Gel Electrophoresis - Genetics

1. A 5 kb plasmid with a single *Xho*I site is digested to linearize it. The reaction product is analyzed on an agarose gel. The lane shows a bright band at the expected 5 kb linear size. However, a second, fainter band is also visible that migrates faster than the 5 kb linear band. This same faster-migrating band is the primary band seen in the uncut plasmid control lane. What is the most likely identity of this second, fainter band in the digested sample?
   Answer: Undigested, supercoiled plasmid from an incomplete reaction. The presence of the expected 5 kb linear band indicates the digestion was successful for most of the plasmid. The key information is that the faster-migrating band matches the band in the uncut control. The predominant form of uncut plasmid is supercoiled, which is more compact than the linear form and thus migrates faster in an agarose gel. The presence of this band in the digested lane indicates that the restriction digest was incomplete and some of the starting supercoiled plasmid was not cut.
2. A researcher loads a DNA ladder in lane 1, which runs clearly with sharp bands. In lane 2, an unpurified PCR sample is loaded directly from the reaction tube. The sample in lane 2 produces a distorted, 'wavy' band that migrates more slowly than its expected size. What is the most likely reason for the anomalous migration of the PCR sample?
   Answer: The high concentration of salts from the PCR buffer in the sample distorted the local electric field. When you encounter gel electrophoresis problems involving distorted or anomalous band migration, think about what factors can disrupt the uniform electric field that drives DNA separation. The key principle is that DNA migrates based on size through the gel matrix under a consistent electric field, but various contaminants can interfere with this process. The wavy, slow-migrating band in lane 2 results from the high salt concentration in the unpurified PCR buffer. PCR reactions contain substantial amounts of potassium chloride and other salts that create localized disruptions in the electric field around the sample. This causes uneven migration patterns and the characteristic "wavy" appearance, while also slowing overall migration compared to purified samples. Answer A is incorrect because while excess dNTPs can chelate magnesium, this wouldn't specifically cause the wavy migration pattern described. Answer B misunderstands the physics - small sample volumes actually concentrate in sharp bands rather than causing diffusion and waviness. Answer C incorrectly suggests Taq polymerase remains bound to DNA during electrophoresis; the denaturing conditions and SDS in loading buffer would disrupt any protein-DNA interactions. The clean, sharp bands from the DNA ladder in lane 1 confirm that the gel and buffer system are working properly, pointing to contamination in lane 2 as the culprit. Study tip: Always remember that unpurified PCR samples contain high salt concentrations that distort electrophoretic migration. When you see questions about band distortion or anomalous migration, immediately consider buffer contaminants as the likely cause.
3. A plasmid digest yields two fragments, Fragment A (2000 bp) and Fragment B (500 bp). The digest is run on an agarose gel and stained with a fluorescent dye that binds DNA proportionally to its mass. Assuming the digest went to completion, the two fragments are present in equimolar amounts. What is the expected relative fluorescent intensity of the two bands?
   Answer: Band A will be approximately 4 times as intense as Band B. Since the fluorescent dye binds proportionally to the mass of DNA, the intensity of a band is proportional to the total mass of DNA in it. The fragments are present in equimolar amounts. The mass of a DNA fragment is proportional to its length (in bp). Fragment A (2000 bp) has 4 times the length of Fragment B (500 bp). Therefore, the total mass of DNA in Band A is 4 times the total mass in Band B, and its fluorescent signal will be approximately 4 times as intense.
4. A gene has a polymorphism that eliminates a *Pst*I restriction site. The allele without the site (Allele A) is dominant and causes a disorder. A study analyzes this RFLP in a family. Genomic DNA from three individuals was digested with *Pst*I and analyzed by Southern blotting. The probe hybridizes to a 500 bp fragment when the *Pst*I site is present (Allele a) and an 800 bp fragment when it is absent (Allele A). Based on the gel shown, which statement is correct?
   Answer: Individual 3 is heterozygous and affected by the disorder. Allele A (disease) = 800 bp band. Allele a (wild-type) = 500 bp band. Individual 1 shows only the 800 bp band, so their genotype is AA (homozygous dominant, affected). Individual 2 shows only the 500 bp band, so their genotype is aa (homozygous recessive, unaffected). Individual 3 shows both the 800 bp and 500 bp bands, so their genotype is Aa (heterozygous). Since Allele A is dominant, the heterozygous individual is affected by the disorder. Therefore, Individual 3 is heterozygous and affected.
5. A technician inadvertently reverses the electrical leads on an electrophoresis chamber, connecting the positive lead to the end with the sample wells and the negative lead to the far end. What will be the result after 30 minutes of running the gel at 100V?
   Answer: The DNA will migrate backwards, out of the wells and into the buffer of the chamber reservoir. DNA has a strong negative charge due to its phosphate backbone. In gel electrophoresis, DNA migrates from the negative electrode (cathode) toward the positive electrode (anode). By reversing the leads, the positive electrode is placed at the same end as the wells. The negatively charged DNA will therefore be drawn toward the nearby positive electrode, causing it to migrate 'up' and out of the wells, into the surrounding buffer, and be lost from the gel.
6. A linear 10 kb DNA fragment has *Hin*dIII restriction sites at 2.0 kb and 7.0 kb from the 5' end. A single *Bam*HI site is located at 4.0 kb from the 5' end. If the fragment is simultaneously and completely digested with both enzymes, which set of fragments will be generated?
   Answer: 2.0 kb, 2.0 kb, 3.0 kb, 3.0 kb The enzymes will cut the 10 kb linear DNA at positions 2.0 kb, 4.0 kb, and 7.0 kb. This creates four fragments. The sizes of these fragments are calculated by the distance between the cut sites (and ends): (2.0 - 0) = 2.0 kb; (4.0 - 2.0) = 2.0 kb; (7.0 - 4.0) = 3.0 kb; and (10.0 - 7.0) = 3.0 kb. Thus, the resulting fragments are 2.0, 2.0, 3.0, and 3.0 kb.
7. A student prepares a sample for agarose gel electrophoresis by mixing their PCR product with a standard 6X loading buffer containing bromophenol blue and glycerol. However, they forget to add a fluorescent DNA stain (like SYBR Safe or ethidium bromide) to the molten agarose when casting the gel and do not use a post-staining protocol. After running the gel and placing it on a UV transilluminator, what will the student most likely observe?
   Answer: The DNA will have separated but will be invisible, while the colored tracking dyes will be seen under ambient light. Loading buffer contains dense agents (like glycerol) to help the sample sink into the well and tracking dyes (like bromophenol blue) that are visible in normal room light to monitor the electrophoresis progress. DNA itself is not visible. Fluorescent stains are required to bind to the DNA so it can be visualized under UV light. Without the stain, the DNA will separate by size but remain invisible. The tracking dyes, however, are colored compounds and will be seen migrating through the gel under ambient (non-UV) light.
8. A 4.0 kb circular plasmid was digested with restriction enzymes *Eco*RI and *Bam*HI, singly and in combination. The results are shown in the agarose gel below. Based on these results, how many restriction sites for *Eco*RI and *Bam*HI are present in the plasmid?
   Answer: One *Eco*RI site and two *Bam*HI sites The *Eco*RI single digest yields one 4.0 kb band, which means it linearized the 4.0 kb circular plasmid; thus, there is one *Eco*RI site. The *Bam*HI single digest yields two bands (1.5 kb and 2.5 kb), and their sizes sum to 4.0 kb; thus, there are two *Bam*HI sites. The double digest results confirm this. The rule for a circular plasmid is that the number of restriction sites equals the number of fragments produced by that enzyme.
9. An agarose gel is run overnight at a low, constant voltage. In the morning, the researcher observes that bands in the top third of the gel are sharp, but bands in the bottom two-thirds are progressively broader and more distorted. What is the most probable cause of this observation?
   Answer: The buffer's ionic strength was exhausted, leading to a drop in conductivity and pH changes in the gel. During prolonged electrophoresis, the buffering capacity of the running buffer can be exhausted due to the electrolysis of water, which generates H+ at the anode and OH- at the cathode. This leads to the formation of a pH gradient and a decrease in ionic strength within the gel and buffer reservoirs. These conditions severely compromise electrophoresis, causing poor conductivity and leading to distorted, smeared, and slow-moving bands, an effect that is more pronounced for bands that have migrated farther down the gel.
10. An unknown DNA fragment (X) was electrophoresed alongside a standard DNA ladder. The fragment X migrated to a position on the gel exactly halfway between the 750 bp and 1000 bp markers. Based on the principle of DNA migration in agarose, what is the best estimate for the size of fragment X?
    Answer: 866 bp The migration distance of linear DNA fragments in an agarose gel is inversely proportional to the logarithm of their size. Therefore, a linear interpolation of fragment size based on migration distance is inaccurate. To estimate the size of fragment X, one must perform a logarithmic interpolation. The log10 of the sizes of the marker bands are log10(1000) = 3.0 and log10(750) ≈ 2.875. The midpoint of these log values is (3.0 + 2.875) / 2 = 2.9375. The estimated size of fragment X is $10^{2.9375}$, which is approximately 866 bp. Choice A (875 bp) represents a common error of using linear interpolation.
11. A researcher digests human genomic DNA with *Eco*RI for a Southern blot. After electrophoresis, the lane shows an expected continuous smear. Before transferring the DNA to a membrane, which of the following treatments is most critical for ensuring the efficient transfer of high-molecular-weight DNA fragments from the gel?
    Answer: Soaking the gel in a dilute acid, such as HCl, prior to denaturation with NaOH. For efficient capillary transfer of large DNA fragments (>10-15 kb) in Southern blotting, a depurination step is crucial. Soaking the gel in a dilute acid like HCl removes purine bases, creating apurinic sites. The phosphodiester backbone at these sites is labile and is cleaved during the subsequent denaturation step with NaOH. This fragmentation of large DNA strands allows them to transfer out of the gel matrix more easily and efficiently.
12. A research team is constructing a physical map of a novel yeast species' chromosomes, which range in size from 225 kb to over 1500 kb. Standard agarose gel electrophoresis fails to resolve these molecules, as they all remain near the well in a single compressed band. Which technique should be employed to achieve separation?
    Answer: Pulsed-field gel electrophoresis (PFGE) Standard gel electrophoresis cannot resolve very large DNA molecules (typically > 50 kb) because they undergo reptation, migrating at a rate independent of size. Pulsed-field gel electrophoresis (PFGE) is specifically designed to separate large DNA molecules (up to megabase sizes) by periodically changing the direction of the electric field. This forces the molecules to reorient themselves, and the time required for reorientation is dependent on molecular size, allowing for effective separation.
13. A researcher is analyzing the products of a dideoxy chain-termination (Sanger) sequencing reaction using slab gel electrophoresis. To accurately determine the sequence, it is critical that the migration rate depends only on fragment length and not on any intrinsic secondary structure of the single-stranded DNA molecules. Which modification to standard electrophoresis is essential to meet these requirements?
    Answer: Including a high concentration of urea or formamide in a polyacrylamide gel. When analyzing Sanger sequencing results, you need DNA fragments to migrate based solely on their length, not their shape or structure. Single-stranded DNA can form secondary structures like hairpins and loops through intramolecular base pairing, which would cause fragments of the same length to migrate differently and scramble your sequence data. The solution is using denaturing conditions that prevent secondary structure formation. Option D is correct because urea and formamide are powerful denaturing agents that disrupt hydrogen bonds between bases. At high concentrations (typically 7-8M urea), these chemicals ensure all DNA fragments remain completely single-stranded and linear, so migration depends only on molecular weight. Option A is wrong because running at 4°C would actually stabilize secondary structures by reducing thermal motion that disrupts base pairing. You want the opposite effect. Option B is incorrect because ethidium bromide intercalates into double-stranded DNA and wouldn't prevent single-strand secondary structures; plus, it's typically used for visualization, not separation optimization. Option C describes a technique for separating very large DNA molecules (>50 kb) by periodically changing the electric field direction, which is irrelevant for the short fragments (100-1000 bp) produced in Sanger sequencing. Study tip: Remember that Sanger sequencing requires denaturing polyacrylamide gels with urea or formamide. Whenever you see sequencing problems involving fragment separation, think "denaturing conditions" to eliminate secondary structure artifacts.
14. A researcher isolates a 4 kb plasmid from E. coli. An aliquot of the untreated plasmid preparation is analyzed on a 1% agarose gel. Three distinct bands are observed: a very bright band that migrated the farthest (Band 1), a faint band with intermediate migration (Band 2), and another faint band that migrated the shortest distance (Band 3). What do these bands most likely represent?
    Answer: Band 1: Supercoiled; Band 2: Linear; Band 3: Nicked circular Plasmid DNA exists in multiple conformations. The most compact form is supercoiled, which migrates fastest through the gel matrix. The linear form, having a more extended shape, migrates at an intermediate rate. The nicked circular (or open circular) form is the least compact and experiences the most drag, causing it to migrate the slowest. The supercoiled form is typically the most abundant in a standard plasmid preparation.
15. A researcher is mapping a 6.0 kb circular plasmid using enzymes *Hin*dIII and *Pst*I. The sizes of the resulting fragments from single and double digests are shown in the table. Based on these data, which of the following statements is true?
    Answer: The single *Hin*dIII site is located within the 2.5 kb fragment generated by *Pst*I. First, determine the number of sites. *Hin*dIII digest gives one 6.0 kb band, linearizing the plasmid, so there is one *Hin*dIII site. *Pst*I digest gives two bands (2.5 kb and 3.5 kb), so there are two *Pst*I sites. To map the sites, compare the single and double digests. The *Pst*I digest creates a 2.5 kb and a 3.5 kb fragment. In the double digest, the 3.5 kb fragment is still present, but the 2.5 kb fragment has been replaced by two smaller fragments, 1.0 kb and 1.5 kb (note that 1.0 + 1.5 = 2.5). This indicates that the single *Hin*dIII site lies within the 2.5 kb fragment originally generated by *Pst*I.
16. A researcher runs a 1.5% agarose gel at 150V for 45 minutes to separate PCR products. Upon visualization, the bands appear distorted as shown in the figure, with bands in the central lanes having migrated farther than those in the outer lanes. This 'smiling' artifact is most likely caused by:
    Answer: Overheating of the gel due to excessive voltage, causing the center to run faster. A 'smiling' effect on an agarose gel is a classic sign of overheating. Running the gel at an excessive voltage generates significant heat. The center of the gel apparatus is less efficient at dissipating this heat than the edges, causing the buffer in the central part of the gel to become warmer. This reduces the local viscosity of the buffer, allowing DNA in the central lanes to migrate faster than the DNA in the cooler, outer lanes, creating a smile-shaped pattern.
17. To accelerate the separation of DNA fragments, a student increases the voltage from the recommended 80 V to 200 V. While the run time is significantly reduced, which of the following is the most likely adverse consequence of this action?
    Answer: The resolution of smaller DNA fragments will be significantly decreased due to frictional heat generation. Increasing the voltage significantly increases the electrical current and leads to substantial Joule heating of the gel and buffer. This heat decreases the viscosity of the buffer, causing all fragments, especially smaller ones, to migrate more quickly and with less effective sieving. The result is a loss of resolution, where bands are compressed together and poorly separated. It can also cause gel melting or 'smiling' artifacts.
18. Ethidium bromide (EtBr) functions as a DNA stain by intercalating between stacked base pairs. How does this intercalation physically affect a linear DNA molecule and its resulting migration during electrophoresis?
    Answer: It lengthens and stiffens the DNA molecule, reducing its electrophoretic mobility by about 15%. The insertion of planar ethidium bromide molecules between base pairs unwinds the DNA helix and increases the distance between base pairs, effectively lengthening the molecule. This process also makes the DNA more rigid. The combined effect of lengthening and increased rigidity increases the frictional drag of the molecule as it moves through the agarose pores, thus decreasing its electrophoretic mobility (i.e., slowing it down) compared to an unstained DNA molecule of the same size.
19. A researcher needs to resolve two linear DNA fragments of 120 bp and 135 bp resulting from a PCR-RFLP analysis. Which of the following agarose gel concentrations would provide the optimal resolution for these two fragments?
    Answer: 3.5% agarose The resolution of an agarose gel depends on its concentration. Higher concentrations of agarose create a matrix with smaller pores, which is more effective at sieving and separating small DNA fragments. For fragments in the range of 100-200 bp, a high-percentage gel is required. While a 2.0% gel would provide some separation, a 3.5% gel is specifically suited for resolving very small fragments (e.g., 20-200 bp) and would offer the best possible resolution to distinguish between 120 bp and 135 bp.
20. A locus is analyzed by PCR. Allele 'W' produces a 200 bp amplicon, while allele 'm' has a 10 bp internal deletion, producing a 190 bp amplicon. The PCR product from a heterozygous (W/m) individual is denatured at 95°C, slowly re-annealed, and then loaded onto a non-denaturing polyacrylamide gel. Which band pattern is expected?
    Answer: Four bands: two bright homoduplex bands (200 bp, 190 bp) and two fainter, slower-migrating heteroduplex bands. After denaturation and re-annealing, four types of duplexes will form: 200 bp/200 bp (homoduplex), 190 bp/190 bp (homoduplex), and two 200 bp/190 bp pairings (heteroduplexes). On a non-denaturing gel, the perfectly matched homoduplexes migrate according to their size. The heteroduplexes contain a 10-base single-stranded loop where the deletion occurred. This mismatch 'bubble' gives the molecule a more open, bulky conformation, which significantly retards its migration through the gel matrix. Therefore, one expects to see the two homoduplex bands and one or two slower-migrating heteroduplex bands.

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