Genetics Quiz: Pcr Polymerase Chain Reaction

What this quiz covers

This quiz focuses on Pcr Polymerase Chain Reaction, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. A long, non-specific smear resulting from random priming events.
2. A product that is approximately twice the length of the expected amplicon due to concatemer formation.
3. Complete failure of the reaction, resulting in no amplification product of any kind on the gel.
4. A short, distinct band corresponding to a primer-dimer, which may consume reagents and inhibit target amplification. (correct answer)

Explanation: When analyzing PCR primer design problems, focus on how primer sequences can interact with each other rather than just with the target DNA. The key issue here is primer complementarity at the 3' ends. When the 3' ends of your forward and reverse primers are complementary (6 nucleotides is significant overlap), they can anneal to each other during the PCR reaction. Since DNA polymerase extends from the 3' end, it will synthesize DNA using one primer as a template for the other. This creates a short, double-stranded product called a primer-dimer. Answer D correctly identifies this artifact - primer-dimers appear as distinct, short bands on gels and are problematic because they consume valuable reagents (primers, dNTPs, and polymerase) that should be used for amplifying your target sequence. Answer A is incorrect because primer-dimers create specific, discrete bands, not random smears. Smears typically result from template degradation or non-specific priming across the genome. Answer B misunderstands the mechanism - concatemers (linked repeated sequences) would form from end-to-end joining of amplicons, not from primer complementarity. The complementary primers don't create products twice the expected length. Answer C is wrong because primer-dimers don't prevent all amplification; they compete with target amplification but don't cause complete reaction failure. For PCR troubleshooting questions, remember that primer-dimer formation is one of the most common artifacts. Always check for 3' end complementarity between primers when designing PCR reactions, as even short complementary sequences can create this problem.

Question 2

1. 10 cycles
2. 20 cycles (correct answer)
3. 25 cycles
4. 30 cycles

Explanation: The number of double-stranded DNA molecules generated by PCR is given by the formula 2ⁿ, where n is the number of cycles. The question asks for the minimum number of cycles to exceed 1,000,000. We need to solve for n in 2ⁿ > 1,000,000. We can test the options: 2¹⁰ = 1,024. 2²⁰ = (2¹⁰)² = 1,024² = 1,048,576. Since 1,048,576 is greater than 1,000,000, 20 cycles is the minimum number required. 19 cycles would yield 2¹⁹ = 524,288, which is not enough.

Question 3

1. Homozygous for Allele A (AA) (correct answer)
2. Homozygous for Allele G (GG)
3. Heterozygous (AG)
4. The genotype is inconclusive without a sequencing result.

Explanation: Allele-specific PCR relies on primers whose 3' ends match one specific allele. Amplification only occurs if the primer perfectly matches the template. In this case, the primer specific for Allele 'A' produced a product, indicating the presence of Allele 'A'. The primer specific for Allele 'G' did not produce a product, indicating the absence of Allele 'G'. Since the individual has Allele 'A' but not Allele 'G', their genotype must be homozygous 'AA'. A heterozygote (AG) would produce a band in both reactions, while a homozygous 'GG' individual would produce a band only in Reaction 2.

Question 4

1. Real-time PCR (qPCR) for precise quantification.
2. Reverse-transcriptase PCR (RT-PCR) to convert RNA to DNA first.
3. Nested PCR, using two sequential amplification reactions with different primer sets. (correct answer)
4. Multiplex PCR to amplify several targets in one reaction.

Explanation: Nested PCR is the ideal technique for this purpose. It involves a first round of PCR using an 'outer' set of primers. The product of this reaction, which is enriched for the target region, is then used as the template for a second round of PCR using an 'inner' set of primers (nested within the first amplicon). This two-step process provides a massive boost in sensitivity. Specificity is also greatly enhanced because the second primer set will only amplify a product if the correct region was amplified in the first round, effectively filtering out any non-specific products from the initial PCR.

Question 5

1. Forward (Tm=60°C): 5'-GATCGCATCGATGCATGCAT-3'; Reverse (Tm=59°C): 5'-AGCTAGCTAGCTACGTAGCTA-3'
2. Forward (Tm=55°C) with a GC-clamp; Reverse (Tm=56°C) with a GC-clamp.
3. Forward (Tm=62°C): 5'-AGTCGACCTGAAAAAAAAAA-3'; Reverse (Tm=61°C): 5'-TCGATCGGATTTTTTTTTTT-3'
4. Forward (Tm=40°C): 5'-ATATATATATATATATATAT-3'; Reverse (Tm=78°C): 5'-GCGCGCGCGCGCGCGCGCGC-3' (correct answer)

Explanation: When designing PCR primers, you need to understand that successful amplification requires primers with compatible melting temperatures (Tm) and appropriate nucleotide composition. The key principle is that both primers should have similar Tm values (ideally within 1-5°C) and avoid extreme sequences that create poor binding conditions. Option D represents a fundamental design failure because the forward and reverse primers have drastically different melting temperatures—40°C versus 78°C. This 38°C difference means you cannot establish optimal annealing conditions during PCR cycling. Additionally, the forward primer consists entirely of AT repeats (very weak hydrogen bonding), while the reverse primer is pure GC repeats (very strong bonding). This creates impossible thermal cycling conditions where one primer won't bind effectively at any chosen annealing temperature. Option A shows well-designed primers with nearly identical Tm values (60°C vs 59°C) and reasonable sequence diversity. Option B describes primers with GC-clamps, which are actually beneficial design features that enhance specificity and stability. Option C presents primers with similar Tm values (62°C vs 61°C); while the poly-A and poly-T tails aren't ideal, the 1°C difference still allows for workable PCR conditions. The critical study tip for PCR primer design questions: always check for compatible melting temperatures first. Primers with Tm differences greater than 5°C signal fundamental design problems. Also watch for extreme sequence compositions like poly-AT or poly-GC repeats, which create binding difficulties and should be avoided in primer design.

Question 6

1. 10 cycles
2. 20 cycles (correct answer)
3. 25 cycles
4. 30 cycles

Explanation: The number of double-stranded DNA molecules generated by PCR is given by the formula 2ⁿ, where n is the number of cycles. The question asks for the minimum number of cycles to exceed 1,000,000. We need to solve for n in 2ⁿ > 1,000,000. We can test the options: 2¹⁰ = 1,024. 2²⁰ = (2¹⁰)² = 1,024² = 1,048,576. Since 1,048,576 is greater than 1,000,000, 20 cycles is the minimum number required. 19 cycles would yield 2¹⁹ = 524,288, which is not enough.

Question 7

1. A gene-specific primer that binds to an exon of Gene X mRNA.
2. Oligo(dT) primers that anneal to the poly(A) tail of mature mRNAs.
3. Random hexamer primers that anneal at various positions along all RNA molecules.
4. A gene-specific primer designed to anneal to a sequence within an intron of Gene X. (correct answer)

Explanation: Mature mRNA has undergone splicing, a process where introns are removed and exons are joined together. Therefore, a primer designed to bind to an intronic sequence will not find a complementary sequence in the mature mRNA template. Consequently, no cDNA corresponding to the Gene X mRNA will be synthesized. A, B, and C are all valid strategies for reverse transcription. Gene-specific primers (A) are highly specific. Oligo(dT) primers (B) target almost all mature mRNAs. Random hexamers (C) will prime synthesis from all types of RNA, including mRNA.

Question 8

1. The annealing temperature remains constant at a very high, stringent level throughout all cycles.
2. The annealing temperature starts low and is gradually increased by 0.5–1°C in each of the initial 10–15 cycles.
3. The annealing temperature starts high, often above the primer Tm, and is decreased by 0.5–1°C in each of the initial 10–15 cycles. (correct answer)
4. The annealing temperature alternates between a high and a low setpoint in every other cycle to 'shock' the reaction.

Explanation: Touchdown PCR is a strategy to increase specificity. The protocol begins with an annealing temperature that is several degrees higher than the calculated Tm of the primers. This high stringency ensures that in the critical early cycles, the primers bind only to their perfect-match target sequence. In each subsequent cycle (for the first 10-15 cycles), the annealing temperature is incrementally lowered. This 'touches down' towards the primers' optimal annealing temperature, eventually allowing for more efficient amplification once the specific target has been enriched.

Question 9

1. High processivity combined with 3' to 5' exonuclease (proofreading) activity. (correct answer)
2. Inherent strand displacement activity to unwind the DNA ahead of synthesis.
3. A complete lack of 3' to 5' exonuclease activity to maximize synthesis speed.
4. An optimal enzymatic activity at a lower temperature (e.g., 60-65°C) to maintain template integrity.

Explanation: Amplifying long fragments requires a polymerase that is both highly processive (can synthesize long stretches of DNA without dissociating from the template) and has high fidelity. High fidelity is achieved through a 3' to 5' exonuclease (proofreading) activity, which allows the enzyme to remove misincorporated nucleotides. Without proofreading, errors would accumulate and cause the polymerase to terminate synthesis, preventing the amplification of long targets. Standard Taq polymerase lacks proofreading and is unsuitable for this task. B: Strand displacement is a feature for other techniques like LAMP, not Long-Range PCR. D: Thermostable polymerases for PCR typically have optimal activity around 72°C.

Question 10

1. The reaction will proceed, but with very low specificity, producing numerous non-target bands.
2. The DNA polymerase will be inactive or have severely crippled function, resulting in little to no amplification. (correct answer)
3. The DNA template strands will fail to denature properly at 95°C, preventing primer access.
4. The primers will degrade rapidly at the high temperatures of the PCR cycle without the stabilizing effect of ions.

Explanation: Thermostable DNA polymerases like Taq polymerase require divalent cations, specifically Mg²⁺, as an essential cofactor. The Mg²⁺ ions are crucial for the catalytic activity of the enzyme, facilitating the correct positioning of the dNTPs for incorporation. Without Mg²⁺, the polymerase is essentially inactive, and no DNA synthesis will occur. A: While Mg²⁺ concentration affects specificity, its complete absence causes enzyme failure, not just non-specificity. C: Denaturation is primarily a function of temperature and is not dependent on Mg²⁺. D: Primers are stable at PCR temperatures; Mg²⁺ is not required for their stability.

Question 11

1. The reaction will be highly efficient, yielding a strong 500 bp band, as the temperature is between the two Tms.
2. No amplification will occur because the annealing temperature is too high for the forward primer to bind effectively.
3. A faint band or no band of the target size will be observed due to significantly reduced binding efficiency of the forward primer. (correct answer)
4. Primarily non-specific bands will be amplified due to the large difference in primer Tms promoting mis-priming.

Explanation: The optimal annealing temperature is typically set 3-5°C below the lower of the two primer Tms. In this case, the annealing temperature of 62°C is 4°C above the Tm of the forward primer (58°C). At a temperature above its Tm, the primer will have very low binding efficiency, leading to poor or no amplification. The reverse primer (Tm=64°C) would bind, but since both primers are required for exponential amplification, the overall reaction will be severely inhibited. Choice B is too absolute ('no amplification'), while C ('faint or no band') more accurately reflects the likely outcome of very low efficiency. Choice D is incorrect because high annealing temperatures generally increase specificity, not decrease it.

Question 12

1. 2.6 µg/mL
2. 26 µg/mL
3. 10.4 µg/mL
4. 104 µg/mL (correct answer)

Explanation: This is a multi-step calculation. First, calculate the moles of amplicon: (6.02 x 10¹² molecules) / (6.02 x 10²³ molecules/mol) = 1.0 x 10⁻¹¹ mol. Second, calculate the molecular weight (MW) of the amplicon: 400 bp * 650 g/mol/bp = 260,000 g/mol = 2.6 x 10⁵ g/mol. Third, calculate the mass of the amplicon: Mass = Moles * MW = (1.0 x 10⁻¹¹ mol) * (2.6 x 10⁵ g/mol) = 2.6 x 10⁻⁶ g, which is 2.6 µg. Finally, calculate the concentration: Concentration = Mass / Volume. The volume is 25 µL, which is 0.025 mL. So, Concentration = 2.6 µg / 0.025 mL = 104 µg/mL.

Question 13

1. Decreasing the annealing temperature to 50°C to ensure primer binding.
2. Increasing the concentration of MgCl₂ significantly to stabilize the polymerase.
3. Adding a co-solvent such as DMSO or betaine to the PCR master mix. (correct answer)
4. Increasing the extension time from 1 minute to 5 minutes to help the polymerase move through the region.

Explanation: GC-rich DNA sequences are difficult to amplify because the strong G-C triple hydrogen bonds can lead to incomplete denaturation and stable secondary structures (like hairpins) that block polymerase progression. Additives like dimethyl sulfoxide (DMSO), betaine, or formamide act as co-solvents that help to disrupt these hydrogen bonds and secondary structures, thereby facilitating both denaturation and primer annealing. A: Lowering the annealing temperature would likely cause non-specific amplification. B: High Mg²⁺ can actually further stabilize GC secondary structures, worsening the problem. D: A longer extension time will not help if the polymerase is physically blocked by secondary structures.

Question 14

1. Expression in treated cells is approximately 8 times higher than in untreated cells.
2. Expression in treated cells is approximately 3 times lower than in untreated cells.
3. Expression in treated cells is approximately 8 times lower than in untreated cells. (correct answer)
4. Expression in treated cells is approximately 3 times higher than in untreated cells.

Explanation: A higher Cq value indicates a lower initial amount of template, as it takes more cycles to reach the fluorescence threshold. The difference in Cq (ΔCq) is 25 (treated) - 22 (untreated) = 3. With 100% efficiency, the amount of DNA doubles each cycle. Therefore, a difference of 3 cycles corresponds to a 2³ = 8-fold difference in initial template concentration. Since the treated cells have the higher Cq value, their initial expression level is 8-fold lower than that of the untreated cells.

Question 15

1. The DNA extract does not contain the target sequence, and the control DNA was likely degraded.
2. The PCR failed due to a fundamental problem with the master mix or thermocycler programming.
3. The DNA extract contains a substance that is inhibiting the DNA polymerase or another aspect of the PCR. (correct answer)
4. The primers used for the control DNA template are not compatible with the PCR buffer system.

Explanation: This experimental design uses an internal positive control (IPC) to check for inhibition. The control DNA is known to amplify under these conditions. Since the control DNA failed to amplify only when mixed with the DNA extract, it strongly implies that something in the extract is preventing the PCR from working. This substance is a PCR inhibitor (e.g., heme from blood, humic acid from soil, or ethanol from purification). B is less likely because an external positive control (control DNA alone) would be needed to confirm a master mix problem. The failure of the internal control is direct evidence of inhibition by the sample.

Question 16

1. The reaction will be highly efficient, yielding a strong 500 bp band, as the temperature is between the two Tms.
2. No amplification will occur because the annealing temperature is too high for the forward primer to bind effectively.
3. A faint band or no band of the target size will be observed due to significantly reduced binding efficiency of the forward primer. (correct answer)
4. Primarily non-specific bands will be amplified due to the large difference in primer Tms promoting mis-priming.

Explanation: The optimal annealing temperature is typically set 3-5°C below the lower of the two primer Tms. In this case, the annealing temperature of 62°C is 4°C above the Tm of the forward primer (58°C). At a temperature above its Tm, the primer will have very low binding efficiency, leading to poor or no amplification. The reverse primer (Tm=64°C) would bind, but since both primers are required for exponential amplification, the overall reaction will be severely inhibited. Choice B is too absolute ('no amplification'), while C ('faint or no band') more accurately reflects the likely outcome of very low efficiency. Choice D is incorrect because high annealing temperatures generally increase specificity, not decrease it.

Question 17

1. The reaction will proceed, but with very low specificity, producing numerous non-target bands.
2. The DNA polymerase will be inactive or have severely crippled function, resulting in little to no amplification. (correct answer)
3. The DNA template strands will fail to denature properly at 95°C, preventing primer access.
4. The primers will degrade rapidly at the high temperatures of the PCR cycle without the stabilizing effect of ions.

Explanation: Thermostable DNA polymerases like Taq polymerase require divalent cations, specifically Mg²⁺, as an essential cofactor. The Mg²⁺ ions are crucial for the catalytic activity of the enzyme, facilitating the correct positioning of the dNTPs for incorporation. Without Mg²⁺, the polymerase is essentially inactive, and no DNA synthesis will occur. A: While Mg²⁺ concentration affects specificity, its complete absence causes enzyme failure, not just non-specificity. C: Denaturation is primarily a function of temperature and is not dependent on Mg²⁺. D: Primers are stable at PCR temperatures; Mg²⁺ is not required for their stability.

Question 18

1. Expression in treated cells is approximately 8 times higher than in untreated cells.
2. Expression in treated cells is approximately 3 times lower than in untreated cells.
3. Expression in treated cells is approximately 8 times lower than in untreated cells. (correct answer)
4. Expression in treated cells is approximately 3 times higher than in untreated cells.

Explanation: A higher Cq value indicates a lower initial amount of template, as it takes more cycles to reach the fluorescence threshold. The difference in Cq (ΔCq) is 25 (treated) - 22 (untreated) = 3. With 100% efficiency, the amount of DNA doubles each cycle. Therefore, a difference of 3 cycles corresponds to a 2³ = 8-fold difference in initial template concentration. Since the treated cells have the higher Cq value, their initial expression level is 8-fold lower than that of the untreated cells.

Question 19

1. 2.6 µg/mL
2. 26 µg/mL
3. 10.4 µg/mL
4. 104 µg/mL (correct answer)

Explanation: This is a multi-step calculation. First, calculate the moles of amplicon: (6.02 x 10¹² molecules) / (6.02 x 10²³ molecules/mol) = 1.0 x 10⁻¹¹ mol. Second, calculate the molecular weight (MW) of the amplicon: 400 bp * 650 g/mol/bp = 260,000 g/mol = 2.6 x 10⁵ g/mol. Third, calculate the mass of the amplicon: Mass = Moles * MW = (1.0 x 10⁻¹¹ mol) * (2.6 x 10⁵ g/mol) = 2.6 x 10⁻⁶ g, which is 2.6 µg. Finally, calculate the concentration: Concentration = Mass / Volume. The volume is 25 µL, which is 0.025 mL. So, Concentration = 2.6 µg / 0.025 mL = 104 µg/mL.

Question 20

1. 25 (correct answer)
2. 32
3. 29
4. 10

Explanation: The number of copies (N) after c cycles is given by N_c = N₀ × (1 + E)ᶜ, where N₀ is the initial copy number and E is the efficiency (expressed as a decimal from 0 to 1). Here, N₀ = 1, E = 0.90, and c = 5. The amplification factor per cycle is (1 + E) = 1.9. Therefore, the number of copies after 5 cycles is 1 × (1.9)⁵. Calculation: 1.9 × 1.9 ≈ 3.6; 3.6 × 1.9 ≈ 6.8; 6.8 × 1.9 ≈ 13; 13 × 1.9 ≈ 24.7. The result is approximately 25 copies. The Cq value is extraneous information designed to mislead.