Ace ASCP CG (Classic Cytogenetics) Certification
Final Exam Questions Comprehensive Resource To Help
You Ace 2026-2027 Exams Includes Frequently Tested
Questions With ELABORATED 100%
Correct COMPLETE SOLUTIONS
Guaranteed Pass First Attempt!! Current Update!!
1. A newborn infant has a WBC count of 37,000/ul with the differential count revealing the
presence of 50 nRBCs per 100 WBCs. What is the corrected WBC count?
a) 15,000/ul
b) 24,700/ul
c) 27,300/ul
d) 12,500/ul - Correct Answer: B. 24,700/ul
Feedback Corrected WBC count= Uncorrected WBC count x100/nRBCs+100: Thus,
(37,000ulx100/50+100= 24,700ul)
2. Gram positive bacilli; non-motile; non-hemolytic; catalase positive; spore forming:
a) B. anthracis
b) B. cereus
c) N. meningitidis
d) N. lactamica - Correct Answer: B. B. anthracis
Feedback Important distinguishing characteristics between B. anthracis and B. cereus is motility,
B. anthracis is non motile while B. cereus is motile.
3. Hair perforation test/Hair baiting test is for identification of which two fungi?
a) Aspergillus niger and A. japonicus
b) Tinea pedis and T. corporis
c) Trichophyton mentagrophyte and T. rubrum
d) C. albicans and C. tropicalis - Correct Answer: C. Trichophyton mentagrophyte and T.
rubrum
,Feedback Trichophyton mentagrophyte is positive and T. rubrum is negative
4. What is the minimum platelet count before you perform apheresis?
a) 500
b) 200
c) 150
d) 250 - Correct Answer: C. 150.
Feedback Apheresis donors with low normal pre-procedure platelet counts (150 - 200 x 10 9/L)
and Hb concentration (12.5 - 13 g/dL) should be examined for post-donation drops in these
haematological parameters
5. Gram negative bacteria, normally found in the oropharyngeal tract of mammals, they are
involved in the pathogenesis of some animal bite wounds as well as periodontal diseases. This
bacteria exhibits GLIDING MOTILITY:
a) Leptospira
b) Actinobacillus
c) Capnocytophaga
d) Chlamydia - Correct Answer: C. Capnocytophaga
Feedback The term Capnocytophaga comes from "Capno" for its dependence on CO2 and
"cytophaga" for its flexibility and mobility shift (" gliding motility ").
6. Urine with an specific gravity consistently between 1.002 and 1.003 indicates:
a) Acute glomerulonephritis
b) Renal tubular failure
c) Addison's disease
d) Diabetes insipidus - Correct Answer: D. Diabetes insipidus
Feedback Low specific gravity (SG) (1.001-1.003) may indicate the presence of diabetes
insipidus, a disease caused by impaired functioning of antidiuretic hormone (ADH). Low SG also
may occur in patients with glomerulonephritis, pyelonephritis, and other renal abnormalities.
,7. Primidone monitoring: suspected to have primidone overdose, but the serum concentration
is within the normal range, what should the MLS do next?
a) Measure procainamide concentration
b) Measure phenobarbital concentration
c) Measure phenytoin concentration
d) Eliminate primidone overdose as possible diagnosis - Correct Answer: B. Measure
phenobarbital concentration
Feedback Primidone is a structural analog of phenobarbital.
8. Postprandial turbidity of the serum is due to the presence of?
a) Fatty acid
b) Cholesterol
c) Phospholipid
d) Lipoproteins - Correct Answer: D. Lipoproteins
Feedback For approximately two to eight hours following a meal there is an additional
LIPOPROTEIN class, called chylomicrons, representing the transport of dietary fat absorbed in
the intestine.
9. A patient with an anti-K and an anti-JKa in her plasma needs 2 units of RBC for surgery. How
many group specific units would need to be screened to find 2 units of RBC? The frequency of
Jka+ is 80% and K+ frequency is 15%. - Correct Answer: B. 12
Feedback Formula: Number of units to crossmatch is equivalent to: # of units need by
patient/(frequency of negativity of antigen #1 x frequency of negativity of antigen #2).
Negativity of Jka+ is .20%, Negativity of K+ is .85%. Using the formula, 2 units/(.20x.85)=11.76
rounded to 12.
10. An alkaline urine refrigerated becomes turbid because of?
a) Amorphous phosphates
b) Amorphous urates
c) WBC
d) Bacteria - Correct Answer: A. Amorphous phosphates
, Feedback Amorphous phosphates are usually formed in alkaline urine and they are without a
color. The precipitate of amorphous phosphates is white while Amorphous urates are yellow,
yellow-brown or pinkish in color. They tend to form in acidic urine and their precipitate is pink
with a cloudy appearance of the mixed urine. Amorphous urates are a normal presence in the
urine.
11. Test for Paroxysmal Cold Hemoglobinuria (PCH):
a) Autohemolysis test
b) Donath-Landsteiner Test
c) Sucrose hemolysis test
d) NOTA - Correct Answer: B. Donath-Landsteiner Test
Feedback Normal= (-) hemolysis on test and control, PCH= (-) hemolysis on control but (+)
hemolysis on test sample.
12. In emphysema, the absence of alpha-1 globulin peak is due the deficiency of:
a) A1-Antitrypsin (AAT)
b) Macroglobulin
c) Hemopexin
d) Transferrin - Correct Answer: A1-Antitrypsin (AAT)
Feedback Alpha-1 antitrypsin deficiency is an inherited disorder that may cause lung disease
and liver disease. Affected individuals often develop emphysema, which is a lung disease caused
by damage to the small air sacs in the lungs (alveoli). Characteristic features of emphysema
include difficulty breathing, a hacking cough, and a barrel-shaped chest.
13. Identify the organism: Fig. 4
a) Clostridium difficile
b) Mycobacterium tuberculosis
c) Listeria monocytogenes
d) Corynebacterium diphtheriae - Correct Answer: A. Mycobacterium tuberculosis
Final Exam Questions Comprehensive Resource To Help
You Ace 2026-2027 Exams Includes Frequently Tested
Questions With ELABORATED 100%
Correct COMPLETE SOLUTIONS
Guaranteed Pass First Attempt!! Current Update!!
1. A newborn infant has a WBC count of 37,000/ul with the differential count revealing the
presence of 50 nRBCs per 100 WBCs. What is the corrected WBC count?
a) 15,000/ul
b) 24,700/ul
c) 27,300/ul
d) 12,500/ul - Correct Answer: B. 24,700/ul
Feedback Corrected WBC count= Uncorrected WBC count x100/nRBCs+100: Thus,
(37,000ulx100/50+100= 24,700ul)
2. Gram positive bacilli; non-motile; non-hemolytic; catalase positive; spore forming:
a) B. anthracis
b) B. cereus
c) N. meningitidis
d) N. lactamica - Correct Answer: B. B. anthracis
Feedback Important distinguishing characteristics between B. anthracis and B. cereus is motility,
B. anthracis is non motile while B. cereus is motile.
3. Hair perforation test/Hair baiting test is for identification of which two fungi?
a) Aspergillus niger and A. japonicus
b) Tinea pedis and T. corporis
c) Trichophyton mentagrophyte and T. rubrum
d) C. albicans and C. tropicalis - Correct Answer: C. Trichophyton mentagrophyte and T.
rubrum
,Feedback Trichophyton mentagrophyte is positive and T. rubrum is negative
4. What is the minimum platelet count before you perform apheresis?
a) 500
b) 200
c) 150
d) 250 - Correct Answer: C. 150.
Feedback Apheresis donors with low normal pre-procedure platelet counts (150 - 200 x 10 9/L)
and Hb concentration (12.5 - 13 g/dL) should be examined for post-donation drops in these
haematological parameters
5. Gram negative bacteria, normally found in the oropharyngeal tract of mammals, they are
involved in the pathogenesis of some animal bite wounds as well as periodontal diseases. This
bacteria exhibits GLIDING MOTILITY:
a) Leptospira
b) Actinobacillus
c) Capnocytophaga
d) Chlamydia - Correct Answer: C. Capnocytophaga
Feedback The term Capnocytophaga comes from "Capno" for its dependence on CO2 and
"cytophaga" for its flexibility and mobility shift (" gliding motility ").
6. Urine with an specific gravity consistently between 1.002 and 1.003 indicates:
a) Acute glomerulonephritis
b) Renal tubular failure
c) Addison's disease
d) Diabetes insipidus - Correct Answer: D. Diabetes insipidus
Feedback Low specific gravity (SG) (1.001-1.003) may indicate the presence of diabetes
insipidus, a disease caused by impaired functioning of antidiuretic hormone (ADH). Low SG also
may occur in patients with glomerulonephritis, pyelonephritis, and other renal abnormalities.
,7. Primidone monitoring: suspected to have primidone overdose, but the serum concentration
is within the normal range, what should the MLS do next?
a) Measure procainamide concentration
b) Measure phenobarbital concentration
c) Measure phenytoin concentration
d) Eliminate primidone overdose as possible diagnosis - Correct Answer: B. Measure
phenobarbital concentration
Feedback Primidone is a structural analog of phenobarbital.
8. Postprandial turbidity of the serum is due to the presence of?
a) Fatty acid
b) Cholesterol
c) Phospholipid
d) Lipoproteins - Correct Answer: D. Lipoproteins
Feedback For approximately two to eight hours following a meal there is an additional
LIPOPROTEIN class, called chylomicrons, representing the transport of dietary fat absorbed in
the intestine.
9. A patient with an anti-K and an anti-JKa in her plasma needs 2 units of RBC for surgery. How
many group specific units would need to be screened to find 2 units of RBC? The frequency of
Jka+ is 80% and K+ frequency is 15%. - Correct Answer: B. 12
Feedback Formula: Number of units to crossmatch is equivalent to: # of units need by
patient/(frequency of negativity of antigen #1 x frequency of negativity of antigen #2).
Negativity of Jka+ is .20%, Negativity of K+ is .85%. Using the formula, 2 units/(.20x.85)=11.76
rounded to 12.
10. An alkaline urine refrigerated becomes turbid because of?
a) Amorphous phosphates
b) Amorphous urates
c) WBC
d) Bacteria - Correct Answer: A. Amorphous phosphates
, Feedback Amorphous phosphates are usually formed in alkaline urine and they are without a
color. The precipitate of amorphous phosphates is white while Amorphous urates are yellow,
yellow-brown or pinkish in color. They tend to form in acidic urine and their precipitate is pink
with a cloudy appearance of the mixed urine. Amorphous urates are a normal presence in the
urine.
11. Test for Paroxysmal Cold Hemoglobinuria (PCH):
a) Autohemolysis test
b) Donath-Landsteiner Test
c) Sucrose hemolysis test
d) NOTA - Correct Answer: B. Donath-Landsteiner Test
Feedback Normal= (-) hemolysis on test and control, PCH= (-) hemolysis on control but (+)
hemolysis on test sample.
12. In emphysema, the absence of alpha-1 globulin peak is due the deficiency of:
a) A1-Antitrypsin (AAT)
b) Macroglobulin
c) Hemopexin
d) Transferrin - Correct Answer: A1-Antitrypsin (AAT)
Feedback Alpha-1 antitrypsin deficiency is an inherited disorder that may cause lung disease
and liver disease. Affected individuals often develop emphysema, which is a lung disease caused
by damage to the small air sacs in the lungs (alveoli). Characteristic features of emphysema
include difficulty breathing, a hacking cough, and a barrel-shaped chest.
13. Identify the organism: Fig. 4
a) Clostridium difficile
b) Mycobacterium tuberculosis
c) Listeria monocytogenes
d) Corynebacterium diphtheriae - Correct Answer: A. Mycobacterium tuberculosis
