CSMLS mock exam questions/past questions Flashcards
Which of the following actions should be taken when the hematoxylin and eosin stain lack 3 distinct shades of pink?
Check the pH of eosin
A dark brown pigment is noted on formalin-fixed, hematoxylin, and eosin-stained sections. What action can prevent this from occurring?
Maintain formalin at a neutral pH
Which of the following situations will cause mast cells to turn blue during metachromatic staining?
Decreased pH of toluidine blue
Which of the following staining problems results from the omission of 1% periodic acid solution in the Jones Methenamine silver technique?
Failure to stain the basement membrane
Which of the following specimen orders prioritizes the processing of microbiology specimens correctly?
CSF, blood culture, urine, swab in a transport medium
Which of the following actions should be taken when tissue sections stained with hematoxylin and eosin show pale cytoplasmic staining?
Check the pH of eosin and adjust with acetic acid if necessary
Explanation: Eosin is an acidic dye, and its pH affects the staining quality. If tissue sections show pale cytoplasmic staining, it indicates potential issues with eosin staining. Checking the pH of eosin and adjusting it with acetic acid if necessary can help restore optimal staining conditions, leading to improved cytoplasmic staining in histological specimens.
Which of the following reagents is used as the mordant and oxidant in the Weigert’s hematoxylin?
Ferric chloride
Explanation: In the Weigert’s hematoxylin staining method, ferric chloride serves as the mordant and oxidant. It helps intensify the staining of nuclei with hematoxylin and promotes the formation of a complex that binds to cellular structures.
Which of the following factors is a possible cause for cocci in chains to appear pink on a gram stain?
The patient is on antibiotics
Explanation: Antibiotic use can affect the cell wall of bacteria, leading to altered staining characteristics. In the case of cocci in chains appearing pink on a Gram stain, it suggests that the bacteria may have lost their ability to retain the crystal violet dye due to antibiotic-induced changes in cell wall structure.
Which naturally occurring antibody(ies) is(are) found in the serum of a group A Rh negative patient
a. Anti-B
b. Anti-D
c. Anti-A,B
d. Anti-D and Anti-B
A
Holes are noted in a cryostat section of muscle. Which of the following is the most likely cause?
a. Too small clearance angle
b. Improper adjustment of the antiroll plate
c. Freezing the section in frozen isopentane in liquid nitrogen
d. Snap freezing the muscle in liquid nitrogen
D
Explanation: Snap-freezing is a process where a tissue is brought down to a temperature below -70C using liquid nitrogen to preserve the tissue morphology and to prevent formation of ice crystals. However, liquid nitrogen will bubble violently when in direct contact with the tissue. Hence, tissue is first submerged into a container isopentane, which then the entire container is lowered into liquid nitrogen. Since isopentane will not form bubbles like liquid nitrogen, this prevents the tissue from getting holes.
Which of the following is true regarding H&E staining using an automated stainer?
a. the staining should be microscopically checked daily using known control tissues
b. the staining should be checked daily only if you are preparing your own staining reagents
c. the staining should always be checked microscopically slide-by-slide
d. microscopic stain checks are not if you are purchasing your reagents
C
Explanation: (A) and (D) are invalid since staining should be checked whether commercial or in-house reagents are used. (B) is too time consuming and impractical. (C) is the proper way to check the quality of the staining. Control tissues will show whether the expected staining features have been stained properly (colour, intensity, and specificity).
Which of the following is true of quality control materials for internal QC?
a. they should be run daily for a week to collect values required to create Levey-Jennings plots
b. they must always have a known value directly traceable to a reference method
c. they should approximate the chemical and physical properties of the sample being tested
d. they should sit on the bench and continue to be used until the bottle is empty or values are > +2SD
C
Explanation: QC materials simulate the composition of patient samples in order to minimize the matrix effect and to correctly reflect the expected performance with patient samples. Matrix effect is the effect on an analytical method caused by all other components of the sample except the except the compound of interest. As opposed to standard reagents, QC reagents do not need a known value, as long as their measured values fall within the acceptable range on the levy-jennings graph. Most QC materials degrade quickly at room temperature; hence they are stored in refrigerators with a few exceptions. The mean and standard deviations of Levy Jennings plots are calculated from QC values collected over a MONTH not a week.
During a routine quality control check, a Kirby-Bauer disc susceptibility test is performed and the resulting zones of inhibition are too large. Which of the following is a possible cause of this result?
a. antibiotics stored at higher temperature than recommended
b. antibiotic discs applied immediately after inoculation of plate
c. inoculum too dense
d.agar depth too thick
B
Explanation: What effect does each answer have on the zone size? An overly dense inoculum means more bugs are added to the plate, which will overcome the effect of the drug causing a falsely small zone size. High temperature can destroy the integrity of the antibiotics making it less effective, hence, a falsely small zone size. Thick agar will prevent the antibiotics from diffusing to bugs that are farther from the disc, causing a falsely small zone size. If the disc is applied too soon before the bugs are settled in the agar, the disc can destroy the bugs immediately, causing a falsely large zone size.
Which antibody could be produced by an R1r individual transfused with R2R2 blood?
a. Anti-C
b. Anti-E
c. Anti-c
d. Anti-e
B
Explanation: Patient is CDe/cde (E-antigen negative RBC). Donor is cDE/cDE (E-antigen positive RBC). Hence the donor red cells can induce the patient to form anti-E.
A good screening test for alcohol ingestion employs comparing the calculated osmolality with what other measurement?
a. freezing point depression osmometer measurement
b. boiling point elevation osmometer measurement
c. vapor-pressure osmometer measurement
d. the osmolar gap
A
Explanation:Freezing point depression = Measured osmolality – Calculated osmolality = osmolar gap (i.e. measured osmo vs. calculated osmo). The more solutes in blood, the lower the freezing point (i.e. freezing point depresses) just like the effect of salt on ice. The solutes in our body are mainly sodium, urea, and glucose (2*Na + urea + gluc) which give us our calculated osmolality. However, our measured osmolality will include the above as well as exogenous substance such as methanol, ethylene glycol, acetone, and isopropyl alcohol. A high osmolar gap (measured – calculated) indicates that these exogenous substances may be present in our body.
When performing a D typing on a patient, a technologist observes that the Rh control is positive. As a result, which of the following would he perform next?
a. Full Rh phenotype
b. Repeat D testing with high protein antiserum
c. Weak D test
d. Direct Antiglobulin Test
D
Explanation: Rh control only consists of a high protein reagent with no anti-D. A positive Rh control indicates that the red cells may already be attached by antibodies in vivo, because the protein in the Rh control may have brought the pre-sensitized red cells together to form an agglutinin. Hence, a DAT is done to prove whether the red cells are pre-sensitized or not.
When entering blood gas results manually into the LIS what should be done if a critical value warning appears when a pH value of 7.14 is entered?
a. immediately consult with the supervisor
b. Check for decimal placement
c. Follow the critical value protocol
d. look at previous results
C
Explanation: Any question that has an answer which tells you to follow the protocol/SOP, it’s the always the safest answer and most likely the right one. Delta check is not a useful tool on pH values, since they can fluctuate immensely depending on the patient’s condition.
Gram positive cocci in chains which grow as small translucent colonies surrounded by a clear zone of hemolysis were isolated from a throat swab. These colonies were found to be catalase negative and resistant to a 0.04 unit bacitracin disc. Which of the following tests would be most useful in identifying this organism?
a. antibody latex agglutination
b. Taxo A disk
c. tube coagulase test
d. susceptibility to optochin
A
Explanation: Catalase neg indicates that this bug is a Strep not Staph. Clear zone indicates that it is a Beta-hem Strep rather than an Alpha-hem Strep (i.e. not S. pneumo and viridans). Amongst the beta-hem strep, Group A Strep is susceptible to bacitracin (Taxo A); hence, this can be either Group C or G. In order to differentiate the two, latex agglutination is the best choice.
What should be done with patient results that are outside reference intervals but are within the percent differences programmed into the delta check?
a. repeat the assay
b. Release the results immediately after review
c. phone the results to the doctor
d. have the specimen recollected
B
Explanation: If the result did not violate delta check, there is no need to question whether any error (technical error, ID error…etc.) is involved. It is unnecessary to phone the doctor since the question didn’t state that the result is critical.
What statement about the dry reagent strip blood test is true?
a. a non-hemolyzed trace is present when there are 1-2 RBC/high power field
b. test for blood is positive for hemoglobin and myoglobin
c. test for blood is based on its reaction with oxidase
d. salicylates cause a false positive reaction
B
Explanation: Both hemoglobin and myoglobin will react with the H2O2 on the pad, which gives a positive result. The colour of the urine (brown vs red) and microscopic examination (red cells) can be used to differentiate haemoglobin from myoglobin. Oxidizing reagents such as bleach can also cause a false positive reaction for blood test, however, salicylate is reducing reagent. Oxidase is used to test for urine glucose.
Which of the following antibodies would not show any reactivity when tested against a panel of reagent red blood cells which had been treated with ficin?
a. Anti-Fya
b. Anti-K
c. Anti-E
d. Anti-Jkb
A
Explanation: Enzymes (e.g. ficin and papain) destroy anti-M, N, S, and Duffy (Fya and Fyb).
What does staining leukemic cells for myeloperoxidase help to distinguish?
a. CML from CLL
b. CLL from ALL
c. AML from ALL
d. AML from CML
C
Explanation: MPO (myeloperoxidase) is an enzyme found in neutrophil primary granules and monocyte lysosomes that catalyze H2O2. Both cells are found in AML, hence very reactive with H2O2 in the stain. ALL is mainly lymphocytes, which has no MPO.
A suspect Enterobacteriaceae isoalte was inoculated to a TSI and decarboxylase test. The TSI showed K/A with gas. The decarboxylase showed:
base: purple, turbid
lysine: purple, turbid
ornithine: purple, turbid
a. the organism is a non-fermentative bacillus
b. the organism is viable but does not decarboxylate the amino acid
c. the test is invalid and must be repeated
d. the organism is negative for lysine and ornithine
C
Explanation:
TSI: 0.1% gluc, 1.0% lactose and sucrose, rest is peptone (protein)
Glucose will first be used up (entire tube turns yellow). If the bug is a lactose fermenter, lactose and sucrose will be utilized next and the peptone will remain untouched (entire tube remains yellow = A/A). However, if the bug does not ferment lactose, it will directly move from glucose to protein. Since deamination of protein occurs aerobically (i.e. the top), only the top will change from yellow to red (top is red, bottom stays yellow due to glucose metabolism = K/A). If the bug is a non-fermenter, it can still metabolize (not ferment) glucose, but only weak acids are produce. Hence, the tube will not turn yellow and will remain a peachy red colour. The protein is then metabolized, turning the top red (top is red, bottom retains original peachy red colour = K/NC).
Summary:
K/A = ferments glucose and utilizes the protein (NLF)
A/A = ferments glucose, lactose, and sucrose (LF)
K/NC = utilizes protein (NFB)
Decarboxylase Test: small amt of glucose and lots of protein (lysine/ornithine); base tube contains glucose only
NLF and LF will first ferment the glucose, producing strong acids that turn the entire tube yellow. If the bugs don’t use ornithine or lysine, the tube will remain yellow. If the bugs metabolize lysine or ornithine or both, they will move on to the protein once glucose is depleted, giving the tube a purple colour. Since the base only contains glucose, NLF and LF will make the base turn yellow. However, since non-fermenters do not ferment glucose, but still utilizes glucose through other pathways, only weak acids are formed. The weak acids cannot turn the tube yellow; hence, the base tube will have a turbid purple colour.
Summary:
NLF and LF ferment the glucose in the base and produce strong acid yellow
NFB utilizes glucose in base, but produces weak acid turbid/purple
Since the TSI result (K/A) tells us that the bug is an NLF, but the decarboxylase test indicates that the bug is an NFB. The conflicting results are deemed invalid.
Given the following patient test results, what is the creatinine clearance?
serum creatinine 100 μmol/L
urine creatinine 8.64 mmol/L
24 hour urine volume 2000 mL
BSA 1.73m^2
a. 120 mm/min
b. 16 mm/min
c. 16 ml/min/1.73m^2
d.120 ml/min/1.73m^2
A
Explanation: **Be careful with the units:
i) Convert umol to mol for serum creatinine
ii) Convert 24 hours to 1440 min (because answer is in minutes)
Ucr (mg/dL) Vur (mL/24hrs) over Pcr (mg/dL) 1440 minutes/24hrs times 1.73 over A
- Cr = creatinine clearance
- Ucr = urine creatinine clearance
- Vur = urine volume excreted in 24 hours
- Pcr = serum creatinine concentration
- 1.73/A= normalization factor for body surface area