Genetics Quiz: Gel Electrophoresis

What this quiz covers

This quiz focuses on Gel Electrophoresis, 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. 0.7% agarose
2. 1.2% agarose
3. 2.0% agarose
4. 3.5% agarose (correct answer)

Explanation: 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.

Question 2

1. Running the electrophoresis at 4°C to stabilize the DNA-polymerase complex.
2. Adding an intercalating agent like ethidium bromide to the gel and running buffer.
3. Using pulsed-field gel electrophoresis with alternating orthogonal fields.
4. Including a high concentration of urea or formamide in a polyacrylamide gel. (correct answer)

Explanation: 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.

Question 3

1. Two distinct bands corresponding to 200 bp and 190 bp fragments.
2. A single broad band centered at approximately 195 bp.
3. Four bands: two bright homoduplex bands (200 bp, 190 bp) and two fainter, slower-migrating heteroduplex bands. (correct answer)
4. Two bands at 200 bp and 190 bp, plus a very fast-migrating 10 bp single-stranded DNA band.

Explanation: 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.

Question 4

1. Two distinct bands corresponding to 200 bp and 190 bp fragments.
2. A single broad band centered at approximately 195 bp.
3. Four bands: two bright homoduplex bands (200 bp, 190 bp) and two fainter, slower-migrating heteroduplex bands. (correct answer)
4. Two bands at 200 bp and 190 bp, plus a very fast-migrating 10 bp single-stranded DNA band.

Explanation: 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.

Question 5

1. The DNA will migrate backwards, out of the wells and into the buffer of the chamber reservoir. (correct answer)
2. The DNA will remain in the wells, as the incorrect electrical field will repel it from entering the gel.
3. The DNA will migrate correctly toward the far end, as the buffer ions will reorient the electrical field.
4. The DNA will precipitate in the wells due to the reversed ionic flow.

Explanation: 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.

Question 6

1. Band A will be approximately 4 times as intense as Band B. (correct answer)
2. Band B will be approximately 4 times as intense as Band A.
3. Band A and Band B will have approximately equal intensity.
4. The relative intensity cannot be determined without knowing the GC content of each fragment.

Explanation: 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.

Question 7

1. The DNA will fail to migrate out of the wells due to rapid denaturation of the fragments.
2. The resolution of smaller DNA fragments will be significantly decreased due to frictional heat generation. (correct answer)
3. The electrophoretic buffer will become depleted, causing the DNA to migrate towards the cathode.
4. The ethidium bromide will dissociate from the DNA, resulting in an inability to visualize any bands.

Explanation: 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.

Question 8

1. 875 bp
2. 866 bp (correct answer)
3. 854 bp
4. 841 bp

Explanation: 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.

Question 9

1. 2.0 kb, 2.0 kb, 3.0 kb, 3.0 kb (correct answer)
2. 2.0 kb, 3.0 kb, 5.0 kb
3. 2.0 kb, 1.0 kb, 3.0 kb, 4.0 kb
4. 4.0 kb, 6.0 kb

Explanation: 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.

Question 10

1. It neutralizes the phosphate backbone's negative charge, causing the DNA to migrate unpredictably.
2. It causes the DNA to become more compact and globular, leading to faster migration through the gel pores.
3. It decreases the net negative charge of the DNA, slowing its migration towards the anode.
4. It lengthens and stiffens the DNA molecule, reducing its electrophoretic mobility by about 15%. (correct answer)

Explanation: 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.

Question 11

1. A 5 kb nicked circular plasmid resulting from nuclease contamination.
2. The XhoI enzyme bound to the DNA, accelerating its mobility.
3. Undigested, supercoiled plasmid from an incomplete reaction. (correct answer)
4. A smaller, unrelated plasmid that co-purified during the preparation.

Explanation: 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.

Question 12

1. Denaturing gradient gel electrophoresis (DGGE)
2. Pulsed-field gel electrophoresis (PFGE) (correct answer)
3. High-resolution capillary electrophoresis
4. Sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE)

Explanation: 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.

Question 13

1. Excess dNTPs in the PCR mix chelated magnesium ions in the buffer, inhibiting migration.
2. The sample volume was too small, causing rapid diffusion and distortion within the well before migration.
3. The Taq polymerase in the sample remained bound to the DNA, creating a large, slow-moving complex.
4. The high concentration of salts from the PCR buffer in the sample distorted the local electric field. (correct answer)

Explanation: 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.

Question 14

1. Band 1: Supercoiled; Band 2: Linear; Band 3: Nicked circular (correct answer)
2. Band 1: Linear; Band 2: Supercoiled; Band 3: Nicked circular
3. Band 1: Nicked circular; Band 2: Linear; Band 3: Supercoiled
4. Band 1: Supercoiled; Band 2: Nicked circular; Band 3: Catenated dimer

Explanation: 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.

Question 15

1. The gel concentration was too low, allowing excessive diffusion of the smaller bands over time.
2. The DNA was progressively degraded by nucleases present in the running buffer.
3. The buffer's ionic strength was exhausted, leading to a drop in conductivity and pH changes in the gel. (correct answer)
4. The ethidium bromide migrated towards the anode faster than the DNA, leaving the lower part of the gel unstained.

Explanation: 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.

Question 16

1. Exposing the gel to high-intensity UV light to crosslink the DNA to the agarose matrix.
2. Soaking the gel in a dilute acid, such as HCl, prior to denaturation with NaOH. (correct answer)
3. Washing the gel extensively with deionized water to remove excess buffer salts.
4. Treating the gel with proteinase K to remove any DNA-binding proteins.

Explanation: 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.

Question 17

1. 2.0 kb, 2.0 kb, 3.0 kb, 3.0 kb (correct answer)
2. 2.0 kb, 3.0 kb, 5.0 kb
3. 2.0 kb, 1.0 kb, 3.0 kb, 4.0 kb
4. 4.0 kb, 6.0 kb

Explanation: 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.

Question 18

1. Band 1: Supercoiled; Band 2: Linear; Band 3: Nicked circular (correct answer)
2. Band 1: Linear; Band 2: Supercoiled; Band 3: Nicked circular
3. Band 1: Nicked circular; Band 2: Linear; Band 3: Supercoiled
4. Band 1: Supercoiled; Band 2: Nicked circular; Band 3: Catenated dimer

Explanation: 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.

Question 19

1. 0.7% agarose
2. 1.2% agarose
3. 2.0% agarose
4. 3.5% agarose (correct answer)

Explanation: 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.

Question 20

1. Denaturing gradient gel electrophoresis (DGGE)
2. Pulsed-field gel electrophoresis (PFGE) (correct answer)
3. High-resolution capillary electrophoresis
4. Sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE)

Explanation: 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.