AP Biology Quiz: Biotechnology

What this quiz covers

This quiz focuses on Biotechnology, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

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. DNA polymerase randomly incorporates ddNTPs at positions where that base is added, terminating strands at multiple sites. (correct answer)
2. Restriction enzymes cut the newly synthesized DNA at multiple recognition sites, generating fragments of varied length.
3. Ligase joins short oligonucleotides into longer products, producing a distribution of fragment sizes.
4. Helicase unwinds the template at multiple points, causing polymerase to start synthesis at many different origins.
5. Ribosomes translate the DNA into peptides of different lengths that are detected as fluorescent fragments.

Explanation: This question examines biotechnology analysis by explaining fragment generation in Sanger sequencing for DNA sequence determination. The correct answer is that DNA polymerase randomly incorporates ddNTPs at positions where that base is needed, terminating strands at multiple sites and producing fragments ending with labeled terminators. This chain-termination method creates a nested set of fragments differing by one nucleotide, separated by electrophoresis to read the sequence from shortest to longest. Fluorescence detection identifies the terminal base, enabling sequence assembly. A tempting distractor is choice B, which is incorrect because restriction enzymes are not used in Sanger synthesis; it's polymerase-driven, reflecting the misconception that sequencing involves cutting rather than synthesis. For sequencing projects, optimize ddNTP ratios to ensure even fragment distribution, a transferable tip for reliable read lengths.

Question 2

1. Approximately 800 bp, because larger fragments migrate farther through agarose matrices.
2. Approximately 600 bp, because fragments near 500 bp cluster and migrate similarly.
3. Approximately 400 bp, because it migrated between the 500-bp and 300-bp ladder bands. (correct answer)
4. Approximately 200 bp, because small fragments remain closer to the wells than larger fragments.
5. Approximately 1,500 bp, because DNA fragments migrate based on base composition, not length.

Explanation: This question evaluates biotechnology analysis by estimating DNA fragment size from agarose gel electrophoresis migration relative to a ladder. The correct answer is approximately 400 bp, as the unknown band migrated between the 500-bp and 300-bp ladder bands, indicating an intermediate size based on the principle that smaller fragments travel farther in the gel matrix under electric field. In agarose gels, DNA separation occurs inversely with size, with larger fragments impeded more by pores, so position interpolation from known ladder bands provides the estimate. Typical conditions ensure linear migration for fragments in this range, supporting accurate sizing. A tempting distractor is choice A, which is incorrect because larger fragments migrate less far, not farther, reflecting the misconception that size and mobility are directly proportional like in some protein gels. To size unknown bands, plot a standard curve of log(size) versus distance using ladder data, a transferable method for precise gel analysis.

Question 3

1. Cell type 1 contains the target protein at detectable levels, while cell type 2 does not under the tested conditions. (correct answer)
2. Cell type 2 contains the target gene, but genes cannot be detected by Western blotting.
3. Cell type 1 has more mRNA, and SDS-PAGE separates mRNA by length to create the 55 kDa band.
4. Cell type 2 proteins migrated off the gel because smaller proteins move toward the wells during electrophoresis.
5. Cell type 1 antibodies replicated the protein, increasing its mass to 55 kDa during transfer.

Explanation: This question evaluates biotechnology analysis by interpreting Western blot banding patterns to detect protein presence in different cell types. The correct answer is that cell type 1 contains the target protein at detectable levels, producing the 55 kDa band via antibody binding, while cell type 2 does not under the tested conditions. SDS-PAGE separates denatured proteins by mass, with transfer to a membrane allowing specific detection using primary and secondary antibodies, where the band indicates antigen presence. The absence in cell type 2 suggests low or no expression, not detection failure, as controls would confirm blot integrity. A tempting distractor is choice D, which is incorrect because smaller proteins migrate farther from wells, not toward them, reflecting the misconception that electrophoresis direction reverses for proteins versus DNA. To interpret blots, compare band positions to molecular weight markers and include loading controls, a transferable approach for quantitative protein analysis.

Question 4

1. Strain Y lacks the primer-binding sites, so no 300 bp amplicon is produced during PCR. (correct answer)
2. Strain Y DNA migrates more slowly, so its 300 bp band remains in the gel well.
3. The primers ligate to form a 300 bp fragment only when template DNA is absent.
4. The no-template control shows no band because restriction enzymes were not added to cut DNA.
5. Lane 2 shows a band because plasmids replicate during electrophoresis, increasing DNA length.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically PCR and gel electrophoresis in distinguishing bacterial strains. The primers are designed to flank a 300 bp region unique to strain X, allowing them to bind and initiate amplification during PCR, resulting in a visible band at 300 bp in lane 2. In strain Y, the absence of these primer-binding sites prevents amplification, explaining the lack of a band in lane 3. The no-template control in lane 4 shows no band because there is no DNA for the primers to amplify, confirming the specificity of the reaction. A tempting distractor is choice B, which incorrectly suggests that strain Y DNA migrates more slowly due to inherent properties, stemming from the misconception that DNA migration in gels varies by source rather than size and charge. To interpret PCR gel results effectively, always compare band positions to expected amplicon sizes and controls to rule out contamination or nonspecific amplification.

Question 5

1. Activation of the transcription factor increased transcription from the promoter, producing more GFP mRNA and protein. (correct answer)
2. The small molecule increased agarose pore size, allowing GFP to enter cells and fluoresce.
3. The solvent-only treatment prevented translation by removing ribosomes from the cytoplasm.
4. The small molecule caused plasmid DNA to mutate into GFP protein without transcription.
5. The transcription factor bound directly to GFP protein, increasing fluorescence without changing expression.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically reporter gene assays for studying transcriptional regulation. The small molecule activates the transcription factor, enabling it to bind the enhancer and promote transcription from the promoter, leading to increased GFP mRNA and subsequent protein production. This results in more fluorescent cells in Treatment 1, as GFP protein accumulates and emits green light. The solvent-only Treatment 2 lacks activation, so minimal transcription occurs, explaining the low fluorescence. A tempting distractor is choice E, which suggests the transcription factor binds directly to GFP protein, arising from the misconception that transcription factors act post-translationally rather than at the DNA level to regulate expression. In reporter assays, compare treated and control groups to quantify regulatory effects and validate molecular mechanisms.

Question 6

1. The corresponding gene has higher expression in treated cells, producing more red-labeled cDNA that hybridizes to that probe. (correct answer)
2. The corresponding gene has equal expression in both conditions, causing red and green signals to cancel into red.
3. The corresponding gene's DNA sequence mutated in treated cells, preventing any hybridization and increasing red intensity.
4. The corresponding gene is translated more in treated cells, and proteins bind the probe to generate red fluorescence.
5. The corresponding gene's probe is longer, so it migrates less during electrophoresis and appears red.

Explanation: This question tests biotechnology analysis by interpreting DNA microarray color signals to compare gene expression levels between conditions. The correct answer is that the gene has higher expression in treated cells, producing more red-labeled cDNA that hybridizes to the probe, resulting in a bright red spot due to dominant red fluorescence. In microarrays, mRNA abundance determines labeled cDNA amounts, and competitive hybridization to probes reveals relative expression, with red indicating upregulation in treated samples. Yellow spots, by contrast, suggest equal expression where green and red signals mix. A tempting distractor is choice B, which is wrong because equal expression produces yellow, not red, stemming from the misconception that color cancellation affects intensity rather than hue. When analyzing microarray data, quantify spot colors using ratios of fluorescence intensities, a strategy applicable to high-throughput expression profiling.

Question 7

1. The treated cells had higher mRNA abundance for the gene, leading to more hybridization to its probe. (correct answer)
2. The treated cells had a deletion of the gene, causing the probe to emit more fluorescence.
3. The chemical increased translation rate, and the microarray directly measured protein abundance.
4. The stronger signal indicates the gene's DNA sequence mutated to match the probe more closely.
5. The stronger signal indicates the treated cells contained more ribosomal RNA, which binds all probes.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically DNA microarrays for comparing gene expression profiles. The stronger fluorescence for the treated-sample dye at the gene spot indicates higher mRNA levels in treated cells, leading to more labeled cDNA hybridizing to the complementary probe. mRNA is converted to fluorescent cDNA, and competitive hybridization allows direct comparison of expression between samples on the same array. The chemical exposure likely upregulated the gene, increasing transcript abundance and thus signal intensity. A tempting distractor is choice C, which claims the microarray measures protein abundance, based on the misconception that arrays detect translation products rather than nucleic acids. When interpreting microarray data, normalize signals and use statistical thresholds to identify differentially expressed genes across conditions.

Question 8

1. Sample A contained more initial target DNA, so fewer amplification cycles were needed to reach the threshold. (correct answer)
2. Sample A had shorter DNA, so it migrated faster during qPCR and crossed the threshold earlier.
3. Sample B contained more initial target DNA, which delayed primer annealing and increased cycle number.
4. Sample B lacked DNA polymerase, so fluorescence increased later due to dye self-activation.
5. Sample A reached threshold earlier because restriction enzymes cut the target into more copies.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically qPCR for quantifying DNA amounts. Sample A, with more initial target DNA, requires fewer amplification cycles to reach the fluorescence threshold because exponential amplification builds on a larger starting template. The fluorescent dye binds to accumulating double-stranded PCR products, and the cycle threshold (Ct) inversely correlates with initial target quantity. Sample B's higher Ct indicates less starting DNA, as it takes more cycles to achieve detectable levels under identical conditions. A tempting distractor is choice C, which incorrectly states Sample B had more DNA delaying primer annealing, stemming from the misconception that higher template inhibits rather than accelerates amplification. In qPCR analysis, use standard curves to convert Ct values to absolute quantities and compare samples for relative expression or copy number.

Question 9

1. The digested sample produces one linear DNA fragment the same length as the plasmid. (correct answer)
2. The digested sample produces two fragments whose sizes add to more than 2,000 bp.
3. The undigested sample produces one band because supercoiled DNA is always linear.
4. The digested sample produces no bands because restriction enzymes remove phosphate groups.
5. The digested sample produces many bands because EcoRI amplifies DNA at its cut site.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically restriction enzyme digestion and gel electrophoresis of plasmids. EcoRI cuts the 2,000 bp plasmid once, converting the circular DNA into a single linear fragment of the same length, which migrates as one band at ~2,000 bp. The undigested plasmid shows multiple bands due to supercoiled, relaxed circular, and linear conformations that migrate differently in the gel. This single band in the digested sample confirms a single cut site, as multiple cuts would produce smaller fragments. A tempting distractor is choice B, which wrongly claims the digested sample produces fragments adding to more than 2,000 bp, based on the misconception that digestion increases total DNA length rather than just linearizing it. When evaluating restriction digests on gels, compare digested and undigested samples to confirm the number of cut sites and resulting fragment sizes.

Question 10

1. Frameshift mutations may occur, altering codons downstream and reducing production of functional protein. (correct answer)
2. All cells will precisely replace the exon with the guide RNA sequence by complementary base pairing.
3. NHEJ will restore the original DNA sequence because it uses RNA primers to correct errors.
4. Cas9 will continue cutting until the entire chromosome is removed from the nucleus.
5. The insertions and deletions will occur only in mRNA because Cas9 targets single-stranded transcripts.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically CRISPR-Cas9 genome editing and DNA repair mechanisms. Cas9, guided by the RNA, creates a double-strand break in the target exon, and NHEJ repair often introduces small insertions or deletions (indels) at the site. These indels can cause frameshift mutations, shifting the reading frame and altering downstream codons, which may lead to nonfunctional proteins. Without a repair template, NHEJ is error-prone, resulting in varied mutations across cells, as confirmed by sequencing. A tempting distractor is choice B, which claims all cells will precisely replace the exon with the guide RNA sequence, stemming from the misconception that NHEJ uses homology-directed repair rather than random end joining. In genome editing, sequence edited regions post-treatment to assess mutation types and efficiencies for reliable outcomes.

Question 11

1. The probe hybridized to a restriction fragment containing a complementary sequence in the sample DNA. (correct answer)
2. The probe translated into protein, which then bound the membrane at the fragment location.
3. The restriction enzyme amplified the 1.2 kb sequence, increasing fluorescence at that size.
4. The membrane converted double-stranded DNA into RNA, which was then detected by the probe.
5. The labeled probe bound nonspecifically to all DNA fragments, but only one migrated in the gel.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically Southern blotting for detecting specific DNA sequences. The labeled probe, being single-stranded and complementary, hybridizes via base pairing to the 1.2 kb restriction fragment on the membrane if the sequence is present in the genomic DNA. After washing away unbound probe, the detected band indicates specific binding to that fragment, confirming the presence of the target sequence. The restriction enzyme cuts the DNA into fragments, and gel electrophoresis separates them by size before transfer, allowing precise detection. A tempting distractor is choice E, which suggests nonspecific binding to all fragments, based on the misconception that probes lack sequence specificity and bind broadly rather than through complementary hybridization. When interpreting blotting results, ensure hybridization conditions promote specificity and use controls to distinguish true signals from background noise.

Question 12

1. It lacks introns because the template mRNA was processed by splicing before reverse transcription. (correct answer)
2. It includes upstream promoter sequences because mRNA contains regulatory DNA elements.
3. It contains all intergenic regions flanking the insulin gene because reverse transcriptase copies whole chromosomes.
4. It includes histone proteins bound to the insulin sequence because plasmids package DNA into nucleosomes.
5. It has a higher mutation rate because reverse transcriptase proofreads more than DNA polymerase.

Explanation: This question assesses biotechnology analysis by comparing features of cDNA and genomic DNA libraries for gene cloning. The correct answer is that the insulin cDNA lacks introns because the template mRNA was processed by splicing in eukaryotic cells before reverse transcription, resulting in a continuous coding sequence. Reverse transcriptase copies the mature mRNA into DNA, excluding non-coding introns that are present in genomic DNA. This makes cDNA libraries useful for expressing functional proteins in bacteria, as they lack splicing machinery. A tempting distractor is choice B, which is wrong because mRNA does not contain promoters, which are DNA elements upstream of genes, based on the misconception that transcription copies regulatory sequences into RNA. When choosing library types, use cDNA for expression studies and genomic for regulatory analysis, a strategy guiding molecular biology resource selection.

Question 13

1. The target mRNA was present in liver cells, providing template for cDNA synthesis and PCR amplification. (correct answer)
2. The gene is absent from the liver genome because PCR products form only from RNA templates.
3. Reverse transcriptase degrades mRNA, so the negative control lacks bands due to RNA loss.
4. The PCR band indicates the liver sample contained restriction fragments of the gene.
5. The negative control lacks a band because agarose gels separate RNA but not DNA.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically RT-PCR for detecting gene expression. The presence of target mRNA in liver cells allows reverse transcriptase to synthesize cDNA, which is then amplified by PCR to produce a visible band, indicating expression. The negative control without reverse transcriptase shows no band, confirming that the signal arises from mRNA-derived cDNA rather than contaminating DNA. RNA isolation and DNA removal steps ensure the assay specifically detects transcribed mRNA, linking the band to active gene expression in the tissue. A tempting distractor is choice B, which incorrectly states the gene is absent from the liver genome, based on the misconception that PCR amplifies only RNA and ignores genomic DNA contamination risks. When performing RT-PCR, include no-reverse-transcriptase controls to verify that amplification originates from mRNA and not residual DNA.

Question 14

1. DNA ligase forms phosphodiester bonds between vector and insert, sealing the sugar-phosphate backbone. (correct answer)
2. Ampicillin induces bacterial DNA polymerase to copy the insert into the chromosome.
3. Restriction enzymes add nucleotides to the insert, increasing its length to match the vector.
4. Transformation converts bacterial mRNA into plasmid DNA by reverse transcription.
5. Gel electrophoresis moves the insert into bacteria by an electric field across the membrane.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically molecular cloning and transformation in creating recombinant DNA. DNA ligase catalyzes the formation of phosphodiester bonds between the complementary sticky ends of the vector and insert, covalently joining them into a stable recombinant plasmid. This ligated plasmid can then be taken up by bacteria during transformation, and the ampicillin resistance gene allows only transformed cells to grow on selective media. Some colonies contain the inserted gene because successful ligation incorporates it into the plasmid, which replicates in the host. A tempting distractor is choice C, which incorrectly states that restriction enzymes add nucleotides to the insert, arising from the misconception that these enzymes synthesize DNA rather than cleave it. In cloning experiments, always verify recombinant plasmids by screening for insert presence, such as through restriction mapping or sequencing.

Question 15

1. Incorporation of a ddNTP terminates strand elongation because no 3' OH is available for phosphodiester bonding. (correct answer)
2. ddNTPs are incorporated only at the first base, forcing polymerase to restart synthesis repeatedly.
3. Capillary electrophoresis cuts DNA at fluorescent bases, producing fragments of different sizes.
4. PCR creates fragments of random length because primers bind nonspecifically after each cycle.
5. DNA polymerase removes nucleotides from the 5' end when ddNTPs are present, shortening strands.

Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically Sanger sequencing for determining DNA sequences. ddNTPs, lacking a 3' OH group, incorporate into the growing strand but prevent further elongation by blocking phosphodiester bond formation with the next nucleotide. This random termination at different positions generates a mixture of fragments of varying lengths, each ending with a labeled ddNTP. Capillary electrophoresis separates these fragments by size, allowing the sequence to be read from the fluorescent signals. A tempting distractor is choice B, which claims ddNTPs are incorporated only at the first base, based on the misconception that ddNTPs inhibit initiation rather than randomly terminate extension. In sequencing experiments, ensure a balanced ratio of dNTPs to ddNTPs to generate a full range of fragment lengths for accurate sequence determination.

Question 16

1. Small indels can shift the reading frame, changing downstream codons and potentially introducing a premature stop codon. (correct answer)
2. NHEJ replaces the entire exon with a homologous exon from a sister chromatid, restoring the original sequence.
3. Cas9 converts the target DNA into RNA, preventing translation of the gene product.
4. Guide RNA binds ribosomes, reducing translation initiation at the edited gene's mRNA.
5. Indels increase DNA methylation at the promoter, which directly changes codons in the mRNA.

Explanation: This question examines biotechnology analysis by explaining how CRISPR-Cas9 editing via NHEJ alters protein coding in a gene. The correct answer is that small indels can shift the reading frame, changing downstream codons and potentially introducing a premature stop codon, which disrupts the protein sequence or truncates it. Cas9, guided by the RNA to the PAM-adjacent site in exon 2, creates a double-strand break, and NHEJ repairs it imprecisely, inserting or deleting nucleotides that affect the triplet code. This frameshift often leads to nonfunctional proteins, as seen in sequencing data showing indels near the cut site. A tempting distractor is choice B, which is incorrect because NHEJ does not use homologous templates and often introduces errors rather than restoring sequences, reflecting the misconception that NHEJ is error-free like homology-directed repair. When designing CRISPR experiments, target exons early in the gene to maximize frameshift disruptions, a strategy for effective knockouts across systems.

Question 17

1. The colony PCR produces a 300-bp band, indicating the plasmid replicated without the insert.
2. The colony PCR produces a 900-bp band, indicating extra DNA is present between the primer sites. (correct answer)
3. No colonies grow on antibiotic agar, indicating the insert disrupted antibiotic resistance.
4. The colony PCR produces no band, indicating the insert was ligated but cannot be amplified.
5. All colonies fluoresce, indicating ligase activity and plasmid uptake occurred in every cell.

Explanation: This question tests biotechnology analysis through interpreting colony PCR results to confirm successful ligation in bacterial transformation. The correct answer is a 900-bp band in colony PCR, which indicates the insert was successfully ligated into the plasmid, increasing the distance between primer sites from 300 bp (empty vector) to 900 bp. Ligation joins the complementary sticky ends, and transformation introduces the recombinant plasmid into bacteria, allowing colony growth on antibiotic agar due to the plasmid's resistance gene. The PCR primers flank the multiple cloning site, so the product size directly reports insert presence without needing sequencing. A tempting distractor is choice A, which is wrong because a 300-bp band would indicate failed ligation and an empty vector, based on the misconception that smaller products confirm insertion rather than the opposite. For verifying cloning success, use PCR product size differences as a quick screen before full sequencing, a transferable approach in molecular cloning workflows.

Question 18

1. The gene's mRNA was present in the RNA sample, enabling cDNA synthesis and subsequent PCR amplification. (correct answer)
2. The gene's protein was present, allowing reverse transcriptase to translate RNA into DNA in the experimental reaction.
3. Genomic DNA contamination caused amplification, but only in the reaction lacking reverse transcriptase.
4. The primers degraded RNA in the control, preventing amplification in the minus–reverse transcriptase reaction.
5. The gel stain binds only to cDNA, so genomic DNA would be invisible even if amplified.

Explanation: This question assesses biotechnology analysis by interpreting RT-PCR gel results to confirm gene expression in liver cells. The correct answer is that the gene's mRNA was present in the RNA sample, enabling reverse transcriptase to synthesize cDNA, which is then amplified by PCR to produce the expected band. The absence of a band in the minus-reverse transcriptase control confirms no genomic DNA contamination, as PCR alone cannot amplify RNA templates without cDNA conversion. This setup distinguishes true expression (mRNA-derived) from artifacts, with the band indicating successful reverse transcription and amplification. A tempting distractor is choice C, which is wrong because genomic contamination would produce a band in the control, not its absence, based on the misconception that controls detect expression rather than contaminants. In RT-PCR experiments, always include a minus-RT control to rule out DNA contamination, a key strategy for validating RNA-based assays.

Question 19

1. Patient 2's DNA lacks primer-binding sites due to sequence changes, preventing amplification of the target region. (correct answer)
2. Patient 2's PCR produced many fragments of different sizes that migrated off the gel during electrophoresis.
3. Patient 2's DNA polymerase degraded the amplified DNA because Taq polymerase has exonuclease activity.
4. Patient 2's sample contained only RNA, which cannot be separated by agarose gel electrophoresis.
5. Patient 2's primers ligated to each other, creating plasmids that cannot be visualized by DNA stains.

Explanation: This question assesses biotechnology analysis skills by interpreting PCR gel electrophoresis results to identify why no band appears for Patient 2. The correct answer is that Patient 2's DNA lacks primer-binding sites due to sequence changes, which prevents the primers from annealing to the target DNA during the PCR annealing step. Without primer binding, DNA polymerase cannot initiate extension, resulting in no amplification of the 220-bp fragment after 30 cycles. Since the negative control shows no bands and conditions were identical, this points to a template-specific issue like mutations in the primer sites rather than reagent or thermocycler errors. A tempting distractor is choice B, which is wrong because PCR typically produces fragments of a specific size rather than many variable sizes that could migrate off the gel, reflecting the misconception that PCR is inherently nonspecific without optimized conditions. To troubleshoot absent PCR bands, always verify primer-template complementarity using sequence alignment tools before assuming other failures.

Question 20

1. Sample Y has a higher GC content, so it migrates as fewer bands during electrophoresis.
2. Sample X contains more recognition sites for the enzyme than sample Y, producing more fragments. (correct answer)
3. Sample Y was amplified by PCR, which prevents restriction enzymes from cutting DNA.
4. Sample X is circular DNA, so enzyme cutting produces more fragments than linear DNA.
5. Sample Y fragments re-annealed after digestion, forming larger DNA that appears as fewer bands.

Explanation: This question evaluates biotechnology analysis by comparing restriction enzyme digestion patterns on an agarose gel to infer differences in DNA sequences. The correct answer is that sample X contains more recognition sites for the enzyme than sample Y, leading to more cuts and thus three fragments instead of two, as each cut increases the number of linear pieces by one. Since both samples are linear DNA of the same initial length, the number of bands directly reflects the number of cut sites, with sample X having two sites (producing three bands) and sample Y having one (producing two bands). This logic relies on the enzyme's specificity for its 6-bp sequence, where sequence variations determine cut site presence. A tempting distractor is choice A, which is incorrect because GC content affects melting temperature but not the number of bands in restriction digestion, stemming from the misconception that base composition influences electrophoretic migration more than fragment count. When analyzing restriction fragment patterns, count the bands and add one to estimate cut sites for linear DNA, a strategy applicable to mapping plasmids or genomes.