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Exam Prep

Molecular Genetics practice questions.

Nucleic acids, mutations and inheritance, extraction, electrophoresis, PCR, sequencing and flow cytometry.

All subjects

Independent practice questions written by Labologics. Not affiliated with DHA, DOH (formerly HAAD), SCFHS, Prometric or any exam body. Not recalled exam questions.

Nucleic acid structure

Q1 Nucleic acid structure · Easy

Which bases are purines?

  1. A: Adenine and guanine
  2. B: Cytosine and thymine
  3. C: Cytosine and uracil
  4. D: Thymine and uracil
Show answer

Answer: A — Adenine and guanine

Purines (A, G) have two rings. Pyrimidines (C, T, U) have one ring. Thymine is found in DNA and uracil replaces it in RNA.

Reference: Thompson & Thompson Genetics in Medicine

Q2 Nucleic acid structure · Easy

How many hydrogen bonds join guanine and cytosine?

  1. A: Two
  2. B: Three
  3. C: Four
  4. D: One
Show answer

Answer: B — Three

G-C pairs have three hydrogen bonds; A-T (DNA) and A-U (RNA) pairs have two. DNA rich in G-C therefore needs a higher temperature to separate (higher melting temperature).

Reference: Thompson & Thompson Genetics in Medicine

Q3 Nucleic acid structure · Easy

In RNA, adenine pairs with:

  1. A: Cytosine
  2. B: Thymine
  3. C: Uracil
  4. D: Guanine
Show answer

Answer: C — Uracil

RNA uses uracil instead of thymine. RNA also contains ribose (with a 2'-OH group) instead of deoxyribose.

Reference: Thompson & Thompson Genetics in Medicine

Q4 Nucleic acid structure · Medium

The two strands of a DNA double helix are:

  1. A: Joined by covalent bonds between bases
  2. B: Both read only 3' to 5'
  3. C: Parallel and identical
  4. D: Antiparallel and complementary
Show answer

Answer: D — Antiparallel and complementary

One strand runs 5' to 3' and the other 3' to 5'. Bases pair by hydrogen bonds; the sugar-phosphate backbone is joined by phosphodiester bonds.

Reference: Thompson & Thompson Genetics in Medicine

Q5 Nucleic acid structure · Medium

A double-stranded DNA sample contains 30% adenine. What percentage is guanine?

  1. A: 20%
  2. B: 30%
  3. C: 40%
  4. D: 70%
Show answer

Answer: A — 20%

A = T = 30%, so A + T = 60%. G + C = 40%, and G = C = 20%.

Reference: Thompson & Thompson Genetics in Medicine

Q6 Nucleic acid structure · Easy

A segment of DNA that codes for a functional product such as a protein is called a:

  1. A: Codon
  2. B: Gene
  3. C: Chromosome
  4. D: Nucleosome
Show answer

Answer: B — Gene

A codon is three bases coding one amino acid; a chromosome carries many genes; a nucleosome is DNA wrapped around histones.

Reference: Thompson & Thompson Genetics in Medicine

Q7 Nucleic acid structure · Medium

The end product of purine breakdown in humans is:

  1. A: Beta-alanine
  2. B: Urea
  3. C: Uric acid
  4. D: Ammonia only
Show answer

Answer: C — Uric acid

Purines (A, G) are broken down to uric acid. Pyrimidines are broken down to soluble products such as beta-alanine.

Reference: Tietz Fundamentals of Clinical Chemistry and Molecular Diagnostics

Replication & gene expression

Q8 Replication & gene expression · Easy

Which enzyme unwinds the DNA double helix during replication?

  1. A: Primase
  2. B: Reverse transcriptase
  3. C: DNA ligase
  4. D: Helicase
Show answer

Answer: D — Helicase

Helicase separates the strands. Primase lays down RNA primers, DNA polymerase extends them, and ligase seals the gaps between Okazaki fragments.

Reference: Thompson & Thompson Genetics in Medicine

Q9 Replication & gene expression · Medium

DNA polymerase synthesizes new DNA in which direction?

  1. A: 5' to 3' only
  2. B: Both directions equally
  3. C: Randomly
  4. D: 3' to 5' only
Show answer

Answer: A — 5' to 3' only

New nucleotides are added to the 3'-OH end, so synthesis is 5' to 3' while the template is read 3' to 5'. This is why the lagging strand is made in Okazaki fragments.

Reference: Thompson & Thompson Genetics in Medicine

Q10 Replication & gene expression · Medium

Which enzyme joins Okazaki fragments on the lagging strand?

  1. A: RNA polymerase
  2. B: DNA ligase
  3. C: Helicase
  4. D: Topoisomerase
Show answer

Answer: B — DNA ligase

DNA ligase forms the final phosphodiester bond. Topoisomerase relieves supercoiling ahead of the replication fork.

Reference: Thompson & Thompson Genetics in Medicine

Q11 Replication & gene expression · Easy

Making messenger RNA from a DNA template is called:

  1. A: Reverse transcription
  2. B: Replication
  3. C: Transcription
  4. D: Translation
Show answer

Answer: C — Transcription

Transcription (by RNA polymerase) happens in the nucleus. Translation of mRNA into protein happens on ribosomes in the cytoplasm.

Reference: Thompson & Thompson Genetics in Medicine

Q12 Replication & gene expression · Medium

Which enzyme makes DNA from an RNA template?

  1. A: DNA ligase
  2. B: Taq polymerase
  3. C: RNA polymerase
  4. D: Reverse transcriptase
Show answer

Answer: D — Reverse transcriptase

Reverse transcriptase is used by retroviruses such as HIV and in the lab to make cDNA, the first step of RT-PCR.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q13 Replication & gene expression · Medium

Mature eukaryotic mRNA has a poly(A) tail. In the lab, this is used to:

  1. A: Prime cDNA synthesis with oligo(dT) primers
  2. B: Cut DNA at specific sites
  3. C: Separate proteins by size
  4. D: Stain RNA on gels
Show answer

Answer: A — Prime cDNA synthesis with oligo(dT) primers

Oligo(dT) primers bind the poly(A) tail, so reverse transcriptase copies mRNA selectively into cDNA.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q14 Replication & gene expression · Easy

A codon consists of:

  1. A: Two nucleotides
  2. B: Three nucleotides
  3. C: Four nucleotides
  4. D: One nucleotide
Show answer

Answer: B — Three nucleotides

Each codon of three bases codes one amino acid or a stop signal. AUG is the start codon (methionine).

Reference: Thompson & Thompson Genetics in Medicine

Mutations & chromosomes

Q15 Mutations & chromosomes · Medium

A single base change that creates a premature stop codon is a:

  1. A: Silent mutation
  2. B: Missense mutation
  3. C: Nonsense mutation
  4. D: Frameshift mutation
Show answer

Answer: C — Nonsense mutation

Nonsense mutations truncate the protein. Missense changes one amino acid (e.g., the sickle mutation); silent mutations do not change the amino acid; frameshifts come from insertions or deletions not divisible by three.

Reference: Thompson & Thompson Genetics in Medicine

Q16 Mutations & chromosomes · Medium

The sickle cell mutation (Glu6Val in HBB) is an example of a:

  1. A: Nonsense mutation
  2. B: Frameshift deletion
  3. C: Trinucleotide repeat expansion
  4. D: Missense point mutation
Show answer

Answer: D — Missense point mutation

A single A-to-T change in codon 6 (GAG to GTG) replaces glutamic acid with valine.

Reference: Thompson & Thompson Genetics in Medicine

Q17 Mutations & chromosomes · Easy

The karyotype 47,XX,+21 describes:

  1. A: Down syndrome in a female
  2. B: Klinefelter syndrome
  3. C: A normal female
  4. D: Turner syndrome
Show answer

Answer: A — Down syndrome in a female

47 chromosomes with an extra chromosome 21. Turner syndrome is 45,X and Klinefelter syndrome is 47,XXY.

Reference: ISCN: International System for Human Cytogenomic Nomenclature

Q18 Mutations & chromosomes · Medium

For a conventional karyotype, cell division is stopped at metaphase by adding:

  1. A: Proteinase K
  2. B: Colchicine (Colcemid)
  3. C: EDTA
  4. D: Ethidium bromide
Show answer

Answer: B — Colchicine (Colcemid)

Colcemid blocks spindle formation, so chromosomes are held at their most condensed stage. G-banding with Giemsa then shows the banding pattern.

Reference: ISCN: International System for Human Cytogenomic Nomenclature

Q19 Mutations & chromosomes · Easy

Autosomes are:

  1. A: Only the chromosomes in gametes
  2. B: The X and Y chromosomes
  3. C: The 22 pairs of non-sex chromosomes
  4. D: Mitochondrial DNA
Show answer

Answer: C — The 22 pairs of non-sex chromosomes

Humans have 22 pairs of autosomes plus one pair of sex chromosomes (46 total).

Reference: Thompson & Thompson Genetics in Medicine

Q20 Mutations & chromosomes · Medium

DiGeorge syndrome (22q11.2 deletion) is most commonly confirmed by:

  1. A: Sanger sequencing of one exon
  2. B: Protein electrophoresis
  3. C: Routine G-banded karyotype
  4. D: FISH or chromosomal microarray
Show answer

Answer: D — FISH or chromosomal microarray

The deletion is too small to be seen reliably on a routine karyotype. FISH with a 22q11.2 probe or a chromosomal microarray detects it.

Reference: Thompson & Thompson Genetics in Medicine

Q21 Mutations & chromosomes · Hard

Which test detects small deletions and duplications (copy number variants) across the whole genome in one assay?

  1. A: Chromosomal microarray
  2. B: Routine karyotype
  3. C: Single-gene PCR
  4. D: Northern blot
Show answer

Answer: A — Chromosomal microarray

Microarrays detect CNVs far smaller than a karyotype can show, but they cannot detect balanced translocations.

Reference: Thompson & Thompson Genetics in Medicine

Inheritance

Q22 Inheritance · Easy

One parent is affected by an autosomal dominant disorder (heterozygous) and the other is unaffected. The chance that each child is affected is:

  1. A: 0%
  2. B: 25%
  3. C: 50%
  4. D: 100%
Show answer

Answer: C — 50%

Autosomal dominant: one abnormal copy causes disease, and an affected heterozygous parent passes it on with 50% probability each pregnancy.

Reference: Thompson & Thompson Genetics in Medicine

Q23 Inheritance · Easy

Both parents are carriers of an autosomal recessive disease. The chance that each child is affected is:

  1. A: 25%
  2. B: 50%
  3. C: 75%
  4. D: 0%
Show answer

Answer: A — 25%

Carrier x carrier gives 25% affected, 50% carriers and 25% unaffected non-carriers.

Reference: Thompson & Thompson Genetics in Medicine

Q24 Inheritance · Medium

The father carries sickle cell trait (HbAS) and the mother is normal (HbAA). Which is correct for each child?

  1. A: 25% affected, 50% carrier
  2. B: 0% affected, 50% carrier
  3. C: 50% affected
  4. D: 100% carrier
Show answer

Answer: B — 0% affected, 50% carrier

AS x AA gives 50% AA and 50% AS. No child can be SS because the mother has no S allele.

Reference: Thompson & Thompson Genetics in Medicine

Q25 Inheritance · Medium

A mother carries an X-linked recessive disorder such as hemophilia A; the father is unaffected. Which is correct?

  1. A: Each son has a 50% chance of being affected
  2. B: Each daughter has a 50% chance of being affected
  3. C: No children can be carriers
  4. D: All sons are affected
Show answer

Answer: A — Each son has a 50% chance of being affected

Each son has a 50% chance of inheriting the affected X and being affected; each daughter has a 50% chance of being a carrier.

Reference: Thompson & Thompson Genetics in Medicine

Q26 Inheritance · Medium

Mitochondrial DNA disorders are inherited from:

  1. A: The father only
  2. B: The mother only
  3. C: Both parents equally
  4. D: Neither; they are always new mutations
Show answer

Answer: B — The mother only

Mitochondria in the embryo come from the egg, so mitochondrial DNA is maternally inherited.

Reference: Thompson & Thompson Genetics in Medicine

Q27 Inheritance · Easy

The observable characteristics of an individual are called the:

  1. A: Haplotype
  2. B: Genotype
  3. C: Phenotype
  4. D: Karyotype
Show answer

Answer: C — Phenotype

The genotype is the genetic makeup (e.g., AO); the phenotype is what is expressed (e.g., blood group A).

Reference: Thompson & Thompson Genetics in Medicine

Q28 Inheritance · Easy

The mother is group O and the father is group AB. Which blood groups are possible in their children?

  1. A: AB only
  2. B: O only
  3. C: A, B, AB or O
  4. D: A or B only (50% each)
Show answer

Answer: D — A or B only (50% each)

OO x AB gives AO (group A) or BO (group B), 50% each. The children cannot be O or AB.

Reference: Thompson & Thompson Genetics in Medicine

Q29 Inheritance · Medium

The mother is group O, the father is group B, and their child is group O. What is the father's genotype?

  1. A: BO
  2. B: OO
  3. C: AB
  4. D: BB
Show answer

Answer: A — BO

The child (OO) must inherit an O allele from each parent, so the group B father must be heterozygous BO.

Reference: Thompson & Thompson Genetics in Medicine

Q30 Inheritance · Medium

Both parents are group A and both are heterozygous (AO). What is the chance that a child is group O?

  1. A: 0%
  2. B: 25%
  3. C: 50%
  4. D: 75%
Show answer

Answer: B — 25%

AO x AO gives 25% AA, 50% AO (both group A) and 25% OO (group O). If either parent were AA, no child could be group O.

Reference: Thompson & Thompson Genetics in Medicine

Extraction & specimens

Q31 Extraction & specimens · Easy

Which blood collection tube is preferred for DNA extraction and PCR?

  1. A: Fluoride oxalate (grey top)
  2. B: Plain serum tube
  3. C: EDTA (lavender top)
  4. D: Heparin (green top)
Show answer

Answer: C — EDTA (lavender top)

EDTA chelates magnesium and inhibits DNases. Heparin inhibits PCR polymerases and should be avoided.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q32 Extraction & specimens · Medium

Which is a standard method of DNA extraction?

  1. A: Gram staining
  2. B: Ethidium bromide staining
  3. C: Reverse transcription
  4. D: Phenol-chloroform (organic) extraction
Show answer

Answer: D — Phenol-chloroform (organic) extraction

Common methods are organic (phenol-chloroform), solid-phase (silica columns or magnetic beads) and chelating resin (e.g., Chelex).

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q33 Extraction & specimens · Medium

A DNA sample has an A260/A280 ratio of about 1.8. This indicates:

  1. A: Reasonably pure DNA
  2. B: Pure RNA
  3. C: Complete degradation
  4. D: Heavy protein contamination
Show answer

Answer: A — Reasonably pure DNA

Pure DNA gives about 1.8 and pure RNA about 2.0. A lower ratio suggests protein or phenol contamination.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q34 Extraction & specimens · Medium

A fresh tissue sample is received for nucleic acid testing. What is the best handling?

  1. A: Leave it at room temperature until the next day
  2. B: Process it immediately or freeze it (e.g., -80 C)
  3. C: Fix it in formalin for one week first
  4. D: Store it in a heparin tube
Show answer

Answer: B — Process it immediately or freeze it (e.g., -80 C)

Nucleases degrade DNA and especially RNA quickly. Prolonged formalin fixation fragments nucleic acids.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q35 Extraction & specimens · Medium

RNA is more difficult to handle than DNA mainly because:

  1. A: RNA does not absorb light at 260 nm
  2. B: RNA is double-stranded
  3. C: RNases are everywhere and very stable, and RNA is chemically less stable
  4. D: RNA cannot be extracted from blood
Show answer

Answer: C — RNases are everywhere and very stable, and RNA is chemically less stable

Use RNase-free reagents, gloves and tubes, and process or stabilize samples quickly.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Electrophoresis

Q36 Electrophoresis · Easy

Electrophoresis separates molecules mainly according to their:

  1. A: Temperature
  2. B: Smell
  3. C: Color
  4. D: Charge and size
Show answer

Answer: D — Charge and size

Charged molecules move in an electric field. For nucleic acids, which have a uniform charge-to-mass ratio, the gel separates mainly by size.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q37 Electrophoresis · Easy

In agarose gel electrophoresis, DNA moves towards the:

  1. A: Anode (positive electrode)
  2. B: Middle of the gel
  3. C: Top of the tank only
  4. D: Cathode (negative electrode)
Show answer

Answer: A — Anode (positive electrode)

The phosphate backbone makes DNA negatively charged. Smaller fragments move faster and further.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q38 Electrophoresis · Medium

Polyacrylamide gel (PAGE) is preferred over agarose for:

  1. A: Very large DNA fragments
  2. B: Proteins and very small DNA fragments
  3. C: Whole chromosomes
  4. D: Live cells
Show answer

Answer: B — Proteins and very small DNA fragments

PAGE has smaller pores and higher resolution; SDS-PAGE separates proteins by molecular weight. Agarose suits larger DNA fragments.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q39 Electrophoresis · Easy

Which dye is commonly used to visualize DNA in an agarose gel under UV light?

  1. A: Ponceau S
  2. B: Coomassie blue
  3. C: Ethidium bromide or SYBR Green
  4. D: Gram stain
Show answer

Answer: C — Ethidium bromide or SYBR Green

These dyes bind nucleic acids and fluoresce. Ethidium bromide is a mutagen, so safer alternatives are often used. Coomassie and Ponceau stain proteins.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q40 Electrophoresis · Medium

Which blotting technique detects specific RNA?

  1. A: Western blot
  2. B: Eastern blot
  3. C: Southern blot
  4. D: Northern blot
Show answer

Answer: D — Northern blot

Southern = DNA, Northern = RNA, Western = protein.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

PCR

Q41 PCR · Easy

What is the correct order of steps in one PCR cycle?

  1. A: Denaturation, annealing, extension
  2. B: Extension, annealing, denaturation
  3. C: Denaturation, extension, annealing
  4. D: Annealing, denaturation, extension
Show answer

Answer: A — Denaturation, annealing, extension

Denaturation at about 94-95 C separates the strands, annealing at about 50-65 C binds the primers, and extension at about 72 C lets Taq polymerase copy the template.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q42 PCR · Easy

Taq polymerase is used in PCR because it:

  1. A: Works only at low temperatures
  2. B: Is heat-stable and survives repeated denaturation steps
  3. C: Copies RNA directly
  4. D: Cuts DNA at specific sites
Show answer

Answer: B — Is heat-stable and survives repeated denaturation steps

It comes from the thermophilic bacterium Thermus aquaticus and survives repeated heating to about 95 C.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q43 PCR · Medium

In PCR, primers:

  1. A: Remove inhibitors from the sample
  2. B: Are long proteins that unwind DNA
  3. C: Are short single-stranded oligonucleotides that define the region to be amplified
  4. D: Are dyes that stain the product
Show answer

Answer: C — Are short single-stranded oligonucleotides that define the region to be amplified

A forward and a reverse primer bind opposite strands and flank the target, giving polymerase a 3'-OH starting point.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q44 PCR · Hard

Which component is an essential cofactor for DNA polymerase in a PCR reaction?

  1. A: Iron
  2. B: Sodium fluoride
  3. C: Calcium
  4. D: Magnesium (Mg2+)
Show answer

Answer: D — Magnesium (Mg2+)

Mg2+ concentration affects enzyme activity, specificity and yield. Too little gives no product; too much gives non-specific bands.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q45 PCR · Easy

Starting from one double-stranded DNA target, how many copies exist after 10 ideal PCR cycles?

  1. A: 20
  2. B: 100
  3. C: 1,024
  4. D: 10,000
Show answer

Answer: C — 1,024

Copies double each cycle: 2^n. After 10 cycles, 2^10 = 1,024. After 30 cycles, about one billion.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q46 PCR · Medium

A PCR no-template control (NTC) shows a band. This most likely means:

  1. A: The test is working perfectly
  2. B: Contamination of reagents or the work area
  3. C: The primers are missing
  4. D: The sample has too much DNA
Show answer

Answer: B — Contamination of reagents or the work area

Any amplification in the NTC means contamination, so patient results cannot be reported. Separate pre- and post-PCR areas with one-way workflow and use dUTP/UNG to reduce carryover.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q47 PCR · Medium

In real-time (quantitative) PCR, a LOWER Ct (cycle threshold) value means:

  1. A: Contamination only
  2. B: Less target in the sample
  3. C: More target in the sample
  4. D: The reaction failed
Show answer

Answer: C — More target in the sample

The more starting template there is, the fewer cycles are needed to cross the fluorescence threshold.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q48 PCR · Hard

Which statement about SYBR Green real-time PCR is correct?

  1. A: It can only detect RNA
  2. B: It cannot be quantified
  3. C: It uses a sequence-specific probe
  4. D: It binds any double-stranded DNA, so a melt curve is used to check specificity
Show answer

Answer: D — It binds any double-stranded DNA, so a melt curve is used to check specificity

SYBR Green also detects primer-dimers and non-specific products. Hydrolysis probes (TaqMan) add sequence specificity.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q49 PCR · Easy

RT-PCR is needed to detect which type of target?

  1. A: RNA, for example RNA viruses or gene expression
  2. B: Proteins
  3. C: Carbohydrates
  4. D: DNA viruses only
Show answer

Answer: A — RNA, for example RNA viruses or gene expression

Reverse transcriptase first converts RNA into cDNA, which PCR then amplifies.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Sequencing & methods

Q50 Sequencing & methods · Medium

Sanger sequencing works by:

  1. A: Cutting DNA with restriction enzymes
  2. B: Chain termination using dideoxynucleotides (ddNTPs)
  3. C: Antibody binding
  4. D: Measuring cell size with a laser
Show answer

Answer: B — Chain termination using dideoxynucleotides (ddNTPs)

A ddNTP lacks the 3'-OH group, so DNA synthesis stops. Fluorescent labels identify the last base, read by capillary electrophoresis.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q51 Sequencing & methods · Medium

The main advantage of next-generation sequencing (NGS) is:

  1. A: It replaces all microscopy
  2. B: It reads only one fragment at a time
  3. C: Massively parallel sequencing of many genes and samples in one run
  4. D: It needs no DNA
Show answer

Answer: C — Massively parallel sequencing of many genes and samples in one run

Barcodes (indexes) allow many samples to be pooled in one run. NGS is used for gene panels, exomes, genomes, tumor profiling and pathogen identification.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q52 Sequencing & methods · Medium

Restriction endonucleases cut DNA at:

  1. A: Only the 3' end
  2. B: RNA only
  3. C: Random positions
  4. D: Specific, often palindromic, recognition sequences
Show answer

Answer: D — Specific, often palindromic, recognition sequences

Bacteria use them as defense. In the lab they are used in cloning and RFLP analysis.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q53 Sequencing & methods · Medium

FISH (fluorescence in situ hybridization) is used to:

  1. A: Detect specific DNA sequences on chromosomes, e.g., BCR::ABL1 fusion or HER2 amplification
  2. B: Measure enzyme activity
  3. C: Sequence whole genomes
  4. D: Count bacteria in urine
Show answer

Answer: A — Detect specific DNA sequences on chromosomes, e.g., BCR::ABL1 fusion or HER2 amplification

Fluorescent probes bind target sequences in cells or tissue, detecting deletions, translocations and amplifications.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q54 Sequencing & methods · Easy

Computer-based analysis of sequence and other biological data is called:

  1. A: Cytogenetics
  2. B: Bioinformatics
  3. C: Histology
  4. D: Immunology
Show answer

Answer: B — Bioinformatics

Bioinformatics pipelines align NGS reads, call variants and interpret them against databases.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Flow cytometry

Q55 Flow cytometry · Easy

In flow cytometry, forward scatter (FSC) mainly reflects:

  1. A: DNA content
  2. B: Antibody binding
  3. C: Cell size
  4. D: Granularity
Show answer

Answer: C — Cell size

Side scatter (SSC) reflects internal complexity or granularity. Fluorescent antibody labels detect CD markers.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q56 Flow cytometry · Medium

The three main systems of a flow cytometer are:

  1. A: Centrifuge, microscope and printer
  2. B: Incubator, shaker and freezer
  3. C: Thermal cycler, gel tank and UV lamp
  4. D: Fluidics, optics and electronics
Show answer

Answer: D — Fluidics, optics and electronics

Fluidics uses hydrodynamic focusing to pass cells one at a time through the laser; optics collects scattered and fluorescent light; electronics converts signals into data.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q57 Flow cytometry · Easy

A common clinical use of flow cytometry is:

  1. A: Immunophenotyping leukemias and lymphomas with CD markers
  2. B: Measuring serum sodium
  3. C: Sequencing DNA
  4. D: Culturing bacteria
Show answer

Answer: A — Immunophenotyping leukemias and lymphomas with CD markers

Other uses include CD4 counts in HIV, PNH testing (CD55/CD59, FLAER), and CD34 stem cell counts.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Clinical applications

Q58 Clinical applications · Easy

Which test confirms MERS-CoV or SARS-CoV-2 infection?

  1. A: Antibody test only
  2. B: RT-PCR (nucleic acid amplification test)
  3. C: Blood culture
  4. D: ESR
Show answer

Answer: B — RT-PCR (nucleic acid amplification test)

Both are RNA viruses, so RT-PCR on respiratory samples is the confirmatory method.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q59 Clinical applications · Medium

Which test confirms active hepatitis C infection after a reactive anti-HCV result?

  1. A: Liver ultrasound
  2. B: Repeat anti-HCV
  3. C: HCV RNA by RT-PCR
  4. D: HBsAg
Show answer

Answer: C — HCV RNA by RT-PCR

Anti-HCV shows exposure; HCV RNA shows current infection. Quantitative HCV RNA is also used to check treatment response.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q60 Clinical applications · Hard

Hepatitis B is a DNA virus. Its viral load is measured by:

  1. A: HBsAg ELISA
  2. B: Western blot
  3. C: RT-PCR for HBV RNA
  4. D: Real-time PCR for HBV DNA
Show answer

Answer: D — Real-time PCR for HBV DNA

HBV DNA quantitation guides treatment and monitoring. RT-PCR is used for RNA viruses such as HCV and HIV.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q61 Clinical applications · Medium

A tissue biopsy shows large cells with 'owl's eye' intranuclear inclusions. Which test best confirms the likely virus?

  1. A: CMV PCR (or immunohistochemistry)
  2. B: HBV DNA
  3. C: HIV antibody
  4. D: EBV serology
Show answer

Answer: A — CMV PCR (or immunohistochemistry)

Owl's eye inclusions are typical of cytomegalovirus. CMV PCR on blood or tissue confirms and quantifies infection.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q62 Clinical applications · Medium

Which molecular test supports the diagnosis of polycythemia vera?

  1. A: BCR::ABL1 RT-PCR
  2. B: JAK2 V617F mutation testing
  3. C: Factor V Leiden PCR
  4. D: CFTR sequencing
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Answer: B — JAK2 V617F mutation testing

About 95% of PV have JAK2 V617F, usually detected by allele-specific or real-time PCR; most of the rest have JAK2 exon 12 mutations.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q63 Clinical applications · Medium

The response of chronic myeloid leukemia to tyrosine kinase inhibitors is monitored by:

  1. A: Bone marrow iron stain
  2. B: Serum LDH only
  3. C: Quantitative RT-PCR for BCR::ABL1 transcripts
  4. D: JAK2 testing
Show answer

Answer: C — Quantitative RT-PCR for BCR::ABL1 transcripts

Results are reported on the International Scale; a major molecular response is BCR::ABL1 of 0.1% or less.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q64 Clinical applications · Medium

Short tandem repeats (STRs) used in paternity and forensic testing are:

  1. A: Whole chromosomes
  2. B: Mitochondrial proteins
  3. C: Blood group antigens
  4. D: Repeats of 2-6 base pairs, analyzed by PCR and capillary electrophoresis
Show answer

Answer: D — Repeats of 2-6 base pairs, analyzed by PCR and capillary electrophoresis

STR loci are highly variable between people. Testing many loci together gives a very high power of discrimination.

Reference: Buckingham: Molecular Diagnostics - Fundamentals, Methods and Clinical Applications

Q65 Clinical applications · Easy

Matching donors and recipients for bone marrow or stem cell transplantation relies mainly on:

  1. A: HLA typing
  2. B: Karyotyping
  3. C: Hemoglobin electrophoresis
  4. D: ABO typing only
Show answer

Answer: A — HLA typing

HLA genes (MHC on chromosome 6) are typed at high resolution, usually by sequencing-based methods.

Reference: Thompson & Thompson Genetics in Medicine

Q66 Clinical applications · Easy

Cystic fibrosis is caused by mutations in which gene?

  1. A: HBB
  2. B: CFTR
  3. C: JAK2
  4. D: F5
Show answer

Answer: B — CFTR

CFTR encodes a chloride channel. F508del is the most common mutation worldwide. Sweat chloride testing supports the diagnosis.

Reference: Thompson & Thompson Genetics in Medicine