MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5c Chromatography Techniques

What this quiz covers

This quiz focuses on 5c Chromatography Techniques, 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. The longer column is likely to provide better resolution because analytes have more opportunity to partition (correct answer)
2. The shorter column is likely to provide better resolution because analytes spend less time diffusing
3. Both columns must give identical resolution because stationary phase chemistry is the same
4. Column length only affects retention time, not resolution, in chromatography

Explanation: This question tests understanding of how column length affects chromatographic resolution. Resolution depends on the number of theoretical plates, which increases with column length when all other factors are equal. The longer column (30 cm) provides more opportunities for repeated partitioning between mobile and stationary phases, leading to better separation of compounds with similar retention properties. While longer columns do increase analysis time, they improve resolution by allowing small differences in partitioning to be amplified over the greater distance. Choice B incorrectly suggests that longer residence time leads to worse resolution due to diffusion - while diffusion does occur, the increased partitioning opportunities outweigh this effect. The key principle is that resolution generally improves with column length, though practical considerations like analysis time and pressure limits must be balanced.

Question 2

1. Compound T, because higher boiling point increases vapor pressure and speeds elution
2. Compound S, because lower boiling point generally corresponds to higher volatility and shorter retention (correct answer)
3. They elute together because polarity is matched to the stationary phase
4. Compound S elutes last because smaller molecules interact more strongly with nonpolar phases

Explanation: This question tests understanding of gas chromatography separation based on volatility and boiling points. In GC, compounds are separated based on their partitioning between the gas phase and the stationary phase, which is strongly influenced by volatility. At a column temperature of 100°C, Compound S (bp 80°C) will be completely vaporized and highly volatile, spending more time in the gas phase and less time interacting with the stationary phase. Compound T (bp 140°C) will be less volatile at 100°C, condensing more on the stationary phase and being retained longer. Choice A incorrectly relates higher boiling point to higher vapor pressure - compounds with higher boiling points have lower vapor pressures at a given temperature. The key principle is that in GC, lower boiling point compounds are more volatile and elute before higher boiling point compounds.

Question 3

1. A new spot appears with higher $R_f$ than the starting alcohol spot (correct answer)
2. A new spot appears with lower $R_f$ than the starting alcohol spot
3. The starting alcohol spot moves to the solvent front because it is polar
4. No spots appear because silica quenches UV absorption

Explanation: This question tests understanding of TLC monitoring for reaction progress based on polarity changes. In normal-phase TLC on silica, less polar compounds have higher Rf values because they interact less with the polar stationary phase. The esterification converts a polar alcohol (with -OH group) to a less polar ester (with -COOR group). Therefore, successful product formation would show a new spot with higher Rf than the starting alcohol. Choice B incorrectly predicts lower Rf for the less polar product, contradicting normal-phase principles. The key concept is that in normal-phase TLC, decreasing polarity leads to increasing Rf values, making this technique ideal for monitoring reactions that change compound polarity.

Question 4

1. Protein Y binds strongly to the cation exchanger and elutes last
2. Protein Z binds to the cation exchanger and elutes at higher salt than Protein Y (correct answer)
3. Both proteins fail to bind because NaCl is required for adsorption
4. Protein Z elutes first because higher $pI$ means more negative charge at pH 7.4

Explanation: This question tests understanding of ion-exchange chromatography and the relationship between protein pI and charge. At pH 7.4, proteins with pI > 7.4 carry net positive charge, while those with pI < 7.4 carry net negative charge. Since Protein Z has pI = 9.2, it is positively charged at pH 7.4 and will bind to the negatively charged cation exchanger. Protein Y with pI = 5.1 is negatively charged at pH 7.4 and will not bind, flowing through immediately. When NaCl concentration increases, the salt ions compete with Protein Z for binding sites, eventually eluting it. Choice D incorrectly states that higher pI means more negative charge, which is the opposite of the correct relationship. The key principle is that proteins bind to ion exchangers when their net charge is opposite to the resin's charge.

Question 5

1. More polar compounds elute earlier because the mobile phase becomes less polar over time
2. Less polar compounds are delayed because increasing ethyl acetate strengthens adsorption to silica
3. More strongly retained (more polar) compounds can be eluted later as the mobile phase becomes more polar (correct answer)
4. All compounds elute at the same time because gradient elution eliminates retention differences

Explanation: This question tests understanding of gradient elution in normal-phase chromatography. In normal-phase silica chromatography, polar compounds are retained more strongly than nonpolar compounds. Starting with pure hexane (nonpolar) will elute the least polar compounds first, but very polar compounds may remain strongly bound. By gradually increasing the ethyl acetate content (making the mobile phase more polar), the eluting power increases, allowing more strongly retained polar compounds to be eluted. The carboxylic acid (O) is most polar and will elute last when sufficient ethyl acetate is present. Choice A incorrectly reverses the elution order, while choice B misunderstands how polarity affects silica adsorption. The key strategy in gradient elution is to start with weak eluent for less retained compounds and gradually increase eluent strength for more retained compounds.

Question 6

1. Increase the ethyl acetate fraction to move both spots upward, with a larger shift for the polar product
2. Decrease the ethyl acetate fraction to reduce overall elution strength and increase retention differences on silica (correct answer)
3. Switch to distillation, since boiling point differences directly increase TLC resolution
4. Add more sample to the baseline, since higher loading increases $R_f$ differences between SM and P

Explanation: This question tests understanding of how to optimize TLC separation by adjusting mobile phase polarity. The current separation shows the nonpolar starting material (SM) with Rf = 0.80 and the polar product (P) with Rf = 0.30, giving a ΔRf of 0.50. To improve separation (increase the distance between spots), we need to increase the difference in how strongly each compound interacts with the stationary phase. Decreasing the ethyl acetate fraction makes the mobile phase less polar, which reduces its ability to compete with silica for both compounds. This causes both compounds to be retained more strongly, lowering both Rf values, but the effect is more pronounced for the polar product, which already has strong interactions with silica. This increases the relative difference in retention between SM and P, improving their separation on the plate. A common misconception is that increasing mobile phase polarity improves separation (choice A), but this would cause both spots to move up the plate with potentially less differentiation. To optimize TLC separation, remember: decreasing mobile phase polarity in normal-phase TLC increases retention differences between compounds of different polarities.

Question 7

1. Aniline (AN), because its amine group is strongly solvated by hexane and moves fastest
2. Ethyl acetate (EA), because it is the least polar solvent and therefore elutes first
3. Hexane (H), because it is least polar and interacts weakest with the polar stationary phase (correct answer)
4. Acetophenone (AP), because aromatic rings have low affinity for silica and elute before alkanes

Explanation: This question tests understanding of normal-phase column chromatography principles, specifically how compound polarity affects elution order. In normal-phase chromatography with a polar stationary phase (silica gel) and nonpolar mobile phase (100% hexane), compounds are separated based on their differential interactions with the stationary phase. Since hexane is the least polar compound in the mixture, it has the weakest interaction with the polar silica gel and will spend the least time adsorbed to the stationary phase. The more polar compounds (EA, AP, and AN) will interact more strongly with silica through hydrogen bonding and dipole-dipole interactions, causing them to be retained longer on the column. A common misconception is that the mobile phase composition determines which compound elutes first (choice B), but in reality, it's the relative affinity for the stationary phase that matters. To predict elution order in normal-phase chromatography, remember: less polar compounds elute first with nonpolar mobile phases, while more polar compounds are retained longer due to stronger interactions with the polar stationary phase.

Question 8

1. Both $R_f$ values decrease because a more polar mobile phase increases adsorption to silica
2. Both $R_f$ values increase, with a larger increase for the more polar compound X (correct answer)
3. Only compound Y's $R_f$ increases because nonpolar compounds respond more to polar solvents
4. Both $R_f$ values remain unchanged because $R_f$ depends only on the stationary phase

Explanation: This question tests understanding of how mobile phase polarity affects Rf values in thin-layer chromatography. In TLC with a polar stationary phase (silica), the Rf value represents the ratio of compound migration distance to solvent front distance, which depends on the balance between compound interactions with the stationary and mobile phases. When switching from a less polar mobile phase (8:2 hexane:ethyl acetate) to a more polar one (2:8 hexane:ethyl acetate), the increased polarity of the mobile phase competes more effectively with the silica for interaction with both compounds. This increased competition causes both compounds to spend less time adsorbed to the stationary phase and more time dissolved in the mobile phase, resulting in higher Rf values for both compounds. The more polar compound X will show a larger increase because the polar mobile phase provides stronger solvation for polar compounds. A common misconception is that increasing mobile phase polarity would decrease Rf values (choice A), but this confuses the effect with increasing stationary phase interactions. To predict Rf changes, remember: increasing mobile phase polarity in normal-phase TLC increases Rf values, with larger increases for more polar compounds.

Question 9

1. The relative migration order is expected to invert: D tends to have lower $R_f$ than E on silica but higher $R_f$ than E on C18 (correct answer)
2. The relative migration order must remain the same because $R_f$ depends only on molecular weight
3. Both compounds will have $R_f \approx 1$ on C18 because nonpolar stationary phases do not retain analytes
4. D will always have lower $R_f$ than E on both plates because polarity always decreases $R_f$

Explanation: This question compares migration order in normal- vs. reverse-phase TLC. Normal-phase on silica retains polar compounds more (lower Rf for D), while reverse-phase on C18 retains nonpolar more (higher Rf for polar D). Switching to C18 in the same solvent inverts the order: D (polar) has higher Rf than E. This reflects reversed polarity affinities. Choice B is incorrect as Rf depends on polarity interactions, not just weight. For similar comparisons, predict order reversal. A key strategy is to note phase type dictates whether polar or nonpolar elutes faster.

Question 10

1. A stronger eluent can reduce differences in retention factors between analytes, compressing elution times and lowering resolution (correct answer)
2. Increasing polarity increases stationary-phase binding, so peaks should spread out and resolve better
3. Resolution decreases because ethyl acetate increases boiling points, causing co-elution
4. Resolution decreases only when the stationary phase is nonpolar, so this cannot occur on silica

Explanation: This question evaluates how mobile phase strength affects resolution in normal-phase chromatography. In silica columns, a stronger (more polar) eluent reduces retention factors and compresses elution ranges, potentially decreasing resolution by minimizing retention differences. Increasing ethyl acetate causes faster elution but overlaps peaks. This occurs as the eluent overpowers selectivity. Choice B fails because higher polarity weakens, not strengthens, binding to silica, leading to compression. In similar optimizations, balance eluent strength. A strategy is to use weaker eluents for better resolution when peaks are close.

Question 11

1. Higher temperature can reduce viscosity and weaken partitioning into the stationary phase, decreasing retention times (correct answer)
2. Higher temperature increases stationary-phase polarity, increasing retention and decreasing retention times
3. Retention time changes indicate the steroids were chemically converted to more polar products
4. Temperature cannot affect HPLC retention times; only flow rate can

Explanation: This question examines temperature effects on retention in reverse-phase HPLC. In reverse-phase HPLC, higher temperature reduces mobile phase viscosity and weakens analyte-stationary phase interactions, often decreasing retention times. Raising from 25°C to 40°C shortens retention for both steroids without changing resolution significantly. This is due to enhanced mass transfer and reduced partitioning. Choice D is incorrect because temperature does affect retention via physical properties, not just flow rate. In similar studies, monitor retention shifts with temperature. A key strategy is to use temperature to fine-tune elution without altering selectivity.

Question 12

1. R has a longer retention time than S because it interacts more strongly with the polar stationary phase (correct answer)
2. R elutes first because polar compounds always have lower boiling points
3. S has a longer retention time because nonpolar compounds bind more strongly to polar stationary phases
4. They cannot be separated because polarity does not affect GC retention

Explanation: This question evaluates retention in gas chromatography with polar stationary phases. In GC, polar stationary phases retain polar compounds longer via stronger interactions, despite similar boiling points. With a polar phase, the more polar R interacts more than S, leading to longer retention. This results in R eluting after S under isothermal conditions. Choice C is incorrect as it claims nonpolar bind more to polar phases, reversing the 'like attracts like' principle. For similar GC analyses, consider polarity match to stationary phase. A key check is to predict longer retention for analytes matching stationary phase polarity.

Question 13

1. The ester is more polar than the alcohol because it traveled farther on silica
2. The ester interacts less strongly with silica than the alcohol in this solvent system (correct answer)
3. The alcohol has a higher $R_f$ than the ester because it is the starting material
4. The ester must have a higher boiling point than the alcohol because it has a higher $R_f$

Explanation: This question tests interpretation of Rf values in normal-phase TLC on silica and their relation to molecular interactions. Thin-layer chromatography on silica, a polar stationary phase, retains compounds capable of hydrogen bonding or polar interactions more strongly, leading to lower Rf in less polar solvents. In this 80:20 hexanes:ethyl acetate system, the ester (Rf 0.55) travels farther than the alcohol (Rf 0.15), indicating weaker silica interactions. The ester interacts less because it lacks the alcohol's strong hydrogen-bonding OH group, making it less polar and more mobile. Choice A is incorrect as it claims the ester is more polar for traveling farther, ignoring that lower polarity reduces retention on silica. For similar TLC analyses, compare functional groups for polarity and H-bonding potential. A transferable check is to ensure Rf increases with decreasing analyte-stationary phase affinity in normal-phase setups.

Question 14

1. 300 Da inhibitor, because small molecules diffuse faster and therefore elute first
2. 60 kDa protein, because it partially enters pores and therefore elutes before excluded species
3. 150 kDa enzyme, because it is excluded from pores and travels the shortest path (correct answer)
4. All components elute together because SEC separates only by charge

Explanation: This question examines separation principles in size-exclusion chromatography based on molecular size. Size-exclusion chromatography separates by size, with larger molecules excluded from resin pores eluting first at the void volume, while smaller ones enter pores and elute later. The resin here excludes >100 kDa, so the 150 kDa enzyme is fully excluded, the 60 kDa protein partially enters, and the 300 Da inhibitor fully enters. Thus, the 150 kDa enzyme elutes first, traveling the shortest path without pore entry. Choice A fails because small molecules like the 300 Da inhibitor elute last in SEC, not first, due to longer paths through pores, not faster diffusion. In similar SEC problems, compare analyte sizes to the exclusion limit. A useful strategy is to visualize elution as inverse to pore accessibility: excluded = early, fully included = late.

Question 15

1. Acidic additives typically decrease tailing by suppressing strong adsorption sites, often causing acids to elute earlier and with sharper bands (correct answer)
2. Acetic acid makes the mobile phase nonpolar, increasing retention of all compounds
3. Acetic acid converts the separation into distillation, so elution order depends on boiling point
4. Acetic acid permanently deactivates silica, preventing any separation

Explanation: This question probes the effect of acidic modifiers in normal-phase chromatography on weak acid separation. In silica column chromatography, polar stationary phases can cause tailing for acids due to strong adsorption or ionization, but acidic additives like acetic acid compete for sites and suppress ionization. Adding 1% acetic acid to the hexanes:ethyl acetate mobile phase reduces tailing by weakening acid-silica interactions. This often leads to earlier elution and sharper peaks for the weak acids. Choice B is wrong because acetic acid increases mobile phase polarity, not decreases it, and would typically decrease retention, not increase it. For similar modifier questions, consider how additives alter analyte-stationary phase interactions. A reasoning strategy is to predict if the additive matches or competes with analyte properties for better band shape.

Question 16

1. P elutes before Q because it spends more time in the gas phase (correct answer)
2. P elutes after Q because low boiling point compounds condense more strongly on the stationary phase
3. P and Q cannot be separated by GC because GC separates only by polarity, not volatility
4. Q elutes first because stronger intermolecular forces reduce retention time

Explanation: This question tests retention principles in gas chromatography based on volatility. In GC with a nonpolar stationary phase, retention depends on analyte volatility and interactions; lower boiling point compounds spend more time in the gas phase and elute earlier under isothermal conditions. Component P, with lower boiling point and weaker forces, interacts less with the stationary phase than Q. Therefore, P elutes before Q, as it partitions less into the liquid phase. Choice B is incorrect because low boiling point compounds actually elute earlier, not later, due to higher volatility, not stronger condensation. For similar GC problems, correlate retention with boiling points for nonpolar phases. A key strategy is to consider time in mobile (gas) vs. stationary phase: more volatile = more gas time = earlier elution.

Question 17

1. Protein B binds more strongly because at pH 7.4 it is positively charged
2. Protein A binds more strongly because at pH 7.4 it is net positively charged (correct answer)
3. Neither protein binds because ion-exchange requires organic solvents
4. Both proteins bind equally because binding depends only on molecular weight

Explanation: This question tests binding in ion-exchange chromatography based on protein charge at given pH. Cation-exchange resins (negatively charged) bind positively charged proteins, with binding strength related to net positive charge; elution uses increasing salt. At pH 7.4, Protein A (pI 9.5) is net positive and binds, while Protein B (pI 5.0) is net negative and does not. Thus, Protein A binds more strongly, requiring higher salt for elution. Choice A is wrong because Protein B is negatively charged and repelled, not bound, by the negative resin. In similar problems, compare pI to buffer pH for net charge. A key strategy is: for cation-exchange, proteins with pI > pH bind; stronger binding means later elution.

Question 18

1. Decreasing mobile-phase polarity decreased $R_f$ because the pigment is less soluble in the mobile phase (correct answer)
2. Decreasing mobile-phase polarity increased $R_f$ because the pigment binds less to paper
3. The change in $R_f$ indicates the pigment's molecular weight changed between runs
4. The solvent front distance determines $R_f$ independent of compound behavior, so $R_f$ should not change

Explanation: This question assesses how solvent polarity affects Rf in paper chromatography. Paper chromatography uses polar cellulose, where less polar solvents reduce compound solubility in the mobile phase, increasing retention and lowering Rf. Switching to a less polar solvent decreases chlorophyll b's Rf from 0.3 to 0.15, as it partitions more into the stationary phase. This occurs because the pigment is less solvated by the nonpolar mobile phase. Choice B errs by claiming decreased polarity increases Rf, ignoring reduced solvation's effect on retention. For similar experiments, note Rf changes with mobile phase polarity. A transferable strategy is to predict Rf decreases when mobile-stationary polarity mismatch increases for the analyte.

Question 19

1. Early fractions near the void volume, where excluded large proteins elute before small ions (correct answer)
2. Late fractions, because large proteins enter pores and are delayed more than ions
3. Middle fractions, because proteins and salts co-elute at the resin's fractionation midpoint
4. No fractions, because SEC cannot separate macromolecules from small ions

Explanation: This question tests desalting via size-exclusion chromatography for macromolecules. SEC separates by size, with large molecules like the 50 kDa enzyme excluded from 1–10 kDa pores, eluting early at void volume, while small salts enter and elute late. The enzyme appears in early fractions, separated from delayed salts. This achieves desalting as the protein travels faster. Choice B is wrong because large proteins are excluded and elute early, not late, contrary to pore entry delaying them. For similar applications, check analyte size vs. fractionation range. A transferable strategy is to collect void volume fractions for excluded species in desalting.

Question 20

1. Retention time decreases because the mobile phase is more aqueous
2. Retention time increases because the mobile phase is more polar and the analyte partitions more into C18 (correct answer)
3. Retention time is unchanged because mobile-phase composition does not affect reverse-phase separations
4. The analyte elutes first because increased water makes C18 more polar

Explanation: This question tests understanding of mobile phase composition effects in reverse-phase HPLC. In reverse-phase chromatography, increasing the water content (from 70:30 to 90:10 water/acetonitrile) makes the mobile phase more polar. This increases the polarity difference between the mobile phase and the nonpolar C18 stationary phase. Hydrophobic analytes will partition more strongly into the C18 phase to avoid the more aqueous mobile phase, resulting in increased retention time. Choice A incorrectly predicts decreased retention with more aqueous conditions, contradicting reverse-phase principles. The key principle is that in reverse-phase HPLC, increasing water content increases retention of hydrophobic compounds, while increasing organic content decreases retention.