Simulate real exam conditions with our Free ASCP MLS Exam Practice Questions: Part 3 – (Laboratory Mathematics) mock test! This timed practice exam features 70 carefully curated MCQs covering the entire scope of laboratory science certifications—including ASCP MLS, AMT MLT/MT, AIMS, CSMLS, IBMS, HAAD/DOH, DHA, and MOH. Track your progress, review performance analytics, and conquer time management challenges. Detailed answer keys help you understand mistakes and boost confidence. Perfect for final readiness checks!
Report a question
ASCP MLS Exam MCQs Chapter 3
Prepare for the ASCP Medical Laboratory Scientist (MLS) Certification Exam with 70 free practice questions focused on Laboratory Mathematics (Sections of the BOC Exam Outline). These multiple-choice questions (MCQs) cover critical topics like preanalytical errors, quality control (QC), regulatory compliance (CLIA, HIPAA), method validation, and laboratory safety protocols. Each question includes detailed rationales to reinforce key concepts and test-taking strategies.
1 / 70
Category:
ASCP Exam Questions
Precision is validated by:
Replicate testing validates precision through repeated measurements of the same sample.
CV
Coefficient of Variation
4 / 70
Category:
ASCP Exam Questions
The formula for the arithmetic mean is:
Sum / number of values
💡 That is:
where:
∑𝑥 = sum of all valuesN = number of values6 / 70
Category:
ASCP Exam Questions
How many mL of 0.25 N NaOH are needed to make 100 mL of 0.05 N NaOH?
✅ Dilution Calculation (C₁V₁ = C₂V₂)
C₁V₁ = C₂V₂
0.25 × V₁ = 0.05 × 100
0.25V₁ = 5
V₁ = 5 ÷ 0.25
V₁ = 20 mL
Protocol:
Measure 20 mL of 0.25N stock solution Add diluent to make 100 mL total volume Mix thoroughly to obtain 0.05N solution 💡 This 5-fold dilution (0.25N → 0.05N) requires stock volume to be 1/5 of final volume
7 / 70
Category:
ASCP Exam Questions
How many grams of H2SO4H2SO4 (MW=98) are in 750 mL of 3N H2SO4H2SO4?
✅ Sulfuric Acid Solution Preparation
Step 1: Calculate equivalent weight
Equivalent weight = MW / valence = 98 / 2 = 49 g/eq
Step 2: Apply formula
grams = N × Eq. weight × Volume(L)
Step 3: Calculate mass
= 3 eq/L × 49 g/eq × 0.75 L
= 3 × 49 × 0.75
= 110.25 g
Required mass of H₂SO₄ = 110.25 g
Preparation Protocol:
Weigh 110.25 g concentrated H₂SO₄ Slowly add to ~500 mL distilled water with stirring Dilute to final volume of 750 mL Allow to cool before use (dilution is exothermic) 10 / 70
Category:
ASCP Exam Questions
Limit of detection (LoD) is best described as:
The lowest concentration of an analyte that can be reliably distinguished from background noise.
Background Noise
Reliable Detection
LoD Threshold
Detects presence (yes/no)
Doesn’t quantify exact value
ExampleA COVID-19 test with LoD of 100 copies/mL can detect the virus at that level, but not at 90 copies/mL.
12 / 70
Category:
ASCP Exam Questions
Reliability of a test to detect disease is its:
The ability of a test to correctly identify those with the disease (true positives)
Measures how reliably a test detects the presence of disease
Also known as the true positive rate
High sensitivity tests rarely miss people with the disease
People with Disease
Correctly Detected (TP)
Missed Cases (FN)
Example:90% sensitivity means the test correctly identifies 90% of people who actually have the disease
15 / 70
Category:
ASCP Exam Questions
The molarity of 18.7 g KCl (MW=74.5) in 500 mL water is:
✅ Molarity Calculation
Step 1: Calculate moles of KCl
Moles =
18.7 g 74.5 g/mol
= 0.25 moles
Step 2: Convert volume to liters
500 mL = 0.5 L
Step 3: Calculate molarity
Molarity =
0.25 moles 0.5 L
= 0.5 M
The molarity of the KCl solution is 0.5 M
16 / 70
Category:
ASCP Exam Questions
The analytical measurement range (AMR) is verified to ensure:
The range of analyte concentrations that can be measured directly without:
Sample pretreatment
Dilution
Concentration
Verification Ensures:Accurate results across the entire range No need for sample modification Method reliability at concentration extremes
17 / 70
Category:
ASCP Exam Questions
The sodium content in 100 g NaCl (Na=23, Cl=35.5) is:
✅ Sodium Content in NaCl
Fraction of sodium in NaCl:
23
58.5
= 39.3 g ≈ 40 g
100 g NaCl contains approximately 40 g sodium
19 / 70
Category:
ASCP Exam Questions
Hemoglobin controls have a mean of 15.2 g/dL and SD of 0.20. The ±2 SD range is:
✅ Statistical Range Calculation (±2 SD)
Step 1: Calculate ±2σ range
±2σ = 2 × 0.20 = 0.40
Step 2: Apply to mean
Range = μ ± 2σ = 15.2 ± 0.40
Step 3: Calculate boundaries
Lower bound = 15.2 – 0.40 = 14.8 g/dL
Upper bound = 15.2 + 0.40 = 15.6 g/dL
The ±2 SD range is 14.8 to 15.6 g/dL
This range includes approximately 95% of data points in a normal distribution
21 / 70
Category:
ASCP Exam Questions
A 1 mL serum + 4 mL water is a:
✅ Serum Dilution Calculation
Dilution = Serum Volume : Total Volume
Serum dilution = 1:5
This means 1 part serum diluted in 4 parts water
22 / 70
Category:
ASCP Exam Questions
A laboratory technician performs a recovery experiment with the following results:
– Mixture 1: 0.9 mL patient serum + 0.1 mL water → 89 mEq/L result
– Mixture 2: 0.9 mL same serum + 0.1 mL 800 mEq/L standard → 161 mEq/L result
What percentage of the added standard was recovered by the testing method?
✅ Standard Addition Recovery Calculation 👉 Original sample (Mixture 1) = 89 mEq/L
👉 After adding standard (Mixture 2) = 161 mEq/L
👉 Amount added = 0.1 mL of 800 mEq/L to 1 mL total volume
👉 Actual measured increase:
161 – 89 = 72 mEq/L
Interpretation: The 90% recovery indicates good accuracy, with 10% potentially lost to matrix effects or measurement error.
25 / 70
Category:
ASCP Exam Questions
How many grams of NaCl are needed to prepare 1 L of 0.9% normal saline?
✅ Let’s calculate the amount of NaCl needed for 0.9% saline:
0.9% NaCl means 0.9 g NaCl per 100 mL of solution.
For 1 L (1000 mL) :
NaCl needed = 0.9% × 1000 mL
= (0.9 ÷ 100) × 1000
= 0.009 × 1000
= 9.0 g
Conclusion: To prepare 1 L of 0.9% saline , dissolve 9.0 g NaCl in water and dilute to 1000 mL.
26 / 70
Category:
ASCP Exam Questions
Limit of quantitation (LoQ) is the:
The lowest concentration of an analyte that can be quantitatively measured with acceptable precision and accuracy .
Background
LoD
LoQ
Quantitation Range
Detection Limit
Quantitation Limit
Measures exact concentration
Requires both precision & accuracy
LoDDetects presence (yes/no)
ExampleA drug test with LoQ of 10 ng/mL can accurately measure concentrations as low as 10 ng/mL.
28 / 70
Category:
ASCP Exam Questions
Predictive value of a positive test is:
Measures the proportion of patients with positive test results who are correctly diagnosed
True Positives (TP)
Patients correctly identified with the condition
False Positives (FP)
Patients incorrectly identified with the condition
Example CalculationPPV = (72 ÷ (72 + 8)) × 100 = (72 ÷ 80) × 100 = 90%
90% of positive test results are true positives
30 / 70
Category:
ASCP Exam Questions
How much stock solution is needed to prepare 8 mL of a 75% solution?
✅ Preparing 8 mL of 75% solution from 100% stock:
Volume of stock = C₂ × V₂ / C₁
Where:
C₂ = 75% (desired concentration) V₂ = 8 mL (desired volume) C₁ = 100% (stock concentration) = (75 × 8) / 100
= 600 / 100
= 6.0 mL of stock needed
Protocol:
Measure 6.0 mL of 100% stock solution Add solvent (e.g., water) to reach 8 mL total volume 31 / 70
Category:
ASCP Exam Questions
How many mL of 30% albumin are needed to make 6 mL of 10% albumin?
1 mL
✅ Dilution Formula Calculation
C₁V₁ = C₂V₂
30 × V₁ = 10 × 6
30V₁ = 60
V₁ = 60 ÷ 30 = 2 mL
Take 2 mL of 30% stock and dilute to 6 mL
32 / 70
Category:
ASCP Exam Questions
A pH of 7.0 represents an H+H+ concentration of:
✅ Hydrogen Ion Concentration Calculation
[H⁺] = 10-pH
Step 1: Calculate [H⁺] in mol/L
[H⁺] = 10-7.0 = 10⁻⁷ mol/L
Step 2: Convert mol/L to nmol/L
1 mol/L = 10⁹ nmol/L
10⁻⁷ mol/L = 10⁻⁷ × 10⁹ nmol/L
= 10(-7+9) = 10² nmol/L
= 100 nmol/L
At pH 7.0, hydrogen ion concentration is 100 nmol/L
💡 Key relationship: pH = -log[H⁺] | 1 mol/L = 10⁹ nmol/L
33 / 70
Category:
ASCP Exam Questions
The statistical term for the average value is the:
Mean
The average value of a dataset
Example: Calculate mean for [5, 7, 3, 9]
Sum = 5 + 7 + 3 + 9 = 24
Number of values = 4
Mean = 24 ÷ 4 = 6
Key Characteristics:
Most common measure of central tendency Sensitive to extreme values (outliers) Appropriate for interval and ratio data Mathematically expressed as x̄ Comparison with Other Measures:
Measure Description Best Used When Mean Average value Data is normally distributed Median Middle value Data has outliers Mode Most frequent value Categorical data
34 / 70
Category:
ASCP Exam Questions
For values (100, 120, 150, 140, 130), the mean is:
1
Calculate the sum:
100 + 120 + 150 + 140 + 130 = 640
The arithmetic mean is 128
35 / 70
Category:
ASCP Exam Questions
Given mean WBC = 12,450/μL and SD = 153.2/μL, the CV is:
✅ Coefficient of Variation (CV) Calculation
Step 1: Calculate SD/Mean
153.2 ÷ 12,450 = 0.0123
Step 2: Convert to percentage
0.0123 × 100% = 1.2%
Interpretation: A CV of 1.2% indicates excellent precision in WBC measurements
Excellent
Good
Moderate
Poor
Unacceptable
💡 In hematology, CV < 3% is generally considered excellent for WBC counts
38 / 70
Category:
ASCP Exam Questions
A serum potassium (MW=39) value of 19.5 mg/100 mL equals:
✅ Concentration Conversion: mg/100mL to mEq/L
Step 1: Convert to mg/L
19.5 mg/100mL × 10 = 195 mg/L
(multiply by 10 since 1L = 1000mL and 1000/100 = 10)
Step 2: Convert mg/L to mmol/L
mmol/L =
195 mg/L 39 mg/mmol
= 5 mmol/L
Step 3: Convert mmol/L to mEq/L
mEq/L = mmol/L × valence = 5 × 1 = 5 mEq/L
Final concentration = 5 mEq/L K⁺
41 / 70
Category:
ASCP Exam Questions
With ±2 SD as the error limit, how many of 100 control values fall outside?
✅ Control Value Analysis (±2 SD Limits)
In a normal distribution:
-2 SD
Mean
+2 SD
Values within ±2 SD: 95%
Values outside ±2 SD: 5%
For 100 control values:
Outliers = 100 × 0.05 = 5 values
Conclusion: Approximately 5 out of 100 control values will fall outside ±2 SD limits in a normal distribution
💡 In quality control, the Westgard rules use this principle to detect systematic errors and random errors in lab tests
44 / 70
Category:
ASCP Exam Questions
Given: Absorbance = 0.500, light path = 1.0 cm, concentration = 0.2 mol/L. The molar absorptivity is:
✅ Beer’s Law Calculation
A = ε × b × c
Rearrange: ε = A / (b × c)
= 0.500 / (1.0 × 0.2)
= 0.500 / 0.2
= 2.5 L·mol⁻¹·cm⁻¹
The molar absorptivity (ε) is 2.5 L·mol⁻¹·cm⁻¹
45 / 70
Category:
ASCP Exam Questions
For a cholesterol QC with mean = 137 mg/dL and SD = 3 mg/dL, the CV is:
Step 1: 3 ÷ 137 = 0.0219
Step 2: 0.0219 × 100 = 2.19%
CV = 2.19%
Excellent precision (CV < 3%)
46 / 70
Category:
ASCP Exam Questions
How many mL of 3% solution can be made from 6 g solute?
✅ Solution Volume Calculation
3 g : 100 mL = 6 g : X mL
3X = 6 × 100
3X = 600
X = 600 ÷ 3
X = 200 mL
200 mL of 3% solution can be made with 6 g solute
47 / 70
Category:
ASCP Exam Questions
To prepare 40 mL of a 3% solution, the volume of stock solution needed is:
👉 Calculating volume of 100% stock needed for 3% solution:
We want to prepare 40 mL of 3% solution from pure stock (100%):
Volume needed = C₂ × V₂
= 3% × 40 mL
= (3 ÷ 100) × 40
= 0.03 × 40
= 1.2 mL of pure stock
Protocol:
1. Measure 1.2 mL of 100% stock solution
2. Dilute with solvent to 40 mL total volume
48 / 70
Category:
ASCP Exam Questions
The most frequent value in a dataset is the:
Mode
The most frequent value in a dataset
No formula – determined by frequency count
Example: Find mode for [3, 7, 3, 9, 2, 3, 5]
Mode = 3 (appears 3 times)
Key Characteristics:
Represents the most common value in a dataset Unaffected by extreme values (outliers) Can be used with nominal, ordinal, interval, and ratio data A dataset can have multiple modes (bimodal, multimodal) May not exist if all values are unique Comparison with Other Measures:
Measure Description Best Used When Mean Average value Data is normally distributed Median Middle value Data has outliers Mode Most frequent value Categorical data or identifying peaks
Real-world Applications:
Identifying most popular product in sales data Finding most common symptom in medical research Determining peak traffic times in analytics Analyzing voter preferences in polls 49 / 70
Category:
ASCP Exam Questions
The middle value in a dataset is the:
Median
The middle value in an ordered dataset
1. Arrange data in order
2. Find the middle position
3. Identify the median value
Example 1: Odd number of values [5, 2, 9, 1, 7]
Ordered: 1, 2, 5, 7, 9
Middle position: 3rd value
Median = 5
Example 2: Even number of values [4, 1, 3, 2]
Ordered: 1, 2, 3, 4
Middle positions: 2nd and 3rd values
Median = (2 + 3) ÷ 2 = 2.5
Key Characteristics:
Robust measure of central tendency Unaffected by extreme values (outliers) Appropriate for ordinal, interval, and ratio data For even n: average of two middle values Better than mean for skewed distributions Comparison with Other Measures:
Measure Description Best Used When Mean Average value Data is normally distributed Median Middle value Data has outliers or is skewed Mode Most frequent value Categorical data
Real-world Applications:
Income distribution analysis Housing price reports Test score interpretations Survival time analysis in medicine Customer wait time statistics Right-Skewed
Mean > Median
50 / 70
Category:
ASCP Exam Questions
Diagnostic specificity is the percentage of:
Measures the proportion of truly negative cases correctly identified by the test
True Negatives (TN)
Actual negatives correctly identified
False Positives (FP)
Actual negatives incorrectly identified as positive
Example Calculation Specificity = (85 ÷ (85 + 15)) × 100 = (85 ÷ 100) × 100 = 85%
A test with 85% specificity correctly identifies 85% of truly negative cases
51 / 70
Category:
ASCP Exam Questions
The normality of 280 g NaOH (MW=40) in 2000 mL solution is:
✅ Normality Calculation for NaOH
Step 1: Calculate moles
Moles = 280 g / 40 g/mol = 7 mol
Step 2: Determine relationship
For NaOH (valence = 1): Normality = Molarity
Step 3: Calculate normality
Normality = moles / liters = 7 mol / 2 L = 3.5 N
The normality of the NaOH solution is 3.5 N
💡 Key point: For monovalent bases like NaOH, normality equals molarity
52 / 70
Category:
ASCP Exam Questions
Adding 25 g NaOH (MW=40) to 0.5 L water, then diluting to 0.75 L, yields:
✅ Molarity Calculation
Step 1: Calculate moles of NaOH
Moles = mass / MW = 25 g / 40 g/mol = 0.625 mol
Step 2: Calculate molarity
Molarity = moles / volume = 0.625 mol / 0.75 L = 0.83 M
The molarity of the NaOH solution is 0.83 M
Preparation: Dissolve 25 g NaOH in water and dilute to 0.75 L total volume
53 / 70
Category:
ASCP Exam Questions
Which parameter varies with disease prevalence?
Varies significantly with disease prevalence
Low Prevalence
Lower Accuracy
High Prevalence
Higher Accuracy
Stable Parameters (Unaffected by Prevalence) Varying Parameters (Affected by Prevalence)Accuracy
Positive Predictive Value (PPV)
Negative Predictive Value (NPV)
Why it matters: In populations with low disease prevalence, even highly specific tests can produce many false positives. Prevalence impacts overall accuracy and predictive values.
54 / 70
Category:
ASCP Exam Questions
To make 1 L of 1.0 N NaOH from 1.025 N NaOH, the volume needed is:
✅ Normality Dilution Calculation
C₁V₁ = C₂V₂
1.025 × V₁ = 1.0 × 1000
1.025V₁ = 1000
V₁ = 1000 ÷ 1.025
V₁ = 975.6 mL
Protocol: Take 975.6 mL of 1.025N solution and dilute to 1000 mL
58 / 70
Category:
ASCP Exam Questions
A spinal fluid protein diluted 1:3 must be:
✅ Let’s break it down:
👉 A 1:3 dilution means: 1 part sample + 2 parts diluent → total 3 parts.
👉 To get the original concentration, you must multiply by the dilution factor (3) .
59 / 70
Category:
ASCP Exam Questions
For a mean of 100 and SD of 1.8 mg/dL, the 95% confidence interval is:
✅ 95% Confidence Interval Calculation
95% CI = μ ± 2σ
Step 1: Calculate ±2σ
2 × σ = 2 × 1.8 = 3.6 mg/dL
Step 2: Apply to mean
Lower bound = μ – 2σ = 100 – 3.6 = 96.4 mg/dL
Upper bound = μ + 2σ = 100 + 3.6 = 103.6 mg/dL
95% Confidence Interval = 96.4 to 103.6 mg/dL
We can be 95% confident that the true population mean lies between 96.4 and 103.6 mg/dL
100
95% Confidence Interval
61 / 70
Category:
ASCP Exam Questions
A solution with 20 g solute in 0.5 L water is:
✅ %(w/v) Calculation
= (20 ÷ 500) × 100
= 0.04 × 100
= 4%
Final concentration = 4% (w/v)
62 / 70
Category:
ASCP Exam Questions
For sodium controls (140, 135, 138, 140, 142 mEq/L), the CV is:
✅ Sodium Control CV Calculation
Mean = (140 + 135 + 138 + 140 + 142) ÷ 5
= 695 ÷ 5 = 139 mEq/L
Value Difference Squared 140 140-139 = 1 1² = 1 135 135-139 = -4 (-4)² = 16 138 138-139 = -1 (-1)² = 1 140 140-139 = 1 1² = 1 142 142-139 = 3 3² = 9 Sum of squares 28
SD = √[Σ(xᵢ – x̄)² ÷ (n-1)] = √[28 ÷ (5-1)]
= √[28 ÷ 4] = √7 ≈ 2.65 mEq/L
CV = (SD ÷ Mean) × 100 = (2.65 ÷ 139) × 100
≈ 1.9%
The coefficient of variation for sodium controls is 1.9%
This indicates excellent precision in sodium measurements
63 / 70
Category:
ASCP Exam Questions
If 0.5 mL of a 1:300 dilution contains 1 antigenic unit, 2 units would be in 0.5 mL of:
👉 0.5 mL of 1:300 dilution = 1 antigenic unit
👉 Target: 2 units in 0.5 mL (double concentration)
1:300 dilution → 1 unit/0.5 mL
For 2 units → Need 2× concentration
Final Dilution: 1:150
Use half the dilution factor (twice as concentrated) to double antigen units
Preparation Steps:
Prepare 1:150 dilution (instead of 1:300) Use same 0.5 mL volume for testing Include positive and negative controls 64 / 70
Category:
ASCP Exam Questions
How many grams of sulfosalicylic acid (MW=254) are needed for 1 L of 3% (w/v) solution?
✅ %(w/v) Solution Preparation
= (3 ÷ 100) × 1000
= 0.03 × 1000
= 30 g
Protocol: Dissolve 30 g sulfosalicylic acid in water and dilute to 1000 mL
70 / 70
Category:
ASCP Exam Questions
A 1 molal solution contains:
Molality = moles solute / kg solvent.
Your score is
The average score is 20%
Follow us on Sicial Media:
Restart quiz
Top 8 Medical Laboratory Scientist (MLS) Exams: Top 8 Medical Laboratory Scientist (MLS) Exams that are recognized globally and can help professionals validate their credentials and enhance their career opportunities:
1. ASCP – American Society for Clinical Pathology (USA) Exam Name: MLS(ASCP)Eligibility: Bachelor’s degree with clinical laboratory experience.Global Recognition: HighPurpose: Certifies Medical Laboratory Scientists in the United States and internationally.2. AMT – American Medical Technologists (USA) Exam Name: MLT(AMT) or MT(AMT)Eligibility: Academic and/or work experience in medical laboratory technology.Global Recognition: ModeratePurpose: Credentialing for medical technologists and technicians.3. AIMS – Australian Institute of Medical and Clinical Scientists Exam Name: AIMS Certification ExamEligibility: Assessment of qualifications and work experience.Recognition: Required for practice in Australia.Purpose: Certification and registration in Australia.4. CSMLS – Canadian Society for Medical Laboratory Science Exam Name: CSMLS General or Subject-specific ExamsEligibility: Graduation from a CSMLS-accredited program or equivalent.Recognition: CanadaPurpose: Entry-to-practice certification in Canada.5. IBMS – Institute of Biomedical Science (UK) Exam Name: Registration and Specialist Portfolio AssessmentEligibility: Accredited degree and lab experience.Recognition: UK and some Commonwealth countries.Purpose: Biomedical Scientist registration with the HCPC (UK).6. HAAD / DOH – Department of Health, Abu Dhabi (UAE) Exam Name: DOH/HAAD License ExamEligibility: Degree in medical laboratory science and experience.Recognition: UAE (Abu Dhabi)Purpose: Licensure for medical laboratory practice in Abu Dhabi.7. DHA – Dubai Health Authority (UAE) Exam Name: DHA License Exam for Medical Laboratory TechnologistsEligibility: Relevant degree and experience.Recognition: Dubai, UAEPurpose: Professional license for clinical laboratory practice in Dubai.8. MOH – Ministry of Health (Gulf Countries like UAE, Saudi Arabia, Kuwait) Exam Name: MOH License ExamEligibility: BSc/Diploma in Medical Laboratory + experience.Recognition: Varies by country.Purpose: Required for practicing in public and private sector labs.Tags: #ASCPMLS #MLSexam #LabTech #MedicalLaboratory #BOCexam #FreePracticeQuestions #QualityControl #LaboratorySafety
Possible References Used
⚠️ Disclaimer: The content on LabTestsGuide.com is for informational and educational purposes only . We do not guarantee the accuracy, completeness, or timeliness of the information provided. Always consult qualified healthcare professionals for medical advice, diagnosis, or treatment. LabTestsGuide.com is not liable for any decisions made based on the information on this site.
ASCP
American Society for Clinical Pathology (USA)
AMT
American Medical Technologists (USA)
AIMS
Australian Institute of Medical and Clinical Scientists
CSMLS
Canadian Society for Medical Laboratory Science
IBMS
Institute of Biomedical Science (UK)
HAAD
Health Authority - Abu Dhabi
MOH
Ministry of Health (UAE)
DHA
Dubai Health Authority
Comprehensive question bank with Answer Key and Mock Test
Performance analytics and get Mock Test Certificate
Mobile-friendly interface for on-the-go studying
Timed mock exams that simulate real test conditions