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ASCP MB practice questions

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Question 1Molecular Biology Principles, Genetics, and Basic Techniques

Which interaction contributes to the stability of the DNA double helix?

  1. Guanine pairs with uracil through three hydrogen bonds
  2. Adenine pairs with thymine through two hydrogen bonds Correct answer.
  3. Cytosine pairs with thymine through two hydrogen bonds
  4. Adenine pairs with guanine through three hydrogen bonds

Explanation

Complementary base pairing holds the two antiparallel strands together: adenine pairs with thymine through two hydrogen bonds and guanine pairs with cytosine through three. Because G–C pairs have an extra hydrogen bond, GC-rich regions are more thermally stable than AT-rich regions.

Question 2Molecular Biology Principles, Genetics, and Basic Techniques

Which two bases are purines?

  1. Thymine and uracil
  2. Guanine and cytosine
  3. Adenine and guanine Correct answer.
  4. Cytosine and thymine

Explanation

Purines have a fused two-ring structure and are adenine and guanine. Pyrimidines have a single ring and include cytosine, thymine, and uracil.

Question 3Molecular Biology Principles, Genetics, and Basic Techniques

In which direction does a DNA polymerase extend a newly synthesized strand?

  1. Bidirectionally from a single origin
  2. 5′ to 3′ along the newly synthesized strand Correct answer.
  3. 3′ to 5′ along the template strand
  4. 3′ to 5′ along the newly synthesized strand

Explanation

DNA polymerases add nucleotides to the free 3′-hydroxyl of the growing chain, so the new strand is made 5′ to 3′. The template strand is therefore read in the opposite, 3′ to 5′, direction.

Question 4Molecular Biology Principles, Genetics, and Basic Techniques

Which step of a standard PCR cycle takes place at approximately 94–95 °C?

  1. Ligation of adjacent amplicons
  2. Primer annealing
  3. Denaturation of the double-stranded template Correct answer.
  4. Extension by Taq polymerase

Explanation

The cycle begins with heating to about 94–95 °C so that hydrogen bonds break and the two template strands separate. Annealing is typically 50–65 °C and extension is usually near 72 °C.

Question 5Molecular Biology Principles, Genetics, and Basic Techniques

During standard agarose gel electrophoresis at neutral pH, DNA migrates toward which electrode, and why?

  1. Neither electrode unless a detergent is added to the buffer
  2. The positive electrode, because DNA becomes positively charged at pH 7
  3. The negative electrode, because the bases carry a net positive charge
  4. The positive electrode, because the phosphate backbone is negatively charged Correct answer.

Explanation

At neutral pH each phosphate group in the sugar-phosphate backbone carries a negative charge, giving DNA a strong net negative charge. The molecule therefore moves through the gel toward the positive electrode (anode).

Question 6Molecular Biology Principles, Genetics, and Basic Techniques

Type II restriction endonucleases characteristically recognize sequences that are:

  1. Palindromic and usually 4–8 base pairs long Correct answer.
  2. Random single-stranded regions
  3. Composed only of adenine and thymine residues
  4. Methylated at every adenine residue

Explanation

Type II enzymes bind a specific recognition site, usually a 4–8 bp inverted repeat (palindrome), and cleave within or near it. This predictable specificity is what makes them useful for mapping and cloning.

Question 7Molecular Biology Principles, Genetics, and Basic Techniques

A Northern blot is designed to detect which type of molecule?

  1. Protein
  2. Double-stranded DNA
  3. RNA Correct answer.
  4. Membrane lipid

Explanation

Northern blotting separates RNA by size, transfers it to a membrane, and detects a specific transcript with a labeled probe. Southern blotting is the DNA counterpart, and western blotting detects protein.

Question 8Molecular Biology Principles, Genetics, and Basic Techniques

How many codons make up the standard genetic code, and how many of them specify an amino acid?

  1. 32 codons; 20 specify amino acids
  2. 64 codons; 61 specify amino acids Correct answer.
  3. 16 codons; 12 specify amino acids
  4. 61 codons; 61 specify amino acids

Explanation

With four bases read as triplets there are 64 possible codons. Three of these (UAA, UAG, UGA) are stop signals, leaving 61 sense codons for the 20 amino acids.

Question 9Molecular Biology Principles, Genetics, and Basic Techniques

The melting temperature (Tm) of a DNA duplex increases when:

  1. The proportion of A–T base pairs increases
  2. The salt concentration is reduced to near zero
  3. The fragment length is shortened to 20 bp
  4. The proportion of G–C base pairs increases Correct answer.

Explanation

G–C pairs share three hydrogen bonds while A–T pairs share only two, so a higher GC content requires more energy to separate the strands. Lower ionic strength destabilizes the duplex and lowers Tm rather than raising it.

Question 10Molecular Biology Principles, Genetics, and Basic Techniques

In Sanger sequencing, chain termination occurs because the incorporated dideoxynucleotide:

  1. Pairs with two template bases simultaneously
  2. Lacks a 2′ hydroxyl group
  3. Cannot be phosphorylated by kinases
  4. Lacks a 3′ hydroxyl group Correct answer.

Explanation

A dideoxynucleotide has no 3′-OH, so the polymerase cannot form the next phosphodiester bond and synthesis stops. Each of the four terminators is labeled, allowing the terminated fragments to be read as a sequence.

Question 11Molecular Biology Principles, Genetics, and Basic Techniques

In a real-time PCR assay, sample A has a Ct of 22 and sample B has a Ct of 29. Which conclusion is best supported?

  1. The assay is invalid because Ct values must be identical
  2. Sample B contains more target than sample A
  3. Sample A contains more target than sample B Correct answer.
  4. Both samples contain equal amounts of target

Explanation

Ct is the cycle at which fluorescence crosses the threshold; a lower Ct means the target was amplified sooner, which reflects a higher starting copy number. A difference of roughly 3.3 cycles corresponds to about a tenfold difference in template.

Question 12Molecular Biology Principles, Genetics, and Basic Techniques

Two unaffected parents have a child with an autosomal recessive disorder. If their next child is unaffected, what is the probability that this child is a carrier?

  1. 1/4
  2. 3/4
  3. 2/3 Correct answer.
  4. 1/2

Explanation

Both parents must be carriers, so each pregnancy is 1/4 affected, 1/2 carrier, and 1/4 homozygous normal. Among the unaffected children (3/4 of all offspring) the carrier proportion is (1/2)/(3/4) = 2/3.

Question 13Molecular Biology Principles, Genetics, and Basic Techniques

An autosomal recessive disorder affects 1 in 10,000 newborns in a population at Hardy–Weinberg equilibrium. What is the approximate carrier frequency?

  1. About 1 in 10,000
  2. About 1 in 50 Correct answer.
  3. About 1 in 5,000
  4. About 1 in 100

Explanation

The affected frequency equals q² = 1/10,000, so q = 0.01 and p ≈ 0.99. Carrier frequency is 2pq ≈ 0.02, or roughly 1 in 50.

Question 14Molecular Biology Principles, Genetics, and Basic Techniques

A man with an X-linked recessive disorder has children with a woman who is homozygous for the normal allele. Which outcome is expected?

  1. All daughters will be carriers Correct answer.
  2. Half of the daughters will be affected
  3. No child will inherit the mutant allele
  4. All sons will be affected

Explanation

A father passes his Y chromosome to sons and his X chromosome to daughters. Daughters therefore receive the mutant X plus a normal X from the mother and are obligate carriers, while sons receive the normal maternal X and are unaffected.

Question 15Molecular Biology Principles, Genetics, and Basic Techniques

Which method most directly detects the BCR-ABL1 fusion transcript in a patient with chronic myeloid leukemia?

  1. Serum protein electrophoresis
  2. Karyotype analysis at 400-band resolution
  3. Reverse-transcription PCR of peripheral blood Correct answer.
  4. Southern blot with a centromeric chromosome 9 probe

Explanation

RT-PCR converts the fusion mRNA to cDNA and amplifies across the junction, detecting the transcript itself with high sensitivity and specificity. Karyotyping may show t(9;22) but cannot define the transcript, and it requires dividing cells.

Question 16Molecular Biology Principles, Genetics, and Basic Techniques

A purified DNA sample has an A260/A280 ratio of 1.4. What does this most likely indicate?

  1. The sample is pure RNA
  2. The sample is highly pure DNA
  3. The sample is contaminated with protein Correct answer.
  4. The DNA is extensively degraded

Explanation

Pure DNA typically gives an A260/A280 ratio near 1.8, while RNA is near 2.0. A ratio of 1.4 suggests protein or phenol carryover, since those contaminants absorb strongly at 280 nm.

Question 17Molecular Biology Principles, Genetics, and Basic Techniques

What is the function of reverse transcriptase in molecular testing?

  1. Synthesizes RNA from a DNA template
  2. Synthesizes a DNA copy from an RNA template Correct answer.
  3. Joins two DNA fragments end to end
  4. Degrades RNA within an RNA–DNA hybrid

Explanation

Reverse transcriptase uses RNA as a template to polymerize a complementary DNA strand. This activity underlies cDNA synthesis and RT-PCR for RNA targets such as fusion transcripts and viral genomes.

Question 18Molecular Biology Principles, Genetics, and Basic Techniques

After ligating a target insert into a plasmid vector, which feature allows selection of bacteria that have taken up the recombinant plasmid?

  1. The origin of replication
  2. The polyadenylation signal
  3. An antibiotic resistance gene carried by the vector Correct answer.
  4. A TATA box within the insert

Explanation

Plating on antibiotic-containing medium kills bacteria that did not acquire the vector, so only transformants grow. The resistance gene must be on the vector so that its presence confers the selectable phenotype.

Question 19Molecular Biology Principles, Genetics, and Basic Techniques

Loss of mismatch repair function in a colorectal tumor most typically produces which molecular phenotype?

  1. Homologous recombination deficiency
  2. Balanced translocations of chromosome 8
  3. Microsatellite instability Correct answer.
  4. Gross chromosomal amplification

Explanation

Mismatch repair proteins correct replication slippage errors within short tandem repeats. When repair is lost, these repeats change length from generation to generation, producing microsatellite instability.

Question 20Molecular Biology Principles, Genetics, and Basic Techniques

Which matrix is best suited to separating DNA fragments that differ in length by a single base pair?

  1. Starch gel
  2. Low-melting agarose at 0.8%
  3. Polyacrylamide gel Correct answer.
  4. Standard agarose at 1.5%

Explanation

Polyacrylamide forms a matrix with small, uniform pores that resolves short fragments differing by one nucleotide. Agarose gels are better suited to larger fragments, typically from a few hundred to several thousand base pairs.

Question 21Molecular Biology Principles, Genetics, and Basic Techniques

Hypermethylation of CpG islands within a gene promoter most commonly leads to:

  1. Loss of the poly-A tail from the mRNA
  2. Increased transcription of that gene
  3. Transcriptional silencing of that gene Correct answer.
  4. A shift in the reading frame during translation

Explanation

Methylated CpG dinucleotides recruit methyl-binding proteins and histone deacetylases, which condense chromatin and restrict transcription factor access. This is a classic epigenetic mechanism of gene silencing in cancer.

Question 22Molecular Biology Principles, Genetics, and Basic Techniques

A woman with a pathogenic mitochondrial DNA variant has children. Which statement best describes the expected inheritance pattern?

  1. No child can inherit the variant because it is not germline
  2. All children may inherit the variant, but expression varies with heteroplasmy Correct answer.
  3. All of her sons will be affected and none of her daughters
  4. Only half of her children will inherit the variant

Explanation

Mitochondrial DNA is transmitted maternally, and each child receives a variable proportion of mutant genomes. The phenotype depends on the heteroplasmy level and the threshold of the affected tissue, so siblings can differ markedly.

Question 23Molecular Biology Principles, Genetics, and Basic Techniques

Which ordering correctly reflects approximate resolution, from lowest to highest, of these techniques?

  1. Karyotype < FISH < Sanger sequencing Correct answer.
  2. Sanger sequencing < karyotype < FISH
  3. Karyotype < Sanger sequencing < FISH
  4. FISH < karyotype < Sanger sequencing

Explanation

G-banded karyotyping resolves imbalances of roughly 5–10 Mb, FISH detects targets in the range of about 100 kb to 1 Mb, and Sanger sequencing resolves single nucleotides. Resolution therefore increases from karyotype to FISH to sequencing.

Question 24Molecular Biology Principles, Genetics, and Basic Techniques

To discriminate between two alleles in an allele-specific PCR assay, the most critical design element is:

  1. A high annealing temperature for all primers
  2. Placement of the variant base at or near the 3′ end of a primer Correct answer.
  3. Use of a proofreading polymerase
  4. A long amplicon of at least 1 kb

Explanation

A mismatch at or near the primer's 3′ terminus strongly impedes extension, so only the perfectly matched allele amplifies efficiently. Stringent annealing conditions amplify that discrimination but cannot substitute for the position of the variant.

Question 25Molecular Biology Principles, Genetics, and Basic Techniques

Sanger sequencing of a tumor sample shows only the wild-type base at a position where a known mutation is suspected. What is the most likely explanation?

  1. The mutation must be homozygous throughout the tumor
  2. Sanger sequencing reliably detects variants at allelic fractions below 1%
  3. The mutant allele is present below the assay's detection limit of roughly 15–20% Correct answer.
  4. Sanger sequencing cannot read through any repeated sequence

Explanation

Sanger chemistry depends on a minor peak being visible above background, which generally requires the variant to represent about 15–20% of the signal. Lower fractions, common in tumors with admixed normal tissue, are missed and require more sensitive methods such as next-generation sequencing.

Question 26Molecular Biology Principles, Genetics, and Basic Techniques

Which nitrogenous base is classified as a purine?

  1. Cytosine
  2. Uracil
  3. Thymine
  4. Adenine Correct answer.

Explanation

Purines have a fused two-ring structure and include adenine and guanine. Pyrimidines have a single ring and include cytosine, thymine, and uracil.

Question 27Molecular Biology Principles, Genetics, and Basic Techniques

During a standard PCR cycle, which temperature typically allows primers to anneal to the template?

  1. 72°C
  2. 55°C Correct answer.
  3. 95°C
  4. 37°C

Explanation

PCR cycling generally uses about 95°C for denaturation, 50–65°C for primer annealing, and about 72°C for extension by a thermostable polymerase. The annealing temperature must be low enough for stable primer–template hybridization but high enough to limit nonspecific binding.

Question 28Molecular Biology Principles, Genetics, and Basic Techniques

In agarose gel electrophoresis at alkaline pH, DNA fragments migrate toward the positive electrode primarily because:

  1. the bases carry a net positive charge at this pH
  2. the DNA is protonated and becomes positively charged
  3. agarose binds DNA and pulls it toward the anode
  4. the phosphate backbone gives DNA a net negative charge Correct answer.

Explanation

Each phosphodiester linkage contributes a negatively charged phosphate group, so DNA has a net negative charge at typical electrophoretic pH. Negatively charged molecules move toward the positive electrode (anode), with smaller fragments migrating faster through the gel matrix.

Question 29Molecular Biology Principles, Genetics, and Basic Techniques

After restriction digestion and agarose gel electrophoresis, a Southern blot is used to detect:

  1. a specific RNA transcript
  2. a specific lipid
  3. a specific protein
  4. a specific DNA sequence Correct answer.

Explanation

Southern blotting transfers size-fractionated DNA from a gel to a membrane and then hybridizes it with a labeled probe complementary to the target sequence. The probe reveals the position of a specific DNA fragment among the separated fragments.

Question 30Molecular Biology Principles, Genetics, and Basic Techniques

A single-base substitution changes a codon specifying glutamine to UAA. This change is best classified as a:

  1. silent mutation
  2. missense mutation
  3. nonsense mutation Correct answer.
  4. frameshift mutation

Explanation

UAA is a stop codon, so replacing an amino acid–specifying codon with UAA introduces premature termination. A missense change would substitute a different amino acid, and a silent change would leave the amino acid unchanged.

Question 31Molecular Biology Principles, Genetics, and Basic Techniques

Which test is most appropriate to identify a balanced reciprocal translocation in a patient with recurrent pregnancy loss?

  1. Allele-specific PCR
  2. Southern blot
  3. Chromosomal microarray
  4. G-banded karyotype Correct answer.

Explanation

Balanced translocations do not change DNA copy number, so copy-number platforms such as chromosomal microarray may miss them. G-banded karyotyping visualizes chromosome structure and can detect balanced rearrangements.

Question 32Molecular Biology Principles, Genetics, and Basic Techniques

Short tandem repeat (STR) loci are useful in identity testing because they:

  1. are uniformly methylated in all tissues
  2. vary in the number of tandem repeats between individuals Correct answer.
  3. are transcribed into mRNA at high levels
  4. contain recognition sites for restriction endonucleases

Explanation

STR polymorphisms arise from differences among individuals in the number of repeated units at a locus. PCR amplification across multiple STR loci produces allele patterns that are highly discriminating for identity and relationship testing.

Question 33Molecular Biology Principles, Genetics, and Basic Techniques

A pathogenic variant is detected in DNA extracted from a breast tumor but is absent in the patient's peripheral blood DNA. What is the most likely interpretation?

  1. Clonal hematopoiesis
  2. Germline pathogenic variant
  3. Somatic tumor variant Correct answer.
  4. Mitochondrial variant

Explanation

A variant present in tumor tissue but absent from normal constitutional DNA is consistent with a somatic change acquired by the tumor. Germline variants are expected to be present in both tumor and normal tissues.

Question 34Molecular Biology Principles, Genetics, and Basic Techniques

A heterozygous 1.5-Mb deletion is suspected. Which method best defines the precise copy-number boundaries of the deletion?

  1. Sanger sequencing
  2. Locus-specific FISH
  3. G-banded karyotype
  4. Chromosomal microarray Correct answer.

Explanation

Chromosomal microarray measures copy number across the genome at kilobase-level resolution, allowing the deleted interval and its approximate breakpoints to be defined. G-banding has much lower resolution, and FISH is targeted rather than genome-wide.

Question 35Specimen Collection, Handling, and Nucleic Acid Extraction

Whole blood for a routine germline DNA PCR assay should be collected in which tube?

  1. Green-top sodium heparin
  2. Lavender-top EDTA Correct answer.
  3. Light blue-top sodium citrate
  4. Gold-top serum separator

Explanation

EDTA chelates magnesium, which shuts down DNases and preserves high-molecular-weight DNA in the cell pellet. Heparin is a well-recognized inhibitor of thermostable polymerases and is avoided for PCR.

Question 36Specimen Collection, Handling, and Nucleic Acid Extraction

During a single venipuncture for multiple tests, which tube is collected first?

  1. Sodium fluoride tube
  2. Sodium citrate tube Correct answer.
  3. EDTA tube
  4. Serum separator tube

Explanation

Order of draw places coagulation tubes containing sodium citrate ahead of serum and other additive tubes. This prevents additive carryover and early clot activation from affecting the citrate tube.

Question 37Specimen Collection, Handling, and Nucleic Acid Extraction

Cell-free DNA for a liquid biopsy is best recovered from which specimen?

  1. Whole blood frozen and thawed once
  2. Buffy coat layer
  3. Plasma separated promptly from EDTA blood Correct answer.
  4. Serum separated after 24 hours at room temperature

Explanation

Cell-free DNA circulates in the plasma fraction, and prompt separation limits lysis of leukocytes that would flood the sample with genomic DNA. Serum is avoided because clotting itself releases cellular DNA.

Question 38Specimen Collection, Handling, and Nucleic Acid Extraction

Sodium heparin is avoided in specimens intended for PCR because it:

  1. It raises the pH of the reaction buffer
  2. It degrades double-stranded DNA directly
  3. It chelates magnesium required by the enzyme
  4. It inhibits thermostable DNA polymerase Correct answer.

Explanation

Heparin binds to and inactivates Taq-family polymerases, suppressing amplification even when DNA is present. Heparinase can sometimes rescue a sample, but choosing another anticoagulant is preferred.

Question 39Specimen Collection, Handling, and Nucleic Acid Extraction

An EDTA whole-blood sample for DNA testing cannot be extracted the same day. What is the best handling?

  1. Freeze at -80 °C after one freeze-thaw
  2. Incubate at 37 °C to prevent clotting
  3. Freeze the whole tube at -20 °C until extraction
  4. Refrigerate at 2–8 °C for a few days Correct answer.

Explanation

Intact whole blood is stable for several days when refrigerated. Freezing unlysed whole blood without a cryoprotectant produces ice crystals that lyse cells and reduce DNA yield and integrity; long-term storage is done on purified DNA.

Question 40Specimen Collection, Handling, and Nucleic Acid Extraction

Compared with a standard cotton swab, a flocked nylon swab improves molecular testing because:

  1. It lyses epithelial cells on contact
  2. It dries the specimen instantly
  3. It contains an RNase inhibitor in the fibers
  4. Cells are released efficiently into medium Correct answer.

Explanation

Flocked fibers project outward and do not trap material in an absorbent core, so collected cells elute efficiently into transport medium. Better release translates into higher nucleic acid yield.

Question 41Specimen Collection, Handling, and Nucleic Acid Extraction

A tissue specimen for RNA analysis is placed in RNAlater. Its purpose is to:

  1. Stabilize RNA by inactivating RNases Correct answer.
  2. Cross-link nucleic acids to the matrix
  3. Dissolve lipids to clarify the specimen
  4. Decalcify mineralized tissue

Explanation

RNAlater is a concentrated salt solution that permeates tissue and inactivates RNases, allowing the sample to be held without freezing. RNA is then recovered by standard lysis and purification.

Question 42Specimen Collection, Handling, and Nucleic Acid Extraction

What is the main consequence of prolonged formalin fixation for molecular testing?

  1. Complete removal of paraffin wax
  2. Progressive loss of tissue architecture
  3. Nucleic acid fragmentation Correct answer.
  4. Increased recovery of intact DNA

Explanation

Formalin cross-links proteins to nucleic acids and over time fragments DNA and RNA, lowering the amount of amplifiable template. Fixation of roughly 24–48 hours is generally optimal before processing.

Question 43Specimen Collection, Handling, and Nucleic Acid Extraction

A blood sample for cell-free DNA is drawn at an outside clinic and will reach the laboratory in 48 hours. Which collection device is most appropriate?

  1. Cell-stabilizing cfDNA tube Correct answer.
  2. Sodium citrate tube
  3. Sodium heparin tube
  4. Plain red-top tube without additive

Explanation

Cell-stabilizing tubes fix nucleated cells so they cannot release genomic DNA during transit, keeping the cell-free fraction representative. This extends acceptable transport time well beyond what plain EDTA allows.

Question 44Specimen Collection, Handling, and Nucleic Acid Extraction

A citrate tube is filled to only half its marked volume. How does this affect a plasma-based molecular assay?

  1. No effect, because the tube still contains anticoagulant
  2. Excess citrate dilutes the plasma Correct answer.
  3. The specimen clots immediately and cannot be tested
  4. The plasma becomes more concentrated

Explanation

Liquid citrate is calibrated for a 1:9 anticoagulant-to-blood ratio. Underfilling leaves excess anticoagulant relative to blood, diluting plasma analytes and altering calcium-dependent reactions.

Question 45Specimen Collection, Handling, and Nucleic Acid Extraction

In silica-column nucleic acid extraction, guanidinium salts serve to:

  1. Denature proteins and aid silica binding Correct answer.
  2. Precipitate DNA in the presence of ethanol
  3. Chelate divalent cations in the lysate
  4. Digest contaminating protein enzymatically

Explanation

Chaotropic agents disrupt hydrogen bonding, denature proteins including nucleases, and promote adsorption of nucleic acid to silica in high salt. Bound nucleic acid is then eluted in a low-salt buffer.

Question 46Specimen Collection, Handling, and Nucleic Acid Extraction

In a phenol-chloroform extraction, the pH of the phenol determines:

  1. The solubility of ethanol in the two phases
  2. The final ionic strength of the aqueous phase
  3. The rate of proteinase K digestion
  4. Whether DNA or RNA enters the aqueous phase Correct answer.

Explanation

At acidic pH, DNA partitions into the organic interphase while RNA remains in the aqueous layer; buffering the phenol to about pH 8 keeps DNA in the aqueous phase. This is why pH 8 phenol is specified for DNA purification.

Question 47Specimen Collection, Handling, and Nucleic Acid Extraction

A dilute nucleic acid solution is precipitated with ethanol. Adding glycogen improves recovery because:

  1. It increases the ionic strength of the solution
  2. It blocks nuclease activity during centrifugation
  3. It acts as an inert carrier for nucleic acid Correct answer.
  4. It neutralizes residual chaotropic salts

Explanation

Glycogen provides a visible, inert matrix that coprecipitates trace nucleic acid and improves pellet formation. Linear acrylamide and tRNA serve the same purpose.

Question 48Specimen Collection, Handling, and Nucleic Acid Extraction

A DNA preparation has an A260/A280 ratio of 1.5. The most likely explanation is:

  1. Residual protein or phenol Correct answer.
  2. Presence of single-stranded DNA
  3. Over-dilution of the sample
  4. Excessive RNA contamination

Explanation

Pure DNA has an A260/A280 of about 1.8; a lower ratio indicates protein or phenol carryover, which absorbs strongly at 280 nm. Additional phenol-chloroform or column cleanup is usually required.

Question 49Specimen Collection, Handling, and Nucleic Acid Extraction

The RNA integrity number (RIN) generated by a bioanalyzer reports:

  1. The 28S-to-18S ratio of DNA fragments
  2. Overall degradation of the RNA sample Correct answer.
  3. Amount of contaminating genomic DNA
  4. Concentration of polyadenylated mRNA only

Explanation

RIN is an algorithm-derived score from 1 to 10 that integrates the entire electropherogram, with 10 indicating intact RNA. A simple 28S:18S ratio is a cruder, older indicator of degradation.

Question 50Specimen Collection, Handling, and Nucleic Acid Extraction

Before reverse transcription of an RNA sample, a DNase step is performed to:

  1. Remove proteins bound to the RNA
  2. Destroy contaminating genomic DNA Correct answer.
  3. Inactivate RNase inhibitors in the buffer
  4. Convert RNA into cDNA

Explanation

Amplification-based RNA assays can amplify residual genomic DNA and produce false-positive results. DNase I digests the DNA and must then be inactivated or removed before reverse transcription.