Page 1 of 401
AMT MLT (MEDICAL LABORATORY TECHNICIANS) EXAM
ACTUAL EXAM TESTBANK COMPLETE 1000+ QUESTIONS AND
VERIFIED SOLUTIONS LATEST UPDATE THIS YEAR
AMT MLT EXAM A
A 56-year-old woman with a history of high blood pressure and cardiovascular disease is
evaluated by her physician due to her recent loss of appetite. The doctor reviews the following
test results:
Plasma creatinine = 2.4mg/dL
Urine creatinine = 65mg/dL
Urine volume (24hrs) = 1400mL
What is the estimated glomerular filtration rate for this patient?
A) 65mL/min
B) 26.32mL/min
C) 51.69mL/min
D) 37,916mL/min - ANSWER-B) 26.32mL/min
The general glomerular filtration rate calculation is as follows:
, Page 2 of 401
(Urine creatinine/plasma creatinine) x (volume/minute) = glomerular filtration rate (mL/min)
*in order to solve this problem, the 24-hr urine must be converted into minutes. 1440 minutes
should be used as the denominator in the volume component.
So...for this patient, the calculation would be:
(65mg/dL/2.4mg/dL) x (1400mL/1440min) = (27.083) x (0.972mL/min) = 26.32mL/min
QUESTION: The vast MAJORITY of would-be invaders are killed or inactivated primarily by
which part of the immune system?
A) Cell mediated immunity
B) Specific immunity
C) Humoral immunity
D) Innate immunity - ANSWER-D) Innate immunity
The innate immunity system is inherent and nonspecific, meaning, that all pathogens are
attacked similarly. The skin, mucus in respiratory tract, acid pH, and others are all examples of
the innate immunity system that the body has to prevent infection upon first exposure.
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Cell mediated immunity is part of the specific or adaptive immunity system that will react when
pathogens breach the initial barriers of the inmate immune system.
Specific immunity, also called adaptive or acquired immunity, is a result of a pathogen
breaching one of the inmate barriers, being processed by antigen processing cells resulting in a
cell mediated, humoral response or both.
Humoral immunity is the aspect of immunity that is mediated by macromolecules found in
extracellular fluids such as secreted antibodies, complement proteins,, and certain
antimicrobial peptides. Humoral immunity is so named because it involves substances found in
the humors, or body fluids.
QUESTION: An increased number if the cells seen in the image, upon microscopic examination
of urine is termed:
A) Glycosuria
B) Hematuria
C) Uremia
D) Normal urine - ANSWER-B) Hematuria
Hematuria indicates the presence of red blood cells in the urine.
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Glycosuria indicates glucose in the urine.
Uremia indicates an increase of urea nitrogen in the blood.
QUESTION: A semen specimen was collected three hours before it was brought to the
laboratory for examination. What course of action should be taken?
A) Complete macroscopic and microscopic examination as quickly as possible.
B) Report the specimen as compromised on the final report
C) Report the macroscopic and morphology procedures only.
D) Perform the wet mount only. - ANSWER-B) Report the specimen as compromised on the final
report
Since the recommended time limit between collection and analysis is one hour, at three hours,
the technologist should report the specimen as compromised on the final report.
The semen specimen is examined under a microscope to determine the concentration, motility
(movement), and morphology (appearance and shape) of the sperm. Since it is important to
observe the sperm while they are still active, samples must be received for analysis within one
hour.
AMT MLT (MEDICAL LABORATORY TECHNICIANS) EXAM
ACTUAL EXAM TESTBANK COMPLETE 1000+ QUESTIONS AND
VERIFIED SOLUTIONS LATEST UPDATE THIS YEAR
AMT MLT EXAM A
A 56-year-old woman with a history of high blood pressure and cardiovascular disease is
evaluated by her physician due to her recent loss of appetite. The doctor reviews the following
test results:
Plasma creatinine = 2.4mg/dL
Urine creatinine = 65mg/dL
Urine volume (24hrs) = 1400mL
What is the estimated glomerular filtration rate for this patient?
A) 65mL/min
B) 26.32mL/min
C) 51.69mL/min
D) 37,916mL/min - ANSWER-B) 26.32mL/min
The general glomerular filtration rate calculation is as follows:
, Page 2 of 401
(Urine creatinine/plasma creatinine) x (volume/minute) = glomerular filtration rate (mL/min)
*in order to solve this problem, the 24-hr urine must be converted into minutes. 1440 minutes
should be used as the denominator in the volume component.
So...for this patient, the calculation would be:
(65mg/dL/2.4mg/dL) x (1400mL/1440min) = (27.083) x (0.972mL/min) = 26.32mL/min
QUESTION: The vast MAJORITY of would-be invaders are killed or inactivated primarily by
which part of the immune system?
A) Cell mediated immunity
B) Specific immunity
C) Humoral immunity
D) Innate immunity - ANSWER-D) Innate immunity
The innate immunity system is inherent and nonspecific, meaning, that all pathogens are
attacked similarly. The skin, mucus in respiratory tract, acid pH, and others are all examples of
the innate immunity system that the body has to prevent infection upon first exposure.
, Page 3 of 401
Cell mediated immunity is part of the specific or adaptive immunity system that will react when
pathogens breach the initial barriers of the inmate immune system.
Specific immunity, also called adaptive or acquired immunity, is a result of a pathogen
breaching one of the inmate barriers, being processed by antigen processing cells resulting in a
cell mediated, humoral response or both.
Humoral immunity is the aspect of immunity that is mediated by macromolecules found in
extracellular fluids such as secreted antibodies, complement proteins,, and certain
antimicrobial peptides. Humoral immunity is so named because it involves substances found in
the humors, or body fluids.
QUESTION: An increased number if the cells seen in the image, upon microscopic examination
of urine is termed:
A) Glycosuria
B) Hematuria
C) Uremia
D) Normal urine - ANSWER-B) Hematuria
Hematuria indicates the presence of red blood cells in the urine.
, Page 4 of 401
Glycosuria indicates glucose in the urine.
Uremia indicates an increase of urea nitrogen in the blood.
QUESTION: A semen specimen was collected three hours before it was brought to the
laboratory for examination. What course of action should be taken?
A) Complete macroscopic and microscopic examination as quickly as possible.
B) Report the specimen as compromised on the final report
C) Report the macroscopic and morphology procedures only.
D) Perform the wet mount only. - ANSWER-B) Report the specimen as compromised on the final
report
Since the recommended time limit between collection and analysis is one hour, at three hours,
the technologist should report the specimen as compromised on the final report.
The semen specimen is examined under a microscope to determine the concentration, motility
(movement), and morphology (appearance and shape) of the sperm. Since it is important to
observe the sperm while they are still active, samples must be received for analysis within one
hour.
