#1 Comp Exam

b) The woman could be pregnant. A low specific gravity could give a false negative result since the concentration of HCG is low in this sample Reason: It is recommended that pregnancy tests (a test for HCG) be done on the first morning specimen when the hormone is most concentrated. This is especially critical if the test is being performed very early in relationship to the regular menstrual cycle (before the first missed menstrual period) when HCG levels are very low.

A 28 year old female had her last regular menstrual period 10 days ago. Her doctor orders a urine pregnancy test. She stops by the laboratory on her way home from the doctor's office and the sample for the pregnancy test is collected and run within 20 minutes of collection. The sample reveals: specific gravity = 1.001 glucose = negative protein - trace pregnancy test results - negative What is your interpretation of the test? a) The woman is not pregnant, doctor should explore other reasons for her missed period. b) The woman could be pregnant. A low specific gravity could give a false negative result since the concentration of HCG is low in this sample c) The woman could be pregnant. A trace protein could interfere with the pregnancy test results causing a false negative. d) The sample was improperly collected and sat too long before testing

c) Positive protein, WBC casts, granular casts Reason: Acute pyelonephritis is an infection of the kidney. It is associate with positive protein, WBCs, and WBC casts. As the WBC casts degenerate, they form coarsely granular casts which are also seen in this disease state.

A 30 year old male has acute pyelonephritis. What would most likely be seen in his urine? a) Positive protein, positive blood, no casts b) Positive protein, RBC casts c) Positive protein, WBC casts, granular casts d) Fatty casts, oval fat bodies

d) barrel shaped arthrospores Reason: With respiratory infections from a fungus, think of the dimorphic fungi. The only one of the dimorphs that grows quickly (usually within 3 days) is Coccidioides immitus. It forms a fluffy white colony that looks like Persian cat fur. It is called the lid-lifter. This culture would have alternating dark-light barrell-shaped arthroconidia. The organism is endemic in the central valley and southern desert of California. It also grows in the arid climates of Arizona and parts of southern Texas. It is easily transmitted in the mold phase and should only be worked on under a biological hood. The organism is really a monomorph rather than a dimorph. The true definition of a dimorphic fungi is that it grows as a mold at room temp and as a yeast at 35-37oC. Coccidiodes differs in that it does not grow in a yeast phase outside the body. It does have two phases: The mold phase which grows outside the body in nature or culture and the spherule phase which grows in the lungs at 35-37oC.

A 46 year old laborer from southern California has respiratory discomfort. Bacteriologic cultures were negative, but fluffy white, cottony growth appeared on Sabouraud Dextrose Agar in less than ten days at room temperature. If you were to mount a teased lactophenol cotton blue mount of this culture, what would you expect to see? a) budding yeast b) pseudohyphae c) chlamydospores d) barrel shaped arthrospores

d) Chronic Lymphocytic Leukemia Reason: In patients with chronic lymphocytic leukemia (chronic lymphoid leukemia, CLL), the complete blood count (CBC) with differential shows absolute lymphocytosis, with more than 5000 B-lymphocytes/µL Microscopic examination of the peripheral blood smear is indicated to confirm lymphocytosis. It usually shows the presence of smudge cells which are artifacts from lymphocytes damaged during the slide preparation. Large atypical cells, cleaved cells, and prolymphocytes are also often seen on the peripheral smear and may account for up to 55% of peripheral lymphocytes. If this percentage is exceeded, prolymphocytic leukemia (B-cell PLL) is a more likely diagnosis. Peripheral blood flow cytometry is the most valuable test to confirm a diagnosis of chronic lymphocytic leukemia (chronic lymphoid leukemia, CLL). It confirms the presence of circulating clonal B-lymphocytes expressing CD5, CD19, CD20(dim), CD 23, and an absence of FMC-7 staining. Although not necessary for the diagnosis or staging of CLL, additional molecular testing now exists that may help predict prognosis or clinical course. Chromosomal evaluation using FISH can identify certain chromosomal abnormalities of CLL with prognostic significance. Patients with a deletion in the short arm of chromosome 17 [del(17p)] tend to have a worse prognosis, as well as resistance to therapy with alkylating agents and purine analogues. Patients with deletions in the long arm of chromosome 11 [del(11q)] also have a worse prognosis and bulky lymphadenopathy at presentation.

A 62 year old male has enlarged lymph nodes. His CBC reveals: WBC = 100,000/cumm RBC = 5.2 million/cumm Hgb = 15.1 g/dl Hct = 47% Differential = 15 % segs, 85% lymphocytes, many smudge cells present Differential smear is shown below: What is the most likely diagnosis? a) Acute viral infection b) Infectious mononucleosis c) Acute Lymphocytic Leukemia d) Chronic Lymphocytic Leukemia

b) 65 mg/dl Reason: The normal CSF glucose, an ultrafiltrate of serum glucose, is about 2/3 of the serum value. Serum value x 2/3 = CSF value CSF value x 3/2 = Serum value 43mg/dl x 3/2 = ~65 mg/dl

A CSF glucose of 43 mg/dl would correspond to a blood glucose of about: a) 110 mg/dl b) 65 mg/dl c) 50 mg/dl d) 120 mg/dl

c) The patient has a cold agglutinin titer, the titer should only be reported out after diluting the patient's serum since the test gives evidence of prozone. Reason: Cold agglutinins are associated with Mycoplasma pneumoniae and usually have anti-I specificity. In this case there is so much antibody present that prozone is present in the initial tubes of the dilution. The titer is the last tube that gives a positive agglutination result which is tube #8 in this case.

A cold agglutinin titer was performed on a patient suspected of having a positive titer. Tubes 1 through 4 showed no agglutination. Tubes 5 through 8 showed agglutination. Tubes 9 through 12 showed no agglutination The positive and negative controls were both fine. What is the problem? a) The patient has positive cold agglutinins. The titer should be reported out as tube #8. b) The patient does not have cold agglutinins since tubes 1 through 4 show no agglutination. c) The patient has a cold agglutinin titer, the titer should only be reported out after diluting the patient's serum since the test gives evidence of prozone. d)The patient has a cold agglutinin titer; the titer should be reported out as tube #4.

d) anti-C Reason: The baby's positive DAT in HDFN is the result of maternal antibodies coating antigens on the baby's cells that were inherited from the father. The father is homozygous for c and e antigens. Because he is type B, his genotype for ABO is either BB or BO. The father's antigenic type is D neg, C neg, E neg, A neg and cannot pass these antigens to the baby but c pos and e pos (both homozygous) which means the baby has c and e antigens on its RBCs. The antibody has to be anti-c.

A cord blood specimen is submitted to the blood bank for routine testing. The following results were obtained: anti-A: 3+; anti-B: neg; anti-D: 2+; Rh Control: neg; Direct Antiglobulin Test: 2+ The baby's father is group B Rh neg with a genotype of cde/cde. Of the following antibodies, which is the mostly likely cause of the baby's + DAT? a) anti-C b) anti-D c) anti-A d) anti-C

a) optochin susceptibility Reason: Perform a "P" disk to test for optochin susceptiblity. S. pneumoniae will be susceptible to optochin whereas Viridans Streptococci will not.

A doctor has sent a sputum specimen to the laboratory. The gram stain shows many gram positive cocci in chains and pairs. Many alpha-hemolytic catalase negative colonies are seen on the blood agar plate. It is not clear, however, whether the organisms are Streptococcus pneumoniae or alpha-hemolytic Viridans Group streptococci. Which of the following tests would make this differentiation? a) optochin susceptibility b) esculin hydrolysis c) bacitracin susceptibility d) coagulase

a) chronic alcohol abuse often associated with alcoholic cirrhosis Reason: Gamma-glutamyltransferase (GGT) is a well-established serum marker for alcohol-related liver disease. It is elevated in chronic alcoholism and is associated with alcoholic cirrhosis.

A doctor orders a gamma glutamyl transferase (GGT). What is she looking for? a) chronic alcohol abuse often associated with alcoholic cirrhosis b) hepatitis especially hepatitis B infection c) obstructive jaundice from gall stones d) myocardial infarction complicated by live stasis

d) Fusobacterium sp. Reason: Fusobacteria are anaerobic gram-negative rods that are a rare cause of serious human disease. F. nucleatum and F. necrophorum, most commonly isolated, may be implicated in infection throughout the body, but with some predilection for the lower respiratory tract, head and neck, peridontium, gingivae, and central nervous system.F. nucleatum is isolated most often. It is an important pathogen, particularly in head and neck and lower respiratory infections. F. necrophorum is a cause of serious infections (necrobacillosis or Lemierre's disease) commonly diagnosed in young adults and also a cause of recurrent sore throats. Fusobacterium species are rods which may be spindle-shaped eg Fusobacterium nucleatum or pleomorphic eg Fusobacterium necrophorum. They exhibit irregular staining. Colonies are variable with an irregular edge. They vary from translucent to granular and opaque. F. necrophorum may be beta-hemolytic, indole +, fluorescent yellow-green under UV light. Isolates are now being identified by MaldiTOF

A gram-negative anaerobic bacillus is isolated from a deep abscess in the abdomen. A note is made on the gram stain that the organism appeared to have tapered ends as seen below. A gas chromatograph report shows butyric acid present. The organism is most likely: a) Peptostreptococcus sp. b) Bacteroides sp. c) Prevotella sp. d) Fusobacterium sp.

c) Antibody to a high titer low avidity (HTLA) antigen Reason: Most weakly reactive antibodies lose reactivity when diluted even modestly but some antibodies that give weak reactions when undiluted continue to react at dilutions as high as 1 in 2048. Such antibodies include anti-Ch, -Rg, -Csa, -Yka, - Kna, -McCa, -JMH, and other specificities. When weak reactions are observed in indirect antiglobulin tests, titrations may be used to indicate specificity within this group. These are characteristics of antibodies to "high titer, low avidity antigens" Antibodies that react to HTLA characteristics cause difficulties in serologic testing because of the weak reactions they produce in the indirect antiglobulin test. Those specificities that are more frequently encountered (anti-Yka, -McCa, -Kna, -Ch) are directed toward antigens of high incidence in both the white and black populations. They have not been shown to cause significant destruction of transfused antigen-positive red cells. The antibodies create problems in serologic tests because the reactions they produce interfere with the identification of reactions due to other, clinically significant antibodies

A medical laboratory scientist performs an antibody workup and finds 1+ and w+ reactions for several of the antibody panel cells. The reactions do not fit any pattern on the panel. Several selected panels and a patient phenotype do not reveal any additional information. The serum is diluted and retested, but the same reactions persist. What type of antibody may be causing these results? a) Antibody to a high-frequency antigen b) Antibody to a low-frequency antigen c) Antibody to a high titer low avidity (HTLA) antigen d) Anti-HLA

c) Trichinella spiralis Reason: It is a common symptom to develop edema in the upper eyelid and fever when first infected with Trichinella spiralis. When the cysts are ingested the individual begins having symptoms 1-2 days after ingestion of the contaminated meat. As the organism migrates through the blood vessels to the various tissues of the body. Frequently infected are the muscles outside the eye that control eye movement; muscles of the jaw, neck and upper and lower back and the diaphragm. The presence of the organism causes swelling and inflammation. This is accompanied by eosinophilia.

A middle aged woman complains of sudden swelling of the upper eyelids and of a fever. She has had nausea and diarrhea in the past few days. A blood examination reveals a high eosinophil count. A biopsy of the eyelid revealed the organism below: Of the following, the most likely causative agent is: a) Enterobius vermicularis b) Taenia solium c) Trichinella spiralis d) Wuchereria bancrofti

a) Cross match using O positive units negative for Jka antigen. Reason: Always check the patient history for previous antibodies that may not be showing up in current testing. Antigen type the units for the historical antibody and give antigen negative type-specific blood. Anti-Jka is a Kidd antibody. They are notorious for titers falling below the detectable level until they are re-stimulated. Can cause delayed transfusion reactions.

A patient admitted to the hospital has a history of anti-Jka. When the type and screen is performed, the patient types as O Rh positive with a negative antibody screen. What is the next best course of action? a) Cross match using O positive units negative for Jka antigen. b) Crossmatch random units, since the antibody is not demonstrating reactivity. c) Repeat the screen with enzyme-treated screening cells. d) Request a new sample.

b) The patient has infectious mononucleosis Reason: While the clinical picture of hepatitis B and mononucleosis can appear similar and both can have a high number of reactive lymphocytes, viral testing can help distinguish between the two. The Epstein Barr Virus is associated with infectious mononucleosis.

A patient diagnosed as having serum sickness has the following lab test performed: HBsAg - negative Epstein Barr virus - positive His CBC shows a reversed differential with an increased number of reactive lymphocytes. What is the most likely problem? a) The patient has Hepatitis A b) The patient has infectious mononucleosis c) The patient has Hepatitis, not A, not B d) The patient has Burkitt's lymphoma

b) Acute Bacterial Infection Reason: Patient has an elevated WBC, left shift is toward immature WBCs which occurs when the bone marrow is under stress to respond, LAP score is elevated. These all point to a leukemoid reaction from acute bacterial infection.

A patient has a WBC = 30,000/cumm. His differential reveals 69 segmented neutrophils; 15 bands; 1 metamyelocyte; 1 myelocyte; 8 lymphocytes; 3 eosinophils; and 3 monocytes. His LAP score is 212. What is the most likely diagnosis? a) Chronic Myelogenous Leukemia b) Acute Bacterial Infection c) Chronic Lymphocytic Leukemia d) Acute Myelogenous Leukemia

d) Wilson's Disease Reason: Wilson's disease (WD) is a genetic disorder in which there is excessive accumulation of copper in the liver and brain because of an inherited defect in the biliary excretion of copper. Copper is an essential cofactor for many enzymes and proteins and plays a role in the mobilization of tissue iron stores. Copper in the diet is absorbed relatively efficiently by the small intestine, is bound relatively loosely by circulating plasma proteins, and is delivered to the liver from the portal circulation. The transport of copper from the hepatocytes to bile is critical in overall copper homeostasis, because biliary excretion undergoes minimal enterohepatic recirculation. In Wilson's disease there is a decrease in biliary copper excretion and ultimately a hepatic accumulation of copper. Along with this failure of biliary excretion, there is also reduced incorporation of copper into ceruloplasmin, a serum glycoprotein that contains six copper atoms per molecule and is synthesized predominantly in the liver. The process of copper incorporation into apoceruloplasmin is absent or diminished in most patients with Wilson's Disease, leading to a reduced circulating level of serum ceruloplasmin in most patients.

A patient has an increased urine copper level and a decreased serum copper and ceruloplasmin level. She most likely has: a) Cushing's Syndrome b) Addison's Disease c) Fanconi's Syndrome d) Wilson's Disease

a) lack of intrinsic factor - (Pernicious Anemia) Reason: Because the Schilling test results were corrected with the addition of the intrinsic factor, this megaloblastic anemia is due to a lack of intrinsic factor and is also called Pernicious Anemia. Pernicious anemia is a condition caused by low B12 levels in the body and is due to the loss of stomach parietal cells that make intrinsic factor which helps the body absorb vitamin B12. Be sure you can distinguish the 3 causes and associated test results for the megaloblastic anemias. Megaloblastic anemia results from inhibition of DNA synthesis during RBC production. When DNA synthesis is impaired, there is cell growth without division which presents as macrocytosis. The two most common causes of megaloblastic anemia are Vitamin B12 or folate deficiency. 1. B12 deficiency - shows low levels of blood Vitamin B12 2. Folate deficiency - shows low levels of blood Folate 3. Lack of intrinsic factor - this impacts the absorption of B12 since intrinsic factor is required for B12 absorption. The Schilling test has been historically used to Diagnose a lack of intrinsic factor but this test is primarily now a historical test only. Since Pernicious Anemia can be due to antibodies against intrinsic factor, the following algorithm for diagnosis is currently used. If vitamin B12 is <150 ng/L, then intrinsic factor blocking antibody (IFBA) is performed. If IFBA is negative or indeterminate, then serum gastrin is performed. Gastrin is usually markedly increased in pernicious anemia and gastrin can be used as a substitute for the Schilling test. If vitamin B12 is 150 to 400 ng/L, then methylmalonic acid (MMA) is performed. If MMA is >0.40 nmol/mL, then IFBA is performed. The most sensitive test for vitamin B12 deficiency at the cellular level is the assay for methylmalonic acid (MMA).

A patient has megaloblastic anemia. A serum gastrin level was performed and was markedly increased. A Schilling test was performed and the following results were obtained: Administration without intrinsic factor - 3% excretion Administration with intrinsic factor - 27% excretion These test results are most compatible with: a) lack of intrinsic factor - (Pernicious Anemia) b) malabsorption syndrome c) folic acid deficiency d) B12 deficiency

c) The patient is showing a false positive result probably from abnormal plasma proteins or autoagglutinins: Cells should be washed with saline and Anti-D saline tube test performed. Reason: In the high protein method for D typing, an Rh control must be run because the protein in the diluent may cause false positive reactions in patients with autoantibodies or abnormal serum proteins. To troubleshoot wash cells with saline and perform a saline Anti-D test

A patient is tested by the Anti-D rapid Slide Modified Tube Method. The results of testing were as follows: Anti-D: + Anti-D Control: + From these results you can conclude: a) The patient is Rho(D) positive and should be crossmatched with Rh positive type specific blood. b) The patients is most likely Rho(D) negative and should be crossmatched with Rh negative type specific blood. c) The patient is showing a false positive result probably from abnormal plasma proteins or autoagglutinins: Cells should be washed with saline and Anti-D saline tube test performed. d) The patient is showing a false positive result most likely from an isoagglutinin from previous sensitization. The isoagglutinin must be identified before correct Rho(D) type can be determined.

c) use prewarmed (37oC) cells and serum for antibody screen and x-match Reason: The effect of cold auto-antibodies on antibody screens and crossmatch results can be eliminated by pre-warming the patient cells and serum to 37oC.

A patient presents in the blood bank with a cold autoantibody in his serum. From the history, it was shown that the patient received three units of blood six weeks ago on a previous admission. The blood bank MLS should: a) perform an autoabsorption to remove the autoantibody, then use the absorbed serum for crossmatch b) eliminate the crossmatch since all the crossmatches will be incompatible, and transfuse ABO and Rh specific units. c) use prewarmed (37oC) cells and serum for antibody screen and x-match d) determine specificity of the cold autoantibody by titration, then give blood negative for that antigen.

c) Factor IX Reason: Mixing studies are tests performed on blood plasma used to distinguish factor deficiencies from factor inhibitors, such as lupus anticoagulant, or specific factor inhibitors, such as antibodies directed against factor VIII. A prolonged aPTT with a normal PT indicate an abnormality in the early part of the intrinsic coagulation pathway which is focused on factors XII, XI, IX, X XII and XI in both serum and plasma so this is not the deficiency IX is only in the aged serum X is in neither the aged serum or the adsorbed plasma Aged serum contains factors: II, VII, IX, XI and XII BaSO4 absorbed plasma contains factors I, V, VIII, XI and XII Since there is correction with aged srum and not with adsorbed plamsa, it is Factor IX in the aged serum that is doing the correction

A patient reveals the following: Lee White clotting time = normal Prothrombin time = normal Bleeding time = normal PTT = prolonged, but corrected with aged serum; BaSO4absorbed plasma showed no correctio This is most probably a deficiency in: a) Factor XII b) Factor VII c) Factor IX d) Factor VIII

a) Dolichos biflorus Reason: The patient forward types as AB and back types as B. This patient may have an anti-A1 in his serum which can occur when the patient has a subtype of A. The plant seed extract from Dolichos biflorus has specificity for A1 and forms complexes with N-acetylgalactosamine and can distinguish between A1 cells and other subgroups of A. A subgroup of A will be negative with Dolichos biflorus and positive when A1 antigens are present. Vicia graminea - has N specificity Aracis hypogaea - has T specificity Ulex europaeus - has H specificity

A patient sample tests as follows: Patient cells with: Patient Serum with: anti-A 3+ A1 cells 2+ anti-B 4+ B cells neg Which typing lectin can be used to resolve this ABO discrepancy? a) Dolichos biflorus b) Vicia graminea c) Arachis hypogaea d) Ulex europaeus

c) paroxysmal nocturnal hemoglobinuria (PNH) Reason: Paroxysmal nocturnal hemoglobinuria is an autoimmune intravascular hemolysis where sensitivity of RBCs to complement causes hemolysis to occur in an acid pH during sleep. The first morning void is red urine containing free hemoglobin. The elevated retic count indicates anemia. The Sugar Water Test and Hams Test show a result that is consistent with RBC hemolysis & sensitivity to complement in acid environment (regardless if the complement comes from the patient or from normal serum) but not when the complement has been inactivated in the serum.

A patient shows the following results: Hematuria which is most pronounced upon waking Hemoglobin in serum with decreased serum haptoglobin Osmotic fragility = normal Sugar Water Test: Increased hemolysis Hams Test results: - Pts RBC + acidified normal serum: + hemolysis - Pts RBC + acidified patient serum: + hemolysis - Pts RBC + inactivated normal serum: no hemolysis Reticulocyte Count = 4% These results suggest: a) methemoglobin disease b) G6PD deficiency c) paroxysmal nocturnal hemoglobinuria (PNH) d) hereditary spherocytosis

a) Perform an auto-control and direct antiglobulin test on the patient Reason: All screening cells and all units are positive at both 37oC and the IAT phase. This indicates the possibility of a high-frequency alloantibody or a warm autoantibody. An auto-control would help to make this distinction. A positive auto-control indicates an autoantibody is present; a negative auto-control and positive screening cells indicates an alloantibody. A DAT should be performed to determine if an antibody has coated the patient's red cells, and is directed against screening cells and donor cells.

A patient types as O positive. All three screening cells and cells from two O-positive donor units show agglutination after incubation at 37oC, and an increase in reactivity at the indirect antiglobulin (IAT) phase of testing. What action should be taken next? a) Perform an auto-control and direct antiglobulin test on the patient b) Perform an enzyme panel c) Perform an elution d) Choose another 2 units and repeat the crossmatch

d) Anti-i Reason: Infectious mononucleosis is associated with the formation of the cold reacting auto-antibody anti-i.

A patient with infectious mononucleosis presents at the blood bank with an auto-cold agglutinin. With what antibody is this most likely associated? a) Anti-I b) Anti-M c) Anti-P1 d) Anti-i

c) Pernicious anemia with hemolysis Reason: The indices and bone marrow point to megaloblastic anemia. The extremely low Hgb and Hct show a hemolytic crisis. The only hint that this is doe to Pernicious anemia is the lack of production of HCl. Patients with Pernicious Anemia produce low levels of HCl The elevated bilirubin which is primarily of the indirect or unconjugated type as seen in hemolysis.

A patient's CBC and lab testing results reveal: Hgb = 6.0 g/dl Hct = 18% RBC = 1.2 million/cumm Platelets = 125,000/cumm WBC = 3,800/cumm Indices indicate macrocytosis Gastric HCl = negative Bone marrow = megaloblastic Bilirubin, total and indirect = increased Serum B12 level = decreased Serum folate level = normal This patient most likely has: a) Iron deficiency anemia with hemolytic crisis b) Folic acid deficiency with hemolytic episode c) Pernicious anemia with hemolysis d) Acute lymphocytic leukemia

a) chronic myelogenous leukemia Reason: In chronic myelogenous leukemia (CML), the total WBC is elevated and there is proliferation of mature granulocytes including neutrophils, eosinophiils and basophils. Yhere is a chromosomal translocation in chromosome 22 called the Philadelphia chromosome. The chromosome is defective and unusually short because of reciprocal translocation of genetic material between chromosome 9 and chromosome 22, and contains a fusion gene called BCR-ABL1. This gene is the ABL1 gene of chromosome 9 juxtaposed onto the breakpoint cluster region BCR gene of chromosome 22, coding for a hybrid protein: a tyrosine kinase signalling protein that is "always on" causing the cell to divide uncontrollably. The LAP score is low in CML and this score can be used to differentiate CML from a leukamoid reaction in which the LAP score is high.

A patient's CBC shows a WBC of 32,000. The RBC count, Hgb and Hct are all normal. The differential reveals: 68 segmented neutrophils, 12 bands, 2 metamyelocytes, 1 myelocyte, 12 lymphocytes, 3 monocytes and 2 eosinophils. The Leukocyte alkaline phosphate (LAP) score is 12 and the patient was positive for the Philadelphia chromosome. The patient most likely has: a) chronic myelogenous leukemia b) an acute bacterial infection c) an acute viral infection d) chronic lymphocytic leukemia

b) Indirect agglutination inhibition Reason: The anti-HCG of the kit will bind with any HCG in the patient serum. When the latex particles containing HCG are added, If the patient is not pregnant (no HCG), there will be agglutination between the kit anti-HCG and the kit HCG If the patient is pregnant (HCG present), the patient HCG will first combine with the kit anti-HCG and be bound up so when latex particles are added, no agglutination occurs. It is indirect, according to serology procedures because it is agglutinating with the latex bead coated with HCG rather than "directly" with the HCG. A direct method uses natural antigens and antibodies (like in blood bank or when you take a bacterial antigen such as in direct agglutination with an antiserum for Neisseria gonorrhoeae from a colony on a plate. It is indirect if the antigen or antibody is attached to something that is artificial so you can see the agglutination. It is agglutination because that is how you are visualizing it. It is inhibition because in the first step you are combining the antigen with antibody to neutralize it so when the indicator system is added, you get no reaction. Go back to your serology course notes if this is unclear.

A pregnancy test in which anti-HCG and and patient urine are mixed together and then added to the HCG coated latex particles in the kit is an example of: a) Passive agglutination b) Indirect agglutination inhibition c) Active precipitation d) Direct agglutination

c) Group AB, Rh-positive Reason: the unit of FFP needs to ABO compatible with the recipient's RBCs. In this case Group AB, Rh-negative could also be used. Always avoid plasma with ABO antibodies to the patient's A or B antigens.

A request for 2 units of Fresh Frozen Plasma is received for a patient who tests group AB, Rh negative. Which of the following units of FFP would be the most acceptable for transfusion? a) Group B, Rh-positive b) Group O, Rh-negative c) Group AB, Rh-positive d) Group A, Rh-negative

b) Falsely depress - fluoride inhibits the enzyme urease Reason: The BUN methodology is enzymatic and the sodium fluoride in a gray top tube inhibits the enzyme urease used in the method.

A sample for blood urea nitrogen (BUN) using the urease methodology and is drawn into a sodium fluoride (NaFl) tube. What effect will this have on the assay? a) Falsely elevate - fluoride acts as an ion activator b) Falsely depress - fluoride inhibits the enzyme urease c) Falsely elevate - fluoride causes turbidity in the sample d) No effect

b) Positive protein, RBC casts Reason: Post streptococcal glomerular nephritis occurs when antibodies to Group A Strep cross react with the basement membrane of the glomerulus. This causes protein albumin - to spill into the urine. This infection is associated with RBC casts.

A six year old male has been over a sore throat for 2 weeks but has developed acute glomerular nephritis. What would you most likely expect to find in his urine? a) Positive protein, positive blood, no casts b) Positive protein, RBC casts c) Positive protein, WBC, casts, granular casts d) Fatty casts, oval fat bodies.

d) Inflammation Reason:

A synovial fluid easily forms small drops when pushed from an aspirating syringe. This viscosity is most likely associated with what condition below? a) Increased viscosity in Multiple myeloma b) Normal synovial fluid c) Hypoparathyroidism d) Inflammation

b) The total CO2 decreases, the pCO2 decreases, the pO2 increases and the pH increases Reason: When the blood for ABGs comes in contact with air, the O2 in the air goes into the sample and the CO2 in the blood comes out into the air because O2 is higher in air than in the blood and CO2 is lower in the air than in the blood due to metabolism and the production of CO2. The pH goes up - becomes more alkaline - because of the loss of CO2 from the sample.

A syringe with blood for Arterial Blood Gases (ABGs) contains a large air bubble. What will be the effect on the ABG results? a) The total CO2 and pCO2 will increase, the pO2 increases and the pH decreases b) The total CO2 decreases, the pCO2 decreases, the pO2 increases and the pH increases c) The total CO2 decreases, the pO2 and pCO2 increase and the pH increases d) The total CO2 increases, the pO2, pCO2 and pH decrease

b) 8:00 p.m. Reason: Platelets prepared from whole blood must be made within 8 hours of the original blood collection.

A unit of whole blood is collected at 12:00 noon and stored at 20oC - 24o C. What is the last hour platelet concentrates may be made from the unit? a) 6:00 p.m. b) 8:00 p.m. c) 9:00 p.m. d) 10:00 p.m.

d) Uric Acid Reason: Uric acid crystals can be seen in normal urine that is concentrated. They are seen in gout or cytotoxic drug use. They appear yellow to yellow-brown but can be colorless. They are pleomorphic in shape including diamond, cube-shaped, rhomic plates, prisms and needles which may be clustered together to form rosettes. Sodium urate is a form of uric acid crystal, appearing as light-yellow slender prisms, single or in clusters; they have no clinical importance.

A urine microscopic on a sample with a pH of 6.0 shows abundant hexagonal yellowish-brown colored crystals. These are most likely: a) Cysteine b) Bilirubin c) Sulfa d) Uric Acid

d) Orientia (Rickettsia) tsutsugamushi Reason: Orientia tsutsugamushi (formerly Richettsia tsutsugamushi) is the etiologic agent of scrub typhus, a chigger-borne zoonosis that is a highly prevalent, life-threatening illness of greatest public health importance in tropical Asia and the islands of the western Pacific Ocean. Scrub typhus is transmitted by some species of trombiculid mites ("chiggers", particularly Leptotrombidium deliense), which are found in areas of heavy scrub vegetation. The bite of this mite leaves a characteristic black eschar that is useful to the doctor for making the diagnosis. The choice of laboratory test is not straightforward, and all currently available tests have their limitations.The cheapest and most easily available serological test is the Weil-Felix test (results given in this question), but this is notoriously unreliable. The gold standard is indirect immunofluorescence, but the main limitation of this method is the availability of fluorescent microscopes, which are not often available in resource-poor settings where scrub typhus is endemic. Indirect immunoperoxidase, a modification of the standard IFA method, can be used with a light microscope, and the results of these tests are comparable to those from IFA. Rapid bedside kits have been described that produce a result within one hour, but the availability of these tests is severely limited by their cost. Serological methods are most reliable when a four-fold rise in antibody titre is found. If the patient is from a nonendemic area, then diagnosis can be made from a single acute serum sample. In patients from endemic areas, this is not possible because antibodies may be found in up to 18% of healthy individuals. Other methods include culture and polymerase chain reaction, but these are not routinely available and the results do not always correlate with serological testing, and are affected by prior antibiotic treatment.

A veteran returning from Southeast Asia is diagnosed as having scrub typhus. The Weil-Felix reaction shows positive agglutination with Proteus OX-K only. What is the organism causing this infection? a) Rickettsia akari b) Rickettsia prowazekii c) Rickettsia rickettsii d) Orientia (Rickettsia) tsutsugamushi

d) Plasmodium Reason: This individual most likely has a malaria infection. The positive syphilis test is due to a false + RPR from the malaria.

A veteran stationed in Vietnam came back from a three month tour of duty with chills, fever and a positive RPR test for syphilis. He most likely has contracted: a) Yaws b) Pinta c) Babesia d) Plasmodium

a) Indirect fluorescence antibody technique (IFA) Reason: In indirect fluorescent antibody techniques, you are assaying for antibody in the patient serum. The patient serum is reacted with an antigen that is fixed on a slide and if the antibody is present in the patient serum, a complex is formed. The slide is washed and an anti-antibody (to the patient antibody) containing a fluorescent tag is then added in the second phase and incubated. Once incubation is complete, the slide is washed again. If the patient was positive for the antibody, the complex can be viewed under a fluorescent scope.

A viral test is performed by incubating tissue culture cells usually in a shell vial with patient serum. The sample is incubated giving time for the virus to grow. The monolayer of cells is then washed followed by the addition of labeled antibody against the viral antibody in the patient's serum. The cell monolayer is viewed under a fluorescent scope. This procedure is an example of which method? a) Indirect fluorescence antibody technique (IFA) b) Fluorescence Inhibition technique (FI) c) Direct fluorescence antibody technique (FA) d) Double indirect fluorescence (DIF)

a) Uncompensated respiratory alkalosis Reason: Patient is alkalotic since pH is elevated. The pCO2 is slightly low which means the patient is hyperventilating so the primary cause is respiratory (it could be lower and probably would be on your board exam). The HCO3- is normal so there is no compensation.

ABG results show the following pH 7.51 pCO2 22 mmHg pO2 103 mm Hg HCO3- 24 meq/l The patient has: a) Uncompensated respiratory alkalosis b) Uncompensated metabolic alkalosis c) Compensated respiratory alkalosis d) Compensated metabolic acidosis

a) Factor V Reason: Vitamin K serves as an essential cofactor for a carboxylase that catalyzes carboxylation of glutamic acid residues on vitamin K-dependent proteins. The key vitamin K-dependent are factors II (prothrombin), VII, IX and X.

All of the following coagulation factors are vitamin K dependent, except: a) Factor V b) Factor VII c) Factor X d) Factor II

c) Quinidine Reason: Qunidine is a heart therapeutic drug. The others listed are used to treat seizures. Review your category of TDMS from your chemistry lecture.

All of the following drugs except one are used to treat epileptics: a) Dilantin b) Phenobarbitol c) Quinidine d) Phenytoin

b) Falsely elevated results Reason: Bananas contain serotonin and tryptophan and can cause false elevations in the test results. Other foods that can elevate catecholamines include: Caffeine, such as coffee, tea, cocoa, and chocolate. Amines. These are found in bananas, walnuts, avocados, fava beans, cheese, beer, and red wine. Any foods or fluids with vanilla. Licorice. Vanillymandelic acid (VMA) is one of the metabolites of epinephrine and norepinephrine, fight or flight hormones. Patients with the tumor pheochromocytoma produce elevated levels of VMA in the urine.

An adult has a 24-hour urine collected for Vanillylmandelic Acid (VMA) determination. The patient had eaten large quantities of bananas and vanilla ice cream in a banana split the night before. What effect will this have on the results? a) None b) Falsely elevated results c) Falsely depressed results d) Bananas depress results and vanilla elevates results falsely, therefore, the two will balance each other out and the results will appear normal

c) Le H Se Reason: Lewis antibodies are usually not clinically significant but may interfere with testing for other clinically significant antibodies. Lewis antibodies are most easily removed by neutralizing them with soluble Lewis substance. The Lewis antigens are secreted into saliva and plasma and are adsorbed onto the red cells. Leb substance is made by adding an L-fucose to both the terminal and next to the last sugar residue on the type 1 precusor chain. This process requires the Le, H, and Se genes. Since some examples of anti-Leb react only with group O and A2 RBCs, neutralization is best achieved if the saliva comes from a person who is group O

An antibody identification panel reveals the presence of anti-Leb and a possible second antibody. Saliva from which person would best neutralize the Leb antibody so the second antibody, if present, can be identified? Genes Lewis ABO Secretor a) Le H sese b) Le hh Se c) Le H Se d) lele hh sese

c) O negative packed cells Reason: For exchange transfusion: 1. Selection of blood for exchange transfusion a. Infant cells must be tested for ABO and Rh (D) b. Mother's blood is used for antibody screen c. Units must be antigen negative for all antibodies in mother's blood (need to use type O since mother potentially have anti-A; unit must be Rh negative so it is compatible with mothers antibodies) 2. FFP is used to reconstitute pRBCs to a hematocrit of approximately 40-50% 3. Any blood products to be transfused must be CMV negative and irradiated on infused with a WBC filter.

An exchange transfusion is required for a baby with hemolytic disease of the fetus and newborn. The mother is O Rho(D) negative. The mother has a positive antibody screen which was identified as being caused by anti-D. The baby is A Rho(D) positive. What type blood do you crossmatch for exchange transfusion? a) A positive whole blood b) O positive whole blood c) O negative packed cells d) A negative packed cells

a) Accept him as a donor Reason: All AABB standards for donation apply to apheresis donors also. However, frequency of donation and additional testing are different for plasmapheresis: every 8 weeks, total protein, IgG and IgM monitored every 4 weeks. This donors screening criteria are normal.

An individual comes to the blood bank as a donor for plasmapheresis. He previously donated plasma 8 weeks ago. His pulse rate is 80 beats per minute; blood pressure 120/80; Hct is 40%. What is the next course of action? a) Accept him as a donor b) Reject; his pulse rate is abnormal c) Reject; his blood pressure is abnormal d) Reject; his Hct is too low

c) pyruvate kinase deficiency Reason: Pyruvate kinase deficiency, one of the most common enzymatic defects of the erythrocyte, second only after G6PD deficiency, manifests clinically as a hemolytic anemia that can range from mild to severe. PKD is a mild to severe normochromic and normocytic anemia with reticulocytosis. Babies with this disorder will have hyperbilirubinemia due to hemolysis. The enzyme activity rate in most patients with pyruvate kinase deficiency is 5-25% of normal, with measurement of the intermediates (2,3-diphosphoglycerol and glucose-6-phosphate) proximal to the enzyme defect helping to confirm the diagnosis. Enzyme assay, as well as deoxyribonucleic acid (DNA) analysis with a polymerase chain reaction (PCR) assay or single-strand conformation polymorphism, can also be used to confirm the diagnosis of pyruvate kinase deficiency. The negative DAT distinguishes this from an autoimmune hemolytic anemia. The normal osmotic fragility test and no spherocytes rules out heriditary spherocytosis. The autohemolysis test is an older test that demonstrates the increased fragility of the RBCs in this deficiency.

An infant with mild splenomegaly and icteric sclera presented with a normochroic, normocytic anemia and reticulocytosis. A review of RBC morphology revealed echinocytes but no spherocytes (see peripheral smear below). Other Test Results: The total and indirect bilirubin were elevated. An osmotic fragility test was normal. The direct antiglobulin test (DAT) on this patient was negative. An auto hemolysis test shows markedly increased auto-hemolysis which is not corrected by the addition of glucose. This pattern is consistent with: a) G6PD deficiency b) hereditary spherocytosis c) pyruvate kinase deficiency d) paroxysmal nocturnal hemoglobinuria (PNH)

c) Ekinella corrodans Reason: The gnr that does not grow on MacConkey Agar and has a bleach smell is Eikinella corrodens, one of the HACEK group. It is often associate with bite wounds or closed fist wounds. E. corrodens must be incubated for 2-3 days before the colonies grow to a size sufficient for counting. When plated the organism is dry, flat and has a light yellowish-pigmented colony. The colony growth has three zones. There is a clear and moist center, a highly visible ring that appears like droplets, and an outer growth ring. The organism can produce either a musty smell or a bleach smell. Some of the colonies seem sightly sunken into the agar and this is called pitting. E. corrodens is small, straight, nonsporeforming, nonencapsulated and nonmotile. It is biochemically inactive for most biochemical tests. It does not produce catalase, urease, indole or H2S. It is oxidase positive and it can reduce nitrate.

An organism isolated from a jaw abscess grows on blood agar with pitting of the agar as seen in the image below. The organism also grew on Chocolate agar but not on MacConkey Agar, The organism is a gram negative rod. When the plate is opened, there is a faint smell of dilute bleach. It is most likely: a) Eschericia coli b) Actinomyces israeli c) Ekinella corrodans d) Veillonella corrodans

d) Moraxella osloensis Reason: Moraxella osloensis, an oxidase-positive, nonmotile, asaccharolytic (does not ferment glucose or other carbohydrates), aerobic gram-negative coccobacilli is a part of the normal flora of the human respiratory tract and the female genital tract. While it is usually a gncb, it can sometimes be mistaken for gndc. This is the rationale behind why Neisseria gonorrhea in a female genital specimen cannot be presumptively identified from a smear as G.C. is identified in males. Some organisms that are normal flora in the female genital tract can mimic gc in gram stain. Clinical significance of M. osloensis isolates even from normally sterile site, such as CSF, may be difficult to determine, because the organism rarely cause invasive infections. It has on very rare occasion been associated with meningitis and septecemia.

An organism isolated from a vaginal swab is oxidase positive and on gram stain appears to be a gram negative diplococci. However, the organism does not ferment glucose. What organism might this be? a) Neisseria gonorrhoeae b) Herella vaginicola c) Pseudomonas aeruginosa d) Moraxella osloensis

a) MCV = 70.3 fl MCH = 20.4 pg MCHC = 29.1% Reason: RBC = 4.45 x 1012/l Hgb = 9.1 g/dl Hct = 31.3% MCV (fL) = Hct(%) x10 / RBC (x 1012/L) = 70.3 fL MCH (pg)= Hgb (g/dl) x 10 / RBC x (1012/L) = 20.4pg MCHC (g/dl) = Hgb (g/dl)/ Hct x100 = 29.1 g/dl

Calculate the indices from the following data and choose the correct answer below: RBC = 4.45 x 1012/l Hgb = 9.1 g/dl Hct = 31.3% a) MCV = 70.3 fl MCH = 20.4 pg MCHC = 29.1% b) MCV = 20.4 fl MCH = 70.3 pg MCHC = 34.3% c) MCV = 70.3 fl MCH = 20.4 pg MCHC = 34.3% d) MCV = 70.3 fl MCH = 20.2 pg MCHC = 25.1%

c) 7.40 Reason: pH = pKa + log of HCO3-/0.03 pCO2 pKa = 6.1 or 6.11 pH = 6.1 + log (27/0.03 x 45) pH = 6.1 + log (20) Log of 20 = 1.30 pH = 7.40

Calculate the pH, given the following information: HCO3 27 mmole/L pCO2 45 mmHg pKa 6.11 a) 7.36 b) 7.38 c) 7.40 d) 7.45

c) Pseudomonas aeruginosa Reason: This is an oxidase + glucose non-fermenter that produced pyocyanin. This pattern of growth fits Pseudomonas aeruginosa.

From a culture of a burn wound, you recover a pure isolate of a fruity-smelling organism which grows well aerobically on blood agar. On MacConkey's agar, the colonies produce both a black-brown pigment - pyomelanin and a reddish pigment - pyorubin as well as a blue-green pigment - pyocyanin. The pigments are more visible after growth on nutrient agar. Carbohydrate fermentation tests reveal the organism does not ferment glucose. The organism gram stains as a gram negative rod. It is motile, positive for catalase and oxidase, negative for indole and urea and positive for citrate. A TSI agar slant shows and AK/no change pattern of growth. What is the identification of this organism? a) Proteus mirabilis b) Salmonella spp. c) Pseudomonas aeruginosa d) Aeromonas hydrophilia

d) 115 fl. 7.5 g/dl 3,500 75,000 Reason: MCV: 115 fl This patient has a high MCV. Normal is 80-100fl Hgb: 7.5 g/dl This patient has a low Hgb. Men Hgb: 13.5-17.5 g/dl women: 12.0=15.5 g/dl WBC: 3,500 This patient has low WBCs. WBC is normall 4,000 - 11,000/cumm Plt: 75,000 This patient has low platelets. Platelet count is normally 150-400 x 109/L.

From the laboratory tests performed on a patient, the smear indicates macrocytosis, anemia, leukopenia and thrombocytopenia. Which of the following sets of data would be consistent with this interpretation? MCV HGB WBC Platelets a) 115 fl. 9.5 g/dl. 6,500 75,000 b) 90 fl. 7.5 g/dl. 3,500. 175,000 c) 75 fl. 9.5 g/dl 6,500. 75,000 d) 115 fl. 7.5 g/dl. 3,500. 75,000

a) Mother's IgG antibodies reacting with baby's antigen Reason: This is a straight forward definition of HDFN. Mother's IgG antibodies cross the placenta and react with the baby's RBC antigens causing RBC destruction that begins in utero.

Hemolytic disease of the fetus and newborn (HDNF) is the result of which of the following? a) Mother's IgG antibodies reacting with baby's antigen b) Mother's IgM antibodies reacting with baby's antigen c) Baby's IgG antibodies reacting with mother's antigen d) Baby's IgM antibodies reacting with mother's antigen

b) The catalase test Reason: Staphylococci are catalase + and Streptococci are catalase neg

How can you BEST and most EASILY differentiate Staphylococcus and Streptococci? a) The coagulase test b) The catalase test c) The oxidase test d) carbohydrate fermentation

b) Staphylococcus aureus Reason: This is the basic definition of Staphylococcus aureus, a very common wound pathogen. GPC in clusters, catalase +, coagulase neg.

If the organism is isolated from a wound and its gram stain shows gram positive cocci in clusters and many polymorphonuclear cells and it is beta hemolytic on BAP and catalase and coagulase positive, it is most likely: a) Staphylococcus epidermidis b) Staphylococcus aureus c) Streptococcus pyogenes d) Micrococcus sp.

c) convert cholesterol esters to free cholesterol Reason: Total Cholesterol Method Cholesterol is measured enzymatically in serum or plasma in a series of coupled reactions that 1) hydrolyze cholesteryl esters to free cholesterol and fatty acids and 2) oxidize the 3-OH group of cholesterol. 3) One of the reaction by products, H2O2 is measured quantitatively in a peroxidase catalyzed reaction that produces a color. In the reaction below, absorbance is measured at 500 nm. The color intensity of the final product is proportional to cholesterol concentration. The reaction sequence is as follows: Cholesteryl ester + H2O -------cholesteryl ester hydrolase---->cholesterol + fatty acid Cholesterol + O2 ------cholesterol oxidase---------> cholest-4-en-3-one + H2O2 2H2O2 + 4-aminophenazone + phenol -----------peroxidase----------> 4-(p-benzoquinonemonoimino)-phenazone + 4 H2O Elevated levels of cholesterol increase the risk for coronary heart disease (CHD). Cholesterol is measured to help assess the patient's risk status and to follow the progress of patient's treatment to lower serum cholesterol concentrations. Desirable cholesterol levels are considered to be those below 200 mg/dL in adults and below 170 mg/dL in children

In cholesterol methodologies the cholesterol in the sample must be first hydrolyzed. The purpose of this step is to: a) split cholesterol into free fatty acids and glycerol b) remove interfering proteins c) convert cholesterol esters to free cholesterol d) precipitate free cholesterol in order to separate it from protein.

c) Antibody to DNA and nucleoprotein Reason: Pattern. Appearance. Antigen Involved. Main Disease Associations. Homogeneous. Solid nucleus staining. dsDNA. High titers: SLE SsDNA. Low titers: SLE or other connective tissue dis. Histones Sd-70 (may give a speckled pattern)

In doing a fluorescent ANA testing, a homogeneous pattern is indicative of: a) Antibody to nucleolar RNA b) Antibody to Sm antigen c) Antibody to DNA and nucleoprotein d) Antibody to DNA and insoluble nucleoprotein.

d) endogenous release of ADP Reason: Platelet aggregation occurs in 2 stages 1. The primary stage is when platelets adhere to one another in the presence of an agonist such as ADP, epinephrine or ristocetin. 2. The secondary stage is characterized by the stage when the platelets are stimulated to release the ADP stored endogenously in their internal organelle.

In platelet aggregation testing, the second phase of aggregation is due to: a) changes in platelet shape b) platelet factor 3 release c) exogenous addition of ADP d) endogenous release of ADP

b) anti-mitochondrial Reason: Primary biliary cirrhosis is an autoimmune disorder caused by the anti-mitochondrial antibodies cross-reacting with the bile duct cells.

In primary biliary cirrhosis, which of the following antibodies is seen in high titers? a) anti-smooth muscle b) anti-mitochondrial c) anti-DNA d) anti-parietal cell

a) The Reiter stain of Treponema pallidium absorbs the non-specific antibodies Reason: Patient serum is pre- treated with the Reiter stain of Treponema pallidium which absorbs the non-specific antibodies out of the patient's serum. The Nichols strain of the Treponema pallidium is specific syphilis and is used in the test as the test antigen. Cardiolipin is the antigen used in the RPR method. Reagin is the anti-body like substance produced during syphilis and is measured by the RPR test.

In the FTA-ABS test for syphilis, false positive results are not a significant problem because: a) The Reiter stain of Treponema pallidium absorbs the non-specific antibodies b) The Nichols strain of T. pallidium absorbs the nonspecific antibodies c) The fluorescein dye is more specific for the reagin than is the cardiolipin of the screening tests. d) The cardiolipin prevents the attachment of nonspecific antibodies.

a) Red Reason: Red rods = + AFB Blue = negative stain for AFB

In the Ziehl-Neilsen acid fast stain, the tubercle bacilli stain: a) Red b) Colorless c) Green d) Purple

a) RBC control = no hemolysis; SO control = complete hemolysis; report out the last patient tube showing no hemolysis. Reason: The controls in the ASO test are important to interpretation of the test results since the QC assesses whether the test components are working correctly. RBC control consists of the kit RBCs and saline. There should be no hemolysis in this control since a + test result is hemolysis. SO control is measuring the strength of the SO reagent in the test kit. The control consists of kit RBCs and kit streptolysin O (SO). The concentration of ASO should be strong enough tocause complete hemolysis of the kit RBCs. If this does not occur in the control, something has destroyed all or part of the ASO activity. Step 1: The anti-streptolysin O (ASO)test is based on mixing patient serum (which would contain the ASO) with the kit SO. If ASO is present in the patient, the ASO will bind with the kit SO in step 1, Step 2: Kit RBCs sensitized to SO are added as an indicator. If patient has ASO, it will be bound to the SO and neutralized. Until the patient ASO is diluted out in serial dilution, there will be no SO available in the test dilution. At higher dilutions, once the patient ASO is diluted out, SO will become available in the reaction mixture. Once SO is available in the reaction mixture, it will lyse the sensitized RBCs and cause hemolysis. Test interpretation: A negative test shows hemolysis in all tubes of the serial dilution. A positive test titer will show no hemolysis in the beginning of the serial dilution. The titer will be the reciprocal of the last tube that shows no hemolysis and the units the test is reported in are "Todd Units".

In the interpretation of the reference method for Anti-streptolysin-O (ASO) titer, which indicates a situation which you could safely report? a) RBC control = no hemolysis; SO control = complete hemolysis; report out the last patient tube showing no hemolysis. b) RBC control = no hemolysis; SO control = complete hemolysis; report out the last patient tube showing hemolysis. c) RBC control = complete hemolysis; SO control = no hemolysis; report out the last patient tube showing no hemolysis. d) RBC control = complete hemolysis; SO control = no hemolysis; report out the last patient tube showing complete hemolysis.

d) Biliary obstruction Reason: In order for urobilinogen to show up in the urine, conjugated bilirubin must be excreted through the common bile duct into the intestines and where it is converted to urobilinogen. Some urobilinogen is reabsorbed from the GI tract through the vessels that line the GI tract and then filtered by the kidneys. The rest of the urobilinogen is converted to urobilin which gives feces its dark color. In obstructive jaundice - caused either by a gall stone or pancreatic cancer - the bile duct is obstructed and conjugated bilirubin does not get into the GI tract. This results in low urine urobilinogen - which can't be measured with any sensitivity by the existing UA dipstick. It is also the reason that stools are very light in obstructive disease.

In which condition would you expect to find a decreased urine urobilinogen? a) Hepatitis b) Hemolytic jaundice c) Alcoholic cirrhosis d) Biliary obstruction

c) methemoglobin Reason: Methemoglobin is a form of the oxygen-carrying metalloprotein hemoglobin, in which the iron in the heme group is in the Fe3+ (ferric) state, not the Fe2+(ferrous) of normal hemoglobin. Methemoglobin cannot bind oxygen, unlike oxyhemoglobin. It is bluish chocolate-brown in color

In which form of hemoglobin is the ferrous ion oxidized to the ferric state? a) carboxyhemoglobin b) sulfhemoglobin c) methemoglobin d) deoxyhemoglobin

a) Providencia: phenylalanine deaminase positive, H2S negative, urease +/= Reason: Both Proteus and Providencia are PDA +. Proteus is H2S + and Providencia is H2S neg The MIO for all Providencia is +/+/=. Most Providencia spp. are urease neg with the exception that P. rettgeri is urease + Proteus spp. are all urease +.

Most Providencia species are closely related to Proteus species, but the Providencia, with the exception of Providencia rettgeri, may be differentiated from the Proteus by which of the following sets of results? a) Providencia: phenylalanine deaminase positive, H2S negative, urease +/= b) Providencia: phenylalanine deaminase positive, H2S positive, urease negative c) Providencia: phenylalanine deaminase negative, H2S positive, urease negative d) Providencia: phenylalanine deaminase positive, H2S negative, always urease positive

a) basophilic stippling Reason: Basophilic stippling is multiple, tiny, fine or coarse inclusions (ribosomal RNA remnants) throughout the cell that stain dark blue to purple. The RBCs containing these inclusions may stain normally in other respects or may be polychromatophilic. The basophilic stippling that is clinically significant is coarse stippling. Fine stippling is often noted in polychromatophilic RBCs and sometimes in other RBCs, and is generally not clinically significant. Coarse basophilic stippling may be seen in sideroblastic anemias, lead poisoning, myelodysplasias, and thalassemias.

Precipitated ribosomes seen in a mature RBC, as in the image below, are called: a) basophilic stippling b) a reticulocyte c) a siderocyte d) a megalocyte

a) decreased TSH, elevated triiodothyronine (T3) and thyroxine (T4) Reason: Primary hyperthyroidism is a condition in which the primary gland, the thyroid gland, makes too much hormone. This feeds back on the pituitary gland that produces thyroid stimulating hormone (TSH) and decreases the production of TSH through a normal negative feedback loop. In secondary hyperthyroidism, the problem is in the pituitary gland where too much TSH is being manufactured thus stimulating the thyroid gland to make too much T3 and T4. In secondary hyperthyroidism, the TSH, T3, and T4 are all elevated

Primary hyperthyroidism is characterized by: a) decreased TSH, elevated triiodothyronine (T3) and thyroxine (T4) b) elevated TSH, elevated triiodothyronine (T3) and thyroxine (T4) c) elevated TSH, decreased triiodothyronine (T3) and thyroxine (T4) d) decreased TSH, decreased triiodothyronine (T3) and thyroxine (T4)

a) are obligate intracellular parasites Reason: Both grow intracellularly inside the host cells.

Rickettsia spp. and Chlamydia spp. are like viruses in that they: a) are obligate intracellular parasites b) are resistant to antibiotics c) multiply by binary fission d) possess either DNA or RNA

c) Yes, if the baby's type is Rh positive. Reason: RhIg is immune anti-D and is given to Rh-negative mothers who give birth to Rh-positive babies and who do not have anti-D already formed from previous pregnancies or transfusion. The baby is Rh positive and the mother has a negative antibody screen so even though the baby has HDFN from anti-A, the mother needs to be protected against forming anti-D in this pregnancy.

Should an O-negative mother receive RhIg if she has a negative antibody screen and the positive DAT on the newborn is caused by immune anti-A? a) No, the mother is not a candidate for RhIg because of the positive DAT b) Yes, if the baby's type is Rh negative c) Yes, if the baby's type is Rh positive. d) No, the baby's problem is unrelated to Rh blood group antibodies.

a) Streptococcus pneumoniae Reason:

Situation: A specimen was sent to the microbiology lab for culture. Source of specimen = sputum culture. The microbiologist is identifying ab isolate with the following characteristics: hemolysis: alpha gram stain: gram positive cocci in pairs and what appear to be short chains catalase negative bacitracin negative optochin disk: 18 mm zone at 24 hr bile esculin negative no growth in 6.5% NaCl What is the most appropriate identification of this organism? a) Streptococcus pneumoniae b) Group D streptococci c) Staphylococcus sp. d) Micrococcus sp.

b) requires at least 2 peptide linkages to react Reason: The Biuret method used for total protein assays, requires at least 2 peptide bonds to react. A single amino acid will not react with this method.

The biuret method for total serum protein: a) is composed of biuret in an acid buffer b) requires at least 2 peptide linkages to react c) is composed of alkaline bromcresol blue d) will measure free amino acids

c) Circulating anticoagulants or inhibitor Reason: When the PT and/or aPTT are abnormal, further testing may be done to identify the specific abnormality. Mixing studies are performed to differentiate a factor deficiency from a circulating inhibitor. The coagulation test with the abnormal result is repeated using several different dilutions of the patient's plasma and normal plasma. The testing is performed immediately and after incubation at 37°C. If the abnormal result is corrected by the addition of normal plasma, a factor deficiency is indicated, whereas no correction of the abnormal result indicates the presence of a circulating inhibitor. This patient has a circulating inhibitor causing the problem because there was no correction of the test results with the addition of normal plasma or serum.

The following substitution studies were performed on a patient's plasma which had a prolonged APTT: Pt plasma + normal plasma Pt plasma + absorbed plasma Pt plasma + normal serum None of the substitutions corrected the prolonged aPTT significantly. What could be causing the abnormal results? a) Abnormal platelet function b) Vitamin K deficiency c) Circulating anticoagulants or inhibitor d) Factor XIII deficiency

a) AB

The genotypes of four children of the same parents are AA, AB, AO, and BO. The mother's genotype is AO. What is the father's genotype? a) AB b) BB c) BO d) AA

b) XI and XII Reason: XII and XI are involved in the initial contact activation of the intrinsic pathway. The intrinsic pathway is activated by trauma inside the vascular system, and is activated by platelets, exposed endothelium, chemicals, or collagen. This pathway is slower than the extrinsic pathway, but more important. It involves factors XII, XI, IX, VIII.

The initial factors necessary for the activation of the intrinsic coagulation system are: a) VIII and XI b) XI and XII c) IX and XII d) IX and XI

b) Factor VII Reason: Stypven Time using Russell's Viper Venom activates factor X (common pathway) directly, thus the Stypven time is normal in factor VII deficiency (extrinsic pathway), but abnormal in factor V, II and most cases of factor X deficiency.

The patient has a prolonged prothrombin time (PT). When you substitute Russell's viper venom for regular thromboplastin, the prothrombin time is corrected. The most likely defect is in: a) Factor II b) Factor VII c) Factor X d) Factor XII

d) XIII Reason: This test is performed as a qualitative screening test for severe FXIII deficiency. It assesses clot solubility in 5M urea or 1% monochloroacetic acid. If the thrombin and Ca2+-induced clot lyses within a few hours, severe FXIII deficiency is suggested provided fibrinogen levels are qualitatively and quantitatively within reference range. Excluding hypofibrinogenemia and dysfibrinogenemia is important, since these conditions cause false-positive results on the 5M urea solubility test. The thrombin-clottable fibrinogen test can be used to exclude hypofibrinogenemia and dysfibrinogenemia.

The patient has an abnormal urea clot solubility. The factor involved is: a) IX b) XII c) III d) XIII

b) 8-20/field Reason: To estimate platelets Use the oil immersion (100X) lens Look at 5-6 fields and take an average Multiply the average by 20,000 Note any macroplatelets Normal = 8-20 platelets/oif < 8 platelets = thrombocytopenia >20 platelets thrombocytosis

Upon completion of a CBC with normal RBC and platelet counts, you use the peripheral smear to estimate the platelet count. How many platelets would you expect to see per oil immersion field? a) 1-8/field b) 8-20/field c) 20-30/field d) 40-50/field

a) Vanillylmandelic acid (VMA) and Homovanillic acid (HVA) Reason: This test measures the levels of catecholamines in the blood. Catecholamines is a term for the collective hormones made by the adrenal glands - the fight or flight hormones. The three catecholamines are epinephrine (adrenalin), norepinephrine (noadrenalin), and dopamine. Epinephrine, norepinephrine and dopamine, hormones in the blood are converted/metabolized by the liver to metanephrine and normetanephrine. These liver metabolites are converted further by the kidney to Vanillylmandelic acid (VMA) and Homovanillic acid (HVA), the urinary metabolites of the catecholamines. In neuroblastoma, HVA is elevated and in pheochromocytoma, VMA is elevated.

What are the major urinary metabolites of the catecholamine? a) Vanillylmandelic acid (VMA) and Homovanillic acid (HVA) b) Epinephrine and norepinephrine c) Dopamine d) Metanephrine and normetanephrine

c) 137 103 34 Reason: Whenever you come across electrolyte results on your boards, calculate the anion gap: Na+ - (Cl- + HCO3-) = 6-18 (the normal anion gap range) - If it is abnormal, especially high >18 you have ketoacidosis, renal failure or lactic acidosis and the lytes are abnormal. Always question a high gap. If the sodium in the 4th option had been higher, say 145, and the chloride had been 98, a normal value, then the A/G would have been 23 and high and you should question these results even though the individual electrolytes were normal since the patient could have one of the conditions above. - The gap in the correct answer is low. This is very unusual and doesn't happen often so puts the results in question. Sometimes myeloma proteins will have a positive change and can cause this situation where the A/G is O or even a negative number.

What electrolyte value in the group below might you question? Sodium meq/l Chloride meq/l Bicarbonate meq/l a) 144 102 31 b) 132 101 26 c) 137 103 34 d) 140 101 24

c) transfers an amino group AST [Aspartate Aminotransferase (AST)] Reason: The enzymes Aspartate Aminotransferase (AST) and Alanine aminotransferase (ALT) are transaminases. They are enzymes that catalyze a transaminationn reaction between an amino acid and an alpha-keto acid. They are important in the synthesis of amino acids which form proteins. Image: Aminotransfer reaction between an amino acid and an alpha-keto acid. The amino (NH2) group and the keto (=O) group are exchanged.

What is the action of a transaminase? Which enzyme is a transaminase? a) transfers a PO4 group LD [Lactate dehydrogenase (LD)] b) transfers a ketone group CK [Creatinine kinase (CK)] c) transfers an amino group AST [Aspartate Aminotransferase (AST)] d) transfers a hydrogen from one amine to another [Gamma-glutamyl transferase (GGT)]

a) Clean catch first morning collection Reason: While UAs are usually performed on random urine collection, the BEST sample is a clean catch first morning so that the formed elements represent what is in the urine vs. other contaminating sources and the sample is most concentrated.

What is the best specimen for a routine urinalysis? a) Clean catch first morning collection b) 24 hour timed specimen c) 2 hour timed urine d) Random urine collection

c) Enzyme Concentration Reason: Enzyme reactions should always be carried out under zero-order kinetics, in which there is enough (excess) substrate to ensure conversion to product. This then makes the reaction enzyme dependent rather than substrate dependent. In patients with very high enzyme concentrations in their serum, the sample can be diluted to bring the enzyme reaction back into zero-order kinetics and then the result multiplies by the dilution factor.

What is the limiting factor in a zero-order kinetics enzyme methodology? a) Substrate Concentration b) Product Concentration c) Enzyme Concentration d) Temperature of the Reaction

c) the protein error of indicators Reason: Testing for protein in the urine is based on the phenomenon called the "protein error of indicators" (ability of protein to alter the color of some acid-base indicators without altering the pH). In a solution that does not contain protein, tetrabromphenol blue, buffered at a pH of 3, is yellow. However, in the presence of protein, particularly albumin, the color changes to green, then blue, depending upon the concentration. This method is more sensitive to albumin than to globulin, detecting as little as 5 mg albumin/dL urine. Bence Jones protein and mucoprotein are examples of globulin components that are sometimes present in urine, but are not distinguishable by the chemical reagent strip method for urine protein. False-positive results can occur when testing for urine protein. A urine specimen that has remained at room temperature for an extended period of time may produce a false-positive protein result on a reagent strip. A false-positive may also occur in the presence of bacterial contamination, alkaline medication, quaternary ammonium compounds, such as disinfectants or drugs, and with skin cleansers containing chlorhexidine.

What is the principle for the method that detects protein on the urine dipstick? a) copper reduction by albumin b) peroxidase converts the protein to amino acids c) the protein error of indicators d) protein reacts with TCA reagent

b) Barium sulfate Reason: Barium sulfate is a granular salt that is weighed into sterile water at various concentrations to make a turbidity standard for susceptibility testing and other kit tests.

What reagent is used in to make the MacFarlandstandard used for Kirby-Bauer sensitivities? a) Calcium sulfate b) Barium sulfate c) Calcium oxalate d) Barium oxalate

b) 1.5% Reason: In untreated iron deficiency anemia, the reticulocyte count is low because the bone marrow is unable to make red cell precursors.

What reticulocyte count would most likely be associated with untreated iron deficiency anemia? a) 10% b) 1.5% c) 18% d) 5%

a) Type II diabetes Reason: This is most likely a case of Type II diabetes that has relative insulin deficiency but is producing some insulin. The patient has a normal fasting glucose, is not back down into the normal range at 2 hours which is the hallmark of diabetes. Diabetics with a relative insulin deficiency can have normal fasting glucose or elevated glucose. A Type I - absolute insulin deficiency - would not return to normal so quickly. By 3 hours this patient is back in the normal reference range.

What type of patient condition would the following values MOST LIKELY indicate? FASTING glucose 70 mg/dl 1 hour glucose 210 mg/dl 2 hour glucose 200 mg/dl 3 hour glucose 110 mg/dl a) Type II diabetes b) Normal patient c) Type I diabetes d) Hypoglycemia

a) Red/pink Reason: Iodine acts as a mordant for gram positive cocci to hold the crystal violet inside the cell. If the iodine is not added, the crystal violet will wash out with the acid alcohol, and the counterstain safranin will make the organism look falsely pink or red.

What will happen in a gram stain if you forget to add the gram's iodine? The organism is a gram + cocci. It will stain: a) Red/pink b) Blue c) Blear - will not stain d) Purple

a) Group A Reason: Lancefield Group A is S. pyogenes and is associated with scarlet fever, pyoderma, rheumatic fever and pharyngitis. Lancefield Group B is S. agalactiae and is associated with neonatal meningitis Lancefield Groups C and G occasionally cause human disease (particularly pharyngitis). Have been less extensively studied but becoming increasingly recognized as important pathogens, especially in Japan (for unclear reasons). Colonize the skin flora, oropharynx, and GI/GU tracts. Appear to cause infections similar to Group A strep as they share many of the same virulence factors. Lancefield Group C includes Streptococcus equi, Streptococcus equi subspecies zooepidemicus, Streptococcus equisimilis, and Streptococcus dysgalactiae. Lancefield Group G includes S. canis. This is normally found in a number of animals but can also cause infection in humans Lancefield Group D Streptococci: These are usually not beta Hemolytic. The group includes Enterococci and Group D Streptococci such as Strep bovis. Lancefield Group F Streptococci: These organisms are usually alpha hemolytic. Organisms of this group have been called S. milleri in the British taxonomic scheme and anginosus (S. anginosus) group in American taxonomic group. Group F Beta-hemolytic streptococci are recognized cause of severe suppurative infections including cellulitis, deep-tissue abscesses, bacteremia, osteomyelitis and endocarditis.

Which Lancefield group comprises the Streptococcus responsible for scarlet fever, rheumatic fever and streptococcal pharyngitis? a) Group A b) Group B c) Group C d) Group G

d) Anti-HBs Reason: Anti-HBs appears later in infection than anti-HBc and is used as a marker for immunity following infection or vaccination rather than for diagnosis of current infection.

Which hepatitis antibody confers immunity against reinfection with hepatitis B? a) Anti-HBc IgM b) Anti-HBc igG c) Anti-HBe d) Anti-HBs

a) Hexokinase and glucose oxidase Reason: Glucose hexokinase and glucose oxidase are enzymatic methods that are most commonly used to test for glucose. The other methods listed here are measuring reducing substance which glucose also is but they are not used in modern chemistry analyzers.

Which methodologies are commonly used to measure glucose levels in serum in the clinical laboratory? a) Hexokinase and glucose oxidase b) Nelson Smogyi and Biuret c) Copper sulfate reduction and Nitroprusside d) Alkaline ferricyanide and alkaline picrate

b) Mycobacterium kansasii Reason: Mycobacterium kansasii is an acid-fast bacillus (AFB) that is readily recognized based on its characteristic photochromogenicity, which produces a yellow carotene pigment when exposed to light. M. kansasii is one of the NTM that causes pulmonary infection.

Which mycobacterium produces a yellow colony from carotene production when exposed to light? a) Mycobacterium bovis b) Mycobacterium kansasii c) Mycobacterium gordonae d) Mycobacterium fortuitum-chelonei complex

a) too heavy of a cell suspension Reason:

Which of the following conditions might cause a false-negative indirect antiglobulin test (IAT)? a) too heavy of a cell suspension b) over-reading c) adding an extra drop of patient serum d) IgG-coated screening cells

a) Methyldopa, MSA (ALDOMET) Reason: There are three types of drugs which can cause + DAT and associated Drug-induced hemolytic anemia: 1. Haptenic type - usually caused by penicillin 2. Immune complex type - usually caused by cephalosporins 3. Those that induce hemolytic anemia - aldomet, methyl dopa and procainamide

Which of the following drugs is frequently associated with a positive direct antiglobulin test? a) Methyldopa, MSA (ALDOMET) b) Anacin c) Diazapan(Valium) d) Bacitracin

d) Anti-K Reason: Unexpected alloantibodies are antibodies other than naturally occurring anti-A or anti-B. Antibodies against Kell and the Rh system are the most frequently encountered antibodies in pre-transfusion testing. No Rh Antibodies are listed so BEST answer is anti-K

Which of the following is the "unexpected" antibody encountered most frequently in pretransfusion testing? a) Anti-Jka b) Anti-S c) Anti-Jsa d) Anti-K

a) Escherichia coli and Salmonella enteritidis Reason: When picking QC organisms, you need to pick a pair that gives a + and a neg reaction across organisms for both tests Lactose Fermentation Citrate Utilization Escherichia coli + neg Salmonella enteritidis neg + Salmonella spp. neg + Morganella morganii neg. neg Klebsiella pneumoniae + + Pseudomonas aeruginosa neg + Shigella boydii neg neg

Which of the following pairs of organisms would give you the expected quality control reactions on MacConkey Agar growth (lactose fermentation and non-lactose fermentation) and Simmons citrate (positive and negative)? a) Escherichia coli and Salmonella enteritidis b) Escherichia coli and Morganella morganii c) Klebsiella pneumoniae and Escherichia coli d) Pseudomonas aeruginosa and Shigella boydii

c) 5.8 g/dl 17% 3.43 million/cumm Reason: Using the rule of 3 in which the Hgb x 3 = Hct and Hct x 3 = RBC, In the set of data: 5.8 g/dl x 3 = 17.4% 17% x 3 = 5.2 million/cumm (doesn't correlate with 3.43. count) 3.43 million/cumm

Which of the following sets of results contains a value which is inconsistent with the other 2 values? HGB HCT RBC a) 11.5 g/dl 36% 3.75 million/cumm b) 15.0 g/dl 44% 4.90 million/cumm c) 5.8 g/dl 17% 3.43 million/cumm d) 10.0 g/dl 29% 3.21 million/cumm

c) An individual with the BB genotype is homozygous for B antigen. Reason: This is a straight forward case of how the alleles are inherited in the ABO blood group system and the definition of homozygous and heterozygous.

Which of the following statements describes inheritance in the ABO system? a) An individual with the OO genotype is heterozygous for O antigen b) An individual with the AB phenotype is homozygous for A and B antigens c) An individual with the BB genotype is homozygous for B antigen. d) An individual with the BO genotype is homozygous for B antigen

d) Anti-Jka Reason: If an antibody gives a stronger reaction with RBCs double-dosed for the target antigen, it shows the dosage effect (or just "dosage"). The dosage effect is most prominently identified in antibodies associated with the Duffy, Kidd, Rh, and MNS blood group systems. Of the options above the Kidd antibodies - Jka - would be the antibody most likely to show dosage that is IgG and warm reacting. Anti-N is rarely encountered as a clinically significant antibody. Anti-M is usually IgM and saline reacting. Anti-S can show dosage and is usually stimulated by transfusion exposure and is IgG but was not an option for this question.

Which of these warm reacting IgG antibodies is most likely to show dosage? a) Anti-N b) Anti-M c) Anti-Vel d) Anti-Jka

d) 500/cumm Reason: To find the absolute count multiply the total WBC count x % cell in the differential. Eosinophil Absolute Count: 10,000 WBCs/cumm x 5% eosinophils = 500 eosinophils/cumm

With a WBC of 10,000/cumm and a differential showing 70% segmented neutrophils 23% lymphocytes, 2% monocytes, and 5% eosinophils, the absolute eosinophil count is: a) 150/cumm b) 250/cumm c) 300/cumm d) 500/cumm

b) 1.5 g/dl Reason: Concentration of the Patient = Abs. Pt/Abs. Std x Conc. of Std .307/.203 x 1.0 = 1.5 g/dl

You are measuring IgG on a nephelometer. You obtain the following results: Sample ID Absorbance 1.0 g/dl IgG standard = .203 Patient unknown sample = .307 What is the patient's IgG concentration? a) 2.2 g/dl b) 1.5 g/dl c) 0.6 g/dl d) 1.0 g/dl

c) The choline chloride in the kit reagent deactivates the complement. Reason: The answer is the reason you need no heat inactivation - choline chloride in the test mixture inactivates complement.

You are performing an RPR test on a patient for syphilis. Why don't you need to heat inactivate the patient's serum? a) The test does not need to be heat inactivated because it is not complement dependent. b) You do need to heat inactivate because complement doesn't interfere with the RPR test; the classical RPR methodology includes heat inactivation for 30 minutes at 56 degrees C. c) The choline chloride in the kit reagent deactivates the complement. d) The sample for testing is drawn into an EDTA tube which inactivates the complement.