INCLUSIVE EXAM 2025 | ACTUAL REAL EXAM ACCURATE
QUESTIONS AND ANSWERS WITH RATIONALES | VERIFIED
AND LATEST UPDATED | CONTAINS A STUDY GUIDE AT THE
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How is the Mean Cell Volume calculated?
A) (Hemoglobin/RBCs) x 10
B) (Hemoglobin/ RBCs) x 100
C) (Hematocrit %/ RBCs) x 10
D) (Hematocrit % / Hemoglobin) x 100
C) (Hematocrit %/ RBCs) x 10
The Mean Cell Volume, or MCV, can be calculated by dividing the hematocrit (%) by the red
blood cell count and multiplying by 10. The MCV describes the volume, or size, of the red blood
cell population.
The Mean Cell Hemoglobin, or the MCH, can be calculated by dividing the hemoglobin (g/dL)
by the red blood cell count and multiplying by 10. The MCH represents the average weight of
hemoglobin in red blood cells.
The Mean Cell Hemoglobin Concentration, or the MCHC can be calculated by dividing the
hemoglobin (g/dL) by the hematocrit (%) and multiplying by 100. The MCHC is the ratio of
hemoglobin mass to volume of RBCs.
Which of the following is characteristic of an exudate effusion?
A) Leukocyte count greater than 1,000 WBC/microliter
B) Clear appearance
C) Protein concentration less than 3.0g/dL
D) Absence of fibrinogen
A) Leukocyte count greater than 1,000 WBC/microliter
Effusion is the increase of fluid between the serous membranes due to disruption of the
mechanisms of formation and reabsorption of serous fluid. Exudates are caused by conditions
,affecting the particular cavity such as infections and malignancies.
Exudate effusion characteristics include:
,- Leukocyte count greater than 1,000WBC/microliters
- Fluid: Serum Protein ratio >0.5
- Yellow or cloudy appearance
- Possible spontaneous clotting
Which bacterial species is most likely represented by the spreading, gray-white, beta-hemolytic
colonies with a "frosted glass" appearance as illustrated in the photograph?
A) Burkholderia pseudomallei
B) Listeria monocytogenes
C) Bacillus cereus
D) Streptomyces anulatus
C) Bacillus cereus
Bacillus cereus colonies are characteristically spreading gray-white, and beta hemolytic with
"frosted glass" appearance.
Burkholderia pseudomallei also produces spreading colonies, but they are more wrinkled in
appearance and without beta hemolysis
Listeria monocytogenes produces small, round, translucent colonies that characteristically have a
narrow zone of beta hemolysis, noted if colonies are removed.
The nocardioform bacteria Streptomyces (formerly gresus) anulatus produces opaque, chalky,
white colonies, This species can also be suspected if the colonies have a pungent, musty
basement odor.
The element that is indicated by the arrow is a:
A) Red blood cell cast
B) Granular cast
C) Hyaline cast
D) Renal tubular epithelial cast
A) Red blood cell cast
This is a red blood cell cast. RBCs can be identified within the matrix of the cast. The
identification of the cast as a red blood cell cast is supported by the appearance of RBCs outside
of the cast.
Granular casts are composed of plasma protein aggregates and cellular remnants. Their presence
in the urine may be of pathologic or onpathologic significane. Granular casts appear as coarse or
fine granules in cast matrix.
, Hyaline cats generally appear colorless, homogeneous, and transparent with rounded ends. They
can also be seen as a result of dehydration, after diuretic therapy, in renal disease, and transiently
as a result of exercise.
Renal tubular epithelial (RTE) casts are cause by the sloughing of necrotic renal tubular cells due
to advance tubular destruction. RTE cells are usually seen as a part of the cast.
A mean value of 100 mg/dL and a standard deviation 1.8mg/dL were obtained from a set of
glucose measurements on a control solution. Based on 95% confidence interval (in mg/dL), what
would be the calculated 2 SD upper and lower limits for this control?
A) 94.6 to 105.4
B) 96.4 to 103.6
C) 97.3 to 102.7
D) 98.2 to 101.8
B) 96.4 to 103.6
The 95% confidence interval can also be described as the probability that 95% of the time, the
quality control will fall within +/- 2 standard deviation of the mean. In this case, 2 standard
deviations would be calculated as 1.8mg/dL multiplied by 2 which equals 3.6mg/dL.
Therefore, the calculated 2SD (3.6mg/dL plus and minus the mean) based on the 95% confidence
interval would be 96.4 - 103.6.
Which of the following sources of error will give a false negative result in antihuman globulin
testing?
A) Low pH of saline
B) Dirty glassware
C) Samples collected in gel separator tubes
D) Refrigerated specimen
C) Samples collected in gel separator tubes
A low pH of saline is indicated as a false negative source of error in antihuman globulin testing.
Dirty glassware, samples collected in gel separator tubes, and refrigerated specimens are all
indicated as false positive sources of error in antihuman globulin testing.
Lymphocyte pleocytosis refers to a(n) in a CSF when compared to a normal shape.
A) Increased number of lymphocytes
B) Decreased number of lymphocytes