levels of all the following EXCEPT: a) Cortisol b) Free catecholamines c) Metanephrines d) Vanillylmandelic acid

a) Cortisol Pheochromocytoma is an adrenal or extraadrenal neoplasm that secretes catecholamines. Patients with pheochromocytoma often exhibit persistent and paroxysmal hypertension. The single best screening test is the measurement of urinary metanephrines. Urinary free catecholamines and vanillylmandelic acid are also elevated.

Which laboratory assay is a highly specific indicator and the most sensitive assay for a diagnosis of rheumatoid arthritis? a) Cyclic citrullinated peptide b) Sensitized sheep cells c) Latex particle agglutination d) C-Reactive Protein agglutination

a) Cyclic citrullinated peptide Cyclic citrullinated peptide antibodies are highly specific indicators for rheumatoid arthritis. Antibodies to CCPs (anti-CCP) were first described in 1998. After the introduction of commercial ELISA products using the so-called second generation peptides, there has been increased interest in using this marker in the diagnosis. Anti-CCP antibodies are detectable in about 69-83% of patients with RA and have specificities ranging from 93% to 95%. Compared with other assays for antibodies in RA, CCP is considered to be more sensitive. Antibodies can be detected up to 16 years before the first clinical symptoms of RA appear. Agglutination tests for RF using a sensitized sheep cell test generally detect IgM rheumatoid factors (RFs). IgM RF is manifested in approximately 70% of adults but is not specific for RA. RF has been associated with some bacterial and viral infections, and some chronic infections, and cancer. Elevated values may be observed in the normal older adult population. Agglutination tests for RF using latex agglutination generally detect IgM RFs. Although a latex agglutination procedure has a 95% correlation with a clinical diagnosis of probable or definite rheumatoid arthritis, rheumatoid factor (RF) is not exclusively limited to patients with rheumatoid arthritis. In using latex tests for the detection of RF, a positive result can be expected in less than 5% of healthy individuals. In patients 60 years old and older, as many as 30% may be seropositive. C-Reactive Protein (C-RP) is a rapidly synthesized, acute phase reactant. It is an indicator of acute inflammation. Traditionally, C-RP has been used clinically for monitoring infection, autoimmune disorders, and, more recently, healing after myocardial infarction. The C-RP rapid latex agglutination test is based on the reaction between patient serum containing C-RP as the antigen and the corresponding antihuman (C-RP) coated to the surface of latex particles. The latex particles allow for easy visual detection of an antigen-antibody reaction. The clinical applications of C-RP are evaluation, including detecting inflammatory disease, particularly infections, screening for inflammatory and malignant disease, and monitoring therapy in inflammatory disease.

Observe the bone marrow sample in the image to the right. All of the descriptors below apply EXCEPT? a) Myeloid precursors predominant b) Inverted M:E ratio - around 1:3 c) Megakaryocyte present d) Erythroid hyperplasia is present

a) Myeloid precursors predominant Myeloid precursors are NOT predominant on this slide. This is a trilinear bone marrow with an erythroid predominance (erythroid hyperplasia). This causes an inverted M:E ratio; meaning there are more erythroid precursors then myeloid precursors present. The normal M:E ratio is 1.5 to 3.1 myeloid cells per erythroid cell. The large cell in the lower left portion of the image is a megakaryocyte.

For which of these conditions or procedures there may be an increased number of megakaryocytes in the bone marrow, but a decreased number of circulating platelets? a) Folic acid deficiency b) Aplastic anemia c) Radiation Therapy d) Wiskott-Aldrich syndrome

a) Folic acid deficiency Pancytopenia is often seen with megaloblastic anemias that are caused by folic acid or vitamin B12 deficiency. Thrombopoiesis (as well as erythropoiesis and granulopoiesis) is ineffective. The bone marrow will contain normal, or even increased megakaryocytes, but the number of platelets entering the peripheral circulation is decreased. In aplastic anemia, megakaryocytes are decreased in number in the bone marrow, leading to a decreased number of circulating platelets. Radiation therapy causes bone marrow hypoplasia. Platelets as well as all other cell lines are depressed. The effect is transient; once the therapy has ended, the marrow will regenerate. In Wiskott-Aldrich syndrome, platelets are very small and thrombocytopenia is present.

Leuconostoc species are streptococcus-like bacteria used in the dairy and pickling industries that have recently caused opportunistic infections in humans. The need to make the laboratory identification is compounded because these bacteria are intrinsically resistant to vancomycin. Which characteristic is most helpful in separating Leuconostoc species from other streptococcus-like organisms? a) Gas from glucose in MRS broth b) Leucine aminopeptidase (LAP) activity c) Ability to grow in 6.5% NaCl (salt tolerance) d) Ability to grow at 10° C

a) Gas from glucose in MRS broth Leuconostoc species may be misidentified as Enterococcus, alpha-hemolytic streptococci, Pediococcus, Lactococcus, or lactobacilli (Facklam R, Elliott JA: Clin Microbiol Rev; 1995 Oct; 8(4):479-95). The detection of gas from glucose in Mann, Rogosa and Sharpe (MRS) broth (available from Difco Laboratories) overlaid with petrolatum is the one characteristics that differentiates Leuconostoc species from the other streptococcal-like organisms. All of the other characteristics listed in this exercise are either variable for Leuconostoc species or are shared by other streptococcal-like organisms and therefore are not discriminatory tests.

One of the most common forms of hereditary hemochromatosis has been found to result from genetic mutations of the HFE gene. This mutation causes hemochromatosis by which of the following mechanisms? a) Impairs hepcidin regulation of ferroportin activity b) Increased binding of ß2 microglobulin c) Impairs transferrin synthesis d) Increases hepcidin synthesis therefore increasing iron absorption.

a) Impairs hepcidin regulation of ferroportin activity All hereditary hemochromatosis involves mutated proteins that impair Hepcidin regulation of ferroportin activity. Hepcidin activity is diminished, and without Hepcidin, ferroportin in the intestinal enterocyte membrane is continually active and absorbing iron regardless of the body's need leading to iron overload. ß2-microglobulin cannot be bound by the mutated HFE molecule. This mutation has no direct effect on transferrin synthesis Hepcidin is an iron regulatory protein, and an increase would decrease iron absorption

Which fungal culture media is used for the primary recovery of pathogenic fungi exclusive of dermatophytes? a) Inhibitory mold agar b) Potato flake agar c) Mycosel agar d) (SABHI) Sabouraud Dextrose with Brain Heart Infusion agar

a) Inhibitory mold agar Inhibitory mold agar, brain-heart infusion agar with antibiotics, and yeast-extract phosphate agar are all used for the primary recovery of pathogenic fungi exclusive of dermatophytes. Potato flake agar is used for the primary recovery of saprobic and pathogenic fungi. Mycosel agar is used for the primary recovery of dermatophytes. Sabouraud Dextrose with Brain Heart Infusion agar is used for the primary recovery of saprobic and pathogenic fungi.

Nitric oxide is associated with the prevention of vaso-occlusion by decreasing cellular adherence to endothelium. Which amino acid is decreased in patients with sickle cell disease and is needed as a substrate to produce nitric oxide? a) L-arginine b) L-glutamine c) L-lysine d) L-tyrosine

a) L-arginine The amino acid L-arginine is a substrate needed to produce nitric oxide. Low arginine bioavailability in sickle cell disease is associated with elevated levels of nitric oxide synthase inhibitor, among other complications.

Which of the following is the most common condition associated with a nonmegaloblastic macrocytic anemia? a) Liver disease b) Aplastic anemia c) Chronic blood loss d) Folic Acid deficiency

a) Liver disease The most common condition associated with nonmegaloblastic macrocytic anemia is liver disease, including cirrhosis due to alcoholism.While aplastic anemia can appear macrocytic, it is often normocytic.Anemia due to chronic blood loss is typically macrocytic.Folic Acid deficiency can lead to macrocytic anemia but is megaloblastic, not nonmegaloblastic.

Which of the following may be seen in a CSF sample of a patient that has a subarachnoid (intracranial) hemorrhage (SAH)? a) Macrophages containing hemosiderin b) Uneven distribution of blood in CSF tubes collected c) Clot is present in the CSF sample d) Supernatant of CSF sample does not have xanthochromia

a) Macrophages containing hemosiderin Characteristics of a subarachnoid hemorrhage (SAH) include even distribution of blood in tubes collected, no clot formation in the sample, as well as a xanthochromic supernatant to the sample. Another characteristic that correlates with a SAH is hemosiderin deposits in macrophages. As red blood cells degenerate, the breakdown products are seen in macrophages as dark, granular, iron-laden hemosiderin deposits.

How does RhIg prevent anti-D production? a) Mainly by suppressing the immune response after exposure to D positive cells. b) Mainly by clearing antibody sensitized D negative RBCs from maternal circulation. c) Mainly by clearing IgG fetal antibodies from maternal circulation. d) Mainly by clearing maternal IgM antibodies from fetal circulation.

a) Mainly by suppressing the immune response after exposure to D positive cells. RhIg suppresses the immune response after exposure to D positive fetal cells and prevents the mother from producing anti-D. The mechanism is not clearly understood, but it may involve the removal of D-positive cells by macrophages, causing the release of cytokines that suppress the immune system response. Since it is the D positive cells that will stimulate the production of anti-D, the D negative cells are not a concern. Also, it is the fetal cells that cause the sensitization, not fetal immunoglobulins. The production of IgG antibodies is considered significant because they cross the placenta. Therefore, maternal IgM antibodies are not of concern.

These illustrations are of parasites. Which of the following does Illustration A represent? a) Flagellate b) Hemoflagellate c) Sporozoa d) Amoeba

d) Amoeba Amoeba is the correct answer because of the nuclear and cytoplasmic features. The illustration shows a nucleus with a centrally located karyosome with evenly distributed chromatin, which represents a possible Entamoeba histolytica trophozoite.

Of the Enterobacteriaceae listed below, which of the following species fails to decarboxylate lysine, ornithine, and arginine? a) Pantoea agglomerans b) Enterobacter aerogenes c) Yersinia enterocolitica d) Proteus mirabilis

a) Pantoea agglomerans From the members of the Enterobacteriaceae family more commonly encountered from clinical specimens, the inability to decarboxylase the amino acids, lysine, arginine, and ornithine, points to Pantoea (Enterobacter) agglomerans. P. agglomerans is primarily found in the environment and is an uncommon cause of infection. Enterobacter aerogenes decarboxylates lysine and ornithine. E. aerogenes is opportunistic, causing various types of infections such as skin and tissue infection and bacteremia. Yersinia enterocolitica and Proteus mirabilis decarboxylate ornithine. Y. enterocolitica is found in a variety of animals, and human infection most often manifests as acute enteritis in young children. Infection may also mimic appendicitis in older populations. Proteus mirabilis is commonly isolated in the clinical laboratory. It causes a variety of infections, especially urinary tract infections.

What procedure utilizes leukapheresis to collect the buffy coat from whole blood? a) Photopheresis b) Plasmapheresis c) Therapeutic apheresis d) Erythrocytapheresis

a) Photopheresis Photopheresis utilizes leukapheresis to collect the buffy coat layer from whole blood. These cells are treated with 8-methoxypsoralen, exposed to ultraviolet A light and then reinfused into the patient. Photopheresis has been shown to be efficacious and has been approved by the Food and Drug Administration for the treatment of cutaneous T-cell lymphoma. Plasmapheresis is the removal and retention of the plasma, with return of all cellular components to the patient. Therapeutic apheresis (TA) involves the removal of a specific blood component, with return of the remaining blood constituents to the patient. However, with TA the component being removed is considered pathological or contributing to the patient's underlying disease state. Erythrocytapheresis, or red cell exchange, removes a large number of red blood cells from the patient and returns the patient's plasma and platelets along with compatible allogeneic donor red blood cells.

An 85-year-old female nursing home patient was treated empirically with a cephalosporin for a urinary tract infection but later failed therapy. The physician then requested a culture and sensitivity on a new urine specimen that grew >100,000 CFU/mL of Escherichia coli. After reviewing the Kirby-Bauer disc susceptibility plate and susceptibility interpretations, what is the resistance mechanism (if any) for this organism? a) Plasmid-mediated Amp b) CCRE - Carbapenem Resistant Enterobacteriaceae c) ESBL - Extended-Spectrum Beta Lactamase d) No resistance mechanism detected

a) Plasmid-mediated Amp Plasmid-mediated AmpC is the correct answer. Cefoxitin and all the first, second, and third-generation cephalosporins are resistant with the exception of Cefepime. Ertapenem and Imipenem are both susceptible. Aztreonam is intermediate. This resistance pattern is indicative of a plasmid-mediated AmpC. AmpC is a cephalosporinase that hydrolyzes all beta-lactam antibiotics except Cefepime and the carbapenems. AmpCs are also not inhibited by clavulanate.

Which of the following species is an oxidase-positive, glucose-fermenting, Gram-negative bacilli? a) Plesiomonas shigelloides b) Serratia marcescens c) Escherichia coli d) Providencia rettgeri

a) Plesiomonas shigelloides All choices are members of the Enterobacteriaceae family, which are Gram-negative bacilli/coccobacilli that ferment glucose. The only genus in the family that produces cytochrome oxidase is Plesiomonas. Plesiomonas shigelloides is the only species in the genus. It is found in aquatic environments and soil. It most frequently causes self-limiting enteric infections but has also been isolated in extraintestinal infections.

In the Coombs phase of a crossmatch, what is the proper procedure to follow if the Check Cells give a negative reaction? a) Repeat procedure with new AHG reagent and check the cell washer. b) Add additional Check Cells and dilute with 100% distilled water. c) Add additional AHG reagent plus proteolytic enzymes to enhance the reaction. d) Accept the crossmatch results as correct - nothing further needs to be done.

a) Repeat procedure with new AHG reagent and check the cell washer. The Check Cells are added when a negative result is obtained at the Combs phase of testing to ensure that the AHG reagent was added and was not neutralized. The addition of Check Cells following a negative result at the Combs phase should yield a positive result. If the Check Cells are negative, then the procedure must be repeated with a new AHG reagent to ensure that the reagent is working properly. It is also recommended to check the cell washer to ensure that the cell buttons are not being under- or over-washed. Adding additional Check Cells diluted with 100% distilled water would hemolyze the cells in the tube. Proteolytic enzymes are used as an enhancement reagent to the reagent red blood cells for antibody detection and identification; they are not used in compatibility testing. A negative result at the Combs phase of compatibility testing is an invalid result and must be resolved before transfusing the patient.

Important applications of polymerase chain reaction (PCR) include the following with the general exception of: a) Search for mutations with large deletions b) Amplification of DNA c) Identification of a target sequence d) Synthesis of an anti-sense probe

a) Search for mutations with large deletions The search for mutations with large deletions in polymerase chain reaction (PCR) product sequence is not an application of the polymerase chain reaction (PCR)technique. In fact, it is a weakness. Only short sequences can be amplified (generally) Amplification of DNA is an important application of polymerase chain reaction (PCR) technique. PCR is a method that amplifies low levels of specific DNA sequences in a sample to higher levels suitable for further analysis. Identification of a target sequence is an important application of polymerase chain reaction (PCR) technique. Qualitative PCR detects the presence or absence of a specific DNA product. Synthesis of an anti-sense probe is an important application of polymerase chain reaction (PCR) technique. Antisense RNA is a single-stranded RNA that is complementary to a messenger RNA (mRNA) strand transcribed within a cell.

This drawing depicts which alpha chain genotype? a) Alpha thalassemia major (Hydrops fetalis) b) Alpha thalassemia intermedia (Hemoglobin H Disease) c) Alpha thalassemia trait d) Silent carrier

b) Alpha thalassemia intermedia (Hemoglobin H Disease) The genotype pictured, --/-a, is known as Alpha thalassemia intermedia or Hemoglobin H disease. That is, three of the four alpha genes are deleted. These patients tend to form tetrads of beta-globin chains, known as Hemoglobin H. Alpha thalassemia major or Hydrops fetalis occurs when all four alpha genes are deleted. These patients produce Hemoglobin Bart which is the formation of gamma globin tetrads. This condition is incompatible with life. Alpha thalassemia trait is the deletion of two of the four alpha genes. The silent carrier has one alpha gene deleted. They tend to have no clinical manifestation of a hemoglobinopathy.

Which antinuclear antibody (ANA) pattern is seen in the image on the right, which represents the result of an ANA test viewed using fluorescent microscopy? a) There is no discernable pattern b) Homogeneous c) Centromere d) Nucleolar

a) There is no discernable pattern In order for the ANA to be positive, there must be a clearly discernible pattern in the nuclei of the interphase cells. Observing the chromosomal area and cytoplasm of the metaphase cells may assist in the identification of the ANA pattern. The cells on this slide demonstrate weak staining in the nuclei of the interphase (nondividing) cells but no clear pattern. In addition, there is an absence of fluorescence in the chromatin region (center of the nucleus) in the metaphase (dividing) cells. In contrast, a homogeneous pattern would show uniform staining of the nuclei in the interphase cells and the presence of staining in the chromatin of the dividing cells. The centromere pattern is characterized by many discrete speckles in both the nuclei of interphase cells and the chromatin of dividing cells. In the nucleolar pattern, there is prominent staining of the nucleoli in the nuclei of the interphase cells and variable staining in the chromatin of the dividing cells.

Aspirin ingestion prevents the synthesis of this signaling molecule in the platelet? a) Thromboxane A2 b) Calcium c) Collagen d) ADP

a) Thromboxane A2 Thromboxane A2 is required to stimulate the activation of new platelets and aids in platelet aggregation. Aspirin ingestion inhibits the COX-1 enzyme that prevents the arachidonic pathway from synthesizing thromboxane A2. Aspirin has no effect on the synthesis of calcium. Aspirin has no effect on the synthesis of collagen. Aspirin has no effect on the synthesis of ADP.

Based on the phenotype of the RBC screening cells, and patient results given below, Anti-Jka and what additional antibody cannot be ruled out? a) Anti-Jkb b) Anti-K c) Anti-Leb d) Anti-M

b) Anti-K Anti-K matches the pattern of reactivity in the temperature at which the reaction occurs and in the screening cell in which it occurs. Anti-Jkb does not match the pattern of reactivity because Jkb is only present in Screen Cell I and not Screen Cell II, whereas the reaction only occurs in Screen Cell II. Anti-Leb does not match the pattern of reactivity since Leb is present in Screen Cells I and II, and Anti-Leb reacts at the immediate spin phase. Anti-M does not match the pattern of reactivity since M is present in Screen Cells I and II, and Anti-M reacts at the immediate spin phase.

All of the following are Lancefield carbohydrate antigens found on beta-hemolytic Streptococcus species. All of the following antigens are found on organisms from the Streptococcus anginosus group EXCEPT? a) A b) B c) F d) G

b) B Group B is the correct answer because the group B Lancefield carbohydrate antigen is found on Streptococcus agalactiae. Group A is the incorrect answer because this Lancefield carbohydrate antigen is commonly found on Streptococcus pyogenes; however, it can also be found on Streptococcus anginosus, which is found in the Streptococcus anginosus group. Streptococcus anginosus can be separated from Streptococcus pyogenes by its small colony size on blood agar and the detection of Group A, C, F, and G carbohydrate group antigens. Group F is the incorrect answer, but this carbohydrate antigen is the most common Lancefield antigen detected in the Streptococcus anginosus group. Streptococcus anginosus and Streptococcus constellatus can both have Group F antigens. Group G is incorrect because it is found on organisms in the Streptococcus anginosus group, specifically Streptococcus anginosus and Streptococcus dysgalactiae. Streptococcus dysgalactiae has a large colony size whereas Streptococcus anginosus has a small colony size.

Which of the following substances can cause a false positive result for blood on the urine chemical reagent strip? a) Ascorbic acid b) Bleach in the collecting bottle c) High levels of nitrite d) High specific gravity

b) Bleach in the collecting bottle The presence of bleach in the urine collection container can lead to a false positive blood result on the chemical reagent strip. Bleach can trigger the color change normally associated with the presence of red blood cells or hemoglobin. It is important to follow up with a microscopic analysis for confirmation of a true positive result. The presence of ascorbic acid, high levels of nitrites, or high specific gravity will not cause false-positive blood results on the chemical reagent strip.

Which of the following phenotypes is most indicative of a natural killer (NK) cell? a) CD2+ CD3+ CD5+ CD7+ b) CD2+ CD3- CD11b+ CD16+ c) CD11b+ CD16+ CD33+ CD56- d) CD19+ CD20+ CD22+ CD57-

b) CD2+ CD3- CD11b+ CD16+ NK cells express CD2, CD11b, and CD16; NK cells do not express CD3, a T cell marker. Other NK cell markers include CD7, CD56, and CD57. CD2, CD3, CD5, and CD7 are all mature T cell markers. CD2 and CD7 are also NK cell markers; however, NK cells are CD3- and CD5-. Although NK cells express CD11b and CD16, CD33 is expressed by monocytes and other myeloid cells. NK cells are CD56+. CD19, CD20, and CD22 are all B cell markers. NK cells are CD57+.

This species of Candida accounts for around 20 percent of urinary yeast isolates. a) Candida albicans b) Candida glabrata c) Candida krusei d) Candida tropicalis

b) Candida glabrata Candida glabrata is the second most common Candida species to cause disease (Candida albicans is the most common). In urinary yeast isolates, it is identified about 20 percent of the time. Infections tend to be aggressive and difficult to treat with traditional antifungal therapy. It has different sugar assimilation patterns than Candida albicans, notably rapid assimilation of trehalose, so it can easily be differentiated. Candida krusei and Candida tropicalis are seen in immunocompromised patients and cause nosocomial infections.

A fusiform-shaped, Gram-negative, oxidase-negative bacillus that produces colonies with marginal finger-like projections was recovered from the oral cavity of a patient with periodontal disease. The most likely identification is: a) Capnocytophaga canimorsus b) Capnocytophaga ochracea c) Actinobacillus actinomycetemcomitans d) Eikenella corrodens

b) Capnocytophaga ochracea Capnocytophaga ochracea, a commensal organism within the oral cavity of humans, may produce a variety of aminopeptidases, neuraminidases, and products that have direct toxic effects on neutrophils resulting in degradation of subgingival and periodontal tissue leading to periodontal disease. The fingerlike projections emanating from the colonies are known as "gliding motility", an additional factor that permits the organism to penetrate deeply into gingival fissures. Capnocytophaga canimorsus has a similar appearance to C. ochracea to culture; however, is indigenous in the oral cavity of dogs and causes human dog bite infections but not an oral disease. Actinobacillus actinomycetemcomitans can also cause periodontal disease; however, the bacterial cells are not fusiform and the colonies do not produce gliding motility. Eikenella corrodens also inhabits the oral cavity of humans but has none of the characteristics for C. ochracea as previously described.

The properties of enzymes are CORRECTLY described by which of the following statements? a) Enzyme activity is not altered by heat denaturation. b) Enzymes are protein catalysts of biological origin. c) Enzymes are stable proteins and unaffected by pH changes. d) Enzymes affect the rate of a chemical reaction by raising the activation energy needed.

b) Enzymes are protein catalysts of biological origin. The only correct choice is choice 2, as enzymes are protein catalysts of biological origin. Enzymes are altered by heat denaturation, are affected by pH changes, and affect the rate of a chemical reaction by decreasing the activation energy needed.

All of the following represent a 2 week temporary deferral from donating blood or blood products EXCEPT: a) Measles (rubeola) vaccine b) German measles (rubella) vaccine c) Mumps vaccine d) Polio vaccine

b) German measles (rubella) vaccine A potential blood donor who has received a German measles (rubella) vaccine is deferred for 4 weeks following the vaccination.Measles (rubella), mumps, and polio vaccines all have a 2 week deferral period following the vaccination.

All of the following are quantitative methods for the determination of albumin EXCEPT? a) Serum protein electrophoresis b) Nephelometry c) Sulfosalicyclic acid precipitation d) Colorimetric spectrophotometry

c) Sulfosalicyclic acid precipitation The sulfosalicylic acid method is a semiquantitative method and not a quantitative method. It is also not specific for albumin. All of the others are quantitative methods for the determination of albumin.

A 7-year-old male presented to the clinic with mild, abdominal pain, diarrhea, and nausea. Patient history revealed that the child lives in a rat and insect infected shack where sanitation practices were poor. A stool sample was collected and submitted for an ova and parasites examination. Numerous suspicious forms were seen in the stool concentrate that measured 60 µm by 75 µm. This child is MOST likely infected with: a) Hymenolepis nana b) Hymenolepis diminuta c) Echinococcus granulosus d) Dipylidium caninum

b) Hymenolepis diminuta The correct answer is Hymenolepis diminuta. Hymenolepis diminuta lacks polar filaments but is otherwise indistinguishable from Hymenolepis nana. Light infections with either species are generally asymptomatic, but large numbers of adult worms can cause abdominal pain, diarrhea, irritability, and headache.

Amoebae stained with this substance may be readily distinguished because it enhances nuclear and structural detail. a) Saline b) Iodine c) Polyvinyl alcohol d) Methylene blue

b) Iodine Iodine is correct. Iodine can be used alone or used in combination with formaldehyde. Iodine can be used to identify Iodamoeba butschlii cyst by being absorbed into the glycogen vacuole. The glycogen vacuole will stain dark brown in color. At times, the stained glycogen vacuole is the only structure that is seen.

All of the following are causes of hypernatremia EXCEPT: a) Excess water loss b) Low aldosterone production c) Decreased water intake d) Increased sodium intake or retention

b) Low aldosterone production Excess water loss, decreased water intake, and increased sodium intake and retention are all causes of high levels of sodium (hypernatremia). Hypernatremia is seen in diabetes insipidus due to excess water loss and in hyperaldosteronism (increased retention of sodium).Aldosterone is a hormone produced by the adrenal cortex. Aldosterone stimulates sodium reabsorption in the distal tubules of the nephron. Low levels of aldosterone would not promote increased levels of sodium.

The presence of an increased number of hypersegmented neutrophils in the peripheral blood, as shown in this image, is an indication of which of the following conditions? a) Preleukemia b) Megaloblastic anemia c) Aplastic anemia d) Myeloproliferative disorder

b) Megaloblastic anemia Hypersegmentation of neutrophils (see image) provides strong morphologic evidence for megaloblastic hematopoiesis. When such cells with six or more lobes are present in the peripheral blood, assays for vitamin B-12 and folate are indicated. These hypersegmented neutrophils are also usually macrocytic as the disruption in normal maturation has resulted in fewer nuclear divisions. In the bone marrow, finding greater than 5% neutrophils with five lobes or greater, or more than 50% with four lobes or greater is evidence that megaloblastic myelopoiesis is to be ruled out. Preleukemia and myeloproliferative disorders are not characterized by hypersegmented neutrophils. Aplastic anemia is characterized by pancytopenia.

You are collecting a blood specimen to be used for forensic (legal) alcohol testing. Which of the following must be done before you can start the specimen collection process? a) Label the tubes in the presence of the patient. b) Must inform the patient that the blood about to be collected is for alcohol testing. c) Seal the collection tubes in front of the patient. d) Collect the specimen, the patient does not need to be informed what the blood is collected for.

b) Must inform the patient that the blood about to be collected is for alcohol testing. Before starting the blood collection process, the patient must be informed that the specimen collected will be tested for blood alcohol. Labeling the blood tubes and sealing them in front of the patient is done after collection is complete.

Which of the following may be found in the cerebral spinal fluid (CSF) and measure approximately 15 µm? a) Acanthamoeba species cyst b) Naegleria fowleri trophozoite c) Toxoplasma gondii bradyzoite d) Entamoeba gingivalis trophozoite

b) Naegleria fowleri trophozoite Naegleria fowleri trophozoite is the correct answer. Naegleria fowleri trophozoites measure 10 - 20 µm and are distinguished from the other atrial protozoa by the presence of a single nucleus lacking peripheral chromatin and blunt pseudopodia (as seen in the image). Naegleria fowleri is found in brackish or fresh waters and can enter through the nasal passage while swimming in this type of water. Once the amoeba invades the nasal mucosa, the amoeba migrates to the olfactory bulb, moves to the olfactory nerve, and invades brain tissue.

This egg measuring 112 µm by 55 µm, was recovered from a sputum sample. What is the correct identification of this organism? a) Schistosoma mansoni egg b) Paragonimus westermani egg c) Pneumocystis jiroveci cyst d) Diphyllobothrium latum egg

b) Paragonimus westermani egg Paragonimus westermani egg is the correct answer because out of all the parasites listed, this species is the only fluke egg known to be found in human sputum. Diagnosis is made by the characteristic size (80-120 µm x 45 - 60 µm), shape (oval), color, (yellow-brown), and the presence of an operculum.

The antiphagocytic capsule of Streptococcus pneumoniae is composed of: a) Proteins b) Polysaccharides c) Lipids d) Lipoproteins

b) Polysaccharides The capsular material of Streptococcus pneumonia is C polysaccharide. This material is antiphagocytic and increases the morbidity and mortality of this organism. Proteins, lipids, and lipoproteins are not major components of the Streptococcus pneumoniae capsule.

Which finding best distinguishes immune hemolytic anemia from other hemolytic anemias? a) Rouleaux b) Positive DAT c) Splenomegaly d) Increased erythrocyte count

b) Positive DAT In the group of disorders referred to as immune hemolytic anemias, erythrocytes are destroyed too early by an immune-mediated process that results from antibodies, complement, or both attaching to the red cell membrane. The presence of immune hemolytic anemia is confirmed by a positive DAT (direct antiglobulin test). Rouleaux is the formation of red cells that are stacked and appear like a stack of coins. This is a characteristic finding in multiple myeloma. Splenomegaly, or an enlarged spleen, may be found in Gaucher's disease or in polycythemia vera. It is not found in immune hemolytic anemia. Increased erythrocyte count is not a finding in immune hemolytic anemia.

Which of the following tests require a 72-hour stool (fecal) collection? a) Occult blood b) Quantitative fecal fat c) Ova and Parasite (O&P) d) Stool culture

b) Quantitative fecal fat Quantitative fecal fat testing requires a 72-hour stool collection. The specimen is collected in a special large container provided by the testing facility and must be refrigerated during the entire 72 hour collection period. The specimen is tested in the Chemistry department as a confirmatory test for a positive qualitative fecal fat test. Occult blood, O & P, and stool culture testing can be performed on a random stool specimen.

All of the following should be done when centrifuging specimens, EXCEPT? a) Specimens without anticoagulant should sit for 30 minutes before centrifugation. b) Specimens with anticoagulant should sit for 20 minutes before centrifugation. c) Cap all tubes for centrifugation. d) Allow the centrifuge to come to a complete stop by itself before opening it.

b) Specimens with anticoagulant should sit for 20 minutes before centrifugation. Specimens with anticoagulants can be spun immediately. There is no reason to delay. Specimens without anticoagulants should sit at least 30 minutes prior to centrifugation to allow for complete clotting of the blood. Proper centrifugation procedure includes making sure all tubes are capped during centrifugation and letting the centrifuge stop completely by itself before opening it.

This antinuclear antibody (ANA) pattern is characterized by granular staining in the nuclei of the interphase cells (a). There is also an absence of staining in the chromosomal area of the metaphase mitotic cells (b). The slide is viewed using fluorescent microscopy. Which pattern is this? a) Homogeneous b) Speckled c) Nucleolar d) Centromere

b) Speckled In order for the ANA to be positive there must be a clearly discernible pattern in the nuclei of the interphase cells. Metaphase mitotic cells are used to assist in identification of the ANA pattern. The image shown displays the features of a speckled ANA pattern: granular/speckled staining in the nuclei of the interphase cells (a) and absence of staining in the chromosomal area of the metaphase mitotic cells (b). The nucleoli do not stain; thus, the pattern is not nucleolar. A homogeneous pattern would show uniform staining of the nuclei in the interphase cells and presence of staining in the chromatin of the dividing cells. The centromere pattern is characterized by many discrete speckles in both the nuclei of interphase cells and the chromatin of dividing cells.

All of the following carbohydrates are considered reducing sugars EXCEPT: a) Lactose b) Sucrose c) Glucose d) Ribose

b) Sucrose Sucrose is not a reducing sugar since the anomeric carbon of both monosaccharides (glucose and fructose) are part of the glycosidic bond, preventing the anomeric carbon of fructose (and therefore its ketone group) from being free and reducing other compounds. Lactose, Glucose and Ribose are all reducing sugars since their anomeric carbons and aldehyde groups are all free to reduce other compounds.

A mother's serologic results are shown above. Her newborn types as group A Rh positive with a (1+) positive direct antiglobulin test (DAT). Which of the following investigative tests would be most useful to resolve the cause of the positive DAT and should be done FIRST? a) Test an eluate prepared from newborn's red cells against an antibody identification panel by IAT. b) Test newborn's plasma against group A1 red cells and group O antibody screen cells by IAT. c) Test newborn's plasma against mother's red cells by IAT. d) Test newborn's plasma against father's red cells by IAT.

b) Test newborn's plasma against group A1 red cells and group O antibody screen cells by IAT. The mother is group O Rh positive with a negative antibody screen and the infant is group A Rh positive. Results are consistent with a possible case of ABO HDFN. ABO HDFN is nearly always limited to A or B infants of group O mothers with potent anti-A,B. The most useful follow-up would be to test the infant's plasma against A1 red cells and group O antibody screen cells (as a control), expecting only the A1 cells to be positive (due to the mother's anti-A,B because she is group O) if the DAT was due to ABO incompatibility.

Which of the following conditions would argue in favor of therapeutic drug monitoring (TDM) for a given drug? a) The drug has a low degree of protein binding. b) The drug is given chronically. c) The drug has low toxicity and few side effects. d) The effective and toxic concentrations are not well defined.

b) The drug is given chronically. Drugs that are given chronically are usually monitored by TDM. Drugs that are highly protein-bound are good candidates for TDM since changes in plasma binding proteins can affect drug levels. A drug that has low toxicity and rare side effects will probably not need TDM. If the effective concentration and toxic concentrations are not well-defined, TDM will not be useful because there is no reference for comparison.

Which cardiac biomarker is a regulator of myocyte contraction? a) Myoglobin b) cTnT c) CK-MB d) CK-MB isoforms

b) cTnT Troponin is a complex of three proteins, TnT, TnI, and TnC, that help regulate striated muscle contraction. TnT and TnI have cardiac-specific isoforms, cTnT and cTnI respectively, which are involved specifically in myocyte contraction. Because of their cardiac specificity, quick release during myocardial infarction (MI), and prolonged elevation in the serum (more than one week), they are great biomarkers for MI.

What is the lower reference limit for sperm concentration according to the World Health Organization, WHO Laboratory Manual for the Examination and Processing of Human Semen, WHO press, Geneva, Switzerland, 2010? a) 5 x 106 spermatozoa/mL b) 10 x 106 spermatozoa/mL c) 15 x 106 spermatozoa/mL d) 30 x 106 spermatozoa/mL

c) 15 x 106 spermatozoa/mL According to the WHO, the reference value for sperm concentration between 10 and 20 million per mL is considered borderline range. The normal reference values are greater than 20 to 250 million per mL.

The normal Myeloid-to-erythroid ratio (M:E ratio) ranges from: a) 2:1 to 7:1 b) 1.2:1 to 5:1 c) 2:1 to 4:1 d) 0.5:1 to 1:1

c) 2:1 to 4:1 The M : E reference range is between 1.5:1 to 3:1. If segmented neutrophils are included, the M : E reference range is between 2:1 to 4:1.

Approximately how many doses are required to obtain a steady-state oscillation allowing for peak and trough levels to be evaluated? a) 1 to 2 b) 3 to 4 c) 5 to 7 d) > 10

c) 5 to 7 Approximately five to seven doses are required before a steady-state oscillation is achieved. After the first dose, absorption and distribution occur, followed only by elimination. Before the concentration of the drug drops significantly, the second dose is given and the peak of the second dose is additive to what remains of the first dose. The third through the seventh scheduled doses all have the same effect, increasing the serum concentration and the amount eliminated. By the end of the seventh dose, the amount of the drug administered is equal to the amount eliminated during the dosage period. At this point, a steady-state is established and peak and trough concentrations can be evaluated.

Which of the following blood tests is used to determine acute pancreatitis? a) Acid phosphatase b) Uric acid c) Amylase d) Alkaline phosphatase (ALP)

c) Amylase Amylase test helps to determine acute pancreatitis. It should be collected in a gel-barrier serum tube (red or gold top). Hemolysis and lipemia should be avoided. Serum should be separated from cells immediately and refrigerated for storage. Acid phosphatase test helps in diagnosing prostate cancer. It should be collected in a gel-separator tube. Serum/plasma should be frozen immediately and transported frozen. Uric acid is a test used to diagnose gout in a patient. It should be collected in a gel-barrier serum tube (red or gold top). Serum should be separated from cells within 45 minutes and stored at room temperature. ALP is the test used to evaluate liver function. It requires to be a fasting collection and collected in a gel-barrier tube. Serum should be separated from cells immediately and refrigerated for storage.

If detected in antibody screen testing, which of the following antibodies is NOT considered clinically significant in prenatal patients? a) Anti-M b) Anti-N c) Anti-Leb d) Anti-Fya

c) Anti-Leb Anti-Leb may be detected in antibody screen testing of prenatal patients; however, this antibody is considered clinically insignificant. It is not indicated in causing hemolytic disease of the fetus and newborn (HDFN). Although rarely seen, both anti-M and anti-N can potentially cause mild to moderate HDFN. The most common clinically significant antibodies noted in prenatal patients include the following: anti-Fya, anti-K, anti-D, anti-E, anti-e, anti-C, and anti-c. These IgG antibodies have been determined to cause moderate to severe HDFN.

Which of the following characteristics of hemolytic disease of the fetus and newborn (HDFN) is NOT different for ABO and Rh HDFN? a) Disease predicted in titers b) Anemia at birth c) Antibody is IgG d) Bilirubin at birth

c) Antibody is IgG The antibodies for ABO HDFN are IgG (anti-A,B) and for Rh HDFN the antibodies are IgG (anti-D). The disease is predicted with titers for the Rh HDFN, but not the ABO HDFN. There is anemia at birth for the Rh HDFN, but not the ABO HDFN. For the ABO HDFN, the bilirubin is in the normal range, but for the Rh HDFN, the bilirubin is elevated.

An increased reticulocyte count MAY be found in all of the following conditions, EXCEPT? a) Hemolytic anemias b) Following acute hemorrhage c) Aplastic anemia d) Satisfactory response to therapy for pernicious anemia

c) Aplastic anemia Aplastic anemia does not cause the bone marrow to release immature red blood cells because it is a pluripotential stem cell disorder that leads to peripheral blood pancytopenia and hypoplastic marrow. Hemolytic anemias, active blood loss, and successful treatment of pernicious anemia all cause an increase in reticulocytes in the peripheral blood, as the bone marrow attempts to replenish the decreased quantity of peripheral RBCs.

Identify the suspicious form (shown in the image) found in a stool specimen, measuring 50 µm. a) Taenia species egg b) Hookworm egg c) Ascaris lumbricoides egg d) Fasciola hepatica egg

c) Ascaris lumbricoides egg The Ascaris (roundworm) egg may be identified by its size of approximately 45-75 µm x 35-50 µm, oval shape, mammillated coat, and semi-organized egg contents. Taenia species (tapeworm) eggs are round, yellow or brown, approximately 30-45 µm, and contain a thick wall with radial striations. A six-hooked oncosphere may be visible in eggs passed in the stool. Hookworm (roundworm) eggs possess a thin, colorless shell. When passed in the stool, the egg contains four to eight cells. Hookworm eggs are oval, and approximately 50-60 µm x 35-40 µm. Fasciola hepatica (fluke) eggs are large in size, measuring approximately 130 x 69-90 µm, ellipsoidal in shape, yellow or brown in color, with an indistinct operculum. Fasciola eggs are unembryonated when found in stool.

Which of the following clinical laboratory departments performs the most tests? a) Urinalysis b) Microbiology c) Clinical Chemistry d) Hematology

c) Clinical Chemistry Clinical Chemistry is the laboratory department that performs most laboratory tests, which include quantitative analytical procedures on various body fluids. This department may have sub-sections such as, Special Chemistry, Toxicology, Therapeutic Drug Monitoring, and Molecular Diagnostics. Urinalysis (UA) is the laboratory department that performs physical, chemical, and microscopic evaluation of urine. Microbiology is the laboratory department that cultures and identifies the presence of microorganisms from body fluids and tissues. This department may have sub-sections such as Bacteriology, Virology, Parasitology, Mycology, and Mycobacteriology. Hematology is the laboratory department which performs tests that identify diseases associated with blood and blood-forming tissues.

The measurement of sodium and chloride in the sweat is the most useful test for the diagnosis of what condition/disease? a) Steatorrhea b) Direct determination of the exocrine secretory capacity of the pancreas. c) Cystic fibrosis d) Zollinger-Ellison syndrome

c) Cystic fibrosis Measurement of the sodium and chloride concentration in sweat is the most useful test in the diagnosis of cystic fibrosis. Significantly elevated concentrations of both ions occur in more than 99% of affected patients.The definitive test for steatorrhea (failure to digest or absorb fats) is the quantitative fecal fat analysis.The secretin/CCK (cholecystokinin) test is the direct determination of the exocrine secretory capacity of the pancreas.Pancreatic cell tumors, which overproduce gastrin, are called gastrinomas; they cause Zollinger-Ellison syndrome and can be duodenal in origin.

Which of the following analytes would be increased due to delay in centrifugation? a) Bicarbonate b) FolateIonized c) Calcium d) ALT

d) ALT The specimen should not be delayed for more than two hours prior to centrifugation because some of the analyte levels (such as glucose, ionized calcium, bicarbonate, folate, etc.) may be falsely decreased due to cellular consumption or falsely increased (such as potassium, ALT, AST, creatinine, etc.) because they are released over time from cells into serum or plasma.

Shown in this photograph are three tubes of Moeller decarboxylase broths, containing from left to right, the amino acids lysine, arginine, and ornithine respectively. A red-blue reaction indicates decarboxylation of the amino acid; yellow is negative. From the bacterial species listed, these reactions are most likely caused by: a) Klebsiella pneumoniae b) Pantoea agglomerans c) Enterobacter aerogenes d) Citrobacter freundii

c) Enterobacter aerogenes From the available choices, Enterobacter aerogenes is the correct answer. The decarboxylase reactions seen in this photograph are produced by Enterobacter aerogenes, Edwardsiella species, and most Serratia species. Within the genus Enterobacter, this profile of decarboxylase reactions is selective for either E. aerogenes or E. gergoviae, as all other species are incapable of decarboxylating lysine. All Klebsiella species, except K. ornithinolytica and a few strains K. terrigena, are ornithine decarboxylase negative. Pantoea agglomerans has as one of its key identifying characteristics negative reactions for the three decarboxylases commonly tested. All Citrobacter species are lysine decarboxylase negative.

A 55-year-old female, who had recently returned from an extensive trip to China, presented to her physician complaining of diarrhea and abdominal cramps. A complete blood count (CBC), comprehensive metabolic panel (CMP), and stool for both culture and parasite examination (O & P) were ordered. The CBC revealed pronounced eosinophilia. The CMP and stool culture were unremarkable. The O & P revealed this suspicious form that measured approximately 140 µm by 80 µm. This patient is most likely infected with: a) Paragonimus westermani eggs b) Diphyllobothrium latum eggs c) Fasciola/Fasciolopsis eggs d) Ascaris lumbricoides eggs

c) Fasciola/Fasciolopsis eggs Fasciola/Fasciolopsis eggs is the correct answer because these eggs measure 130-140 µm x 63-90 µm. Fasciola/Fasciolopsis are indistinguishable and recovery of an adult worm is necessary to speciate these organisms. Paragonimus westermani eggs are incorrect because patients infected with Paragonimus westermani typically suffer from respiratory difficulties. The organism produces eggs similar to those shown here that are recovered in sputum specimens. Eggs from this organism measure 80-120 µm x 39-67 µm and even though it has an operculum, the operculum is flattened with raised opercular shoulders. Diphyllobothrium latum eggs is incorrect because eggs are generally smaller in size (measuring 58-76 µm x 40-51 µm) than those depicted here and possess a small terminal knob opposite the operculum. Infected patients complain of several symptoms including digestive difficulties and abdominal discomfort. Ascaris lumbricoides eggs is incorrect because these eggs typically measure 45-95 µm x 35-50 µm and do not have an operculum. The egg in this image is much larger and has an operculum.

Which of these methods measures fetal hemoglobin or D positive red cells or both to evaluate fetomaternal hemorrhage? a) Rosette test b) Kleihauer-Betke test c) Flow cytometry d) AHG testing

c) Flow cytometry Flow cytometry measures fetal hemoglobin or D positive red cells or both. In patients with a positive rosette test (a screening for fetomaternal hemorrhage), a quantitative test such as Kleihauer-Betke test or flow cytometry is performed to calculate the dose of Rh immune globulin. The Kleihauer-Betke acid elution is based on the fact that fetal hemoglobin is resistant to acid elution and adult hemoglobin is not resistant to it. Examples of polyclonal antiserum produced for blood bank testing are known as antihuman globulin (AHG) reagents. These products contain multiple antibody specificities.

How does hydroxyurea aid in the treatment of sickle cell disease? a) Acts as an analgesic in pain management. b) Prevents sickle cells from clumping together. c) Induces increased production of HbF. d) Reduces the number of sickle cells that form.

c) Induces increased production of HbF. Hydroxyurea induces increased production of HbF. Most sickle cell patients who have increased levels of HbF experience milder forms of the disease than do patients with normal or low levels of HbF. Therefore, the focus of molecular treatments for sickle cell disease is to increase fetal hemoglobin (HbF). Anti-inflammatory agents are given to serve as an analgesic in pain management for sickle cell patients. Nitric oxide treatment prevents sickle cells from clumping together. Clotrimazole is an over-the-counter antifungal medication. This drug prevents water loss from the red blood cells, which helps prevent the formation of sickle cells.

Which of the following tests on amniotic fluid would be included when assessing fetal lung maturity: a) Alpha Fetoprotein b) Bilirubin c) L/S Ratio d) Fetal hemoglobin

c) L/S Ratio The correct answer choice here is L/S ratio. The lecithin/sphingomyelin ratio is determined by examining the amniotic fluid of a pregnant mother. Because amniotic fluid is continuously swallowed or inhaled and then replaced or exhaled in uterus the ratio can be measured to determine how much surfactant is in the fetal lungs. This ratio will help determine how well the infant's lungs will perform breathing at birth. Alpha-fetoprotein is used in assessing the potential for genetic disorders. Bilirubin is used in diagnosing the cause of jaundice and other liver problems.Fetal hemoglobin is not used to assess fetal lung maturity.

A variety of additives are used in blood collection tubes. Which of the following additives prevents clotting by inhibiting thrombin and thromboplastin? a) EDTA b) Gel c) Lithium or sodium heparin d) Sodium fluoride

c) Lithium or sodium heparin Lithium/sodium heparin is an additive used to prevent clotting by inhibiting thrombin and thromboplastin. It is found in green and light green tubes. EDTA prevents blood from clotting by binding calcium, which in turn inhibits the coagulation cascade. EDTA is found in lavender, pink, white, royal blue, and tan top tubes. The gel in the tube is used to form a barrier between plasma/serum and blood cells upon centrifugation. Sodium fluoride is found in gray top tubes and its purpose is to inhibit glycolysis (metabolism of glucose by red blood cells).

A three-year-old boy was brought unconscious to the emergency room. Arterial blood gas results were as follows: pH = 7.07p CO2 = 90 mm Hg HCO3- = 23 mmol/L This patient is most likely suffering from: a) Metabolic alkalosis b) Metabolic acidosis c) Respiratory acidosis d) Respiratory alkalosis

c) Respiratory acidosis Since the blood pH is only 7.07 (normal 7.35-7.45), it is acidosis. In addition, the increased level of pCO2 90 mm Hg (normal 34-45 mm Hg) demonstrates that there is a build-up of acid from the respiratory system, while the bicarbonate is within the normal range (22-29 mmol/L). Therefore, the patient is suffering from respiratory acidosis. Judging by the pH, which is only 7.07 (normal 7.35-7.45), this patient is experiencing acidosis. The patient's carbon dioxide (pCO2) level is increased by 90 mm Hg (normal 35-45 mmol/L), while the bicarbonate (HCO3-) level is 23 mmol/L, still within the reference range (22-29 mmlo/L). This indicates that the acidosis is respiratory in nature. The bicarbonate is not increased in response to the increase in acid, indicating that the respiratory acidosis is uncompensated. Reference Range Recap: pH = 7.35-7.45 pCO2 = 35-45 mm Hg HCO3- = 22-29 mmol/L pO2 = 85-105 mm Hg

A D-zone test is performed to determine if Staphylococcus aureus or Streptococcus agalactiae has an inducible resistance to clindamycin. This inducible resistance is due to which of the following mechanisms? a) vanC gene b) msrA gene c) erm gene d) mecA gene

c) erm gene erm gene is the correct answer because when performing a D-zone test, a "D" pattern will occur when clindamycin and erythromycin disks are placed next to each other on a 5% sheep blood agar for Streptococcus agalactiae or Mueller Hinton agar plate for Staphylococcus aureus (see photo to the right). Clindamycin inducible resistance occurs due to iMLSB (macrolide-lincosamide-streptogramin-B) mechanism caused by either the erm A gene or erm C gene. van C gene is incorrect as this causes an intrinsic resistance to vancomycin for Enterococcus gallinarium and Enterococcus casseliflavus, which are both pigmented Enterococcus species. msrA gene is incorrect because this gene causes the organisms Staphylococcus aureus or Streptococcus agalactiae to be sensitive to clindamycin and resistant to erythromycin by way of an efflux mechanism. A "D" pattern will not occur when a D-Test is performed. mecA gene is incorrect because this gene can be detected by using a cefoxitin disk to determine oxacillin resistance. For example, if Staphylococcus aureus grows around the disk, the organism is resistant to oxacillin and will be determined to be MRSA (Methicillin-Resistant Staphylococcus aureus).

Which marker is most useful for the detection of gestational trophoblastic disease? a) CEA (Carcinoembryonic antigen) b) AFP (a-fetoprotein) c) hCG (human chorionic gonadotropin) d) CA-125 (cancer antigen 125)

c) hCG (human chorionic gonadotropin) hCG is used for pregnancy testing, but it is also the most useful marker for the detection of gestational trophoblastic disease.The main clinical use of CEA is a marker for colorectal cancer.AFP is often elevated in patients with hepatocellular carcinoma and germ cell tumors.CA-125 is a serological marker of ovarian cancer.

A manual white blood cell count was performed by the hematology technologist. The cell counts for each of two sides was 38 and 42 respectively. All nine large squares were counted on each side. The dilution for this kit was pre-measured at 1:10. What should the technologist report as the white cell count? a) 4.8 x 109/L b) 4.4 x 109/L c) 0.48 x 109/L d) 0.44 x 109/L

d) 0.44 x 109/L Calculation: Cells Counted (in this case the average of both sides) X dilution factor (in this case 10) / # of sqaures counted (in this case 9) X 0.1mm (depth of solution) X area of each square (1mm2) So, in this problem: (40 x 10) / (9 x 1mm2 x 0.1mm) = 444.4/mm3 (can be converted to 0.44 x 109/L)

A computer monitor should be set approximately how many inches away from the eyes? a) 1 - 3 inches b) 5 - 10 inches c) 10 - 15 inches d) 20 - 40 inches

d) 20-40 inches Computer monitors should be approximately 20 - 40 inches away from the eyes. The top of the monitor is best set at eye level so that the eyes gravitate toward the center of the screen. OSHA recommends: Ensuring adequate desk space between the user and the monitor (table depth). If there is not enough desk space, consider doing the following:Make more room for the back of the monitor by pulling the desk away from the wall or divider; orProvide a flat-panel display, which is not as deep as a conventional monitor and requires less desk space, orPlace monitor in the corner of a work area. Corners often provide more desk depth than a straight run of desk top. Move back and install an adjustable keyboard tray to create a deeper working surface.

What is the increase in the risk for developing antibodies against red cell antigens (RBC alloimmunization) for patients who are characterized as chronically transfused patients? a) 1% - 4% b) 2% - 8% c) 5% - 10% d) 30% or greater

d) 30% or greater In chronically transfused patients, such as those with thalassemia, autoimmune hemolytic anemia, and sickle cell disease, the risk of developing antibodies against red cell antigens (RBC alloimmunization) increases by 30% or more.

What is the correct identification for each egg illustrated? a) A: Hookworm egg B: Trichuris trichiura egg C: Enterobius vermicularis egg b) A: Ascaris lumbricoides egg B: Enterobius vermicularis egg C: Hookworm egg c) A: Diphyllobothrium latum egg B: Hookworm egg C: Trichuris trichiura egg d) A: Trichuris trichiura egg B: Hookworm egg C: Enterobius vermicularis egg

d) A: Trichuris trichiura egg B: Hookworm egg C: Enterobius vermicularis egg Trichuris trichiura eggs have a thick shell with hyaline polar plugs. Hookworm eggs have a thin shell that is broadly oval and between four to eight cells inside when passed in the stool. Enterobius vermicularis eggs have a colorless shell that is flattened on one side and a c-shaped larva inside. Ascaris lumbricoides have a bumpy (mammillated) outer shell. Diphyllobothrium latum eggs have an inconspicuous operculum with a small knob at the opposite end.

On a quiet evening shift at a small hospital, you encounter a specimen with a positive antibody screen. As per your current laboratory protocol, you check for agglutination at the immediate spin phase of testing; and then again at the antihuman gloubulin (AHG) phase of 37°C. According to your laboratory guidelines, a single homozygous cell may be used to rule out an antibody. Based on the following 3-cell screen performed by tube, which of the following clinically significant antibodies are you unable to rule out? a) Anti-Fya, -D b) Anti-Lua, -Lea, -Fya, -C c) Anti-Lea, -Fya, -C d) Anti-Fya, -C, -Lub

d) Anti-Fya, -C, -Lub The correct answer is anti-Fya, anti-C, and anti-Lub. This is a fairly basic panel, but it involves further thought. First, it asks you to determine which antibodies are not ruled out. Upon completing the rule-outs on screen cell #3, one must then evaluate those remaining antibodies that have not been ruled out for clinical significance.

Which of the following protozoa has largest trophozoite? a) Entamoeba histolytica trophozoite b) Endolimax nana trophozoite c) Iodamoeba butschlii trophozoite d) Balantidium coli trophozoite

d) Balantidium coli trophozoite The correct answer is Balantidium coli trophozoite. Balantidium coli trophozoites can be up to 100 µm. It is the only member of the ciliates known to cause disease in humans. It is found worldwide, but most reports of infection come from Latin America, the Far East, and New Guinea.Entamoeba histolytica trophozoites range from 12 - 40 µm. It is known to cause amebic dysentery and is transmitted via the fecal-oral route. Iodamoeba butschlii trophozoites range from 6-25 µm. These protozoa are not considered pathogenic by the CDC but are necessary to identify in order to differentiate from protozoa that are pathogenic.Endolimax nana trophozoites range from 5-8 µm. These protozoa are also not considered to be pathogenic by the CDC.

To assess drug concentrations during the trough phase: a) Blood should be drawn about one hour after the administration of an oral dose of the drug. b) Blood should be drawn about half an hour before the next dose is given. c) Blood should be drawn about two hours after the administration of an oral dose of the drug. d) Blood should be drawn immediately before the next dose is given.

d) Blood should be drawn immediately before the next dose is given. To assess drug concentrations during the trough phase, blood should be drawn immediately before the next dose is given. If the blood specimen was collected at one or two hours after the administration of an oral dose of the medication, the result would most likely reflect the peak level. If the blood specimen was collected half an hour before the dose is given, it will not be representing the real trough phase.

Homogentisic acid in urine will cause urine to be which of the following colors? a) Green b) Pink c) Port-wine d) Brown/black

d) Brown/black The presence of homogentisic acid in urine will cause urine to be brown/black. Homogentisic acid in the urine is found in Alkaptonuria, an inherited amino acid disorder. These individuals lack an enzyme that breaks down homogentisic acid. Green-colored urine can be associated with Pseudomonas infections. Pink urine is most commonly associated with the presence of a small amount of blood. Port-wine-colored urine is associated with porphyrins.

Which of the following D variants has the best likelihood to receive D-positive RBCs without any adverse effects? a) Del b) Partial D c) Partial weak D d) C in Trans to RHD

d) C in Trans to RHD Individuals with C in Trans to RHD possess complete D antigen structures. The allele that carries C is in trans (or opposite position) from the allele carrying RHD. As a result, the position of the C antigen in relationship to the D antigen interferes with the expression of the D antigen. This can be a weakened expression of D antigen; however, D antigen is still present. These patients can receive D-positive RBCs without the implication of an Rh-mediated transfusion reaction. In partial D and partial weak D individuals, they can produce Rh antibodies to the portion of the RhD gene that is missing. There is a risk for potential adverse effects to occur when these individuals are recipients of D-positive RBCs. Individuals with Del phenotype possess a low number of D antigen sites that can go undetected in routine serological testing. This testing discrepancy can lead to misclassification of Rh type for the recipient (i.e., classified RhD-negative but maybe RhD-positive). Molecular studies are needed to detect the mutant RHD gene responsible for altering the expression of the RhD protein.

Which parasitic infection may result from human ingestion of contaminated fish? a) Dipylidium caninum b) Taenia saginata c) Taenia solium d) Diphyllobothrium latum

d) Diphyllobothrium latum The life cycle of Diphyllobothrium latum requires two intermediate hosts that both reside in fresh water: copepods and fish. In this life cycle, fish eat infected copepods. Ingestion of contaminated fish initiates human infection. The organisms then settle in the human intestine where they mature. Adult worms lay eggs that may be passed out of the body via feces. Egg contact with freshwater allows for the emergence of the coracidium, which is later ingested by a copepod and the cycle repeats itself. Dipylidium caninum is known as the dog tapeworm. Human transmission may occur through accidental ingestion of fleas containing infective cysticercoid. Taeniasis is caused by eating infected, undercooked beef (Taeniasaginata), or pork (Taenia solium).Taenia soliummay cause a serious complication called cysticercosis, through ingestion of eggs.

A 21-year-old male presented to the emergency room complaining of abdominal pain, fever, cough, nausea, vomiting, and constipation alternating with diarrhea. Radiologic studies revealed the presence of a hepatic abscess. The form in this image, measuring 20 µm, was seen in the microscopic examination of a sample taken from the abscess. What condition is the patient most likely suffering from? a) Intestinal amebiasis b) Meningoencephalitis c) Giardiasis d) Extraintestinal amebiasis

d) Extraintestinal amebiasis The organism in the image is an Entamoeba histolytica trophozoite. The size ranges from 10 to 50 µm. The nucleus contains a small central karyosome with fine even peripheral chromatin. This organism is the only intestinal protozoan that can cause extraintestinal infections. In severe infections, a flask-shaped ulcer forms in the intestinal mucosa. The organisms may completely erode the intestinal wall gaining access to the circulation. The organism can then colonize the liver, typically in the right lobe. Patients with hepatic abscesses may have symptoms such as fever, chills, and upper right quadrant pain. The form found in the abscess sample and the patient's symptoms indicate this is extraintestinal amebiasis caused by Entamoeba histolytica.

Which FAB designation is called the "true" monocytic leukemia and is characterized by monoblasts, promonocytes, and monocytes? a) FAB M1 b) FAB M3 c) FAB M4 d) FAB M5

d) FAB M5 FAB M5 is acute monoblastic leukemia. Most cells are monocytes. There are 2 subtypes. Subtype a is characterized by large blasts in the bone marrow and peripheral blood. Subtype b is characterized by monoblasts, promonocytes, and monocytes. FAB M1 is acute myeloblastic leukemia without maturation. Blasts and promyelocytes predominate with minimal maturation of myeloid cells. FAB M3 is acute promyelocytic leukemia. Promyelocytes predominate in the bone marrow. FAB M4 is acute myelomonocytic leukemia. Myeloid and monocytic cells represent at least 20% of the total leukocytes.

This organism, identified as a Klebsiella oxytoca, was recovered from the blood of an elderly patient with an infected anal fissure . The Vitek identified this organism as an ESBL. After reviewing the Kirby-Bauer plate and susceptibility interpretations, what is the resistance mechanism demonstrated by this organism? a) Klebsiella pneumoniae carbapenemase (KPC) b) Extended-spectrum beta-lactamase (ESBL) c) AmpC beta-lactamase d) K-1 beta-lactamase

d) K-1 beta-lactamase The correct answer is K-1 beta-lactamase. The K-1 beta-lactamse (not an ESBL) found in Klebsiella oxytoca, is primarily a penicillinase that also hydroloyzes aztreonam, cefuroxime, and ceftriaxone. It demonstrates weak activity against cefotaxime and ceftazidime. Low level beta-lactamase production is responsible for the penicillin resistance. Hyperproduction of K-1 beta-lactamase causes resistance for aztreonam and the labile cephalosporins. These hyperproducers may be identified because there is greater activity against ceftriaxone than cefotaxime and also greater activity against aztreonam than ceftazidime. Most commercial systems incorrectly call the K-1 an ESBL.No post-analytic comment required: Patient does not require contact isolation This Klebsiella oxytoca is aztreonam resistant and ceftazidine susceptible which classifies the resistance mechanism as K-1 beta-lactamase. Because the isolate is sensitive to cefoxitin, it is thus negative for AmpC. There is no clavulanic effect, thus it is negative for ESBL. Because all carbapenems are susceptible, the isolate is negative for KPC.

Various methods have been employed for the detection of Clostridioides difficile disease, which method is the new gold standard for detection? a) Culture on cycloserine-cefoxitin-fructose agar (CCFA) b) Cell cytotoxicity neutralization assay (CCNA) c) Enzyme immunoassays (EIA) detecting toxins d) Nucleic acid amplification tests (NAATs)

d) Nucleic acid amplification tests (NAATs) Nucleic acid amplification tests (NAATs) are becoming the new gold standard for detection. They determine the presence of toxin A and B genes in a stool sample. It is becoming the new gold standard since it does not need confirmation like other test methods. Cell cytotoxicity neutralization assays (CCNA) were considered for a long time the gold standard for Clostridioides difficile detection because they detect the toxins in the stool sample. However, the assay is labor-intensive and requires 48 of incubation for the cells to grow. The culture of stool on cycloserine-cefoxitin-fructose agar (CCFA) is very effective in detecting C. difficile, although it requires up to four days to isolate the organism. Isolates must also be subsequently tested for toxin production to prove their causal relationship to disease. Enzyme immunoassays (EIA) provide a more efficient and rapid means of diagnostic testing. Sensitivity of detection is enhanced by choosing a method that allows for the detection of both toxins A and B.

A patient has a history of repeated spontaneous abortion. Coagulation studies reveal an elevated APTT, normal PT, normal platelet function, and normal thrombin time. Schistocytes were seen on the peripheral blood smear. Which test should be performed to determine if the patient has lupus anticoagulant? a) Factor VIII assay b) Mixing studies with factor-deficient plasmas c) Antinuclear antibody test d) Platelet neutralization test

d) Platelet neutralization test Platelet neutralization tests are one of the confirmatory tests that can be used to determine if a patient has a circulating lupus antibody, or lupus anticoagulant. The principle in this test involves the use of a freeze-thawed platelet suspension that, when mixed with the patient plasma, will neutralize the anti-phospholipid antibodies (lupus anticoagulant) present and allow for a corrected APTT result upon re-testing. A Factor VIII assay would be abnormal due to the interference of the lupus anticoagulant with the phospholipid in the APTT test that is used to measure Factor VIII activity. Mixing studies are useful as screening tests for lupus anticoagulant and phospholipid antibodies. Mixing studies with factor deficient plasmas and the APTT test would all be abnormal due to the interference of the lupus anticoagulant with the phospholipid in the APTT test that is used to perform the mixing studies. A mixing study using the patient's plasma incubated 1:1 with normal plasma would also be abnormal (uncorrected). The antinuclear antibody test is a screening test for autoimmune diseases such as systemic lupus erythematosus (SLE). Although some patients with SLE may have a lupus anticoagulant, not all patients with lupus anticoagulant have SLE.

Fluorometers are designed so that the path of the excitation light is at a right angle to the path of the emitted light. What is the purpose of this design? a) Prevent the loss of emitted light b) Prevent the loss of the excitation light c) Focus emitted and excitation light upon the detector d) Prevent incident light from striking the detector

d) Prevent incident light from striking the detector The angle is set at 90 degrees in fluorometers in order to only detect the emitted light from the sample substance after it has been excited with a light beam. This helps to avoid measuring the ultraviolet light emitted from the light beam itself, or any other incident light. Excitation is the light that is absorbed by the sample (higher energy, shorter wavelength) while fluorescent light is that light that is emitted as energy leaves the sample (lower energy, longer wavelength). By having the photodetector at a right angle, only the emitted/fluoresced light will be detected.

Observation of colonies producing brick red fluorescence under long-wave ultraviolet light on KVLB agar allows for presumptive identification of which anaerobic bacterial genus? a) Fusobacterium b) Peptostreptococcus c) Bacteroides d) Prevotella

d) Prevotella Of the anaerobic bacteria listed, Prevotella grow on KVLB, which contains kanamycin and vancomycin. Prevotella may produce brick red fluorescence under UV light, such as a Wood's lamp. Some species of Prevotella may not fluoresce, and some may fluoresce other colors such as orange or yellow.

Each of the following organisms may possess extended-spectrum beta-lactamase (ESBL) activity except: a) Escherichia coli b) Klebsiella oxytoca c) Proteus mirabilis d) Pseudomonas aeruginosa

d) Pseudomonas aeruginosa Pseudomonas aeruginosa is not tested for ESBL production. The organism is typically resistant to ampicillin, cefazolin, and other first and second generations cephalosporins. Escherichia coli, Klebsiella oxytoca, and Proteus mirabilis all can possess ESBL activity. These enzymes have the ability to inactivate penicillins, extended-spectrum cephalosporins, and aztreonam. Klebsiella pneumoniae is also a very common organism that can display ESBL activity, and other genera including Salmonella species and Enterobacter species also have been identified with ESBL activity. ESBL enzymatic activity is confirmed in the lab by using clavulanic acid, which inhibits the enzyme activity. The organism is tested typically with cefotaxime and ceftazidime alone and then with each drug plus clavulanic acid. If the organism is resistant without clavulanic acid but becomes sensitive with the addition of clavulanic acid, the resistance is due to the ESBL enzyme activity. Any identified ESBL organism should be reported resistant to cephalosporins, penicillins, and aztreonam, regardless of test results.

The qualitative differences between A1 and A2 phenotypes includes all of the following EXCEPT: a) The formation of anti-A1 in A subgroups. b) The amount of transferase enzymes. c) The length of the precursor oligosaccharide chains. d) The lack of agglutination of patient red cells with anti-A reagent.

d) The lack of agglutination of patient red cells with anti-A reagent. Qualitative differences for A1 and A2 phenotypes include the following: differences in the precursor oligosaccharide chains (in length and complexity of branching), small differences in transferase enzymes (decreased in A2 subgroup), and the formation of anti-A1 in the serum of A2 phenotype individuals. Both A1 and A2 patient red cells react with the anti-A reagent. Dolichos biflorus or anti-A1 lectin reagent is used to differentiate between A1 and A2 phenotypes. This lectin reagent agglutinates with A1 patient red cells but does NOT agglutinate with A2 patient red cells.