MLT ASCP BOARD EXAM LATEST ACTUAL EXAM 100 QUESTIONS AND CORRECT DETAILED ANSWERS WITH RATIONALES (VERIFIED ANSWERS) |ALREADY GRADED A

MLT ASCP BOARD EXAM LATEST 2023-2024 ACTUAL

EXAM 100 QUESTIONS AND CORRECT DETAILED

ANSWERS WITH RATIONALES (VERIFIED ANSWERS)

|ALREADY GRADED A

For which of the following antibodies is the DAT most likely to be negative when

testing a newborn for possible HDFN?

A. anti-A

B. anti-c

C. anti-D

D. anti-K

E. anti-Fya - ANSWER- A;

Rationale: The DAT is most likely to be negative in ABO HDFN. It's possible

that the washing done as part of the DAT may break the bonds between antiA (or anti-B) and the newborn's poorly developed A (or B) antigens.

General Hospital is considering the addition of a new chemistry panel containing

12 tests. The laboratory is asked to calculate the total cost of quality control per

new chemistry test panel. Quality control must be performed 3 times per day

(every 8 hours). The labor cost per quality control test for this panel is $2.63. A

month's worth of quality control reagent costs $354.00. What is the total quality

control cost per new chemistry test panel if 76,000 of these new panels are

performed each y - ANSWER- B;

This scenario's answer can be calculated by first deciding what the total

quality control labor costs are as well as what the total consumable costs are.

In this case, if quality control is run 3 times per day, a total of 1095 quality

control runs are performed each year. The direct labor cost of $2.63

multiplied by 1095 quality control runs equals $2879.85 per year in quality

control direct labor. The hospital pays $354.00 per month on quality control

consumables, which equals $4248.00 per y

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;

Rationale: 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.

Which of the following immunoglobulin classes is chiefly responsible for the

degranulation of mast cells and basophils:

A. IgG

B. IgA

C. IgM

D. IgE - ANSWER- D;

Rationale: IgE levels are often increased in patients with allergic disease. IgE

binds to the membranes of mast cells and basophils, and if specific antigen is

present to react with the IgE molecule, degranulation of these cells occurs,

releasing histamines, and other substances into the blood or tissues.

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;

Rationale: Ammonium biurate crystalsare typically round, irregularly spiked

and yellow-brown in color.

Electrophoretic separation of hemoglobin fundamentally relies on:

A. Size differences of molecules

B. Net charge differences of molecules

C. Concentration differences of molecules

D. Shape variations of molecules - ANSWER- B;

Rationale: Hemoglobin electrophoresis uses an electric field to separate

hemoglobin molecules based on differences in net electrical charge. The rate

of electrophoretic migration is also dependent on the ionic radius of the

molecule, the viscosity of the solution through which it is migrating, the

electrical field strength, temperature, and the type of supporting medium

used.

True/False

For those facilities that in the interest of safety use a special calculation for RhIg

dosage, regardless if they round up or round down, they always add one vial. -

ANSWER- True

Rationale: Regardless of rounding up or down, when calculating RhIg dosage,

such facilities always add one vial. For example:

1.4 = 1 +1 = 2 vials

1.5 = 2 +1 = 3 vials

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

Using the standard hemocytometer formula shown on the right, what is the

nucleated cell count per microliter (µL)?

A. 1.03 x 10^4

B. 2.06 x 10^4

C. 4.62 x 10^4

D. 9.25 x 10^4 - ANSWER- B;

Using the formula on the right,

Cells/µL = 370 x 100 / 18 x 0.1

Cells/µL = 37000 / 1.8

Cells/µL = 20556 or 2.06 x 104