Biochemistry Quiz: Redox Chemistry And Biological Electron Carriers

What this quiz covers

This quiz focuses on Redox Chemistry And Biological Electron Carriers, giving you a quick way to practice the rules, question types, and explanations that matter most for Biochemistry.

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. A standard reduction potential that is precisely midway between that of NAD⁺ and the first iron-sulfur cluster, ensuring a smooth energy drop.
2. A chemical structure nearly identical to nicotinamide, allowing it to accept a hydride ion (H⁻) directly from NADH before donating single electrons.
3. The permanent covalent attachment of the flavin to the enzyme, which prevents the coenzyme from diffusing away during the multi-step electron transfer.
4. The ability of the isoalloxazine ring to accept or donate electrons either one or two at a time, permitting the stable formation of a semiquinone radical. (correct answer)

Explanation: When you encounter questions about electron transport chains, focus on how different coenzymes bridge the gap between donors and acceptors with different electron transfer mechanisms. Flavin coenzymes (FMN and FAD) are uniquely suited to connect two-electron donors like NADH with one-electron acceptors like iron-sulfur clusters because of their flexible redox chemistry. The isoalloxazine ring system in flavins can exist in three stable oxidation states: fully oxidized (quinone), one-electron reduced (semiquinone radical), and two-electron reduced (hydroquinone). This allows flavoproteins to accept two electrons from NADH simultaneously, then donate them sequentially as single electrons to downstream acceptors like iron-sulfur centers. Option A is incorrect because the standard reduction potential alone doesn't determine electron transfer mechanisms - it's about thermodynamic favorability, not whether electrons are transferred singly or in pairs. Option B misrepresents flavin chemistry; while flavins can accept hydride ions, their structure is not nearly identical to nicotinamide, and this similarity isn't what enables the electron transfer bridging function. Option C describes a real property of many flavoproteins, but covalent attachment prevents diffusion rather than enabling the one/two-electron flexibility that's crucial for bridging different types of redox centers. The correct answer is D because the semiquinone radical intermediate is the key feature that allows flavins to "split" two-electron transfers into sequential one-electron steps. Remember: flavoproteins are the "universal adapters" of electron transport - their ability to stabilize odd-electron species makes them essential intermediates between different classes of redox partners.

Question 2

1. It activates citrate synthase by an allosteric mechanism, committing more acetyl-CoA to the cycle to store the excess energy.
2. It causes potent product inhibition of NAD⁺-dependent dehydrogenases, particularly isocitrate dehydrogenase and α-ketoglutarate dehydrogenase. (correct answer)
3. It enhances the activity of succinate dehydrogenase (Complex II), bypassing the NADH-producing steps to maintain some cycle flux.
4. It reverses the malate dehydrogenase reaction, favoring malate production to consume excess NADH and alleviate inhibition.

Explanation: A high NADH/NAD⁺ ratio signals a high energy state where the cell has abundant reducing power. NADH acts as an allosteric inhibitor and a product inhibitor for several key regulatory enzymes in the TCA cycle. Specifically, high levels of NADH directly inhibit isocitrate dehydrogenase and α-ketoglutarate dehydrogenase, two major control points, thus slowing down the entire flux through the cycle.

Question 3

1. It stimulates the conversion of pyruvate to acetyl-CoA via the pyruvate dehydrogenase complex to increase TCA cycle flux.
2. It drives the reduction of pyruvate to lactate, a process which regenerates the NAD⁺ required to sustain glycolysis. (correct answer)
3. It allosterically activates pyruvate carboxylase, shunting the excess pyruvate towards gluconeogenesis to store energy.
4. It inhibits phosphofructokinase-1, the main regulatory enzyme of glycolysis, to prevent further production of pyruvate.

Explanation: A high NADH/NAD⁺ ratio indicates a deficit of NAD⁺, which is an essential substrate for the glyceraldehyde-3-phosphate dehydrogenase step of glycolysis. To continue ATP production via glycolysis under anaerobic conditions, NAD⁺ must be regenerated. The reduction of pyruvate to lactate by lactate dehydrogenase oxidizes NADH to NAD⁺, thus replenishing the NAD⁺ pool and allowing glycolysis to proceed.

Question 4

1. NADH, because it is produced in large quantities by glycolysis in the cytosol, making it readily available for biosynthetic reactions.
2. FADH₂, because its prosthetic group is tightly bound to the fatty acid synthase complex, allowing for efficient, localized electron transfer.
3. NADPH, because the cell maintains a high NADPH/NADP⁺ ratio, which creates a strong reducing environment favorable for anabolic pathways. (correct answer)
4. NADPH, because its standard reduction potential is significantly more positive than that of NADH, allowing it to donate electrons more readily.

Explanation: Cells partition their nicotinamide nucleotide pools. The NADH/NAD⁺ ratio is typically kept low, favoring oxidation (catabolism). In contrast, the NADPH/NADP⁺ ratio is kept very high, favoring reduction (anabolism). This high ratio provides the necessary thermodynamic driving force for reductive biosynthetic pathways like fatty acid synthesis. The primary source of this NADPH is the pentose phosphate pathway.

Question 5

1. FAD is a stronger oxidizing agent than NAD⁺, and its reduction potential is sufficient to drive the thermodynamically challenging oxidation of an alkane to an alkene. (correct answer)
2. NAD⁺ is restricted to the mitochondrial matrix, while FAD is a cytosolic coenzyme, which is where β-oxidation primarily occurs.
3. The reaction involves the transfer of a single electron, and only FAD can form the stable semiquinone radical intermediate required for such a transfer.
4. The enzyme's active site has a specific binding pocket for the isoalloxazine ring of FAD, which physically excludes the bulkier nicotinamide ring of NAD⁺.

Explanation: The standard reduction potential of FAD (in most flavoproteins) is higher (more positive) than that of NAD⁺. This makes FAD a stronger oxidizing agent. The oxidation of a C-C single bond in an alkane to a C=C double bond in an alkene is energetically more difficult than oxidizing an alcohol to a ketone. The greater oxidizing power of FAD is necessary to make this dehydrogenation reaction thermodynamically favorable.

Question 6

1. O₂, Cytochrome c (Fe³⁺), FADH₂, NADH
2. NADH, FADH₂, Coenzyme Q (ubiquinone), O₂ (correct answer)
3. Glucose, FADH₂, Coenzyme Q (ubiquinone), NADH
4. Cytochrome c (Fe³⁺), O₂, NADH, FADH₂

Explanation: Electrons with higher energy are better reducing agents and are associated with more negative standard reduction potentials (E°'). The order of electron flow in the electron transport chain reflects a decrease in electron energy. NADH (E°' ≈ -0.32 V) has the highest energy electrons. It passes them to carriers like FMN and then Coenzyme Q. FADH₂ (E°' in complex II is ≈ 0 V) also passes electrons to Coenzyme Q. From there, electrons flow through cytochromes to the final electron acceptor, O₂ (E°' = +0.82 V), which has the lowest energy state for these electrons.

Question 7

1. The reaction is non-spontaneous because the overall change in reduction potential (ΔE°') is negative.
2. The reaction is spontaneous, with NAD⁺ acting as the primary reducing agent in the forward direction.
3. The reaction is spontaneous with a standard free energy change (ΔG°') of approximately -23.7 kJ/mol. (correct answer)
4. The reaction is spontaneous but requires energy input, resulting in a standard free energy change (ΔG°') of +23.7 kJ/mol.

Explanation: To find the overall reaction potential (ΔE°'), we identify the electron donor (NADH) and acceptor (Acetaldehyde). The reaction for NADH oxidation is the reverse of its reduction, so its potential is +0.320 V. The overall ΔE°' = E°'(acceptor) - E°'(donor) = (-0.197 V) - (-0.320 V) = +0.123 V. A positive ΔE°' indicates a spontaneous reaction. We then use the formula ΔG°' = -nFΔE°', where n=2 moles of electrons and F ≈ 96.5 kJ·V⁻¹·mol⁻¹. ΔG°' = -2 * (96.5) * (0.123) ≈ -23.7 kJ/mol. The negative ΔG°' confirms spontaneity.

Question 8

1. Carboxylation reactions that require a biotin coenzyme for one-carbon transfers, such as the first step of gluconeogenesis.
2. Dehydrogenase reactions that utilize a soluble coenzyme to oxidize substrates such as alcohols, aldehydes, or α-hydroxy acids. (correct answer)
3. Transamination reactions that depend on pyridoxal phosphate (PLP) for the transfer of amino groups in amino acid metabolism.
4. Electron transfer reactions involving tightly bound flavoproteins that are responsible for oxidizing C-C single bonds to C=C double bonds.

Explanation: Niacin (vitamin B3) is the precursor for nicotinamide adenine dinucleotide (NAD⁺ and NADP⁺). NAD⁺ is a crucial coenzyme that acts as a soluble electron acceptor in a wide range of dehydrogenase-catalyzed oxidation reactions, including key steps in glycolysis, the TCA cycle, and fatty acid oxidation. A deficiency impairs these central energy-yielding pathways, leading to symptoms like fatigue.

Question 9

1. Cyto-C → Cyto-A → Cyto-D → Cyto-B
2. Cyto-B → Cyto-D → Cyto-A → Cyto-C (correct answer)
3. Cyto-B → Cyto-A → Cyto-D → Cyto-C
4. Cyto-A → Cyto-C → Cyto-B → Cyto-D

Explanation: In a thermodynamically favorable electron transport chain, electrons flow spontaneously from carriers with a lower (more negative) standard reduction potential (E°') to carriers with a higher (more positive) E°'. Ordering the given carriers by their E°' from most negative to most positive gives: Cyto-B (-0.05 V), Cyto-D (+0.08 V), Cyto-A (+0.15 V), and Cyto-C (+0.25 V). Therefore, the favored direction of electron flow is B → D → A → C.

Question 10

1. -41.7 kJ/mol (correct answer)
2. +41.7 kJ/mol
3. -29.7 kJ/mol
4. +29.7 kJ/mol

Explanation: The spontaneous reaction involves the oxidation of the species with the more negative E°' (lactate) and the reduction of the species with the more positive E°' (fumarate). The overall reaction is Lactate + Fumarate → Pyruvate + Succinate. The change in potential is ΔE°' = E°'(acceptor) - E°'(donor) = (+0.031 V) - (-0.185 V) = +0.216 V. The free energy change is ΔG°' = -nFΔE°' = -2 * (96.5 kJ·V⁻¹·mol⁻¹) * (0.216 V) ≈ -41.7 kJ/mol.

Question 11

1. The substrate is oxidized by losing two electrons and one proton, which are accepted by the NAD⁺ coenzyme as a single chemical entity. (correct answer)
2. The substrate is reduced by donating a proton, while NAD⁺ accepts two electrons from a separate source within the enzyme active site.
3. NAD⁺ accepts two distinct electrons sequentially to form a radical intermediate, followed by the subsequent binding of a free proton from solution.
4. A proton (H⁺) is transferred from the substrate to NAD⁺, and two electrons are then tunneled through the enzyme backbone to neutralize the charge.

Explanation: Dehydrogenase reactions using NAD⁺ involve the removal of two hydrogen atoms from the substrate. One of these is transferred as a hydride ion (H⁻, which is a proton plus two electrons) to the nicotinamide ring of NAD⁺, forming NADH. The other is released into the solvent as a proton (H⁺). Therefore, the substrate is oxidized by losing two electrons and one proton, which are accepted by NAD⁺.

Question 12

1. E becomes significantly more positive than E°' because the high concentration of the oxidant (NAD⁺) favors reduction. (correct answer)
2. E becomes significantly more negative than E°' because the presence of the product (NADH) shifts the equilibrium to the left.
3. E remains equal to E°' because cellular conditions are buffered to maintain the standard state for thermodynamic consistency.
4. E approaches zero, as the reaction is near equilibrium under these mitochondrial conditions, preventing large net electron flow.

Explanation: The Nernst equation relates the actual potential (E) to the standard potential (E°') and the concentrations of the redox species. The term ln([reduced]/[oxidized]) is ln([NADH]/[NAD⁺]), which is ln(1/10). Since ln(1/10) is a negative number, the term -(RT/nF)ln([reduced]/[oxidized]) becomes a positive value. Thus, E = E°' + (a positive value), meaning the actual reduction potential is more positive than the standard potential, making the couple a stronger oxidizing agent under these conditions.

Question 13

1. The reaction directly generates one molecule of GTP via substrate-level phosphorylation, similar to the succinyl-CoA synthetase step.
2. The resulting NADH molecule donates two electrons to Complex I of the electron transport chain, contributing to the synthesis of approximately 2.5 ATP. (correct answer)
3. The electrons from the newly formed NADH are immediately transferred to FAD within Complex II, leading to the synthesis of approximately 1.5 ATP.
4. The oxidation of malate is highly exergonic, and the large release of free energy is directly used to pump protons across the mitochondrial membrane.

Explanation: The malate dehydrogenase reaction generates NADH by reducing NAD⁺. This NADH is a mobile electron carrier in the mitochondrial matrix. It subsequently binds to Complex I of the electron transport chain and donates its two high-energy electrons. This initiates a series of redox reactions that power the pumping of protons, ultimately leading to the synthesis of approximately 2.5 molecules of ATP per NADH molecule oxidized.

Question 14

1. A substrate-level phosphorylation occurs where a phosphate group from ATP is used to create the high-energy intermediate.
2. The enzyme utilizes the energy from FADH₂ oxidation to phosphorylate the substrate, using inorganic phosphate to form the product.
3. The reduction of the aldehyde to an alcohol is coupled to the oxidation of NADH, consuming one of the cell's reducing equivalents.
4. The oxidation of an aldehyde functional group to a high-energy acyl phosphate is coupled to the reduction of one molecule of NAD⁺ to NADH. (correct answer)

Explanation: The GAPDH reaction is a crucial step in glycolysis that demonstrates how cells couple oxidation-reduction reactions with energy conservation. This reaction involves both redox chemistry and the formation of a high-energy phosphate bond in a single enzymatic step. In this reaction, glyceraldehyde-3-phosphate (which contains an aldehyde group) undergoes oxidation to form an acyl phosphate intermediate. This oxidation provides the energy needed to incorporate inorganic phosphate (Pi) into the product, creating 1,3-bisphosphoglycerate with its high-energy acyl phosphate bond. Simultaneously, NAD⁺ accepts electrons from this oxidation, becoming reduced to NADH. This makes option D correct - it accurately describes both the oxidation of the aldehyde to a high-energy acyl phosphate and the coupled reduction of NAD⁺. Option A incorrectly suggests substrate-level phosphorylation using ATP, but no ATP is consumed in this reaction - instead, the energy comes from aldehyde oxidation. Option B wrongly identifies FADH₂ as the electron carrier; GAPDH specifically uses NAD⁺/NADH, not FAD/FADH₂. Option C has the redox direction backwards - it claims the aldehyde is reduced to an alcohol and NADH is oxidized, when actually the aldehyde is oxidized and NAD⁺ is reduced. Remember that GAPDH is unique because it's the only glycolytic enzyme that directly couples substrate oxidation with NAD⁺ reduction while simultaneously creating a high-energy phosphate bond. This "pay-off" phase reaction sets up the subsequent substrate-level phosphorylation that generates ATP.

Question 15

1. NADH produced during glycolysis transfers a hydride ion directly to GSSG, splitting the disulfide bond in a single, direct reaction step.
2. Reduced cytochrome c leaks from the mitochondria and acts as the direct, single-electron donor for the glutathione reductase enzyme.
3. FADH₂, shuttled from the mitochondria, directly reduces GSSG in the cytosol, thus linking fatty acid catabolism to antioxidant defense.
4. NADPH, primarily from the pentose phosphate pathway, provides the reducing power to the FAD cofactor of glutathione reductase. (correct answer)

Explanation: When you encounter questions about cellular antioxidant systems, focus on the specific electron carriers and cofactors involved in each pathway. The glutathione system is a major cellular defense against oxidative stress, and understanding its mechanism requires knowing the electron flow. Glutathione reductase catalyzes the reduction of oxidized glutathione (GSSG) back to its reduced form (GSH). This enzyme contains FAD as a prosthetic group, which must be reduced to FADH₂ to transfer electrons to the disulfide bond in GSSG. The reducing power comes from NADPH, primarily generated by the pentose phosphate pathway in the cytosol. NADPH transfers electrons to the FAD cofactor, creating the reduced enzyme form that can then reduce GSSG to two GSH molecules. Option A is incorrect because NADH, while an important electron carrier, is not the primary reductant for glutathione reductase—NADPH is specifically required. Option B misrepresents the role of cytochrome c, which functions in the electron transport chain within mitochondria and doesn't directly participate in cytosolic glutathione reduction. Option C incorrectly suggests FADH₂ shuttling from mitochondria, but the FAD is bound to glutathione reductase itself and gets reduced in situ by NADPH. Remember that NADPH (not NADH) is the key reducing agent for biosynthetic and antioxidant processes in cells. When you see questions about cellular antioxidant systems, immediately consider the pentose phosphate pathway as the primary source of NADPH for these protective mechanisms.

Question 16

1. The large driving force would cause uncontrolled radical formation and cellular damage if water oxidation occurred directly at the chlorophyll special pair
2. Water molecules cannot directly interact with chlorophyll due to hydrophobic effects in the photosystem membrane environment, requiring intermediate protein-bound cofactors for electron transfer
3. The thermodynamic driving force is actually insufficient under physiological conditions due to pH effects and the need to pump protons across the thylakoid membrane simultaneously
4. Direct four-electron oxidation of water has an extremely high activation energy that cannot be overcome without sequential one-electron transfer steps through intermediate carriers (correct answer)

Explanation: When you encounter questions about photosystem II water oxidation, focus on the fundamental challenge: breaking water's strong O-H bonds requires removing four electrons and four protons simultaneously, which is kinetically prohibitive in a single step. The oxygen-evolving complex solves this through sequential one-electron oxidations using a Mn₄CaO₅ cluster that cycles through different oxidation states (S₀ through S₄). Each photochemical event advances the cycle by one electron, accumulating oxidizing equivalents until the S₄ state can finally oxidize water and release O₂. This stepwise mechanism dramatically lowers the activation energy compared to direct four-electron oxidation. Answer D correctly identifies that direct water oxidation has an insurmountable activation energy barrier. The $\Delta G°'$ tells us the reaction is thermodynamically favorable, but says nothing about the kinetic pathway. Answer A incorrectly suggests the issue is uncontrolled radical formation. While P680•⁺ is indeed a strong oxidant, the problem isn't radical damage but rather the impossibility of simultaneous four-electron transfer. Answer B misunderstands the mechanism. Water molecules do access the active site, and hydrophobic effects aren't the limiting factor—the Mn₄CaO₅ cluster is specifically positioned to interact with water. Answer C incorrectly claims insufficient thermodynamic driving force. The 1.62 V potential difference provides ample energy; pH effects and proton pumping don't negate this thermodynamic favorability. Remember: favorable thermodynamics doesn't guarantee a feasible kinetic pathway. Multi-electron reactions typically require sophisticated mechanisms to overcome activation barriers through sequential single-electron steps.

Question 17

1. Succinate directly donates electrons to cytochrome c oxidase, bypassing the entire electron transport chain upstream of Complex IV
2. Succinate is oxidized by succinate dehydrogenase, which transfers electrons to ubiquinone downstream of the rotenone-sensitive site at Complex I (correct answer)
3. Succinate can be converted to NADH by reverse electron flow, allowing normal electron transport through an alternative pathway around Complex I
4. Succinate acts as an uncoupler that allows electron flow through Complex I despite rotenone inhibition by providing an alternative proton translocation mechanism

Explanation: Succinate dehydrogenase (Complex II) oxidizes succinate to fumarate and reduces FAD to FADH₂. The electrons from FADH₂ enter the electron transport chain at ubiquinone (CoQ), which is downstream of Complex I where rotenone acts. This allows electrons to flow through Complexes III and IV, generating some ATP synthesis, though less than with NADH since Complex I proton pumping is bypassed. Choice A is incorrect because succinate doesn't directly interact with Complex IV. Choice C is incorrect because succinate oxidation doesn't produce NADH, and reverse electron flow would require energy input. Choice D is incorrect because succinate is not an uncoupler and doesn't affect proton translocation mechanisms.

Question 18

1. Hypoxic conditions directly alter the standard reduction potential of NAD⁺, making it a weaker oxidizing agent and shifting the equilibrium toward NADH
2. The absence of oxygen as the terminal electron acceptor prevents NADH reoxidation through the electron transport chain, leading to NADH accumulation (correct answer)
3. Hypoxic stress increases the cellular concentration of reducing enzymes that convert NAD⁺ to NADH more rapidly than under normal conditions
4. Low oxygen levels cause compartmental redistribution of pyridine nucleotides, concentrating NADH in the cytosol while depleting mitochondrial NAD⁺ pools

Explanation: Under aerobic conditions, NADH is continuously reoxidized to NAD⁺ through the electron transport chain, with oxygen serving as the terminal electron acceptor. This maintains a high [NAD⁺]/[NADH] ratio. In hypoxic conditions, oxygen availability limits electron transport chain activity, creating a bottleneck for NADH reoxidation. Consequently, NADH accumulates and NAD⁺ is depleted, dramatically lowering the ratio. Choice A is incorrect because standard reduction potentials are constants that don't change with oxygen availability. Choice C is incorrect because the change is due to impaired NADH oxidation, not increased NADH production. Choice D is incorrect because compartmental redistribution alone wouldn't explain the magnitude of the ratio change observed.

Question 19

1. Electron flow will reverse direction because decreasing lactate concentration makes the pyruvate/lactate couple more reducing than the NAD⁺/NADH couple
2. Electron flow will continue in the same direction but with decreased driving force because the lower lactate concentration decreases the reduction potential of the pyruvate/lactate couple
3. Electron flow will continue in the same direction but with increased driving force because the lower lactate concentration increases the reduction potential of the pyruvate/lactate couple (correct answer)
4. The direction of electron flow will remain unchanged because standard reduction potentials are independent of concentration changes in biological systems

Explanation: When you encounter redox problems involving concentration changes, you need to apply the Nernst equation to determine how shifting concentrations affect reduction potentials and electron flow direction. First, let's establish the initial conditions. Under standard biochemical conditions, the NAD⁺/NADH couple has E°' = -0.32 V and pyruvate/lactate has E°' = -0.19 V. Since pyruvate/lactate has the higher (less negative) reduction potential, it acts as the electron acceptor, while NADH donates electrons. This means electrons flow from the NAD⁺/NADH half-cell to the pyruvate/lactate half-cell. Now, when lactate concentration drops to 10 mM (0.01 M), we apply the Nernst equation: $E = E°' + \frac{RT}{nF}\ln\frac{[oxidized]}{[reduced]}$. For the pyruvate/lactate couple, decreasing [lactate] increases the ratio [pyruvate]/[lactate], making the reduction potential more positive. This creates a larger potential difference between the half-cells, increasing the driving force for electron flow in the same direction. Option A incorrectly suggests the pyruvate/lactate couple becomes more reducing - actually, it becomes more oxidizing. Option B has the right direction but wrong magnitude effect - the driving force increases, not decreases. Option D wrongly claims standard potentials are concentration-independent, confusing E°' (which is constant) with actual reduction potentials under non-standard conditions. Study tip: Remember that decreasing the concentration of the reduced form (product) in a redox couple makes it more oxidizing, while decreasing the oxidized form makes it more reducing. Always consider how concentration ratios affect the Nernst equation.

Question 20

1. The reduction of NAD⁺ to NADH provides additional favorable free energy change that makes the overall reaction thermodynamically favorable under standard conditions
2. The cellular concentrations of reactants and products deviate significantly from standard conditions, making the actual ΔG more favorable than the calculated ΔG°'
3. The enzyme couples these reactions mechanistically through a covalent intermediate, allowing the favorable oxidation to drive the unfavorable phosphorylation reaction (correct answer)
4. The high-energy acyl phosphate bond formation is thermodynamically favorable in the cellular environment due to the presence of inorganic phosphate and metal ion cofactors

Explanation: The glyceraldehyde-3-phosphate dehydrogenase reaction works by mechanistic coupling through a covalent thioester intermediate between the enzyme and the substrate. The favorable oxidation reaction (-43 kJ/mol) drives the formation of the high-energy acyl phosphate (+49 kJ/mol), but without additional thermodynamic input, the overall ΔG°' would be +6 kJ/mol. However, the enzyme mechanism allows the energy from aldehyde oxidation to be captured in the thioester intermediate, which then drives acyl phosphate formation. Choice A is incorrect because NAD⁺ reduction alone doesn't provide enough favorable energy. Choice B, while potentially contributing, doesn't explain the fundamental mechanism. Choice D is incorrect because acyl phosphate formation remains energetically unfavorable regardless of cofactors without mechanistic coupling.