Challenge your understanding of Nucleic Acids and Molecular Diagnostics with this Biochemistry Mock Test (Part 50) . This quiz simulates real ASCP MLS exam questions, helping you test your knowledge of DNA/RNA chemistry, replication, PCR, and other key molecular diagnostic methods used in clinical laboratories. It’s ideal for MLS students, medical technologists, and clinical biochemistry learners aiming for exam mastery.
📘 Topics Included DNA and RNA structure and function Molecular replication and transcription PCR and hybridization techniques Clinical molecular applications (infectious, genetic, oncologic) Laboratory safety and quality in molecular testing 🧠 Why Take This Mock Test? Reinforce key concepts in molecular biochemistry. Evaluate your preparation for the ASCP MLS molecular section. Practice molecular-based questions in a clinical context. Improve interpretation of molecular test results.
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ASCP MLS Exam MCQs Chapter 50
Why Take This Mock Test? Strengthens exam confidence Highlights areas for improvement Provides practice with clinically relevant scenarios This mock test (60 MCQs (3941 – 4000) ) is part of our ongoing ASCP MLS Exam Practice Series , giving you structured preparation for all major immunology topics.
Our Biochemistry – Nucleic Acids & Molecular Diagnostics Mock Test is specifically designed for candidates appearing in ASCP MLS, AMT MLT/MT, AIMS, CSMLS, IBMS, HAAD/DOH, DHA, and MOH exams. This mock test mirrors the structure, difficulty level, and question style you can expect in the actual examination.
Take this test to: ✅ Strengthen your Pathology exam preparation. ✅ Boost confidence before the ASCP MLS Exam .
Who Should Use This Mock Test? Medical Laboratory Scientists and Technicians
Pathology Students
Professionals preparing for international laboratory certification exams
Anyone seeking to strengthen their knowledge of Biochemistry – Nucleic Acids & Molecular Diagnostics
How to Use This Mock Test Effectively Simulate Exam Conditions: Attempt the test in one sitting without referring to notes.
Track Your Time: Keep within the allotted time limit to build speed.
Review Explanations: Study the answer explanations to strengthen understanding.
Repeat for Retention: Re-attempt after revision to measure improvement.
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ASCP Exam Questions
The laboratory test most useful for monitoring the course of a patient with testicular cancer is:
Testicular cancer (especially non-seminomatous germ cell tumors ) produces tumor markers that help in diagnosis, monitoring treatment response, and detecting recurrence.
Alpha-fetoprotein (AFP) is elevated in yolk sac tumors and some embryonal carcinomas .
Beta-hCG (not listed here) is another marker elevated in choriocarcinoma and some seminomas.
Other options:
CEA → mainly used for colorectal and some other gastrointestinal cancers.
Prolactin → related to pituitary disorders.
Testosterone → not a tumor marker; levels can vary but do not track testicular cancer.
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ASCP Exam Questions
The tumor marker CA 125 is most strongly associated with which type of carcinoma?
CA 125 (Cancer Antigen 125) is a tumor marker that is most strongly associated with epithelial ovarian carcinoma . It is elevated in approximately 80% of women with advanced epithelial ovarian cancer. It is also frequently elevated in endometrial (uterine) cancer .
While CA 125 can be elevated in other malignancies (such as pancreatic, breast, or lung cancer) and in various benign conditions (like endometriosis, pelvic inflammatory disease, pregnancy, and liver cirrhosis), its primary and strongest clinical association is with ovarian cancer.
Why the other options are incorrect:
a) Breast: The primary tumor markers for breast cancer are CA 15-3 and CA 27.29 . CEA can also be used.
b) Colon: The primary tumor marker for colorectal cancer is Carcinoembryonic Antigen (CEA) .
c) Lung: No single tumor marker is definitive for lung cancer, but common ones include CEA (for adenocarcinoma), SCC (for squamous cell), and NSE or ProGRP (for small cell lung cancer). CA 125 is not a primary marker for lung cancer.
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ASCP Exam Questions
In the context of heme synthesis, ALA (aminolevulinic acid) is produced by the enzyme:
In the heme synthesis pathway, the first and rate-limiting step is the condensation of glycine and succinyl-CoA to form δ-aminolevulinic acid (ALA) . This reaction is catalyzed by the enzyme ALA synthase (also known as δ-aminolevulinic acid synthase).
Here is the role of the other enzymes listed:
a) ALA dehydratase: This enzyme uses the product (ALA) to perform the next step: it condenses two molecules of ALA to form porphobilinogen (PBG).
c) Porphobilinogen deaminase: This enzyme acts on porphobilinogen (PBG) to form hydroxymethylbilane.
d) Ferrochelatase: This is the final enzyme in the pathway; it inserts iron into protoporphyrin IX to form heme.
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ASCP Exam Questions
A 9-month-old boy of Ashkenazi Jewish descent presents with a loss of motor skills and seizures. Laboratory analysis reveals an accumulation of GM2-ganglioside in neurons and a deficiency of hexosaminidase A. These findings are most consistent with:
The presentation is classic for Tay-Sachs disease:
Demographics: The patient is of Ashkenazi Jewish descent , a population with a high carrier frequency for this condition.
Age and Symptoms: Onset around 6-10 months with loss of motor skills (regression) and seizures is characteristic.
Pathophysiology: The accumulation of GM2-ganglioside in neurons is the direct result of a deficiency of the enzyme hexosaminidase A , which is necessary for breaking down this lipid.
Let’s review why the other options are incorrect:
a) Niemann-Pick disease: This involves the accumulation of sphingomyelin (not GM2-ganglioside) due to a deficiency of sphingomyelinase. It can cause neurodegeneration, but the specific enzyme and stored material do not match.
c) Phenylketonuria (PKU): This is caused by a deficiency of phenylalanine hydroxylase, leading to the accumulation of phenylalanine (not a lipid). It causes intellectual disability but not the specific neurodegenerative course or the lipid accumulation described.
d) Hurler syndrome: This is a mucopolysaccharidosis caused by a deficiency of alpha-L-iduronidase, leading to the accumulation of heparan sulfate and dermatan sulfate (glycosaminoglycans, not GM2-ganglioside).
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ASCP Exam Questions
The main advantage of real-time PCR (qPCR) over conventional PCR is:
Real-time PCR (qPCR) allows the quantitative measurement of DNA as it is being amplified — in real time .
It uses fluorescent dyes or probes that emit light proportional to the amount of DNA produced during each PCR cycle.
This makes it possible to measure starting DNA quantities and monitor gene expression accurately.
Other options:
a) Faster gel electrophoresis → qPCR does not require gel electrophoresis at all.
c) Reduced sample requirement → may be true, but not the main advantage.
d) Color detection only → color (fluorescence) is used, but the key feature is quantification , not just detection.
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ASCP Exam Questions
The Southern blot technique is used to detect:
The Southern blot technique is used to detect specific DNA sequences in a sample.
It involves transferring DNA fragments (separated by gel electrophoresis) onto a membrane, then using a labeled DNA probe that hybridizes to the complementary sequence.
Why the other options are incorrect:
a) RNA: The corresponding technique for detecting RNA is called Northern blot .
c) Proteins: The technique for detecting specific proteins is called Western blot .
d) Antibodies: Antibodies are used as the detection probes in a Western blot to identify proteins; they are not the target molecule of a Southern blot.
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ASCP Exam Questions
The term “amplicon” refers to:
In molecular biology, an amplicon specifically refers to the product of an amplification reaction , most commonly the piece of DNA that is amplified (copied) by the Polymerase Chain Reaction (PCR) . It is the specific, targeted DNA fragment that accumulates exponentially during the PCR process.
Why the other options are incorrect:
a) A restriction enzyme: This is an enzyme that cuts DNA at specific sequences.
c) A DNA polymerase inhibitor: This is a substance that inhibits the activity of DNA polymerase enzymes.
d) A transcription factor: This is a protein that controls the rate of transcription of genetic information from DNA to mRNA.
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The enzyme responsible for unwinding the DNA double helix is:
DNA helicase is the enzyme that unwinds and separates the two strands of the DNA double helix by breaking the hydrogen bonds between the base pairs.
This creates the replication fork , allowing other enzymes to synthesize new DNA strands.
Other options:
a) DNA polymerase → synthesizes the new DNA strand by adding nucleotides to the template.
b) DNA ligase → joins Okazaki fragments on the lagging strand by forming phosphodiester bonds.
d) RNA polymerase → synthesizes RNA during transcription , not DNA replication.
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ASCP Exam Questions
The screening test for fetal neural tube defects involves measuring which substance in maternal serum or amniotic fluid?
Alpha-fetoprotein (AFP) is produced by the fetal liver , yolk sac, and GI tract.
It normally crosses into the amniotic fluid and then into maternal serum .
Elevated AFP levels in maternal serum or amniotic fluid suggest:
Open neural tube defects (NTDs) such as spina bifida or anencephaly
Abdominal wall defects (e.g., gastroschisis, omphalocele)
Multiple pregnancies (due to more fetuses producing AFP)
Low AFP may indicate:
Other options:
a) Estriol → part of triple/quad screen; assesses fetal well-being, not NTDs specifically.
b) hCG → elevated in Down syndrome; not for NTDs.
d) Progesterone → reflects corpus luteum or placental function, not NTD screening.
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ASCP Exam Questions
The enzyme that synthesizes RNA using a DNA template is:
RNA polymerase is the central enzyme of transcription. Its primary function is to synthesize a complementary RNA strand using a DNA template. It reads the DNA sequence and builds the RNA molecule by adding ribonucleotides, following the base-pairing rules (A-U, T-A, G-C).
Why the other options are incorrect:
a) DNA polymerase: This enzyme synthesizes new DNA strands during DNA replication, not RNA.
c) Ligase: This enzyme joins DNA fragments together by forming phosphodiester bonds (e.g., Okazaki fragments on the lagging strand). It does not synthesize RNA.
d) Primase: This is a specialized type of RNA polymerase that synthesizes short RNA primers to initiate DNA replication. However, when referring to the enzyme that synthesizes RNA using a DNA template in the general sense (like for mRNA, tRNA, rRNA), the correct and direct answer is RNA polymerase .
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ASCP Exam Questions
The basic structural unit of DNA is called a:
Other options:
a) Nucleoside → base + sugar (no phosphate group)
c) Amino acid → building block of proteins , not DNA
d) Ribosome → site of protein synthesis , not part of DNA structure
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The method used to determine the sequence of nucleotides in DNA is:
Sanger sequencing (also known as the dideoxy chain termination method ) is used to determine the exact sequence of nucleotides (A, T, G, C) in a DNA molecule.
It uses dideoxynucleotides (ddNTPs) that terminate DNA chain elongation at specific bases, producing fragments that can be separated and read to reveal the sequence.
Other options:
a) PCR → amplifies DNA, but does not determine its sequence.
b) ELISA → detects antigens or antibodies , not nucleotides.
d) Immunofluorescence → used to detect proteins or structures using fluorescent antibodies.
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ASCP Exam Questions
The metabolite 11-nor-tetrahydrocannabinol-9-COOH is detected in the urine following the use of:
11-nor-Δ⁹-tetrahydrocannabinol-9-carboxylic acid (11-nor-THC-COOH) is the major inactive metabolite of Δ⁹-tetrahydrocannabinol (THC) — the psychoactive component of marijuana (cannabis) .
It is lipid-soluble , stored in body fat, and excreted slowly, so it can be detected in urine for days to weeks after use (depending on frequency).
Other options:
Methamphetamine → detected as amphetamine/methamphetamine .
Cocaine → metabolized to benzoylecgonine and ecgonine methyl ester .
Benzodiazepine → detected as oxazepam or related metabolites depending on the drug.
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In PCR, the process of separating DNA strands is called:
In PCR, the process of separating the double-stranded DNA into two single strands is called denaturation . This is achieved by heating the reaction mixture to a high temperature (typically 94-98°C), which breaks the hydrogen bonds between the complementary base pairs.
Why the other options are incorrect:
a) Annealing: This is the step following denaturation, where the temperature is lowered to allow short DNA primers to bind (anneal) to their complementary sequences on the single-stranded DNA templates.
c) Extension (or Elongation): This is the step following annealing, where the temperature is raised to the optimal temperature for Taq polymerase (around 72°C) to synthesize new DNA strands by extending the primers.
d) Amplification: This is the overall goal and result of the entire PCR process, not a specific step for separating DNA strands.
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ASCP Exam Questions
The recommended specimen for forensic toxicology to detect drugs of abuse over a longer time period is:
Hair testing provides a long detection window , often up to months , because drugs and their metabolites are incorporated into growing hair.
This makes it ideal for forensic cases , monitoring chronic use , or establishing historical patterns of drug use.
Other options:
Random urine → detects recent use , typically hours to a few days.
Serum → short detection window; useful for recent ingestion or acute toxicity.
Whole blood → also reflects recent use , mainly for immediate post-ingestion detection.
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The different forms of hemoglobin (e.g., HbA, HbS, HbC) can be separated and identified based on their charge using:
Hemoglobin electrophoresis is the standard laboratory method used to separate and identify different hemoglobin variants (e.g., HbA, HbS, HbC, HbF ).
The technique works by applying an electric field to a medium (such as cellulose acetate or agar gel).
Each hemoglobin type moves at a different rate based on its charge and structure , allowing clear separation.
Other options:
a) Immunoassay → detects specific antigens or antibodies; not used for separating hemoglobin variants.
c) Mass spectrometry → can identify hemoglobin variants at the molecular level but is not the routine clinical method.
d) Ion-selective electrode → used to measure electrolytes (e.g., Na⁺, K⁺, Cl⁻), not proteins.
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ASCP Exam Questions
The temperature range for DNA denaturation in PCR is approximately:
In the Polymerase Chain Reaction (PCR), the denaturation step requires a high temperature to break the hydrogen bonds holding the double-stranded DNA together, separating it into two single strands. The standard temperature range used for this step is 90–95°C .
This high temperature is necessary to fully denature the DNA and ensure the reaction starts with single-stranded templates for the primers to bind to in the next step (annealing).
The specific temperature within this range (e.g., 94°C or 95°C) is a common choice in many protocols.
Why the other options are incorrect:
a) 30–40°C: This range is far too low for denaturation; it is more typical for some enzyme activities or bacterial growth.
b) 50–60°C: This is the standard temperature range for the annealing step, where primers bind to their complementary sequences on the single-stranded DNA.
c) 70–80°C: This is the typical range for the extension/elongation step, where the DNA polymerase (like Taq polymerase) synthesizes the new DNA strand.
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ASCP Exam Questions
The method of choice for confirming a positive immunoassay screen for barbiturates in a urine drugs of abuse panel is:
Immunoassays are commonly used for initial urine drug screens because they are fast and easy.
However, immunoassays can give false positives due to cross-reactivity.
Confirmation of a positive screen requires a highly specific and sensitive method , which is GC/MS :
Other options:
Nephelometry → measures light scattering; used for proteins, not drugs.
Thin-layer chromatography (TLC) → can separate compounds but is less specific than GC/MS.
UV absorption spectroscopy → not specific enough for confirmation of drugs.
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The initial analysis for a urine sample received for “porphyrins” without further specification should include:
When a urine sample is received for “porphyrins” without specific instructions, the initial analysis should be a broad screening approach to detect the two major categories of heme precursors that can appear in urine:
Porphobilinogen (PBG) Screen: This is critical for identifying the acute porphyrias (e.g., Acute Intermittent Porphyria). A marked increase in PBG is a medical emergency.
Porphyrin Screen: This detects the presence of elevated total porphyrins. If this screen is positive, further fractionation can be performed to identify the specific types of porphyrins (e.g., uroporphyrin, coproporphyrin) to help diagnose specific porphyrias like Porphyria Cutanea Tarda.
This combined screening approach ensures that both acute and cutaneous porphyrias are initially evaluated.
Why the other options are less correct:
a) Porphyrin screen and quantitative total porphyrin: This misses the crucial screen for porphobilinogen (PBG), which is essential for diagnosing acute attacks.
b) Quantitative total porphyrin and porphobilinogen screen: While this includes the necessary PBG screen, performing a quantitative total porphyrin test is more specific and labor-intensive. A simple, rapid porphyrin screen (often by fluorescence) is a more appropriate and cost-effective first step. Quantification can follow if the screen is positive.
d) Porphobilinogen screen and ion-exchange analysis: Ion-exchange analysis is a specific method for quantifying individual porphyrins. It is a follow-up test, not an appropriate initial screen when the request is non-specific.
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ASCP Exam Questions
A patient’s serum is tested for a tumor marker using a chemiluminescent EIA. The reaction is quantified by measuring the:
In a chemiluminescent enzyme immunoassay (EIA) , the detection system relies on a chemical reaction that emits light (chemiluminescence).
The intensity of the emitted light is directly proportional to the amount of antigen or antibody present in the sample.
The emitted light is measured by a luminometer or photomultiplier tube , providing a quantitative result.
Other options:
a) Absorption of light by the product → describes colorimetric (spectrophotometric) assays , not chemiluminescent.
c) Change in electrical potential → describes electrochemical assays .
d) Scatter of light by microparticles → describes nephelometry or turbidimetry .
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ASCP Exam Questions
A patient with a microcytic, hypochromic anemia has a low serum iron and low total iron-binding capacity (TIBC). This pattern is most consistent with:
The key to distinguishing these causes of microcytic anemia lies in the combination of serum iron and total iron-binding capacity (TIBC) .
Why the other options are incorrect:
a) Iron Deficiency Anemia (IDA): This is characterized by low serum iron but a high TIBC . The body tries to compensate for the lack of iron by producing more transferrin to maximize iron uptake and transport.
c) Hemochromatosis: This is an iron overload disorder, characterized by high serum iron and low TIBC (as transferrin saturation is extremely high). It does not typically cause a microcytic anemia.
d) Sideroblastic Anemia: This can have a variable pattern, but it is often characterized by normal or high serum iron and normal TIBC . The defining feature is the presence of ringed sideroblasts in the bone marrow, reflecting ineffective iron utilization within erythroid precursors.
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ASCP Exam Questions
The process of protein synthesis from mRNA is called:
The process of protein synthesis, where the genetic code in a messenger RNA (mRNA) molecule is decoded to produce a specific sequence of amino acids (a polypeptide chain), is called translation . This process occurs on ribosomes in the cytoplasm.
Why the other options are incorrect:
a) Transcription: This is the process of synthesizing RNA from a DNA template.
c) Replication: This is the process of synthesizing a new DNA molecule from an existing DNA template.
d) Recombination: This is the process by which genetic material is rearranged, such as during the crossing over of chromosomes in meiosis.
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ASCP Exam Questions
The principle of the test for occult blood in feces depends on the:
The fecal occult blood test (FOBT) detects hidden (occult) blood in stool.
The test is based on the peroxidase-like (pseudoperoxidase) activity of hemoglobin , which catalyzes the oxidation of a colorless indicator (like guaiac) by hydrogen peroxide → producing a blue color .
Reaction principle:
Hemoglobin+H2O2+Guaiac→Blue oxidized guaiac\text{Hemoglobin} + H_2O_2 + \text{Guaiac} \rightarrow \text{Blue oxidized guaiac} Hemoglobin + H 2 O 2 + Guaiac → Blue oxidized guaiac
Other options:
a) Coagulase ability of blood → irrelevant; coagulase relates to bacteria (e.g., Staphylococcus aureus ).
b) Oxidative power of atmospheric oxygen → not involved in the test.
c) Hydrogen peroxide in hemoglobin → hemoglobin doesn’t contain hydrogen peroxide; H₂O₂ is a reagent in the test.
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ASCP Exam Questions
In a patient with suspected hemolytic anemia, which of the following laboratory results would be expected?
Haptoglobin is a plasma protein that binds free hemoglobin released during red blood cell (RBC) destruction.
In hemolytic anemia , RBCs break down → free hemoglobin increases → haptoglobin binds it → the complex is removed from circulation , leading to low serum haptoglobin levels .
Other options:
a) Increased haptoglobin → opposite of what happens; would indicate inflammation, not hemolysis.
c) Increased serum iron → may occur secondarily but not diagnostic.
d) Decreased unconjugated bilirubin → incorrect; it actually increases due to heme breakdown.
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ASCP Exam Questions
The sugar present in DNA is:
The name DNA stands for Deoxyribo nucleic Acid. The “deoxyribo” part specifically refers to its sugar component, deoxyribose .
Why the other options are incorrect:
a) Ribose: This is the sugar found in RNA (Ribonucleic Acid).
c) Fructose & d) Glucose: These are simple 6-carbon sugars (hexoses) that are involved in cellular metabolism and energy production, but they are not components of nucleic acids.
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ASCP Exam Questions
The complementary base of adenine in DNA is:
In DNA, the base pairs are held together by specific hydrogen bonding, following the rule of complementary base pairing:
This specific pairing (A-T and G-C) is crucial for the accurate replication of DNA and the transmission of genetic information.
Why the other options are incorrect:
b) Cytosine: Cytosine pairs with Guanine (G).
c) Uracil: Uracil is not found in DNA; it replaces Thymine in RNA , where it pairs with Adenine.
d) Guanine: Guanine pairs with Cytosine (C).
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24-hour homovanillic acid (HVA) is a useful test to aid in the diagnosis of:
Homovanillic acid (HVA) is the major urinary metabolite of dopamine .
Measuring 24-hour urinary HVA (often along with VMA , vanillylmandelic acid) helps in the diagnosis and monitoring of neuroblastoma , a catecholamine-secreting tumor that arises from the adrenal medulla or sympathetic nervous system —most commonly in children.
Other options:
a) Cushing disease → involves excess cortisol; diagnosed by dexamethasone suppression or cortisol levels, not HVA.
c) Conn disease → primary hyperaldosteronism; diagnosed by aldosterone and renin ratio.
d) Graves disease → autoimmune hyperthyroidism; diagnosed by thyroid hormones and TSH, not catecholamine metabolites.
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ASCP Exam Questions
The enzyme responsible for DNA synthesis in PCR is:
Taq polymerase is a heat-stable DNA polymerase isolated from the thermophilic bacterium Thermus aquaticus . It is the key enzyme used in the Polymerase Chain Reaction (PCR) because it can withstand the high temperatures (around 95-98°C) required to denature the DNA double helix during each PCR cycle without being denatured itself. Its function is to synthesize new DNA strands by adding nucleotides to the primers, using the single-stranded DNA as a template.
Why the other options are incorrect:
a) RNA polymerase: This enzyme synthesizes RNA from a DNA template during transcription. It is not used in PCR.
c) DNA ligase: This enzyme joins DNA fragments together by sealing nicks in the DNA backbone. It is not involved in the synthesis of new DNA strands in PCR.
d) Helicase: This enzyme unwinds the DNA double helix in in vivo DNA replication. In PCR, the high-temperature denaturation step replaces the need for a helicase enzyme.
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In eukaryotes, DNA is located in the:
In eukaryotic cells, the vast majority of the cell’s DNA is stored within the nucleus , which is a membrane-bound organelle. The DNA is organized into chromosomes and is separated from the rest of the cellular machinery by the nuclear envelope.
Why the other options are incorrect:
a) Cytoplasm: This is where DNA is located in prokaryotic cells (like bacteria), which lack a nucleus.
b) Ribosome: Ribosomes are structures in the cytoplasm that synthesize proteins using the genetic code from mRNA; they do not contain DNA.
d) Endoplasmic reticulum: This is an organelle involved in the synthesis and transport of proteins and lipids; it does not store DNA.
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In molecular separation techniques, High-Performance Liquid Chromatography (HPLC) involves:
High-Performance Liquid Chromatography (HPLC) is defined by its use of a high-pressure pump to force a liquid mobile phase through a tightly packed column containing a stationary phase. Components of the sample separate based on their different interactions with these two phases.
Here is why the other options are incorrect:
a) Separation in a gel medium under an electric field: This describes gel electrophoresis , not HPLC.
c) Vaporization of the sample with a carrier gas: This describes Gas Chromatography (GC) , where the mobile phase is an inert gas.
d) Ionization and separation based on mass-to-charge ratio: This describes Mass Spectrometry (MS) , which is often coupled with HPLC (as LC-MS) but is not the principle of HPLC itself.
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ASCP Exam Questions
The laboratory investigation of a comatose patient with suspected methanol poisoning would most likely reveal:
Methanol (parent alcohol) raises the measured serum osmolality (giving a high osmolar/osmolal gap ). As it’s metabolized to formic acid , that produces a high anion gap metabolic acidosis . Clinically you may also see visual disturbances and an increased serum methanol level.
Why the other options are incorrect:
a) A low osmolal gap and a low anion gap: This pattern is not characteristic of any common poisoning. A low anion gap is rare and can be seen in conditions like hypoalbuminemia.
c) A low osmolal gap and a high anion gap: This pattern might be seen in other causes of a high anion gap metabolic acidosis (e.g., lactic acidosis, ketoacidosis) where no unmeasured osmoles are present. However, in methanol poisoning, the osmolal gap is high.
d) A high osmolal gap and a low anion gap: This is an inconsistent pattern. A high osmolal gap suggests the presence of unmeasured osmoles, which are often alcohols or glycols that metabolize into acids, causing a high anion gap. A low anion gap would not be expected.
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ASCP Exam Questions
The Western blot technique identifies:
The Western blot (also called immunoblot) is a widely used analytical technique specifically for detecting specific proteins in a sample.
The general process involves:
Separation: Proteins are separated by size using gel electrophoresis (usually SDS-PAGE).
Transfer: The separated proteins are transferred (blotted) from the gel onto a solid membrane.
Detection: The membrane is probed with antibodies specific to the target protein. A labeled secondary antibody is then used to visualize the protein-antibody complex.
Why the other options are incorrect:
a) DNA: Detection of specific DNA sequences is done by Southern blot .
b) RNA: Detection of specific RNA sequences is done by Northern blot .
d) Lipids: There is no standard “blotting” technique for lipids. Lipids are typically analyzed using methods like thin-layer chromatography (TLC) or mass spectrometry.
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ASCP Exam Questions
A patient with Wilson’s disease would be expected to have which of the following laboratory findings?
Other options:
a) Incorrect — both serum copper and urine copper cannot be high simultaneously in Wilson’s.
c) Incorrect — ceruloplasmin is decreased , not increased.
d) Incorrect — urine copper is increased , not decreased.
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ASCP Exam Questions
A nucleotide is composed of:
A nucleotide is the basic building block of nucleic acids (DNA and RNA) and consists of three components:
Nitrogenous base — purine (adenine, guanine) or pyrimidine (cytosine, thymine, uracil)
Pentose sugar —
Deoxyribose in DNA
Ribose in RNA
Phosphate group(s) — one or more attached to the sugar
Other options:
a) Sugar + phosphate → missing the base → that’s not a complete nucleotide.
c) Phosphate + base → missing the sugar.
d) Base + amino group → incorrect; amino groups are part of the bases but not separate components.
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ASCP Exam Questions
The genetic code is said to be:
The genetic code has two key characteristics:
Universal: With very few minor exceptions, the same codons specify the same amino acids in almost all living organisms, from bacteria to humans. This universality is powerful evidence for a common ancestry of all life.
Degenerate (or Redundant): Most amino acids are encoded by more than one codon. For example, the amino acid leucine is specified by six different codons (UUA, UUG, CUU, CUC, CUA, CUG). This redundancy provides a buffer against harmful effects of mutations.
Why the other options are incorrect:
a) Ambiguous: Ambiguous would mean that a single codon could code for more than one amino acid. The genetic code is not ambiguous; each codon specifies only one amino acid (or a stop signal).
c) Random: The code is not random; it follows a specific, set pattern. While the specific assignments may seem arbitrary, they are consistent and non-random.
d) Specific to humans: The code is universal, not specific to humans. The same genetic code is used by bacteria, plants, animals, and all other forms of life.
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The Northern blot technique is used for:
The Northern blot technique is specifically used for the detection of specific RNA molecules. The name is a play on the original Southern blot (for DNA), with “Northern” being assigned for RNA.
The process involves:
Separating RNA fragments by size using gel electrophoresis.
Transferring (blotting) the RNA onto a membrane.
Hybridizing the membrane with a labeled DNA or RNA probe complementary to the target RNA sequence.
Detecting the bound probe to identify the presence and size of the specific RNA.
Why the other options are incorrect:
a) DNA detection: This is the purpose of the Southern blot .
c) Protein detection: This is the purpose of the Western blot .
d) Enzyme activity: This is measured by specific enzymatic assays, not by blotting techniques.
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The anticodon is located on:
The anticodon is a crucial sequence of three nucleotides found on a transfer RNA (tRNA) molecule. Its function is to base-pair with a complementary codon on the messenger RNA (mRNA) during translation. This ensures that the correct amino acid, which is attached to the 3′ end of the same tRNA, is added to the growing polypeptide chain.
Why the other options are incorrect:
a) mRNA: mRNA contains codons , which are the three-nucleotide sequences that the anticodons recognize.
b) rRNA: Ribosomal RNA (rRNA) is a structural and catalytic component of the ribosome; it does not have anticodons.
d) DNA: DNA contains genes that are transcribed into mRNA; it does not have anticodons.
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ASCP Exam Questions
The primary urinary metabolite measured to assess epinephrine and norepinephrine production is:
The catecholamine metabolic pathway ends with VMA as the major final metabolite. Here is a brief overview:
Norepinephrine and Epinephrine are metabolized primarily to Normetanephrine and Metanephrine , respectively.
These metanephrines are then further broken down to Vanillylmandelic Acid (VMA) .
Therefore, measuring urinary VMA provides a good assessment of the total production of epinephrine and norepinephrine over a 24-hour period. It has been a traditional screening test for catecholamine-secreting tumors like pheochromocytoma .
Why the other options are incorrect:
a) Dopamine: Dopamine is a precursor to norepinephrine, not a primary metabolite of it. Its own major metabolite is Homovanillic Acid (HVA) .
b) Dihydroxyphenylalanine (DOPA): This is the amino acid precursor to all catecholamines, not a metabolite.
c) Homovanillic acid (HVA): This is the primary end metabolite of dopamine , not of epinephrine or norepinephrine.
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ASCP Exam Questions
DNA replication is described as:
Other options:
a) Conservative → would mean the entire parent DNA stays intact and a completely new double helix is made
c) Dispersive → would mean DNA is a random mix of old and new segments
d) Random → not a recognized model of replication.
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ASCP Exam Questions
Which of the following enzymes in the heme biosynthesis pathway is inhibited by lead?
Lead poisoning has a well-known inhibitory effect on several enzymes in the heme biosynthesis pathway. The most sensitive and classically tested enzyme is δ-Aminolevulinate dehydratase (also known as porphobilinogen synthase).
why the other options are incorrect:
a) δ-Aminolevulinate dehydratase (ALAD): ✅ Yes, this is the classic enzyme inhibited by lead. ALA accumulates in blood and urine.
b) Porphobilinogen synthase: Another name for ALAD, so sometimes considered the same.
c) Uroporphyrinogen synthase: Not affected by lead; defect here causes congenital porphyria.
d) Bilirubin synthetase: Not part of heme synthesis; bilirubin is a breakdown product, not a biosynthesis enzyme.
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ASCP Exam Questions
Which of the following is used to separate DNA fragments by size?
Gel electrophoresis (usually agarose gel electrophoresis ) is used to separate DNA fragments by size .
When an electric current is applied, negatively charged DNA fragments move toward the positive electrode .
Smaller fragments move faster and farther through the gel than larger ones.
Other options:
a) Spectrophotometer → measures DNA concentration and purity , not size.
c) Chromatography → separates molecules based on chemical properties , not typically used for DNA size.
d) Centrifuge → separates substances based on density , not fragment length.
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ASCP Exam Questions
PCR (Polymerase Chain Reaction) is used to:
PCR (Polymerase Chain Reaction) is a fundamental technique in molecular biology. Its primary purpose is to exponentially amplify a specific segment of DNA, creating millions to billions of copies from a very small initial sample.
Why the other options are incorrect:
a) Sequence amino acids: Amino acid sequencing is determined by protein analysis techniques (like Edman degradation or mass spectrometry), not PCR. DNA sequencing determines the order of nucleotides.
c) Transcribe RNA: Transcription is the cellular process of creating RNA from a DNA template, carried out by the enzyme RNA polymerase. While a related technique called RT-PCR uses RNA as a starting material, it first converts the RNA to DNA before the amplification process.
d) Measure enzyme activity: PCR itself is a process that uses a DNA polymerase enzyme, but it is not a tool for measuring the activity of various enzymes.
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ASCP Exam Questions
A mutation that replaces one nucleotide with another is called a:
A substitution mutation occurs when a single nucleotide base is replaced by a different base. For example, an adenine (A) might be replaced by a guanine (G) in the DNA sequence.
Why the other options are incorrect:
a) Deletion: This is the loss of one or more nucleotides from the DNA sequence.
b) Insertion: This is the addition of one or more nucleotides into the DNA sequence.
d) Frameshift: A frameshift mutation is not a specific type of mutation itself, but rather a consequence of an insertion or deletion (that is not a multiple of three). It alters the reading frame of the gene.
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ASCP Exam Questions
Which type of RNA carries amino acids to the ribosome?
Transfer RNA (tRNA) is specifically responsible for carrying amino acids to the ribosome during protein synthesis. Each tRNA molecule has an anticodon that base-pairs with a complementary codon on the messenger RNA (mRNA), ensuring the correct amino acid is added to the growing polypeptide chain.
Why the other options are incorrect:
a) mRNA (Messenger RNA): This type of RNA carries the genetic code from DNA in the nucleus to the ribosome, serving as a template for protein synthesis. It does not carry amino acids.
c) rRNA (Ribosomal RNA): This is a structural and catalytic component of the ribosome itself, where protein synthesis occurs. It does not transport amino acids.
d) snRNA (Small Nuclear RNA): This is involved in the processing of pre-mRNA (e.g., splicing) within the nucleus and is not involved in delivering amino acids to the ribosome.
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ASCP Exam Questions
The complementary base of adenine in RNA is:
In RNA , the base uracil (U) replaces thymine (T) found in DNA.
Therefore, adenine (A) pairs with uracil (U) through two hydrogen bonds .
The other base pairing remains the same: guanine (G) pairs with cytosine (C) .
Other options:
a) Thymine → found only in DNA , not RNA.
b) Cytosine → pairs with guanine , not adenine.
d) Guanine → pairs with cytosine , not adenine.
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ASCP Exam Questions
The process by which a drug is metabolized can be studied using a technique where separation occurs in a column with a liquid pumped under high pressure. This describes:
Other options:
a) High-pressure focused electrophoresis → not a standard analytical technique; electrophoresis uses an electric field, not pressure.
c) Gas-liquid chromatography (GLC) → uses a gas as the mobile phase and requires volatile compounds .
d) Mass spectrophotometry–mass spectrometry (MS–MS) → identifies and quantifies compounds based on their mass-to-charge ratio , but it’s not a separation-by-liquid-pressure technique.
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ASCP Exam Questions
The presence of basophilic stippling in RBCs on a peripheral smear and an elevated erythrocyte protoporphyrin level would lead to suspicion of poisoning by which heavy metal?
Basophilic stippling of red blood cells occurs because lead inhibits delta-aminolevulinic acid dehydratase (ALAD) and ferrochelatase , enzymes involved in heme synthesis. This causes accumulation of ribosomal RNA , which appears as stippling in RBCs.
Elevated erythrocyte protoporphyrin is another hallmark of lead poisoning, due to impaired incorporation of iron into protoporphyrin IX.
Why the other options are incorrect:
a) Arsenic: Arsenic poisoning can cause basophilic stippling, but it does not characteristically cause an elevation in erythrocyte protoporphyrin. Its primary hematological effect is often pancytopenia.
b) Iron: Iron is not a toxic heavy metal in this context; it is an essential nutrient. Iron deficiency is a common cause of elevated erythrocyte protoporphyrin, but it does not cause basophilic stippling.
c) Mercury: Mercury poisoning does not typically cause either of these two specific findings. Its effects are primarily neurological and renal.
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ASCP Exam Questions
The process of RNA synthesis from a DNA template is called:
Transcription is the process by which RNA is synthesized from a DNA template .
During transcription, RNA polymerase reads the DNA template strand and produces a complementary RNA molecule (mRNA, tRNA, or rRNA).
Other options:
a) Translation → process where mRNA is used to make proteins on ribosomes.
b) Replication → process of making a new DNA molecule from an existing one.
d) Transformation → process by which a cell takes up foreign DNA from its surroundings.
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ASCP Exam Questions
The start codon in mRNA is:
The start codon in mRNA is AUG , which signals the beginning of translation .
It codes for the amino acid methionine (in eukaryotes) or formyl-methionine (fMet) in prokaryotes.
Other options (a, b, d) are stop codons :
UAA → Stop
UAG → Stop
UGA → Stop
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ASCP Exam Questions
Agarose gel electrophoresis separates nucleic acids based on:
Agarose gel electrophoresis separates DNA or RNA fragments based on their size (length) and negative charge .
Nucleic acids are negatively charged due to their phosphate backbone and move toward the positive electrode (anode) when an electric current is applied.
Smaller fragments move faster and farther through the gel pores than larger ones.
Other options:
a) Shape → has little effect; mainly important in protein electrophoresis.
c) Sequence → determined by DNA sequencing, not electrophoresis.
d) Color → not a factor; dyes are used only to visualize bands.
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ASCP Exam Questions
Which of the following statements correctly describes the utility of clinical laboratory assays for tumor markers?
Tumor markers (like AFP, β-hCG, CEA, CA-125, PSA) are not highly specific or sensitive enough for general population screening .
Their main clinical utility is to:
Monitor treatment response – levels should decrease if therapy is effective.
Detect recurrence – rising levels can indicate relapse.
Other options:
a) Highly specific for screening asymptomatic patients → False; markers can be elevated in benign conditions.
c) Confirm absence of disease → False; a normal marker does not guarantee absence of cancer.
d) Identify patients at risk for developing cancer → False; markers do not predict risk reliably.
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ASCP Exam Questions
The definitive diagnosis of an active myocardial infarction is best supported by an elevation of which of the following?
Other options:
a) Total CK → nonspecific; can be elevated in skeletal muscle injury too.
b) CK-MB → more specific to cardiac tissue than total CK, but less specific and sensitive than troponin; now used mainly as a secondary marker .
c) LD (Lactate dehydrogenase) → older, nonspecific marker; replaced by troponin testing.
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ASCP Exam Questions
The definitive diagnosis for hereditary coproporphyria (HCP) is a marked increase in:
Hereditary Coproporphyria (HCP) is caused by a deficiency in the enzyme coproporphyrinogen oxidase. This defect leads to the specific accumulation of its substrate, coproporphyrinogen III (which auto-oxidizes to coproporphyrin III), primarily in the feces.
Let’s review why the other options are incorrect:
a) Urine δ-aminolevulinic acid (ALA) & b) Urine porphobilinogen (PBG): While ALA and PBG are often increased in the urine during acute attacks of HCP, this is not the definitive or unique diagnostic feature. A marked increase in urinary ALA and PBG is more characteristic of Acute Intermittent Porphyria (AIP) .
d) Erythrocyte protoporphyrin: This is characteristically increased in Iron Deficiency Anemia and Protoporphyria , not in HCP.
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ASCP Exam Questions
Which biosafety practice is most important when working with molecular samples?
In molecular biology, especially in sensitive techniques like PCR, preventing contamination is the single most critical biosafety and quality assurance practice. Even minute amounts of contaminating DNA, RNase, or other molecules can ruin experiments, leading to false positives or degraded samples.
Gloves prevent contamination from the user’s skin (e.g., nucleases, foreign DNA).
Filtered (barrier) pipette tips prevent aerosols from contaminating the pipette shaft and subsequent samples.
Why the other options are incorrect:
a) Use of dry heat sterilization: While sterilizing equipment is important, it is a general lab practice and not the most specific critical practice for handling molecular samples themselves.
b) Avoidance of UV exposure: UV light is actually used to sterilize surfaces and crosslink DNA in gels. While prolonged UV exposure should be avoided by personnel, it is not the primary focus for sample integrity.
d) Running PCR near open flames: Open flames (like Bunsen burners) are rarely used in modern molecular biology and can create air currents that increase the risk of contamination. They are not a recommended practice for PCR setup.
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ASCP Exam Questions
The most common form of congenital adrenal hyperplasia (21-hydroxylase deficiency) is detected by elevated plasma levels of:
The most common form of congenital adrenal hyperplasia (CAH) is due to 21-hydroxylase deficiency (≈90–95% of cases).
21-hydroxylase is required for the conversion of 17-hydroxyprogesterone → 11-deoxycortisol in the cortisol synthesis pathway.
When this enzyme is deficient:
Cortisol and aldosterone ↓ (decreased synthesis)
ACTH ↑ (due to loss of feedback inhibition)
17-hydroxyprogesterone accumulates → key diagnostic marker
Other options:
a) Cortisol → decreased, not increased.
b) Aldosterone → decreased (causes salt-wasting form).
d) 11-deoxycortisol → decreased because its precursor can’t be converted without 21-hydroxylase.
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ASCP Exam Questions
The backbone of DNA is formed by:
The backbone of a DNA strand is composed of alternating sugar (deoxyribose) and phosphate molecules. These are linked together by strong phosphodiester bonds , which form between the 5′ phosphate of one nucleotide and the 3′ hydroxyl group of the next nucleotide’s sugar. This creates a continuous, stable chain.
Why the other options are incorrect:
a) Base pair hydrogen bonds: Hydrogen bonds hold the two strands of the DNA double helix together by connecting complementary base pairs (A-T, G-C). However, these bonds are between the strands and do not form the backbone of a single strand.
c) Peptide bonds: These are the covalent bonds that link amino acids together in proteins. They are not involved in DNA structure.
d) Sulfhydryl groups: These are -SH groups found in the amino acid cysteine, and they form disulfide bridges that help stabilize protein structure. They are not part of DNA.
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ASCP Exam Questions
A condition in which erythrocyte protoporphyrin is increased is:
In iron deficiency anemia , the lack of iron disrupts the final step of heme synthesis, where iron is incorporated into protoporphyrin IX to form heme. This causes a buildup of the precursor, free erythrocyte protoporphyrin (FEP).
The other options are types of porphyria, but they involve defects in different steps of the heme synthesis pathway:
a) Acute intermittent porphyria and d) Acute porphyric attack are characterized by a buildup of porphobilinogen and delta-aminolevulinic acid, not protoporphyrin.
c) Porphyria cutanea tarda is characterized by a buildup of uroporphyrinogen, not protoporphyrin.
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ASCP Exam Questions
A “Hook effect” in a tumor marker immunoassay, resulting in a falsely low value, can be suspected when:
The “Hook effect” (also called the prozone effect ) occurs in immunometric (sandwich) assays when the antigen concentration is extremely high .
In this situation, excess antigen saturates both the capture and detection antibodies , preventing proper “sandwich” formation.
As a result, the measured signal is falsely low , even though the true antigen level is very high.
Other options:
a) African American male → unrelated to assay interference.
c) Specimen after digital rectal exam → may cause transient PSA elevation, not the hook effect.
d) Smoker → may elevate CEA slightly but not cause the hook effect.
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