For a week, a real-time PCR assay's no-template control has been positive at Ct 35 to 37, with the same melt peak as the target. New reagent lots did not help. What is the most likely cause, and how is it best prevented?

Amplicon carryover; use dUTP with UNG

Primer dimers; use a hot-start polymerase

Mispriming; raise the annealing temperature

Reagent contamination; use a new reagent lot

Sanger sequencing of the HFE gene from a patient's blood DNA shows two overlapping peaks of about equal height, G and A, at c.845, with clean single peaks elsewhere. The patient is most likely:

Showing a dye artifact, not a true variant

Carrying a frameshift deletion at c.845

During each PCR cycle, the reaction is cooled to about 55 to 65 °C. What happens at this step?

Primers bind to their target sequences

RNA is copied into complementary DNA

Polymerase extends the new strands

What makes the fluorescent signal rise with each cycle in a hydrolysis (TaqMan) probe assay?

Polymerase cuts the probe, separating reporter from quencher

A dye binds to all double-stranded DNA as it forms

The probe folds into a hairpin once it binds target

More quencher dye is made with each new copy

Molecular methods

8 ASCP MLS practice questions on Molecular methods, part of Laboratory Operations. The first 8 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Hard
A laboratory-developed real-time PCR for BK polyomavirus DNA in plasma uses a standard curve of Ct = -3.32 × log10(copies per reaction) + 39.5, with an efficiency of about 100%. The extraction takes 0.50 mL of plasma and elutes the DNA in 60 µL, and 10 µL of eluate goes into each reaction. A kidney transplant recipient's sample gives a Ct of 26.2 in duplicate; the internal control, positive control and no-template control are all acceptable. What viral load should be reported?
1. AAbout 10,000 copies/mL
2. BAbout 30,000 copies/mL
3. CAbout 120,000 copies/mL
4. DAbout 60,000 copies/mL
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Answer: C · About 120,000 copies/mL
From the curve, log10(copies per reaction) = (39.5 - 26.2) ÷ 3.32 = 4.0, so each reaction held about 10,000 copies. That reaction used 10 µL of a 60 µL eluate, so the whole eluate held 6 × 10,000 = 60,000 copies. These came from 0.50 mL of plasma, so the viral load is 60,000 ÷ 0.50 = about 120,000 copies/mL (log10 about 5.1). The value can be released only because the internal, positive and no-template controls are valid and the Ct lies within the validated range of the standard curve.
- A: About 10,000 is the number of copies in one reaction, before any volume correction.
- B: About 30,000 multiplies by 0.50 mL instead of dividing by it.
- D: About 60,000 is the total in the eluate, which came from 0.50 mL of plasma, not per mL.
Question 2 · Hard
For a week, a real-time PCR assay's no-template control has been positive at Ct 35 to 37, with the same melt peak as the target. New reagent lots did not help. What is the most likely cause, and how is it best prevented?
1. APrimer dimers; use a hot-start polymerase
2. BMispriming; raise the annealing temperature
3. CReagent contamination; use a new reagent lot
4. DAmplicon carryover; use dUTP with UNG
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Answer: D · Amplicon carryover; use dUTP with UNG
A positive no-template control with the target's own melt peak means target sequence is getting into the tubes, and new reagent lots rule out a contaminated lot. The usual source is amplified product (amplicon) from earlier runs, spread as aerosols. If dUTP replaces dTTP in every run, all products contain uracil, and uracil-N-glycosylase (UNG) in the next reaction destroys them before cycling while leaving thymine-containing patient DNA intact. UNG acts only on products made with dUTP, so existing contamination must still be cleaned away with bleach and pre- and post-amplification areas kept separate.
- A: Primer dimers give a smaller product with a lower melt peak than the target.
- B: With no template in the tube, mispriming cannot make a product with the target's melt peak.
- C: New reagent lots were tried and did not clear the problem.
Question 3 · Medium
Sanger sequencing of the HFE gene from a patient's blood DNA shows two overlapping peaks of about equal height, G and A, at c.845, with clean single peaks elsewhere. The patient is most likely:
1. AHomozygous for c.845G>A
2. BHeterozygous for c.845G>A
3. CShowing a dye artifact, not a true variant
4. DCarrying a frameshift deletion at c.845
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Answer: B · Heterozygous for c.845G>A
Both copies of the gene are sequenced together. When one copy carries G and the other A at the same position, the trace shows two peaks of about equal height there, which is a heterozygous single-nucleotide variant. HFE c.845G>A is the change that gives p.Cys282Tyr (C282Y). Seeing the same double peak on the reverse strand helps confirm it is real.
- A: A homozygous variant gives one clean peak of the new base, A.
- C: Dye artifacts are broad blobs, not two sharp peaks of equal height at one base.
- D: A heterozygous frameshift gives mixed peaks from that point to the end of the read.
Question 4 · Easy
During each PCR cycle, the reaction is cooled to about 55 to 65 °C. What happens at this step?
1. APrimers bind to their target sequences
2. BRNA is copied into complementary DNA
3. CThe two DNA strands separate
4. DPolymerase extends the new strands
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Answer: A · Primers bind to their target sequences
A PCR cycle has three steps. Heating to about 94 to 95 °C separates (denatures) the double-stranded DNA. Cooling to the annealing temperature, usually 55 to 65 °C depending on the primers, lets each primer bind its complementary sequence. At about 72 °C the heat-stable polymerase extends the primers, and the cycle repeats, roughly doubling the target each time.
- B: Reverse transcription is a single step before cycling begins, used only for RNA targets.
- C: Strand separation (denaturation) needs about 94 to 95 °C.
- D: Extension takes place at about 72 °C, the best temperature for Taq polymerase.
Question 5 · Medium
What makes the fluorescent signal rise with each cycle in a hydrolysis (TaqMan) probe assay?
1. AA dye binds to all double-stranded DNA as it forms
2. BThe probe folds into a hairpin once it binds target
3. CPolymerase cuts the probe, separating reporter from quencher
4. DMore quencher dye is made with each new copy
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Answer: C · Polymerase cuts the probe, separating reporter from quencher
The probe binds the target between the two primers and carries a reporter dye at one end and a quencher at the other. While the probe is intact, the quencher absorbs the reporter's energy. As Taq polymerase extends the primer, its 5' to 3' exonuclease activity cuts the bound probe and frees the reporter, which then fluoresces. Each new copy releases more reporter, so fluorescence tracks the amount of product.
- A: A dye that binds all double-stranded DNA describes SYBR Green, which needs no probe.
- B: Molecular beacons open out of a hairpin when they bind; TaqMan probes are cut, not folded.
- D: The quencher is part of the probe and is not made during the reaction.
Question 6 · Easy
What makes each new strand stop at a known base in Sanger sequencing?
1. ARestriction enzymes that cut at set sequences
2. BLabeled dideoxynucleotides that end extension
3. CProbes that give a signal only when matched
4. DAn enzyme that makes light as each base is added
Answer and explanation in LearnFree account
Question 7 · Medium
A transplant patient's plasma CMV PCR gives a Ct of 31 one week and a Ct of 22 the next, in the same assay with near 100% efficiency. Compared with the first sample, the second contained:
1. AAbout 9 times more target
2. BAbout 9 times less target
3. CAbout 500 times less target
4. DAbout 500 times more target
Answer and explanation in LearnFree account
Question 8 · Hard
A kidney transplant recipient's plasma for CMV DNA PCR was collected in a lithium heparin tube. Results: CMV target, no amplification; internal control, no amplification (expected Ct about 30); positive control, Ct 27; negative control, no amplification. What is the correct action?
1. ARepeat the run with a new positive control
2. BReport CMV DNA as below the quantitation limit
3. CReport CMV DNA as not detected
4. DReport invalid and request an EDTA sample
Answer and explanation in LearnFree account

Molecular methods

For a week, a real-time PCR assay's no-template control has been positive at Ct 35 to 37, with the same melt peak as the target. New reagent lots did not help. What is the most likely cause, and how is it best prevented?

Sanger sequencing of the HFE gene from a patient's blood DNA shows two overlapping peaks of about equal height, G and A, at c.845, with clean single peaks elsewhere. The patient is most likely:

During each PCR cycle, the reaction is cooled to about 55 to 65 °C. What happens at this step?

What makes the fluorescent signal rise with each cycle in a hydrolysis (TaqMan) probe assay?

What makes each new strand stop at a known base in Sanger sequencing?

A transplant patient's plasma CMV PCR gives a Ct of 31 one week and a Ct of 22 the next, in the same assay with near 100% efficiency. Compared with the first sample, the second contained:

A kidney transplant recipient's plasma for CMV DNA PCR was collected in a lithium heparin tube. Results: CMV target, no amplification; internal control, no amplification (expected Ct about 30); positive control, Ct 27; negative control, no amplification. What is the correct action?