Genetics Quiz: Dna Sequencing

What this quiz covers

This quiz focuses on Dna Sequencing, 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. Covalently attaching fluorescent dyes to the 3' end of each DNA fragment.
2. Ligating specific adapter sequences to both ends of the fragmented DNA. (correct answer)
3. Denaturing the double-stranded DNA fragments into single strands using a high-temperature step.
4. Using DNA polymerase and dNTPs to create blunt ends on all fragmented DNA.

Explanation: The adapters ligated to the ends of DNA fragments contain specific sequences (e.g., P5 and P7) that are complementary to the lawn of oligonucleotides covalently bound to the flow cell surface. This complementarity is required for the library fragments to hybridize (or 'bind') to the flow cell, which is the first step of cluster generation by bridge PCR.

Question 2

1. Illumina NGS is most prone to substitution errors, while Sanger sequencing has higher error rates when calling the length of homopolymer regions. (correct answer)
2. Sanger sequencing is highly prone to substitution errors, while Illumina NGS is more prone to insertion/deletion errors in homopolymer regions.
3. Both methods have uniformly low error rates across the entire read length, with NGS simply providing higher throughput.
4. Sanger sequencing errors are primarily due to adapter dimer formation, while Illumina NGS errors result from premature chain termination.

Explanation: When evaluating DNA sequencing technologies, you need to understand that different sequencing methods have distinct error profiles based on their underlying chemistry and detection mechanisms. Sanger sequencing uses chain-terminating ddNTPs and capillary electrophoresis, making it highly accurate for individual base calls with error rates around 0.001%. Its main weakness occurs in homopolymer regions (stretches of identical bases like AAAAA) where the electrophoretic separation becomes difficult to resolve, leading to uncertainties in determining the exact length of these runs. Illumina NGS uses reversible terminators and bridge amplification on a flow cell. While highly efficient for throughput, it's more prone to substitution errors due to several factors: incomplete removal of fluorescent labels, carry-forward of signal from previous cycles, and occasional incorporation of incorrect bases. The error rate increases toward the end of reads as these effects accumulate. Looking at the options: Answer A correctly identifies this key difference - Illumina's primary issue is substitutions, while Sanger struggles with homopolymer length determination. Answer B reverses these error types incorrectly. Answer C ignores the distinct error profiles entirely, falsely claiming uniform low error rates. Answer D mentions completely irrelevant error sources - adapter dimers are library preparation issues, not sequencing errors, and premature chain termination isn't a major Illumina problem. Study tip: Remember the acronym "SIL" - Sanger struggles with Identical bases (homopolymers), Illumina makes Letter substitutions. This captures the core error profile difference you'll need to recognize on genetics exams.

Question 3

1. dATP, dTTP, dCTP, and ddGTP only.
2. dATP, dTTP, dCTP, dGTP, and ddGTP. (correct answer)
3. ddATP, ddTTP, ddCTP, and ddGTP only.
4. dGTP and ddGTP only, along with the other required reaction components.

Explanation: For a Sanger sequencing reaction to generate a ladder of products, DNA polymerase must be able to extend the primer. This requires all four standard deoxynucleotides (dATP, dCTP, dGTP, dTTP) to be available as building blocks. The chain termination that creates the sequence-specific fragments is caused by the occasional, stochastic incorporation of a single type of dideoxynucleotide (in this case, ddGTP). Therefore, the 'ddG' tube must contain a large amount of all four dNTPs plus a small, limiting amount of ddGTP.

Question 4

1. The sequencing primer had a low melting temperature and dissociated from the template after 600 bases.
2. An insufficient concentration of all four ddNTPs was added, preventing termination in longer fragments.
3. The physical limits of electrophoretic separation and decreasing enzyme processivity for long fragments. (correct answer)
4. A strong hairpin loop formed in the template DNA, causing a hard stop for the DNA polymerase at the 600 bp mark.

Explanation: Sanger sequencing via capillary electrophoresis has an inherent read length limit. As DNA fragments become longer, the resolution of the electrophoresis decreases, meaning the ability to separate fragments that differ by only one base (e.g., 601 vs 602 bp) diminishes. This causes the fluorescent peaks to broaden and overlap. Additionally, factors like DNA polymerase falling off the template (processivity) contribute to a weaker signal for longer products, leading to a practical read length limit.

Question 5

1. To correct the random sequencing errors that are present in the raw FASTQ read data.
2. To quantify the number of reads mapping to each gene to infer its expression level.
3. To assemble the short reads into a single, contiguous sequence representing each chromosome.
4. To systematically identify genomic positions where the sequenced sample differs from the reference sequence. (correct answer)

Explanation: When you encounter questions about bioinformatics workflows, focus on understanding what each step accomplishes in the pipeline from raw sequencing data to biological insights. Variant calling is fundamentally about comparison—it systematically examines each position in your aligned reads against the reference genome to identify differences. These differences (variants) include single nucleotide polymorphisms (SNPs), insertions, deletions, and other genomic variations that make each individual's genome unique. The algorithm analyzes read depth, base quality scores, and statistical confidence to determine where your sample genuinely differs from the reference sequence. Option A confuses variant calling with quality control steps. While sequencing errors exist, variant calling doesn't correct them—it distinguishes real genetic variants from technical artifacts using statistical methods and quality filters. Option B describes RNA-seq analysis, not DNA variant calling. Gene expression quantification requires counting reads mapping to transcripts, which isn't the goal in whole-exome sequencing focused on identifying genetic variants. Option C describes genome assembly, a completely different process. Assembly reconstructs genomes from scratch using overlapping reads, while variant calling works with reads already aligned to an existing reference genome. Option D correctly captures variant calling's core purpose: identifying positions where your sample's DNA sequence differs from the reference genome. Study tip: Remember the bioinformatics pipeline order: raw reads → quality control → alignment → variant calling → annotation. Each step has a distinct purpose, and variant calling always comes after alignment because you need a reference point for comparison.

Question 6

1. Sequencing will terminate randomly at all G positions, producing a full range of fragment lengths representative of the template.
2. The resulting fragments will be heavily skewed towards shorter lengths, terminating at the first G encountered in the synthesized strand. (correct answer)
3. The reaction will produce predominantly full-length strands with no termination, as DNA polymerase preferentially incorporates dNTPs.
4. Fragments of all possible lengths will be generated, but only those ending in G will be fluorescently labeled and detected.

Explanation: In Sanger sequencing, dNTPs and ddNTPs compete for incorporation by DNA polymerase. A 1:1 ratio of ddGTP to dGTP means there is a 50% chance of termination each time a guanine is to be incorporated. This high probability will cause most chains to terminate at the very first 'G' position encountered after the primer, leading to a strong bias towards very short fragments. This corresponds to the first cytosine on the template strand. A normal reaction uses a much lower ratio of ddNTP:dNTP to generate a wider range of fragment lengths.

Question 7

1. The 3' carbon, which prevents DNA polymerase from adding the next nucleotide to the growing chain. (correct answer)
2. The 2' carbon, which prevents the formation of the phosphodiester bond with the subsequent nucleotide.
3. The 5' carbon, which prevents the attachment of the triphosphate group necessary for incorporation.
4. The 1' carbon, which destabilizes the N-glycosidic bond attaching the nitrogenous base.

Explanation: When you encounter questions about Sanger sequencing and chain termination, focus on the mechanism of DNA synthesis and what's required for chain elongation to continue. DNA polymerase synthesizes new DNA by forming phosphodiester bonds between nucleotides. This reaction specifically occurs between the 3'-OH group of the growing chain and the 5'-phosphate of the incoming nucleotide. For chain elongation to continue, the newly added nucleotide must have a free 3'-OH group available for the next nucleotide to attach. Dideoxynucleotides (ddNTPs) function as chain terminators because they lack the 3'-OH group on their deoxyribose sugar. When DNA polymerase incorporates a ddNTP into the growing chain, there's no 3'-OH available for the next nucleotide to bond with, effectively stopping synthesis at that point. This is exactly what answer choice A describes. Answer B incorrectly identifies the 2' carbon - normal DNA already lacks a hydroxyl group at the 2' position (that's what makes it "deoxy"), so this wouldn't create termination. Answer C is wrong because ddNTPs do have the 5'-triphosphate group necessary for incorporation - they're successfully added to the chain. Answer D misunderstands the termination mechanism entirely; the N-glycosidic bond connecting the base to the sugar remains stable in ddNTPs. Remember this key principle: DNA synthesis always requires a free 3'-OH group to continue. Any modification that eliminates this group will terminate chain elongation, which is exactly how Sanger sequencing generates the different-length fragments needed for sequence determination.

Question 8

1. It doubles the total throughput (number of bases) for the same cost, providing higher overall coverage.
2. It simplifies the library preparation workflow by eliminating the need for DNA fragmentation.
3. It inherently reduces the per-base error rate by sequencing each base twice, once from each direction.
4. It provides spatial information about the distance and orientation of two reads from the same DNA fragment. (correct answer)

Explanation: When tackling questions about NGS sequencing strategies, focus on what unique information each approach provides rather than just technical specifications. Paired-end sequencing's key advantage lies in generating spatial information about DNA fragments. In this method, both ends of the same DNA fragment are sequenced, creating two reads with a known distance between them (the insert size) and defined orientation. This spatial relationship is crucial for genome assembly because it helps you span repetitive regions, resolve ambiguous assemblies, and detect structural variants like insertions, deletions, or rearrangements that would be invisible with single reads alone. Choice A is incorrect because while paired-end does generate more total bases, the primary advantage isn't simply doubled throughput—it's the structured information those additional bases provide. Choice B misrepresents the workflow; paired-end sequencing still requires DNA fragmentation, just like single-read methods. The fragments are simply sequenced from both ends rather than one. Choice C contains a fundamental error—paired-end doesn't sequence the same base twice. Instead, it sequences different bases at opposite ends of the same fragment, so there's no inherent error reduction from redundant base calling. The spatial information from paired-end reads acts like molecular rulers, telling you not just what sequences are present, but how far apart they are and in what orientation. This makes it invaluable for detecting structural variants and improving assembly quality. Study tip: Remember that "paired-end" means paired in space, not paired for redundancy. The power comes from knowing the relationship between the two reads, not from sequencing the same region twice.

Question 9

1. The variant is a heterozygous somatic mutation present in approximately 50% of the tumor cells. (correct answer)
2. The variant is a germline homozygous SNP.
3. The variant is likely a sequencing error because its frequency (25%) is low.
4. The variant is a heterozygous somatic mutation present in all tumor cells.

Explanation: When analyzing sequencing data from tumor samples, you need to interpret variant allele frequencies in the context of tumor biology. The key insight is that tumors are clonal populations that can be heterogeneous, meaning not all cells necessarily carry the same mutations. In this case, you have 25% of reads showing variant 'A' and 75% showing reference 'G'. If this were a simple heterozygous variant present in all diploid cells, you'd expect roughly 50% of each allele. The 25% frequency suggests that only about half the tumor cells carry this heterozygous mutation (50% of cells × 50% of alleles = 25% variant frequency). Looking at the answer choices: Option A correctly identifies this as a heterozygous somatic mutation present in approximately 50% of tumor cells, which explains the 25% allele frequency. Option B is wrong because a germline homozygous SNP would show nearly 100% variant frequency, not 25%. Option C incorrectly dismisses the variant as sequencing error—25% is far too high to be technical noise, which typically occurs at <5% frequency. Option D suggests the mutation is in all tumor cells, but this would produce a 50% allele frequency for a heterozygous variant, not 25%. Remember that in cancer genomics, variant allele frequency reflects both the mutation's zygosity and the fraction of cells carrying it. Always consider tumor heterogeneity when interpreting sequencing data—not every mutation is present in every cancer cell.

Question 10

1. The presence of a heterozygous single-nucleotide polymorphism (SNP) at the position of the sharp peak.
2. A heterozygous insertion or deletion mutation just upstream of the observed region of overlapping peaks. (correct answer)
3. Contamination of the sequencing reaction with a second, unrelated DNA template of similar concentration.
4. A strong secondary structure in the DNA template that caused the polymerase to dissociate prematurely.

Explanation: A heterozygous insertion or deletion (indel) creates two populations of template molecules that differ in length. From the point of the indel onwards, the sequencing reactions for the two alleles are out of frame with each other. This results in the superimposition of two different sequences on the chromatogram, creating a jumble of overlapping peaks that are unreadable. A heterozygous SNP would show two clean, overlapping peaks at a single position, followed by a return to a clean sequence.

Question 11

1. Illumina sequencing (sequencing-by-synthesis)
2. PacBio SMRT sequencing (single-molecule real-time)
3. Ion Torrent sequencing (semiconductor sequencing) (correct answer)
4. Sanger sequencing (dideoxy chain termination)

Explanation: The defining principle of Ion Torrent technology is the detection of proton release upon nucleotide incorporation. The formation of a phosphodiester bond releases a hydrogen ion, which causes a localized change in pH. This change is detected by an ion-sensitive field-effect transistor (ISFET) sensor integrated into the bottom of each microwell. Other technologies use optical (fluorescence) detection.

Question 12

1. Unique molecular identifiers (UMIs) attached to each cDNA molecule prior to PCR amplification.
2. Biotin labels attached to the adapters to allow for purification of the pooled library.
3. Unique DNA barcode (index) sequences within the adapters ligated to each sample's library. (correct answer)
4. Random hexamer primers used during the initial reverse transcription step for each sample.

Explanation: This process is called multiplexing. To distinguish between reads from different samples that have been pooled together, each sample's library is prepared using adapters that contain a unique DNA sequence, known as a barcode or index. After the sequencing run is complete, the reads are sorted based on their barcode sequence in a computational step called demultiplexing, assigning them to their sample of origin.

Question 13

1. Next-generation sequencing of the whole genome to ensure all potential causative variants are identified comprehensively.
2. Sanger sequencing of the 5 kb candidate gene after PCR amplification from each family member. (correct answer)
3. Next-generation sequencing of the exome to identify all potential coding variants across the family's genomes.
4. Sanger sequencing using random primers to sequence the entire genome of one affected individual.

Explanation: For a targeted analysis of a small, specific region like a single 5 kb gene, Sanger sequencing is the gold standard. It is highly accurate, relatively inexpensive for a small number of targets, and provides straightforward data analysis. Whole-genome or exome NGS would be unnecessarily expensive, generate vast amounts of off-target data, and be computationally more complex for this focused question.

Question 14

1. Generating highly accurate 150 bp paired-end reads, as high accuracy is crucial for distinguishing between repeat copies.
2. Generating extremely high coverage (e.g., 500X) of 150 bp single-end reads to statistically resolve the repeats.
3. Using Sanger sequencing to 'walk' through the repetitive regions, as it has the lowest error rate.
4. Generating 15 kb reads with a higher random error rate, as the read length can span entire repetitive elements. (correct answer)

Explanation: When you encounter questions about genome assembly, especially with repetitive sequences, focus on how different sequencing technologies handle structural complexity rather than just accuracy or coverage depth. Long, nearly identical repetitive sequences create a fundamental assembly challenge called the "repeat problem." Short reads, even highly accurate ones, cannot span these repetitive regions, making it impossible for assembly algorithms to determine how many copies exist or how they're arranged. This is where read length becomes the critical factor. Option D is correct because 15 kb reads can span entire repetitive elements that might be several kilobases long. Even with higher random error rates, these long reads provide the structural context needed to resolve repeat organization. Assembly algorithms can use overlapping long reads to bridge repetitive regions and determine the correct genome structure. Option A fails because 150 bp paired-end reads, despite high accuracy, are too short to span long repeats. The assembly software cannot determine where each short read originated within the repetitive regions. Option B compounds this same fundamental problem - no amount of coverage can compensate for reads that are inherently too short to provide structural information across repeats. Option C misses the point entirely; while Sanger sequencing is accurate, the "walking" approach is extremely labor-intensive and impractical for whole genome assembly of complex repeat structures. Remember: for repetitive genome regions, read length trumps accuracy. Long reads provide the structural scaffolding that high-accuracy short reads simply cannot deliver, regardless of coverage depth.

Question 15

1. Sanger sequencing is unable to sequence DNA from tumor samples due to their inherent instability.
2. The base-calling algorithm in NGS is inherently more sensitive to minor signals than Sanger sequencing's algorithm.
3. Sanger sequencing is a bulk, analog method with a practical detection limit of around 15-20% for a minor allele. (correct answer)
4. NGS library preparation specifically enriches for mutant alleles, making them easier to detect than wild-type alleles.

Explanation: Sanger sequencing measures a mixed signal from all molecules in the sample. A variant at 1% frequency would generate a fluorescent peak that is only 1% the height of the main peak, making it indistinguishable from background noise. The practical limit of detection is thus around 15-20%. Deep NGS is a digital method; it sequences individual molecules. By generating thousands of reads (high depth), it can count the number of molecules with the variant, allowing for confident detection of alleles at frequencies of 1% or lower.

Question 16

1. (6.7 \times 10^7) (correct answer)
2. (1.3 \times 10^8)
3. (3.3 \times 10^6)
4. (2.0 \times 10^{10})

Explanation: First, calculate the total number of bases required: Total Bases = Genome Size × Depth = (500 × 10^6 bases) × 40 = 20,000 × 10^6 bases = 2 × 10^10 bases. Second, determine the number of bases sequenced per fragment. A 150 bp paired-end read generates 2 × 150 = 300 bases of sequence data per fragment. Finally, divide the total bases needed by the bases per fragment: Number of fragments = (2 × 10^10 bases) / (300 bases/fragment) = 0.0667 × 10^9 fragments = 6.67 × 10^7 fragments.

Question 17

1. It allows the DNA polymerase to reverse direction and perform proofreading after each incorporation event.
2. It allows the unincorporated nucleotides in the flow cell to be chemically reactivated for use in a later cycle.
3. It allows the chemical cleavage of the terminating group and the dye, regenerating a 3'-OH for the next cycle. (correct answer)
4. It allows the entire newly synthesized DNA strand to be stripped off the template, which is then sequenced again.

Explanation: The 'reversible terminator' concept is key to SBS. Each modified dNTP has a blocking group on the 3'-OH position that 'terminates' synthesis, and a fluorescent dye. In each cycle, only one nucleotide can be added. After the fluorescent signal is recorded, a chemical reaction removes both the blocking group and the dye. This cleavage 'reverses' the termination, leaving a standard 3'-OH group, which allows the next nucleotide to be incorporated in the following cycle.

Question 18

1. Next-generation sequencing of the whole genome to ensure all potential causative variants are identified comprehensively.
2. Sanger sequencing of the 5 kb candidate gene after PCR amplification from each family member. (correct answer)
3. Next-generation sequencing of the exome to identify all potential coding variants across the family's genomes.
4. Sanger sequencing using random primers to sequence the entire genome of one affected individual.

Explanation: For a targeted analysis of a small, specific region like a single 5 kb gene, Sanger sequencing is the gold standard. It is highly accurate, relatively inexpensive for a small number of targets, and provides straightforward data analysis. Whole-genome or exome NGS would be unnecessarily expensive, generate vast amounts of off-target data, and be computationally more complex for this focused question.

Question 19

1. Covalently attaching fluorescent dyes to the 3' end of each DNA fragment.
2. Ligating specific adapter sequences to both ends of the fragmented DNA. (correct answer)
3. Denaturing the double-stranded DNA fragments into single strands using a high-temperature step.
4. Using DNA polymerase and dNTPs to create blunt ends on all fragmented DNA.

Explanation: The adapters ligated to the ends of DNA fragments contain specific sequences (e.g., P5 and P7) that are complementary to the lawn of oligonucleotides covalently bound to the flow cell surface. This complementarity is required for the library fragments to hybridize (or 'bind') to the flow cell, which is the first step of cluster generation by bridge PCR.

Question 20

1. Illumina NGS is most prone to substitution errors, while Sanger sequencing has higher error rates when calling the length of homopolymer regions. (correct answer)
2. Sanger sequencing is highly prone to substitution errors, while Illumina NGS is more prone to insertion/deletion errors in homopolymer regions.
3. Both methods have uniformly low error rates across the entire read length, with NGS simply providing higher throughput.
4. Sanger sequencing errors are primarily due to adapter dimer formation, while Illumina NGS errors result from premature chain termination.

Explanation: When evaluating DNA sequencing technologies, you need to understand that different sequencing methods have distinct error profiles based on their underlying chemistry and detection mechanisms. Sanger sequencing uses chain-terminating ddNTPs and capillary electrophoresis, making it highly accurate for individual base calls with error rates around 0.001%. Its main weakness occurs in homopolymer regions (stretches of identical bases like AAAAA) where the electrophoretic separation becomes difficult to resolve, leading to uncertainties in determining the exact length of these runs. Illumina NGS uses reversible terminators and bridge amplification on a flow cell. While highly efficient for throughput, it's more prone to substitution errors due to several factors: incomplete removal of fluorescent labels, carry-forward of signal from previous cycles, and occasional incorporation of incorrect bases. The error rate increases toward the end of reads as these effects accumulate. Looking at the options: Answer A correctly identifies this key difference - Illumina's primary issue is substitutions, while Sanger struggles with homopolymer length determination. Answer B reverses these error types incorrectly. Answer C ignores the distinct error profiles entirely, falsely claiming uniform low error rates. Answer D mentions completely irrelevant error sources - adapter dimers are library preparation issues, not sequencing errors, and premature chain termination isn't a major Illumina problem. Study tip: Remember the acronym "SIL" - Sanger struggles with Identical bases (homopolymers), Illumina makes Letter substitutions. This captures the core error profile difference you'll need to recognize on genetics exams.