ASCP MLT Practice Test Questions
board practice
A 46-year old known alcoholic with liver damage is brought in the ER unconscious.
One would expect his lipid values to be affected in what way?
A. Increased
B. Decreased
C. Normal
D. Unaffected by the alcoholism - -------------ANSWER: A;
High triglycerides may be caused by disorders such as type 2 diabetes,
hypothyroidism, Cushing's sydnrome, liver disease, uremia, dysglobulinemia,
nephrotic syndrome, and alcoholism can cause hypertriglyceridemia.
A dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400 - -------------ANSWER: B;
A dilution commonly used for a routine sperm count is a 1:20.
A laboratory manager wants to evaluate the timeliness of patient services in order to
prevent medical errors caused by delay in treatment. Measuring the turnaround time
for which of these analytes would provide the most valuable information?
A. Routine glucose tests from patients on medical floors
B. Lipid panels from the outpatient clinic
C. Troponin tests from the emergency department - -------------ANSWER: C;
Measuring the turnaround times of troponin tests from the emergency department
would provide valuable information. This data represents circumstances in which
patient safety may be compromised if results are delayed and treatment is not
started as soon as possible.
A manual white blood cell count was performed by the hematology technologist. The
cell counts for both sides were 152 and 164 respectively. All nine large squares were
counted on each side. The dilution for this kit was pre-measured at 1:100. What
should the technologist report as the white cell count?
A. 177.5 x 10^9/L
B. 17.5 x 10^9/L
,C. 1.75 x 10^9/L
D. 175 x 10^9/L - -------------ANSWER: B;
Calculation:
Cells Counted (in this case the average of both sides) X dilution factor (in this case
100) / # of squares counted (in this case 9) X area of each square (1mm2) X 0.1mm
(depth factor)
So, in this problem:
158 x x 1 x 0.1mm = 17555.55/mm3 (can be converted to 17.5 x 109/L*)
*There are 1,000,000 mm3 in a liter (L). So 17555.55 X 1,000,000 = 17.5 x 109/L
A microscopic examination of a normal urine pH 8.0 shows 2+ yellow-brown thorny
spheres which are MOST probably:
A. ammonium biurate crystals
B. ampicillin crystals
C. amorphous urate crystals
D. crenated red cells
E. waxy casts - -------------ANSWER: A;
Ammonium biurate crystalsare typically round, irregularly spiked and yellow-brown in
color.
A normal hemoglobin molecule is comprised of the following:
A. Ferrous iron and four globin chains
B. Four heme and four globin chains
C. Four heme and one globin chains
D. One heme and four globin chains - -------------ANSWER: B;
A normal hemoglobin molecule consists of a tetramer, of four heme and four globin
chains, with a molecular weight of 64,500 daltons. Each of the four units can bind a
molecule of oxygen for transport to the body's tissues. In the image shown below,
there are four monomers (2 red globin chains and 2 blue globin chains) which form
the entire hemoglobin tetramer structure. The green portions represent the 4 heme
groups.
A patient has been characterized as a CYP2D6 poor metabolizer (PM) after
genotyping. Which of the following statements is not true?
A. The patient will likely need lower doses of CYP2D6-metabolized drugs.
B. The patient is less likely to require therapeutic drug monitoring (TDM) since the
genotype is known.
,C. CYP2D6 metabolizes many drugs, and so attention must be given to the doses of
drugs from different classes. - -------------ANSWER: B;
Though it may not be required, TDM should still be used to confirm adequate dosing.
Genotyping does not make TDM redundant.
A PM will metabolize the drug more slowly and therefore will need lower doses.
CYP2D6 metabolizes many different drugs; it is not associated with just one class of
drugs. Anytime a drug is taken that competes for the same metabolizing enzyme as
another drug, there is potential for the concentrations of both drugs to be increased.
A patient is admitted to the emergency room with lethargy and pallor. The CBC
results are as follows:
RBC = 4.1 x 1012/L
Hemoglobin = 7.9 g/cL
Hematocrit = 29%
How would you classify this anemia?
A. microcytic, hypochromic
B. normocytic, normochromic
C. macrocytic, normochromic
D. microcytic, hyperchromic - -------------ANSWER: A;
First, the RBC indices must be calculated. The MCV ((Hct/RBC) x 10) = 71 fL. Since
the reference range for the MCV is 80-100 fL, this anemia would be classified as
microcytic. The MCH ((Hgb/RBC) x 10) = 19.3 pg. Since the reference range for the
MCH is 27-33 pg, this would be considered hypochromic. Finally, the MCHC
((Hgb/Hct) X 100) = 27%. Since the normal range for the MCHC is 33%-36%, this
would indicate hypochromia which correlates with the MCH findings. The correct
answer is therefore microcytic, hypochromic anemia.
A patient is taking cimetidine for a stomach ulcer. This drug inhibits CYP2D6. The
patient is now prescribed amphetamine for narcolepsy. Amphetamine is metabolized
by CYP2D6. What would you predict?
A. The dose for the amphetamine needs to be lower than normal.
B. The dose for amphetamine needs to be higher than normal.
C. Nothing can be assumed until you know the patient's status (PM, EM, UM).
D. The two drugs can never be given together since they interact. -
-------------ANSWER: A;
Since cimetidine inhibits CYP2D6, less amphetamine will probably need to be given
since it will not be able to be metabolized as readily.
, Most drug interactions are like this: one drug inhibits or competes with the same
CYP450 as another drug. The end result is that higher concentrations of one, or
both, drugs are present, leading to potential toxicity.
A patient who has a primarily vegetarian diet will most likely have an acid urine pH. -
-------------ANSWER: False
A patient who has a primarily vegetarian diet will most likely have an alkaline urine
pH. A low-carbohydrate diet as well as the ingestion of citrus fruits can also lead to a
more alkaline urine sample.
A patient with an infectious mononucleosis infection presents in the emergency
room. Physicians order a spinal tap which is immediately sent to the laboratory for
review. Please identify the cell in the image below from this patient's cerebrospinal
fluid sample.
A. Reactive Lymphocyte
B. Monocyte
C. Macrophage
D. Mesothelial Cell - -------------ANSWER: A;
The cell depicted with the arrow in this image is an atypical (reactive) lymphocyte.
These cells are common found in certain viral infections, especially infectious
mononucleosis. Notice the larger size and abundant cytoplasm present in this
lymphocyte. There is also apparent vacuoliation which is a key feature of atypical
lymphocytes. The chromatin pattern of this cell as well as the overall shape, color
and size rules out the monocyte, macrophage, and mesothelial cell choices.
A refrigerator used to store whole blood must be able to maintain a temperature in
the ranges of:
A. 0 - 4 degrees Celsius
B. 2 - 4 degrees Celsius
C. 2 - 8 degrees Celsius
D. 1 - 6 degrees Celsius - -------------ANSWER: D;
After collection all blood should be stored at 1 - 6oC, unless it is going to be used as
a source of platelets.
A sample of cerebrospinal fluid is diluted 1:100; the standard 9 squares of a
hemocytometer are counted on each side for a total of 18 large squares.
Side 1-- 186 nucleated cells counted
Side 2-- 184 nucleated cells counted
total nucleated cells = 370
board practice
A 46-year old known alcoholic with liver damage is brought in the ER unconscious.
One would expect his lipid values to be affected in what way?
A. Increased
B. Decreased
C. Normal
D. Unaffected by the alcoholism - -------------ANSWER: A;
High triglycerides may be caused by disorders such as type 2 diabetes,
hypothyroidism, Cushing's sydnrome, liver disease, uremia, dysglobulinemia,
nephrotic syndrome, and alcoholism can cause hypertriglyceridemia.
A dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400 - -------------ANSWER: B;
A dilution commonly used for a routine sperm count is a 1:20.
A laboratory manager wants to evaluate the timeliness of patient services in order to
prevent medical errors caused by delay in treatment. Measuring the turnaround time
for which of these analytes would provide the most valuable information?
A. Routine glucose tests from patients on medical floors
B. Lipid panels from the outpatient clinic
C. Troponin tests from the emergency department - -------------ANSWER: C;
Measuring the turnaround times of troponin tests from the emergency department
would provide valuable information. This data represents circumstances in which
patient safety may be compromised if results are delayed and treatment is not
started as soon as possible.
A manual white blood cell count was performed by the hematology technologist. The
cell counts for both sides were 152 and 164 respectively. All nine large squares were
counted on each side. The dilution for this kit was pre-measured at 1:100. What
should the technologist report as the white cell count?
A. 177.5 x 10^9/L
B. 17.5 x 10^9/L
,C. 1.75 x 10^9/L
D. 175 x 10^9/L - -------------ANSWER: B;
Calculation:
Cells Counted (in this case the average of both sides) X dilution factor (in this case
100) / # of squares counted (in this case 9) X area of each square (1mm2) X 0.1mm
(depth factor)
So, in this problem:
158 x x 1 x 0.1mm = 17555.55/mm3 (can be converted to 17.5 x 109/L*)
*There are 1,000,000 mm3 in a liter (L). So 17555.55 X 1,000,000 = 17.5 x 109/L
A microscopic examination of a normal urine pH 8.0 shows 2+ yellow-brown thorny
spheres which are MOST probably:
A. ammonium biurate crystals
B. ampicillin crystals
C. amorphous urate crystals
D. crenated red cells
E. waxy casts - -------------ANSWER: A;
Ammonium biurate crystalsare typically round, irregularly spiked and yellow-brown in
color.
A normal hemoglobin molecule is comprised of the following:
A. Ferrous iron and four globin chains
B. Four heme and four globin chains
C. Four heme and one globin chains
D. One heme and four globin chains - -------------ANSWER: B;
A normal hemoglobin molecule consists of a tetramer, of four heme and four globin
chains, with a molecular weight of 64,500 daltons. Each of the four units can bind a
molecule of oxygen for transport to the body's tissues. In the image shown below,
there are four monomers (2 red globin chains and 2 blue globin chains) which form
the entire hemoglobin tetramer structure. The green portions represent the 4 heme
groups.
A patient has been characterized as a CYP2D6 poor metabolizer (PM) after
genotyping. Which of the following statements is not true?
A. The patient will likely need lower doses of CYP2D6-metabolized drugs.
B. The patient is less likely to require therapeutic drug monitoring (TDM) since the
genotype is known.
,C. CYP2D6 metabolizes many drugs, and so attention must be given to the doses of
drugs from different classes. - -------------ANSWER: B;
Though it may not be required, TDM should still be used to confirm adequate dosing.
Genotyping does not make TDM redundant.
A PM will metabolize the drug more slowly and therefore will need lower doses.
CYP2D6 metabolizes many different drugs; it is not associated with just one class of
drugs. Anytime a drug is taken that competes for the same metabolizing enzyme as
another drug, there is potential for the concentrations of both drugs to be increased.
A patient is admitted to the emergency room with lethargy and pallor. The CBC
results are as follows:
RBC = 4.1 x 1012/L
Hemoglobin = 7.9 g/cL
Hematocrit = 29%
How would you classify this anemia?
A. microcytic, hypochromic
B. normocytic, normochromic
C. macrocytic, normochromic
D. microcytic, hyperchromic - -------------ANSWER: A;
First, the RBC indices must be calculated. The MCV ((Hct/RBC) x 10) = 71 fL. Since
the reference range for the MCV is 80-100 fL, this anemia would be classified as
microcytic. The MCH ((Hgb/RBC) x 10) = 19.3 pg. Since the reference range for the
MCH is 27-33 pg, this would be considered hypochromic. Finally, the MCHC
((Hgb/Hct) X 100) = 27%. Since the normal range for the MCHC is 33%-36%, this
would indicate hypochromia which correlates with the MCH findings. The correct
answer is therefore microcytic, hypochromic anemia.
A patient is taking cimetidine for a stomach ulcer. This drug inhibits CYP2D6. The
patient is now prescribed amphetamine for narcolepsy. Amphetamine is metabolized
by CYP2D6. What would you predict?
A. The dose for the amphetamine needs to be lower than normal.
B. The dose for amphetamine needs to be higher than normal.
C. Nothing can be assumed until you know the patient's status (PM, EM, UM).
D. The two drugs can never be given together since they interact. -
-------------ANSWER: A;
Since cimetidine inhibits CYP2D6, less amphetamine will probably need to be given
since it will not be able to be metabolized as readily.
, Most drug interactions are like this: one drug inhibits or competes with the same
CYP450 as another drug. The end result is that higher concentrations of one, or
both, drugs are present, leading to potential toxicity.
A patient who has a primarily vegetarian diet will most likely have an acid urine pH. -
-------------ANSWER: False
A patient who has a primarily vegetarian diet will most likely have an alkaline urine
pH. A low-carbohydrate diet as well as the ingestion of citrus fruits can also lead to a
more alkaline urine sample.
A patient with an infectious mononucleosis infection presents in the emergency
room. Physicians order a spinal tap which is immediately sent to the laboratory for
review. Please identify the cell in the image below from this patient's cerebrospinal
fluid sample.
A. Reactive Lymphocyte
B. Monocyte
C. Macrophage
D. Mesothelial Cell - -------------ANSWER: A;
The cell depicted with the arrow in this image is an atypical (reactive) lymphocyte.
These cells are common found in certain viral infections, especially infectious
mononucleosis. Notice the larger size and abundant cytoplasm present in this
lymphocyte. There is also apparent vacuoliation which is a key feature of atypical
lymphocytes. The chromatin pattern of this cell as well as the overall shape, color
and size rules out the monocyte, macrophage, and mesothelial cell choices.
A refrigerator used to store whole blood must be able to maintain a temperature in
the ranges of:
A. 0 - 4 degrees Celsius
B. 2 - 4 degrees Celsius
C. 2 - 8 degrees Celsius
D. 1 - 6 degrees Celsius - -------------ANSWER: D;
After collection all blood should be stored at 1 - 6oC, unless it is going to be used as
a source of platelets.
A sample of cerebrospinal fluid is diluted 1:100; the standard 9 squares of a
hemocytometer are counted on each side for a total of 18 large squares.
Side 1-- 186 nucleated cells counted
Side 2-- 184 nucleated cells counted
total nucleated cells = 370
