Microbiology Quiz: Transport Mechanisms
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Transport MechanismsQuestion 1 of 20

Consider the transport of a small, uncharged molecule across a bacterial cell membrane. Which of the following changes would most likely increase the rate of simple diffusion of this molecule into the cell?

Adding a non-competitive inhibitor that binds to protein transporters.
Increasing the number of specific carrier proteins for the molecule in the membrane.
Increasing the degree of saturation of the fatty acid tails in the membrane phospholipids.
Increasing the concentration of the molecule in the external medium.
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Microbiology Quiz

Microbiology Quiz: Transport Mechanisms

Practice Transport Mechanisms in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transport Mechanisms, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

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.

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Question 1

Consider the transport of a small, uncharged molecule across a bacterial cell membrane. Which of the following changes would most likely increase the rate of simple diffusion of this molecule into the cell?

  1. Adding a non-competitive inhibitor that binds to protein transporters.
  2. Increasing the number of specific carrier proteins for the molecule in the membrane.
  3. Increasing the degree of saturation of the fatty acid tails in the membrane phospholipids.
  4. Increasing the concentration of the molecule in the external medium. (correct answer)
Explanation: The rate of simple diffusion is governed by Fick's Law, which states that the rate is directly proportional to the concentration gradient across the membrane. By increasing the external concentration of the molecule, the concentration gradient becomes steeper, driving more molecules into the cell per unit of time. A and B are incorrect because simple diffusion does not involve protein transporters. C is incorrect because increasing the saturation of fatty acid tails makes the membrane more viscous and less permeable, which would decrease the rate of simple diffusion.

Question 2

A bacterium possesses a membrane transporter for the amino acid leucine. This transporter is competitively inhibited by the presence of isoleucine. In an environment with a low concentration of leucine, how would the addition of a moderate concentration of isoleucine affect the kinetics of leucine transport?

  1. It would decrease the Vmax of leucine transport but not affect the Km.
  2. It would increase the Km of leucine transport but not affect the Vmax. (correct answer)
  3. It would decrease both the Vmax and the Km of leucine transport.
  4. It would have no effect on transport kinetics as it is a different amino acid.
Explanation: This question applies the principles of enzyme kinetics to transport proteins. Competitive inhibitors bind to the same active site as the substrate. In this case, isoleucine competes with leucine for the transporter's binding site. This increases the apparent substrate concentration required to reach half-maximal velocity, meaning the Km (a measure of affinity) increases. However, if the leucine concentration is raised high enough, it can outcompete the inhibitor and the transporter can still reach its normal maximum velocity (Vmax). Therefore, competitive inhibition increases Km but does not change Vmax. A describes non-competitive inhibition. C describes uncompetitive inhibition. D is incorrect as competitive inhibition is a well-established phenomenon for transporters with specificity for similar substrates.

Question 3

The transport of substance P into a bacterial cell is coupled to the transport of substance Q out of the cell. The concentration of P is higher outside the cell than inside, while the concentration of Q is higher inside the cell than outside. The transport process requires no direct input of metabolic energy such as ATP. What type of transport mechanism is described?

  1. A symporter using secondary active transport.
  2. An antiporter functioning by facilitated diffusion. (correct answer)
  3. A primary active transport pump.
  4. Two separate channels for P and Q.
Explanation: The transporter moves two substances in opposite directions, so it is an antiporter. Both substances are moving down their respective concentration gradients (P from high outside to low inside; Q from high inside to low outside). Since the movement for both substrates is energetically favorable and no metabolic energy (like ATP) is required, the mechanism is a form of facilitated diffusion. It is not active transport because no substance is being moved against its gradient. A is incorrect because a symporter moves substances in the same direction. C is incorrect because there is no direct energy input. D is incorrect because the transport of P and Q are described as being coupled, implying a single protein.

Question 4

A halophilic (salt-loving) bacterium maintains osmotic balance in a high-salt environment by accumulating high internal concentrations of a compatible solute, such as glycine betaine. This import process is inhibited by compounds that dissipate the membrane's sodium gradient.

Given this information, the uptake of glycine betaine is most likely mediated by which of the following?

  1. A Na+/glycine betaine symporter. (correct answer)
  2. A K+/glycine betaine antiporter.
  3. An ATP-dependent primary transporter.
  4. A group translocation system specific for glycine betaine.
Explanation: The bacterium is accumulating glycine betaine against a concentration gradient, which indicates active transport. The finding that this process is inhibited by dissipating the sodium gradient strongly implicates this gradient as the energy source. This is the defining feature of secondary active transport. Since glycine betaine is being imported, and marine bacteria typically have a strong Na+ gradient directed inwards, the most plausible mechanism is a symporter that couples the favorable influx of Na+ to the unfavorable influx of glycine betaine. B is incorrect because a K+ gradient is not implicated. C is incorrect because the energy source is the Na+ gradient, not ATP directly. D is incorrect as there is no evidence of chemical modification.

Question 5

An experiment measuring the uptake of an uncharged molecule into a bacterium shows that the transport rate reaches a maximum velocity (Vmax) at high external concentrations. This observation alone is insufficient to distinguish between facilitated diffusion and active transport. Which of the following additional findings would definitively point to active transport?

  1. The transport rate decreases significantly when the temperature is lowered from 37°C to 20°C.
  2. A structurally analogous molecule is found to competitively inhibit the transport of the original molecule.
  3. The steady-state intracellular concentration of the molecule exceeds the extracellular concentration. (correct answer)
  4. The transport protein is found to be an integral membrane protein spanning the lipid bilayer.
Explanation: Saturation kinetics (reaching a Vmax) is a property of both facilitated diffusion and active transport, as both rely on a finite number of carrier proteins. The defining difference is that active transport can move substances against their concentration gradient, while facilitated diffusion cannot. Therefore, observing that the molecule is accumulated inside the cell to a concentration higher than the outside medium is conclusive evidence for active transport. A is not definitive, as both processes are enzyme-catalyzed and temperature-dependent. B is also not definitive, as competitive inhibition is characteristic of any specific binding site on a protein, common to both mechanisms. D is true for both types of transporters and thus does not distinguish them.

Question 6

A bacterium has two different transport systems for an essential nutrient, Nutrient X. Transporter A has a high affinity (Km = 5 µM) and a low maximum velocity (Vmax). Transporter B has a low affinity (Km = 500 µM) and a high maximum velocity. Which statement accurately describes the physiological roles of these transporters?

  1. Transporter A is most effective in nutrient-rich environments, while Transporter B is essential in nutrient-poor environments.
  2. Transporter B is an active transport system, while Transporter A functions by facilitated diffusion, explaining the difference in Vmax.
  3. Transporter A is likely the primary system used for scavenging the nutrient in oligotrophic (low-nutrient) conditions. (correct answer)
  4. Both transporters would be equally effective regardless of the external concentration of Nutrient X.
Explanation: A low Km value indicates a high affinity of the transporter for its substrate. This means Transporter A can bind and transport Nutrient X efficiently even when it is present at very low concentrations. This makes it ideal for scavenging nutrients in oligotrophic (nutrient-poor) environments. Transporter B, with its high Km, requires a much higher concentration of the nutrient to function effectively, but its high Vmax allows for rapid uptake when the nutrient is abundant. Therefore, C is the correct statement. A is the reverse of the correct situation. B makes an unsubstantiated claim about the transport mechanism; kinetic parameters alone do not distinguish between active transport and facilitated diffusion. D is incorrect as their effectiveness is clearly concentration-dependent.

Question 7

The bacterial phosphotransferase system (PTS) is responsible for transporting many carbohydrates, including glucose. A key feature of this system is that the transported sugar is chemically altered as it crosses the cell membrane. Which of the following statements most accurately describes the advantage of this modification?

  1. The modification allows the sugar to be transported by simple diffusion out of the cell if its internal concentration becomes too high.
  2. The phosphorylation of the sugar traps it inside the cell, maintaining a steep concentration gradient for the unmodified sugar. (correct answer)
  3. The chemical alteration of the sugar makes it unrecognizable to other transport proteins, ensuring a single pathway for its metabolism.
  4. The addition of a phosphate group directly energizes the transport protein, functioning similarly to ATP hydrolysis in primary active transport.
Explanation: In the PTS, glucose is converted to glucose-6-phosphate during transport. Because the transporter is specific for glucose, not glucose-6-phosphate, the internal concentration of the transported substrate (glucose) remains very low. This maintains a steep concentration gradient favoring the continued influx of glucose from the outside. The phosphorylation effectively 'traps' the sugar inside in a modified form. A is incorrect because the phosphorylated sugar is highly charged and cannot exit via simple diffusion. C is a potential side-effect but not the primary advantage. D is incorrect because the energy for the process comes from the high-energy phosphate bond of phosphoenolpyruvate (PEP), not from the modification energizing the protein in the manner of ATP hydrolysis.

Question 8

A bacterium possesses a membrane transporter for the amino acid leucine. This transporter is competitively inhibited by the presence of isoleucine. In an environment with a low concentration of leucine, how would the addition of a moderate concentration of isoleucine affect the kinetics of leucine transport?

  1. It would decrease the Vmax of leucine transport but not affect the Km.
  2. It would increase the Km of leucine transport but not affect the Vmax. (correct answer)
  3. It would decrease both the Vmax and the Km of leucine transport.
  4. It would have no effect on transport kinetics as it is a different amino acid.
Explanation: This question applies the principles of enzyme kinetics to transport proteins. Competitive inhibitors bind to the same active site as the substrate. In this case, isoleucine competes with leucine for the transporter's binding site. This increases the apparent substrate concentration required to reach half-maximal velocity, meaning the Km (a measure of affinity) increases. However, if the leucine concentration is raised high enough, it can outcompete the inhibitor and the transporter can still reach its normal maximum velocity (Vmax). Therefore, competitive inhibition increases Km but does not change Vmax. A describes non-competitive inhibition. C describes uncompetitive inhibition. D is incorrect as competitive inhibition is a well-established phenomenon for transporters with specificity for similar substrates.

Question 9

The transport of substance P into a bacterial cell is coupled to the transport of substance Q out of the cell. The concentration of P is higher outside the cell than inside, while the concentration of Q is higher inside the cell than outside. The transport process requires no direct input of metabolic energy such as ATP. What type of transport mechanism is described?

  1. A symporter using secondary active transport.
  2. An antiporter functioning by facilitated diffusion. (correct answer)
  3. A primary active transport pump.
  4. Two separate channels for P and Q.
Explanation: The transporter moves two substances in opposite directions, so it is an antiporter. Both substances are moving down their respective concentration gradients (P from high outside to low inside; Q from high inside to low outside). Since the movement for both substrates is energetically favorable and no metabolic energy (like ATP) is required, the mechanism is a form of facilitated diffusion. It is not active transport because no substance is being moved against its gradient. A is incorrect because a symporter moves substances in the same direction. C is incorrect because there is no direct energy input. D is incorrect because the transport of P and Q are described as being coupled, implying a single protein.

Question 10

Valinomycin is an ionophore that specifically binds K+ and facilitates its diffusion across biological membranes. A typical bacterial cell maintains a high internal concentration of K+ relative to the external environment. What is the immediate effect of adding valinomycin to a suspension of these bacteria?

  1. A rapid influx of K+ into the cells, leading to an increase in membrane potential (hyperpolarization).
  2. A rapid efflux of K+ from the cells, leading to a decrease in membrane potential (depolarization). (correct answer)
  3. The cell will actively pump valinomycin out using an ABC transporter, preventing any effect on K+ levels.
  4. The intracellular pH will decrease as K+ is exchanged for H+ ions to maintain charge balance.
Explanation: The bacterium has a high internal K+ concentration. Valinomycin creates a new pathway for K+ to move across the membrane via facilitated diffusion. K+ will therefore move down its concentration gradient, from inside the cell to the outside. Since K+ is a positive ion, this efflux of positive charge will make the inside of the cell more negative relative to the outside, causing a decrease in membrane potential (depolarization). A is incorrect because K+ will move out, not in. C is irrelevant to the direct physicochemical effect of the ionophore. D is incorrect; while charge balance must be maintained, the primary immediate effect is on the membrane potential, not necessarily a 1:1 exchange with H+.

Question 11

A bacterial Na+/Ca2+ exchanger protein moves Na+ ions into the cell and Ca2+ ions out of the cell. The concentration of Na+ is much higher outside the cell than inside, while the concentration of Ca2+ is much lower outside the cell than inside. Which statement best describes this transport process?

  1. This is a primary active transport process where ATP is used to pump both ions.
  2. This is a form of facilitated diffusion, as the movement of Na+ drives the movement of Ca2+.
  3. This is a symport mechanism because two different cations are being transported simultaneously.
  4. This is an example of secondary active transport, where the Ca2+ is moved against its gradient. (correct answer)
Explanation: When you encounter transport proteins moving multiple substances across membranes, focus on the energy source and direction relative to concentration gradients to classify the mechanism correctly. This Na⁺/Ca²⁺ exchanger represents secondary active transport. The key insight is recognizing that Na⁺ moves down its concentration gradient (high outside → low inside), releasing energy that drives Ca²⁺ movement against its gradient (high inside → low outside). The Na⁺ gradient, established by primary active transport elsewhere, provides the driving force. Since Ca²⁺ moves from low to high concentration, energy is required, making this active transport. Because the energy comes from an ion gradient rather than direct ATP hydrolysis, it's secondary active transport. Answer A is incorrect because no ATP is directly used in this exchanger's operation. Primary active transport requires direct ATP hydrolysis, which isn't happening here. Answer B misclassifies this as facilitated diffusion. While Na⁺ does move down its gradient, Ca²⁺ moves against its gradient, which requires energy input and cannot be passive transport. Answer C incorrectly identifies this as symport. This is actually antiport (countertransport) because the ions move in opposite directions - Na⁺ enters while Ca²⁺ exits. Symport would require both ions moving in the same direction. Remember: Secondary active transport always involves one substance moving down its gradient to power another substance's movement against its gradient. Look for the coupling of favorable and unfavorable movements to identify these mechanisms on exams.

Question 12

Consider the transport of a small, uncharged molecule across a bacterial cell membrane. Which of the following changes would most likely increase the rate of simple diffusion of this molecule into the cell?

  1. Adding a non-competitive inhibitor that binds to protein transporters.
  2. Increasing the number of specific carrier proteins for the molecule in the membrane.
  3. Increasing the degree of saturation of the fatty acid tails in the membrane phospholipids.
  4. Increasing the concentration of the molecule in the external medium. (correct answer)
Explanation: The rate of simple diffusion is governed by Fick's Law, which states that the rate is directly proportional to the concentration gradient across the membrane. By increasing the external concentration of the molecule, the concentration gradient becomes steeper, driving more molecules into the cell per unit of time. A and B are incorrect because simple diffusion does not involve protein transporters. C is incorrect because increasing the saturation of fatty acid tails makes the membrane more viscous and less permeable, which would decrease the rate of simple diffusion.

Question 13

The LacY permease of E. coli is a symporter that uses the proton motive force to drive the uptake of lactose against its concentration gradient. A mutation in the gene encoding the primary proton pump of the cell leads to a significantly reduced proton gradient across the membrane. What is the most likely immediate consequence for lactose transport via LacY?

  1. Lactose transport will reverse direction, actively pumping lactose out of the cell to restore the proton gradient.
  2. Lactose transport will cease entirely, as the transporter is irreversibly damaged by the change in membrane potential.
  3. The transporter will continue to import lactose, but will now hydrolyze ATP to compensate for the loss of the proton gradient.
  4. The rate of lactose accumulation against its concentration gradient will be severely reduced or eliminated. (correct answer)
Explanation: Secondary active transporters like LacY are directly powered by an ion gradient (in this case, protons). If the primary pump that generates this gradient is compromised, the energy source for LacY is diminished. Consequently, the ability of LacY to pump lactose against a concentration gradient will be severely reduced or abolished. The transporter may still facilitate lactose movement down a gradient, but the 'active' component of its function will be lost. A is incorrect because the transporter is not designed to run in reverse to create a proton gradient. B is an overstatement; the protein is not damaged, its energy source is just missing. C is incorrect as the transporter cannot switch its energy source from PMF to ATP.

Question 14

A researcher is studying an ABC transporter responsible for drug efflux in a pathogenic bacterium. The protein has two nucleotide-binding domains (NBDs) and two transmembrane domains (TMDs). A mutation in one of the NBDs prevents it from binding ATP, but does not alter its overall structure. What is the most likely consequence of this mutation?

  1. The transporter will now function as a facilitated diffusion channel, allowing the drug to flow out down its concentration gradient.
  2. The transporter will be unable to bind the drug substrate at the TMDs, preventing recognition of the molecule to be exported.
  3. The transporter will bind the drug and ATP at the functional NBD, but will be unable to complete the conformational changes required for translocation. (correct answer)
  4. The transporter will work in reverse, hydrolyzing ATP to actively import the drug into the cell against its concentration gradient.
Explanation: ABC transporters function via a cycle of conformational changes powered by ATP binding and hydrolysis at the NBDs. Binding of the substrate (drug) to the TMDs is typically followed by the binding of two ATP molecules to the NBDs. This triggers a large conformational change that exposes the drug to the outside. ATP hydrolysis resets the transporter. If one NBD cannot bind ATP, the full conformational change powered by the binding of two ATP molecules cannot occur, and the transport cycle will be arrested after the drug is bound. A is incorrect because the fundamental mechanism is not converted to passive transport. B is incorrect because substrate binding occurs at the TMDs, which are unaffected by the NBD mutation. D is incorrect; a loss-of-function mutation will not reverse the direction of transport.

Question 15

A bacterium has two different transport systems for an essential nutrient, Nutrient X. Transporter A has a high affinity (Km = 5 µM) and a low maximum velocity (Vmax). Transporter B has a low affinity (Km = 500 µM) and a high maximum velocity. Which statement accurately describes the physiological roles of these transporters?

  1. Transporter A is most effective in nutrient-rich environments, while Transporter B is essential in nutrient-poor environments.
  2. Transporter B is an active transport system, while Transporter A functions by facilitated diffusion, explaining the difference in Vmax.
  3. Transporter A is likely the primary system used for scavenging the nutrient in oligotrophic (low-nutrient) conditions. (correct answer)
  4. Both transporters would be equally effective regardless of the external concentration of Nutrient X.
Explanation: A low Km value indicates a high affinity of the transporter for its substrate. This means Transporter A can bind and transport Nutrient X efficiently even when it is present at very low concentrations. This makes it ideal for scavenging nutrients in oligotrophic (nutrient-poor) environments. Transporter B, with its high Km, requires a much higher concentration of the nutrient to function effectively, but its high Vmax allows for rapid uptake when the nutrient is abundant. Therefore, C is the correct statement. A is the reverse of the correct situation. B makes an unsubstantiated claim about the transport mechanism; kinetic parameters alone do not distinguish between active transport and facilitated diffusion. D is incorrect as their effectiveness is clearly concentration-dependent.

Question 16

A halophilic (salt-loving) bacterium maintains osmotic balance in a high-salt environment by accumulating high internal concentrations of a compatible solute, such as glycine betaine. This import process is inhibited by compounds that dissipate the membrane's sodium gradient.

Given this information, the uptake of glycine betaine is most likely mediated by which of the following?

  1. A Na+/glycine betaine symporter. (correct answer)
  2. A K+/glycine betaine antiporter.
  3. An ATP-dependent primary transporter.
  4. A group translocation system specific for glycine betaine.
Explanation: The bacterium is accumulating glycine betaine against a concentration gradient, which indicates active transport. The finding that this process is inhibited by dissipating the sodium gradient strongly implicates this gradient as the energy source. This is the defining feature of secondary active transport. Since glycine betaine is being imported, and marine bacteria typically have a strong Na+ gradient directed inwards, the most plausible mechanism is a symporter that couples the favorable influx of Na+ to the unfavorable influx of glycine betaine. B is incorrect because a K+ gradient is not implicated. C is incorrect because the energy source is the Na+ gradient, not ATP directly. D is incorrect as there is no evidence of chemical modification.

Question 17

An experiment measuring the uptake of an uncharged molecule into a bacterium shows that the transport rate reaches a maximum velocity (Vmax) at high external concentrations. This observation alone is insufficient to distinguish between facilitated diffusion and active transport. Which of the following additional findings would definitively point to active transport?

  1. The transport rate decreases significantly when the temperature is lowered from 37°C to 20°C.
  2. A structurally analogous molecule is found to competitively inhibit the transport of the original molecule.
  3. The steady-state intracellular concentration of the molecule exceeds the extracellular concentration. (correct answer)
  4. The transport protein is found to be an integral membrane protein spanning the lipid bilayer.
Explanation: Saturation kinetics (reaching a Vmax) is a property of both facilitated diffusion and active transport, as both rely on a finite number of carrier proteins. The defining difference is that active transport can move substances against their concentration gradient, while facilitated diffusion cannot. Therefore, observing that the molecule is accumulated inside the cell to a concentration higher than the outside medium is conclusive evidence for active transport. A is not definitive, as both processes are enzyme-catalyzed and temperature-dependent. B is also not definitive, as competitive inhibition is characteristic of any specific binding site on a protein, common to both mechanisms. D is true for both types of transporters and thus does not distinguish them.

Question 18

A bacterial Na+/Ca2+ exchanger protein moves Na+ ions into the cell and Ca2+ ions out of the cell. The concentration of Na+ is much higher outside the cell than inside, while the concentration of Ca2+ is much lower outside the cell than inside. Which statement best describes this transport process?

  1. This is a primary active transport process where ATP is used to pump both ions.
  2. This is a form of facilitated diffusion, as the movement of Na+ drives the movement of Ca2+.
  3. This is a symport mechanism because two different cations are being transported simultaneously.
  4. This is an example of secondary active transport, where the Ca2+ is moved against its gradient. (correct answer)
Explanation: When you encounter transport proteins moving multiple substances across membranes, focus on the energy source and direction relative to concentration gradients to classify the mechanism correctly. This Na⁺/Ca²⁺ exchanger represents secondary active transport. The key insight is recognizing that Na⁺ moves down its concentration gradient (high outside → low inside), releasing energy that drives Ca²⁺ movement against its gradient (high inside → low outside). The Na⁺ gradient, established by primary active transport elsewhere, provides the driving force. Since Ca²⁺ moves from low to high concentration, energy is required, making this active transport. Because the energy comes from an ion gradient rather than direct ATP hydrolysis, it's secondary active transport. Answer A is incorrect because no ATP is directly used in this exchanger's operation. Primary active transport requires direct ATP hydrolysis, which isn't happening here. Answer B misclassifies this as facilitated diffusion. While Na⁺ does move down its gradient, Ca²⁺ moves against its gradient, which requires energy input and cannot be passive transport. Answer C incorrectly identifies this as symport. This is actually antiport (countertransport) because the ions move in opposite directions - Na⁺ enters while Ca²⁺ exits. Symport would require both ions moving in the same direction. Remember: Secondary active transport always involves one substance moving down its gradient to power another substance's movement against its gradient. Look for the coupling of favorable and unfavorable movements to identify these mechanisms on exams.

Question 19

A microbiologist is studying the uptake of the amino acid proline by a bacterial species. They observe that proline is accumulated against a significant concentration gradient. When the protonophore carbonyl cyanide m-chlorophenyl hydrazone (CCCP), a chemical that dissipates the proton motive force (PMF), is added to the culture, proline transport ceases almost immediately.

Based on these observations, what is the most likely mechanism of proline transport in this bacterium?

  1. An ATP-binding cassette (ABC) transporter that hydrolyzes ATP to pump proline into the cell.
  2. A phosphotransferase system (PTS) where proline is chemically modified during transport.
  3. A secondary active transport system, likely a proton-proline symporter. (correct answer)
  4. A facilitated diffusion channel that allows proline to move down its electrochemical gradient.
Explanation: The accumulation of proline against its concentration gradient indicates an active transport mechanism. The immediate cessation of transport upon addition of CCCP, a protonophore that eliminates the proton motive force (PMF), strongly suggests that the energy source for this transport is the PMF, not ATP directly. This is the defining characteristic of secondary active transport. A proton-proline symporter would use the influx of protons down their electrochemical gradient to drive the influx of proline against its gradient. A is incorrect because ABC transporters use ATP, and their inhibition by CCCP would be indirect and slower. B is incorrect as there is no mention of proline modification. D is incorrect because facilitated diffusion does not allow for transport against a concentration gradient.

Question 20

A microbiologist is studying the uptake of the amino acid proline by a bacterial species. They observe that proline is accumulated against a significant concentration gradient. When the protonophore carbonyl cyanide m-chlorophenyl hydrazone (CCCP), a chemical that dissipates the proton motive force (PMF), is added to the culture, proline transport ceases almost immediately.

Based on these observations, what is the most likely mechanism of proline transport in this bacterium?

  1. An ATP-binding cassette (ABC) transporter that hydrolyzes ATP to pump proline into the cell.
  2. A phosphotransferase system (PTS) where proline is chemically modified during transport.
  3. A secondary active transport system, likely a proton-proline symporter. (correct answer)
  4. A facilitated diffusion channel that allows proline to move down its electrochemical gradient.
Explanation: The accumulation of proline against its concentration gradient indicates an active transport mechanism. The immediate cessation of transport upon addition of CCCP, a protonophore that eliminates the proton motive force (PMF), strongly suggests that the energy source for this transport is the PMF, not ATP directly. This is the defining characteristic of secondary active transport. A proton-proline symporter would use the influx of protons down their electrochemical gradient to drive the influx of proline against its gradient. A is incorrect because ABC transporters use ATP, and their inhibition by CCCP would be indirect and slower. B is incorrect as there is no mention of proline modification. D is incorrect because facilitated diffusion does not allow for transport against a concentration gradient.