MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5c Electrophoresis Protein Separation

What this quiz covers

This quiz focuses on 5c Electrophoresis Protein Separation, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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. Increased heating and diffusion at higher voltage broadened bands and compressed apparent spacing. (correct answer)
2. Higher voltage decreased the negative charge provided by SDS, equalizing mobilities.
3. Higher voltage caused large proteins to become positively charged and reverse direction.
4. Higher voltage eliminated the sieving effect by enlarging gel pores during the run.

Explanation: This question evaluates voltage's effect on band spacing in SDS-PAGE, considering heating and diffusion. In electrophoresis, higher voltage increases speed but causes heating, broadening bands and compressing apparent spacing, especially for disparate MW. In this pH 8.3 setup, doubled voltage decreased spacing between 15 and 150 kDa despite farther migration. The correct answer A is consistent as heating broadened bands, reducing effective separation. Distractor D is incorrect, claiming voltage enlarges pores and eliminates sieving, a misconception since voltage doesn't alter gel structure. A check is to run with constant current to minimize heating and preserve spacing. Monitoring band widths at varying voltages can verify diffusion's role in resolution loss.

Question 2

1. Spots 1–3 likely represent proteins of identical MW but different pI values (e.g., isoforms or post-translational modifications). (correct answer)
2. Spots 1–3 must be the same protein because identical MW implies identical pI.
3. Spot 4 must have the highest pI because it has the lowest MW.
4. Spots 1–3 differ only in size because IEF separates primarily by MW.

Explanation: This question tests interpretation of 2D electrophoresis data, combining isoelectric focusing (IEF) and SDS-PAGE for separation by pI and MW. Electrophoresis separates by charge in IEF (proteins stop at their pI) and by size in SDS-PAGE, allowing identification of isoforms with same MW but different pI. In this 2D setup, horizontal IEF positions reflect pI, and vertical SDS-PAGE reflects MW. The correct answer A is supported as spots 1–3 share MW (66 kDa) but differ in pI (4.9, 6.8, 8.9), likely isoforms from modifications. Distractor D fails by stating IEF separates by MW, a misconception since IEF is charge-based, not size-based. To verify, re-run individual spots in 1D IEF to confirm pI differences. Comparing with unmodified proteins can check if modifications alter pI without changing MW.

Question 3

1. At pH 6.8 the enzyme is near neutral net charge; at pH 9.0 it is net negative and migrates toward the anode. (correct answer)
2. At pH 6.8 the enzyme is net negative; at pH 9.0 it becomes net positive and migrates toward the anode.
3. At pH 6.8 the enzyme is net positive and is repelled by the cathode; at pH 9.0 it is neutral.
4. The change in migration is primarily due to MW changing with pH rather than charge.

Explanation: This question assesses how pH affects protein charge and migration in native PAGE, testing understanding of isoelectric point (pI) influence. In electrophoresis, proteins' net charge depends on pH relative to pI: negative above pI, positive below, with charge magnitude affecting migration speed and direction. In this native PAGE setup loaded at the cathode, migration toward the anode at higher pH indicates net negative charge. The correct answer A is consistent because at pH 6.8 (below pI 7.4), the enzyme is near neutral or positive with little migration (0.5 cm), while at pH 9.0 (above pI), it is net negative and migrates far (4.7 cm). Distractor D is incorrect as it attributes migration change to MW variation with pH, a misconception since MW is constant and charge drives the difference. To verify, test at pH equal to pI to confirm zero migration. Comparing migration at multiple pH values can map charge transitions around pI for similar proteins.

Question 4

1. Increased Joule heating at higher voltage increased diffusion and band broadening. (correct answer)
2. Higher voltage decreased the electric field strength, lowering protein velocity and increasing overlap.
3. Higher voltage increased protein molecular mass by promoting aggregation, reducing migration differences.
4. At higher voltage, SDS no longer binds uniformly, making smaller proteins less negatively charged than larger proteins.

Explanation: This question tests understanding of how voltage affects band resolution in SDS-PAGE through Joule heating effects. In gel electrophoresis, applying voltage generates heat (Joule heating) proportional to the square of the voltage, and excessive heating can reduce separation quality by increasing molecular diffusion and band broadening. The data shows that at 200V, despite similar total migration distance, band separation decreased (0.35 to 0.30 cm) while band width nearly doubled (0.18 to 0.34 cm), indicating significant band broadening that reduced resolution between the two proteins. Increased Joule heating at higher voltage causes uneven temperature distribution in the gel, leading to increased diffusion of protein bands and loss of sharpness, which explains the observed reduction in separation efficiency. Choice B incorrectly states higher voltage decreases electric field strength, when voltage directly increases field strength; the issue is the thermal effects, not field strength. For optimal SDS-PAGE resolution, balance voltage to achieve reasonable run times while minimizing Joule heating - lower voltage with longer run times often provides sharper bands than high voltage with shorter times.

Question 5

1. Voltage, because increasing voltage changes the proteins' pI values and increases their charge difference.
2. Net charge at the running pH, because E1 is farther from its pI than E2 and migrates more. (correct answer)
3. Molecular size, because proteins of identical mass separate primarily by sieving through the gel.
4. Direction of the electric field, because reversing the leads would increase the distance between E1 and E2.

Explanation: This question tests understanding of factors affecting protein separation in native PAGE, specifically the role of net charge versus other parameters. In native electrophoresis, proteins separate based on their charge-to-mass ratio and size, with net charge being the primary factor when proteins have similar sizes. At pH 8.0, E1 (pI 5.0) is 3.0 pH units above its pI, giving it a large negative charge, while E2 (pI 7.8) is only 0.2 pH units above its pI, resulting in a very small negative charge - this large difference in net charge explains why E1 migrates much farther (3.6 cm at 150V) than E2 (0.6 cm). The separation efficiency between E1 and E2 is determined by their different net charges at the running pH, not by voltage (which affects migration speed but not relative separation), molecular size (they're identical), or field direction. Choice A incorrectly suggests voltage changes pI values, when pI is an intrinsic protein property independent of applied voltage. To optimize protein separation in native PAGE, choose a pH that maximizes the difference in net charge between proteins by considering their pI values - proteins far from their pI at the running pH will have larger net charges and migrate more.

Question 6

1. P1 has a higher $pI$ than P2.
2. P2 has a higher $pI$ than P1. (correct answer)
3. Both proteins have the same $pI$ because they swap migration distances.
4. Neither protein has a $pI$ because only nucleic acids have $pI$ values.

Explanation: This question tests understanding of how protein migration changes with pH relative to pI in native PAGE. In electrophoresis, proteins are positively charged below their pI and negatively charged above their pI. At pH 6.0, P2 migrates farther toward the anode (24 mm vs 6 mm), indicating it has more negative charge. At pH 8.5, P1 migrates farther (20 mm vs 8 mm), showing P1 is more negative at higher pH. This reversal indicates P2's pI lies between pH 6.0 and 8.5, while P1's pI is below pH 6.0. Since P2 transitions from more negative to less negative as pH increases from 6.0 to 8.5, P2 must have the higher pI. A useful check is that proteins become less negative as pH approaches their pI from above.

Question 7

1. Protein R likely has a more negative net charge than Protein Q at pH 7.0.
2. Protein Q likely has a more negative net charge than Protein R at pH 7.0. (correct answer)
3. Protein Q likely has a larger hydrodynamic radius, causing faster migration.
4. Protein R must be positively charged because it migrated toward the anode.

Explanation: This question tests understanding of how net charge affects migration in native PAGE. In native electrophoresis at pH 7.0, proteins migrate toward the anode based on their net negative charge - more negative proteins migrate farther. Since Protein Q migrated 26 mm while Protein R only migrated 5 mm (both toward the anode), Protein Q must have a more negative net charge at pH 7.0. The similar masses (~50 kDa) rule out size as the primary factor differentiating their migration. Choice A reverses the charge relationship, while Choice D incorrectly states that anode-migrating proteins are positive (they're actually negative). A key check is that in native PAGE, migration distance directly correlates with net charge magnitude when sizes are similar.

Question 8

1. Smaller pore size increases sieving, enhancing size-based differences in mobility. (correct answer)
2. Higher gel percentage increases protein net charge, enhancing charge-based differences.
3. Higher gel percentage decreases voltage, slowing all proteins equally.
4. Smaller pore size reverses migration direction for larger proteins.

Explanation: This question tests understanding of gel percentage effects on protein separation in PAGE. Higher percentage gels have smaller pore sizes due to increased acrylamide concentration, creating more sieving effect. In the 6% gel, proteins U and V migrated similarly (34 vs 30 mm), but in the 12% gel, their separation increased dramatically (18 vs 9 mm). The smaller pores in the 12% gel create greater friction for larger proteins, enhancing size-based differences in mobility. This improved separation results from differential sieving - the larger protein experiences proportionally more resistance. Choice B incorrectly suggests gel percentage affects protein charge, which is determined by amino acid composition and pH, not gel concentration.

Question 9

1. The 100 kDa species is likely a disulfide-linked complex that separates into two subunits under reducing conditions. (correct answer)
2. BME increases SDS binding, causing proteins to migrate toward the cathode.
3. Reducing conditions increase protein $pI$, causing two bands to appear.
4. The two smaller bands indicate proteolysis during electrophoresis due to higher voltage.

Explanation: This question tests understanding of reducing versus non-reducing SDS-PAGE conditions. β-mercaptoethanol (BME) is a reducing agent that breaks disulfide bonds between cysteine residues. The observation of one 100 kDa band without BME transforming into two bands (60 kDa and 40 kDa) with BME strongly suggests the 100 kDa species was a disulfide-linked complex. Under reducing conditions, the disulfide bonds break, releasing the individual subunits which migrate according to their separate molecular weights. Choice D incorrectly attributes the band pattern to proteolysis, but proteolytic cleavage would occur in both lanes, not just the reducing lane. The key principle is that disulfide bonds maintain protein complexes that separate into subunits when reduced.

Question 10

1. Lowering pH increased the protein's net negative charge, reducing migration.
2. Lowering pH decreased the protein's net negative charge, reducing migration toward the anode. (correct answer)
3. Lowering pH increased gel pore size, reducing migration.
4. Lowering pH increased voltage across the gel, reducing migration.

Explanation: This question tests understanding of pH effects on protein charge in native PAGE. In electrophoresis, proteins migrate toward the anode when negatively charged, with migration distance proportional to net charge. At pH 7.2, the protein migrated 22 mm toward the anode, indicating substantial negative charge. When pH was lowered to 5.8, migration decreased dramatically to 2 mm, indicating much less negative charge. Lowering pH protonates acidic residues (reducing negative charges) and maintains positive charges on basic residues, resulting in decreased net negative charge. Choice A incorrectly states that lowering pH increases negative charge - the opposite is true. The key principle is that proteins become less negative (more positive) as pH decreases.

Question 11

1. Replacing a surface Asp with Lys, increasing net positive charge and speeding migration to the anode.
2. Replacing a surface Lys with Glu, increasing net negative charge and speeding migration to the anode. (correct answer)
3. Adding a disulfide bond, increasing net negative charge and speeding migration.
4. Removing a glycosylation site, decreasing net charge and causing faster migration to the anode.

Explanation: This question tests understanding of how amino acid substitutions affect protein migration in native PAGE. At pH 8.2, proteins migrate toward the anode based on net negative charge. The mutant (Mut) migrated much farther than wild-type (29 mm vs 14 mm), indicating increased net negative charge. Since both variants have identical mass, the difference must be due to charge. Replacing a positively charged Lys with negatively charged Glu removes one positive charge and adds one negative charge, creating a net change of -2 in charge number. This makes the protein more negative and increases anode migration. Choice A would decrease negative charge (Asp to Lys adds positive charge), causing less migration, not more.

Question 12

1. The enzyme's $pI$ is likely above 10.0.
2. The enzyme's $pI$ is likely between 7.0 and 10.0.
3. The enzyme's $pI$ is likely below 7.0. (correct answer)
4. The enzyme has no $pI$ because it is an enzyme rather than a structural protein.

Explanation: This question tests understanding of how protein migration changes with pH relative to pI. In native PAGE, proteins are positively charged below their pI and negatively charged above their pI. The enzyme migrated toward the anode at both pH values, indicating negative charge, but migrated much farther at pH 10.0 (26 mm) than at pH 7.0 (8 mm). This shows the enzyme becomes more negative as pH increases from 7.0 to 10.0, which only occurs when the pI is below the pH range tested. If the pI were above 7.0, the protein would be positive at pH 7.0 and migrate toward the cathode. Choice A incorrectly suggests pI > 10.0, which would make the protein positive at both pH values tested.

Question 13

1. Increased ionic strength screens electric fields around proteins, reducing effective electrophoretic mobility. (correct answer)
2. Added NaCl increases protein size by hydration, increasing mobility.
3. Added NaCl reverses the sign of protein charge at pH 6.5.
4. Added NaCl decreases gel pore size, making proteins migrate farther.

Explanation: This question tests understanding of ionic strength effects on electrophoresis. In electrophoresis, proteins migrate in an electric field, but dissolved ions create an ionic atmosphere that screens the electric field around charged molecules. Adding NaCl increases ionic strength, which enhances this screening effect and reduces the effective electric field experienced by proteins. This decreases electrophoretic mobility for all proteins, causing shorter migration distances and reduced separation between bands. The screening effect follows Debye-Hückel theory - higher ionic strength compresses the electrical double layer around proteins. Choice B incorrectly suggests NaCl increases protein size significantly enough to increase mobility, but any size increase would actually decrease mobility.

Question 14

1. It likely had a net positive charge at pH 8.6. (correct answer)
2. It likely had the smallest molecular weight at pH 8.6.
3. It likely had the most negative charge at pH 8.6.
4. It likely bound SDS in the sample buffer, reversing its migration.

Explanation: This question tests understanding of protein charge and migration direction in electrophoresis. In native agarose electrophoresis, proteins migrate based on their net charge - negative proteins toward the anode, positive proteins toward the cathode. Since this band migrated toward the cathode (opposite most proteins), it must have a net positive charge at pH 8.6. This is unusual because pH 8.6 is above most protein pI values, typically rendering proteins negative. However, some proteins with very high pI values (>8.6) remain positively charged at this pH. Choice C incorrectly assigns negative charge to a cathode-migrating protein, contradicting fundamental electrophoresis principles. The key verification is migration direction: cathode = positive charge.

Question 15

1. Protein P becomes less negatively charged as pH increases, increasing anode migration.
2. Protein P is likely near its $pI$ around pH 5.0. (correct answer)
3. Protein P is net positive above pH 8.0, explaining the plateau.
4. Protein P's size decreases with pH, causing greater migration.

Explanation: This question tests understanding of protein migration patterns near the pI. In native PAGE, proteins migrate based on net charge, with minimal migration at their pI where net charge is zero. The data shows Protein P has very low migration at pH 5.0 (2 mm), then increasing migration as pH rises to 8.0 (23 mm), with little further increase at pH 9.5 (25 mm). This pattern indicates the pI is near pH 5.0 - at this pH, the protein has minimal charge and migration. As pH increases above the pI, the protein becomes increasingly negative, explaining increased anode migration. The plateau above pH 8.0 suggests most ionizable groups are already deprotonated. Choice A incorrectly states proteins become less negative as pH increases.

Question 16

1. A2 likely has a smaller hydrodynamic radius than A1, reducing friction and increasing mobility. (correct answer)
2. A2 likely has a larger hydrodynamic radius than A1, increasing mobility.
3. A1 likely has a more negative net charge than A2, increasing mobility.
4. A1 migrated less because it was loaded closer to the anode.

Explanation: This question tests understanding of how both charge and size affect migration in native PAGE. In native electrophoresis, migration depends on the balance of electrophoretic force (proportional to charge) and frictional resistance (proportional to size). Since A1 and A2 have identical net charge but different migration (A2 migrates farther), the difference must be due to size. A2's greater migration distance indicates it experiences less friction, meaning it has a smaller hydrodynamic radius than A1. The smaller size allows A2 to move more easily through the gel matrix despite having the same driving force (charge). Choice B incorrectly suggests larger proteins migrate faster - in reality, larger proteins experience more friction and migrate slower when charge is equal.

Question 17

1. Protein Bdp becomes more positively charged as pH decreases, increasing migration toward the cathode. (correct answer)
2. Protein Bdp becomes more negatively charged as pH decreases, increasing migration toward the cathode.
3. Protein Bdp becomes uncharged at lower pH, increasing migration toward the cathode.
4. Protein Bdp's size decreases as pH decreases, reversing its migration direction.

Explanation: This question tests understanding of how pH affects protein charge and migration in native PAGE electrophoresis. In electrophoresis, charged molecules migrate through a gel matrix when an electric field is applied - positively charged molecules migrate toward the cathode (negative electrode) while negatively charged molecules migrate toward the anode (positive electrode). Since Protein Bdp is described as a basic DNA-binding protein, it has a high isoelectric point (pI) and carries a net positive charge at both pH 8.8 and pH 6.0. As pH decreases from 8.8 to 6.0, more acidic groups on the protein become protonated, increasing the net positive charge on Protein Bdp. The increased positive charge at pH 6.0 causes stronger electrostatic attraction to the cathode, explaining why the protein migrates even farther toward the cathode at the lower pH. Choice B incorrectly states that the protein becomes more negatively charged at lower pH, which contradicts fundamental acid-base chemistry where proteins gain positive charge as pH decreases below their pI. To verify protein charge in electrophoresis, remember that migration direction indicates charge sign (cathode = positive, anode = negative) and migration distance reflects charge magnitude.

Question 18

1. C1 is most net negative, C2 is near neutral, and C3 is net positive or near neutral. (correct answer)
2. C3 is most net negative because its pI is highest.
3. C2 is most net negative because its pI is closest to pH 6.5.
4. All three are net negative because they migrated at least slightly toward the anode.

Explanation: This question tests net charge deduction from native PAGE data, using pI relative to pH and migration. In electrophoresis, at pH near pI, charge is low; below pI positive, above negative, with distance reflecting charge magnitude for similar MW. In this pH 6.5 setup, farther anode migration indicates more negative charge. The correct answer A is supported as C1 (pI 5.0 < 6.5, most negative, 4.8 cm), C2 (pI 6.4 ≈ 6.5, near neutral, 1.0 cm), C3 (pI 8.2 > 6.5, positive or neutral, 0.2 cm). Distractor B fails by claiming C3 (highest pI) most negative, misunderstanding high pI means less negative at low pH. To verify, adjust pH to 9.0 and check reversed order. Calculating charge differences can confirm migration patterns match expected net charges.

Question 19

1. H1 is more net negative than H2 at pH 8.6. (correct answer)
2. H2 is more net negative than H1 at pH 8.6.
3. H1 is net positive at pH 8.6, explaining its greater migration toward the anode.
4. H1 and H2 have identical net charge because they are both hemoglobin.

Explanation: This question assesses relative charge comparison in native PAGE for hemoglobin variants, based on pI and migration. In electrophoresis, at pH above pI, lower pI means more negative charge, leading to farther anode migration. In this pH 8.6 setup, H1 (pI 7.1 < 8.6) migrated farther (3.9 cm) than H2 (pI 8.9 > 8.6, 0.6 cm). The correct answer A follows as H1 is more net negative than H2. Distractor C fails by claiming H1 net positive drives anode migration, confusing positive charge's cathode direction. To verify, test at pH 10.0 to enhance both negativities and check distances. Comparing with pI standards can confirm lower pI correlates with greater negative charge at high pH.

Question 20

1. The 120 kDa species likely contained disulfide-linked subunits that separated into ~60 kDa monomers upon reduction. (correct answer)
2. Reduction increases net positive charge, slowing migration and splitting bands.
3. SDS-PAGE separates by pI, so reduction changed pI and created two bands.
4. β-mercaptoethanol polymerized the proteins, doubling their MW and causing band splitting.

Explanation: This question tests interpretation of reducing agents' effects in SDS-PAGE, focusing on disulfide bond disruption. In electrophoresis, SDS-PAGE separates by MW, and reducing agents like β-mercaptoethanol break disulfides, dissociating multimers into subunits. In this setup, the shift from 120 kDa to 60 kDa bands upon reduction indicates dimer dissociation. The correct answer A is supported as the 120 kDa species was likely disulfide-linked dimers that separated into 60 kDa monomers. Distractor D fails by claiming β-mercaptoethanol polymerizes proteins, a misconception since it reduces, not forms, disulfides. To verify, analyze non-reduced samples with cross-linkers to confirm multimer status. Comparing reduced vs. non-reduced MW can identify disulfide-dependent structures in other proteins.