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301. The lock-and-key model mainly proposes that the enzyme active site is:
ⓐ. Rigid and already complementary to the substrate
ⓑ. Always changing shape before substrate arrives
ⓒ. Made of lipid bilayer to trap substrate
ⓓ. Formed only after product is released
Correct Answer: Rigid and already complementary to the substrate
Explanation: The lock-and-key model describes the active site as a pre-formed structure whose shape and chemical features match the substrate closely. Because of this matching, the substrate can bind directly with minimal structural adjustment, similar to a key fitting a lock. The model highlights specificity as a direct result of complementarity between the substrate and a relatively fixed active site. It is especially useful for explaining why enzymes discriminate among similar molecules. Although modern views include flexibility, lock-and-key remains a standard conceptual model in exam questions. Therefore, it proposes a rigid, pre-complementary active site.
302. The induced-fit model mainly proposes that:
ⓐ. Substrate binds only after enzyme is denatured
ⓑ. Enzyme changes conformation to fit the substrate
ⓒ. Substrate changes into enzyme during binding
ⓓ. Active site is always identical in all enzymes
Correct Answer: Enzyme changes conformation to fit the substrate
Explanation: The induced-fit model explains enzyme specificity and efficiency by allowing flexibility in the active site. When the substrate approaches, binding interactions trigger a conformational change in the enzyme that improves the fit. This adjustment aligns catalytic residues properly, strengthens binding interactions, and can better stabilize the transition state. Induced fit therefore explains why some enzymes can accommodate substrates that are similar but still maintain strong specificity. The model emphasizes dynamic structure rather than a permanently rigid pocket. Hence, induced fit means the enzyme changes conformation to fit the substrate.
303. A key advantage of induced fit over lock-and-key in explaining catalysis is that induced fit:
ⓐ. Eliminates the need for substrate binding
ⓑ. States that enzymes never change shape
ⓒ. Explains active-site adjustment that improves transition-state stabilization
ⓓ. Requires covalent attachment of substrate always
Correct Answer: Explains active-site adjustment that improves transition-state stabilization
Explanation: Induced fit provides a mechanism for how enzymes become catalytically efficient after substrate binding by reshaping the active site. This adjustment can bring catalytic amino acids into the correct positions and create an environment that stabilizes the transition state more effectively. By stabilizing the transition state, the activation energy is lowered and the reaction rate increases. Lock-and-key focuses mainly on pre-existing fit and does not emphasize this dynamic alignment and transition-state tuning. Induced fit therefore better explains the link between binding and catalysis in many enzymes. Hence, it explains active-site adjustment that improves transition-state stabilization.
304. Which observation most strongly supports the induced-fit model?
ⓐ. Enzymes bind substrates with no structural change ever
ⓑ. Active site residues move to enclose the substrate on binding
ⓒ. Enzymes are made of amino acids only
ⓓ. Enzymes increase activation energy for reactions
Correct Answer: Active site residues move to enclose the substrate on binding
Explanation: Induced fit predicts that substrate binding triggers conformational changes in the enzyme, especially in and around the active site. If experimental evidence shows that active site residues shift position and the enzyme “closes” around the substrate, that directly supports induced fit. Such movements improve contact, orientation, and catalytic alignment, which explains higher efficiency after binding. This dynamic behavior is inconsistent with a strictly rigid lock-and-key view. The presence of amino acids alone does not distinguish the models, but observed shape changes during binding do. Therefore, movement of active site residues to enclose the substrate supports induced fit.
305. In lock-and-key description, specificity is mainly due to:
ⓐ. Substrate forming peptide bonds with enzyme
ⓑ. Enzyme changing shape only after product release
ⓒ. Random binding that becomes specific later
ⓓ. Active site and substrate having matching shape/chemistry before binding
Correct Answer: Active site and substrate having matching shape/chemistry before binding
Explanation: The lock-and-key model emphasizes that the active site already has the proper shape and chemical groups to bind its substrate. Because the fit is pre-existing, only the correct substrate can bind efficiently, leading to high specificity. This model explains discrimination among similar molecules by focusing on geometric and chemical complementarity. It does not require significant enzyme reshaping to explain binding. The model therefore treats specificity as a direct consequence of a rigid active-site structure that matches the substrate. Hence, specificity is mainly due to pre-existing matching shape and chemistry.
306. Which statement correctly compares the two models?
ⓐ. Lock-and-key emphasizes rigid active site; induced fit emphasizes flexibility
ⓑ. Lock-and-key emphasizes flexibility; induced fit emphasizes rigidity
ⓒ. Both models deny any role of active site in specificity
ⓓ. Both models require covalent enzyme–substrate bonding always
Correct Answer: Lock-and-key emphasizes rigid active site; induced fit emphasizes flexibility
Explanation: The lock-and-key model treats the active site as largely rigid and already complementary to the substrate, so binding occurs without major changes. The induced-fit model allows the active site to be flexible and to adjust its shape after the substrate begins interacting with it. This flexibility can enhance binding strength and catalytic alignment by optimizing interactions and stabilizing the transition state. Both models still rely on the active site as the central region of specificity and catalysis, but they differ in whether the site is fixed or adaptable. Therefore, the correct comparison is rigid versus flexible emphasis. Hence, lock-and-key is rigid and induced fit is flexible.
307. A scenario where induced fit is especially helpful is when:
ⓐ. The substrate fits perfectly without any rearrangement
ⓑ. Slight active-site rearrangement improves binding of a similar substrate
ⓒ. The enzyme has no defined active site
ⓓ. The enzyme catalyzes no reaction at all
Correct Answer: Slight active-site rearrangement improves binding of a similar substrate
Explanation: Induced fit is particularly useful for explaining enzymes that bind substrates requiring minor adjustments in the active site for optimal interaction. Many biological substrates are flexible or have closely related analogs, and a small conformational change can allow the enzyme to grip the correct substrate and align it for catalysis. This model explains how binding can become tighter and more specific after initial contact. It also accounts for improved transition-state stabilization through dynamic repositioning of catalytic residues. A perfectly fitting substrate without changes matches lock-and-key more closely, not induced fit. Therefore, induced fit is helpful when slight rearrangement improves binding and catalysis.
308. Which model most directly explains that substrate binding can “activate” the enzyme’s catalytic geometry?
ⓐ. Lock-and-key model
ⓑ. Denaturation model
ⓒ. Induced-fit model
ⓓ. Diffusion-only model
Correct Answer: Induced-fit model
Explanation: Induced fit explains that the enzyme’s active site may not be perfectly arranged for catalysis until the substrate binds and triggers a conformational adjustment. This change can reposition key amino acids, strengthen interactions, and create the correct catalytic environment, effectively “activating” the geometry needed for reaction. The lock-and-key model does not focus on activation via shape change; it assumes the site is already prepared. Induced fit therefore links binding to catalysis more explicitly by describing structural rearrangements. This is a central competitive-style concept in enzyme mechanism questions. Hence, induced fit most directly explains activation of catalytic geometry upon binding.
309. The lock-and-key model is most closely associated with explaining:
ⓐ. Absolute specificity with a highly complementary rigid site
ⓑ. Lipid bilayer formation in membranes
ⓒ. Phosphodiester bond formation in DNA
ⓓ. Protein denaturation by heat only
Correct Answer: Absolute specificity with a highly complementary rigid site
Explanation: Lock-and-key is classically used to illustrate enzymes that show high specificity because the substrate fits the active site with strong pre-existing complementarity. This is especially aligned with absolute specificity, where only one substrate type is recognized. The model helps explain why even small differences in substrate structure can prevent binding if the rigid site cannot accommodate them. It is not a model for membrane assembly, nucleic acid backbone formation, or denaturation processes. Its key conceptual use is in specificity via a rigid active site. Therefore, lock-and-key is associated with absolute specificity and a highly complementary rigid site.
310. If an enzyme clearly changes shape upon substrate binding, the best-supported model is:
ⓐ. Lock-and-key only
ⓑ. Substrate-only model
ⓒ. No specificity model
ⓓ. Induced fit
Correct Answer: Induced fit
Explanation: Induced fit explicitly predicts conformational change in the enzyme during binding, so direct evidence of shape change supports this model. When the enzyme adjusts its active site, it can improve substrate alignment and strengthen interactions that promote catalysis. This dynamic behavior is a core feature distinguishing induced fit from a strictly rigid lock-and-key concept. Such changes can also help explain transition-state stabilization and improved reaction rates after binding begins. Therefore, observed enzyme shape change upon substrate binding is best explained by induced fit. Hence, induced fit is the best-supported model.
311. For most enzymes from human cells, the highest activity is typically observed near:
ⓐ. $0^\circ C$ (ice-point conditions)
ⓑ. Around $37^\circ C$ (body temperature)
ⓒ. Around $100^\circ C$ (boiling conditions)
ⓓ. Around $-10^\circ C$ (subzero conditions)
Correct Answer: Around $37^\circ C$ (body temperature)
Explanation: Enzymes have an optimum temperature at which their active site geometry and molecular flexibility best support substrate binding and catalysis. In humans, cellular enzymes evolved to function most efficiently at normal body temperature, so activity generally peaks near $37^\circ C$. Below this temperature, molecular collisions and conformational motions slow down, reducing reaction rate. Far above this range, the enzyme’s structure becomes unstable and activity falls due to loss of functional shape. The result is a characteristic rise to an optimum followed by a decline at higher temperatures. Therefore, human enzymes typically show maximum activity around $37^\circ C$.
312. The primary reason enzyme activity decreases sharply at high temperatures is:
ⓐ. Irreversible denaturation of the enzyme protein
ⓑ. Complete freezing of the reaction mixture
ⓒ. Permanent removal of water from the enzyme
ⓓ. Conversion of enzymes into nucleic acids
Correct Answer: Irreversible denaturation of the enzyme protein
Explanation: At high temperatures, the weak bonds and interactions that maintain an enzyme’s secondary and tertiary structure become disrupted. When this happens, the active site loses its precise shape and chemical arrangement, so the substrate can no longer bind and react effectively. This structural loss is called denaturation and it commonly leads to a large, sudden drop in activity. Unlike low-temperature effects, which mainly slow molecular motion, denaturation directly destroys the functional conformation. Because the active site is highly structure-dependent, even modest unfolding can significantly reduce catalysis. Hence, sharp decline at high temperature is mainly due to irreversible protein denaturation.
313. At low temperatures, enzyme-catalyzed reactions usually slow down mainly because:
ⓐ. The enzyme is always permanently denatured at low temperature
ⓑ. Substrate becomes chemically inactive in all cases
ⓒ. Molecular collisions and kinetic energy decrease
ⓓ. The active site forms new peptide bonds rapidly
Correct Answer: Molecular collisions and kinetic energy decrease
Explanation: Lower temperature reduces the kinetic energy of molecules, so both enzyme and substrate move more slowly. This decreases the frequency of effective collisions and lowers the probability that reacting molecules reach the activation state. Importantly, the enzyme’s structure generally remains intact at low temperature, so the active site is still present but less frequently engaged. As a result, the reaction rate drops, but activity can often return when temperature is raised again. This is a classic “rate-limiting motion” explanation rather than a structural destruction explanation. Therefore, low temperature slows enzyme reactions mainly by reducing molecular collisions and kinetic energy.
314. The term “optimum temperature” for an enzyme refers to the temperature at which:
ⓐ. The enzyme is completely denatured
ⓑ. The enzyme shows maximum catalytic activity
ⓒ. The substrate becomes insoluble in water
ⓓ. The enzyme is converted into a cofactor
Correct Answer: The enzyme shows maximum catalytic activity
Explanation: Enzyme activity typically increases with temperature up to a point because molecular motion and collision frequency improve. Beyond a certain temperature, the enzyme’s structure begins to destabilize and the active site loses its functional shape, causing activity to decline. The peak of this activity–temperature curve is called the optimum temperature, representing the best balance between increased kinetic energy and structural stability. It is not the point of denaturation; rather, it is the point of maximum catalytic rate under given conditions. This optimum can differ among organisms and enzyme types. Hence, optimum temperature is the temperature at which the enzyme shows maximum activity.
315. If temperature rises moderately (but still below the denaturation range), enzyme activity generally increases because:
ⓐ. Substrate concentration becomes zero instantly
ⓑ. Enzyme molecules become smaller in size permanently
ⓒ. Enzyme changes into a lipid to trap substrate
ⓓ. Collision frequency and activation events increase
Correct Answer: Collision frequency and activation events increase
Explanation: As temperature increases within a safe range, molecules move faster and collide more frequently. This raises the chance that substrate molecules collide with the active site in the correct orientation and with enough energy to cross the activation barrier. The enzyme remains properly folded in this moderate range, so its catalytic machinery is still intact. Therefore, the rate increases primarily due to kinetic effects rather than structural changes. Only when temperature becomes too high does denaturation dominate and reverse the trend. Thus, moderate warming increases activity by increasing collision frequency and activation events.
316. When an enzyme is heated beyond its stability limit and then cooled, the activity often remains low mainly because:
ⓐ. Denaturation may not readily reverse to the native structure
ⓑ. Low temperature permanently increases enzyme activity
ⓒ. The substrate becomes a cofactor after heating
ⓓ. Phosphodiester bonds form in the active site
Correct Answer: Denaturation may not readily reverse to the native structure
Explanation: High temperature can disrupt the interactions that maintain an enzyme’s precise 3-D shape, causing unfolding or misfolding. After such denaturation, the protein may aggregate or settle into incorrect conformations that do not recreate the original active site. Cooling reduces molecular motion but does not necessarily restore the correct folding pattern automatically. Because catalytic function depends on exact structure, failure to regain the native conformation means activity stays low. This explains why heat damage is often effectively irreversible for many enzymes. Therefore, activity may remain low after cooling because denaturation may not readily reverse.
317. Enzymes from thermophilic organisms typically show higher optimum temperatures because they:
ⓐ. Have more stable structures resistant to heat disruption
ⓑ. Lack any tertiary structure and therefore cannot denature
ⓒ. Are made only of carbohydrates and not proteins
ⓓ. Work only at subzero temperatures in nature
Correct Answer: Have more stable structures resistant to heat disruption
Explanation: Thermophilic organisms live in high-temperature environments, so their enzymes are adapted to remain folded and functional under heat stress. These enzymes often have structural features that increase stability of the active site and overall fold, allowing catalysis at temperatures that would denature typical enzymes. Because stability is higher, the activity–temperature curve shifts so that maximum activity occurs at a higher temperature. This is an adaptation of protein structure–function, not a change in the basic principle of denaturation. The enzymes still require correct folding; they are simply better at maintaining it. Hence, thermophilic enzymes have higher optima because their structures are more heat-stable.
318. In a typical enzyme vs temperature graph, the decline after the optimum is mainly due to:
ⓐ. Reduced collision rate at higher temperature
ⓑ. Enzyme denaturation reducing active site integrity
ⓒ. Increased substrate solubility in water
ⓓ. Formation of extra cofactors at high temperature
Correct Answer: Enzyme denaturation reducing active site integrity
Explanation: Before the optimum, rising temperature increases reaction rate because molecules move faster and collide more often. After the optimum, the dominant effect becomes structural damage to the enzyme, where folding interactions weaken and the active site loses its precise shape. This reduces substrate binding and catalytic efficiency sharply, producing the characteristic drop in activity. The decline is therefore not explained by fewer collisions, which actually increase with temperature, but by loss of functional conformation. This concept is a common exam trap: the curve falls at high temperature due to denaturation, not kinetics. Therefore, the post-optimum decline is mainly due to denaturation reducing active site integrity.
319. If a reaction mixture is cooled from $37^\circ C$ to $10^\circ C$, the enzyme activity usually:
ⓐ. Increases sharply due to faster collisions
ⓑ. Becomes permanently zero due to instant denaturation
ⓒ. Decreases but can recover when warmed again
ⓓ. Converts substrate into product without enzyme
Correct Answer: Decreases but can recover when warmed again
Explanation: Cooling lowers kinetic energy, so enzyme and substrate molecules move more slowly and collide less effectively, reducing reaction rate. However, low temperature generally does not destroy the enzyme’s 3-D structure, so the active site remains intact. When the temperature is raised back toward the optimum, molecular motion and effective collisions increase again, and activity typically returns. This reversibility distinguishes low-temperature slowing from high-temperature denaturation, which often causes lasting loss of function. Thus, the expected result is reduced observed activity with recovery on warming. Therefore, activity decreases on cooling but can recover when warmed again.
320. A practical laboratory reason to store enzymes at low temperature is that it:
ⓐ. Increases random denaturation speed
ⓑ. Stops all chemical reactions forever
ⓒ. Makes enzyme proteins polymerize into DNA
ⓓ. Slows degradation and unwanted side reactions
Correct Answer: Slows degradation and unwanted side reactions
Explanation: Lower temperature reduces molecular motion and reaction rates, which helps slow processes that can damage enzymes over time. This includes unwanted chemical changes, proteolytic breakdown, and other degradation pathways that proceed faster at warmer temperatures. Storing enzymes cold helps preserve the native structure and active site integrity for longer periods. While activity is lower at low temperature, the goal of storage is stability rather than immediate catalysis. The enzyme can often be warmed to working temperature when needed for reactions. Therefore, enzymes are stored cold because it slows degradation and unwanted side reactions.
321. The main reason pH affects enzyme activity is that pH changes the:
ⓐ. Number of peptide bonds in enzyme
ⓑ. Molecular mass of the substrate
ⓒ. Ionization of active-site residues
ⓓ. Total count of amino acids in enzyme
Correct Answer: Ionization of active-site residues
Explanation: Enzyme active sites contain amino acid side chains that must have specific charges to bind substrate and perform catalysis. When pH changes, the ionization state of these side chains changes, altering charge-based interactions like ionic bonding and hydrogen bonding. This can reduce substrate binding, disrupt catalytic steps, or change active-site shape indirectly. Because the required charge pattern is precise, even moderate pH shifts can reduce the reaction rate. The effect is therefore mechanistic, linked to how the active site chemistry works. Hence, pH affects enzyme activity mainly by changing ionization of active-site residues.
322. The “optimum pH” of an enzyme refers to the pH at which the enzyme shows:
ⓐ. Maximum catalytic activity
ⓑ. Complete permanent denaturation
ⓒ. Zero seeable reaction rate
ⓓ. Only substrate binding, no product
Correct Answer: Maximum catalytic activity
Explanation: Enzymes typically show peak activity at a particular pH where their active site has the most favorable charge state and structure for catalysis. At this pH, substrate binding and transition-state stabilization are most efficient. When pH moves away from the optimum, key residues may gain or lose protons, weakening essential interactions. This reduces the rate even if temperature and substrate concentration remain constant. The optimum pH is therefore a functional peak point on an activity–pH curve. Thus, optimum pH is the pH of maximum catalytic activity.
323. Extremely acidic or extremely basic pH can reduce enzyme activity mainly by:
ⓐ. Increasing collision frequency too much
ⓑ. Converting enzymes into carbohydrates
ⓒ. Raising substrate concentration instantly
ⓓ. Disrupting bonds that maintain folding
Correct Answer: Disrupting bonds that maintain folding
Explanation: Very low or very high pH can alter many ionizable groups across the protein, not only at the active site. This widespread charge disruption weakens ionic interactions and hydrogen bonding that stabilize secondary and tertiary structure. As folding integrity is lost, the active site geometry is distorted and catalysis falls sharply. In many cases, this structural damage behaves like denaturation and may not fully reverse. The decline is therefore primarily structural rather than due to simple collision changes. Hence, extreme pH reduces activity by disrupting bonds that maintain folding.
324. Pepsin (a stomach enzyme) shows highest activity in a medium that is:
ⓐ. Neutral, around $pH\ 7$
ⓑ. Strongly acidic, around $pH\ 2$
ⓒ. Mildly basic, around $pH\ 8$
ⓓ. Strongly basic, around $pH\ 12$
Correct Answer: Strongly acidic, around $pH\ 2$
Explanation: Pepsin functions in the stomach, where the environment is highly acidic due to gastric acid. Its active site residues and overall structure are adapted to remain stable and catalytically effective at low pH. At neutral or basic pH, the ionization pattern needed for catalysis is disturbed and activity drops markedly. This is a classic example showing that different enzymes have different optimum pH values based on their physiological location. The acidic optimum supports efficient protein digestion in the stomach. Therefore, pepsin works best around $pH\ 2$.
325. Trypsin (a small-intestine enzyme) typically shows highest activity near:
ⓐ. Mildly basic, around $pH\ 8$
ⓑ. Strongly acidic, around $pH\ 2$
ⓒ. Strongly basic, around $pH\ 12$
ⓓ. Neutral, around $pH\ 7$
Correct Answer: Mildly basic, around $pH\ 8$
Explanation: Trypsin acts in the small intestine, where conditions are slightly alkaline due to bicarbonate-rich secretions. Its catalytic residues are tuned to the charge state found at mildly basic pH, enabling efficient substrate binding and peptide bond cleavage. If the pH becomes too acidic, those residues shift protonation states and catalysis weakens. If the pH becomes too high, structural stability can also be compromised. The optimum therefore aligns with the enzyme’s functional environment in digestion. Hence, trypsin works best around $pH\ 8$.
326. A buffer helps maintain enzyme activity because it:
ⓐ. Removes the enzyme from the solution
ⓑ. Makes enzyme reactions independent of pH
ⓒ. Resists sudden changes in pH
ⓓ. Permanently locks the active site
Correct Answer: Resists sudden changes in pH
Explanation: Enzymes require a suitable pH range to keep the correct ionization state of active-site residues and overall folding stability. A buffer system limits large pH shifts by neutralizing added acids or bases, so the reaction mixture stays near a target pH. This helps preserve the active site’s charge pattern and prevents rapid loss of activity. Buffers do not eliminate pH dependence; they simply keep pH stable during the experiment or in cells. Stable pH supports consistent enzyme performance and reproducible rates. Therefore, buffers support enzyme activity by resisting sudden pH changes.
327. If the pH is changed far from an enzyme’s optimum, the most direct immediate effect is often:
ⓐ. A change in amino acid sequence
ⓑ. Formation of new peptide bonds in enzyme
ⓒ. A sudden increase in enzyme quantity
ⓓ. Altered charge on key active-site groups
Correct Answer: Altered charge on key active-site groups
Explanation: pH directly changes the protonation state of ionizable groups, especially side chains located in or near the active site. When these groups gain or lose $H^+$, their charge and ability to form ionic or hydrogen-bond interactions changes. This can reduce substrate binding, disrupt catalytic acid–base steps, or misalign reactive groups needed for the mechanism. These effects occur quickly because they depend on equilibrium of protonation, not on slow changes like sequence alteration. The immediate outcome is therefore functional loss at the active site. Hence, the most direct effect is altered charge on key active-site groups.
328. A typical enzyme activity vs pH graph is expected to show:
ⓐ. A peak around optimum pH with decline on both sides
ⓑ. A constant flat line at all pH values
ⓒ. A steady rise with increasing pH only
ⓓ. A steady fall with increasing pH only
Correct Answer: A peak around optimum pH with decline on both sides
Explanation: Enzymes usually have an optimum pH where their active site is correctly ionized and the protein structure is most supportive of catalysis. As pH becomes more acidic or more basic than this optimum, the charge state of essential residues shifts and activity falls. At extreme pH, broader structural disruption can occur, further reducing activity. This produces a characteristic curve with a maximum at the optimum and lower activity on both sides. The shape reflects loss of catalytic efficiency when the chemical environment becomes unfavorable. Therefore, a peak at optimum pH with decline on both sides is expected.
329. Salivary amylase generally shows highest activity close to:
ⓐ. Strongly acidic, around $pH\ 2$
ⓑ. Strongly basic, around $pH\ 12$
ⓒ. Near-neutral, around $pH\ 7$
ⓓ. Mildly basic, around $pH\ 9$
Correct Answer: Near-neutral, around $pH\ 7$
Explanation: Salivary amylase functions in the mouth where the pH is usually near neutral. Its active site is adapted to maintain the correct charge distribution and shape under these conditions, allowing efficient breakdown of starch. In highly acidic conditions, its catalytic residues become improperly protonated and activity decreases sharply. In strongly basic conditions, stability and key interactions can also be disrupted. The optimum therefore aligns with the physiological environment of saliva. Hence, salivary amylase works best near $pH\ 7$.
330. The optimum pH of an enzyme is best predicted by considering the:
ⓐ. Enzyme’s color in solution
ⓑ. Enzyme’s location and functional environment
ⓒ. Number of carbon atoms in substrate
ⓓ. Total water percentage in tissue
Correct Answer: Enzyme’s location and functional environment
Explanation: Different body compartments and cellular locations maintain different pH conditions, and enzymes adapt to function best in their usual environment. The active site residues must have the correct protonation state at that local pH to support binding and catalysis. For example, digestive enzymes in acidic or basic regions show optima matching those conditions, while cytosolic enzymes often peak near neutral pH. This is a structure–function adaptation rather than a random property. Knowing where an enzyme operates helps predict its optimum pH in exam questions. Therefore, optimum pH is best predicted from the enzyme’s location and functional environment.
331. When substrate concentration increases while enzyme concentration is fixed, the reaction rate typically:
ⓐ. Decreases continuously due to crowding
ⓑ. Remains constant at all substrate levels
ⓒ. Becomes zero after a small increase in substrate
ⓓ. Increases then approaches a maximum rate
Correct Answer: Increases then approaches a maximum rate
Explanation: As $[S]$ increases, more substrate molecules collide with enzyme active sites, so the rate rises because more enzyme–substrate complexes form per unit time. At low to moderate $[S]$, many active sites are still available, so adding substrate noticeably increases the reaction rate. Eventually, nearly all enzyme molecules have their active sites occupied most of the time, meaning the enzyme becomes “saturated.” Beyond this point, adding more substrate cannot increase the number of active sites, so the rate approaches a limiting maximum called $V_{max}$. This plateau reflects enzyme limitation, not substrate shortage. Therefore, the rate increases and then approaches a maximum.
332. At very low substrate concentration (with fixed enzyme), the reaction rate is generally:
ⓐ. Independent of $[S]$
ⓑ. Directly proportional to $[S]$
ⓒ. Equal to $V_{max}$ immediately
ⓓ. Greater than $V_{max}$
Correct Answer: Directly proportional to $[S]$
Explanation: When $[S]$ is very low, most enzyme active sites are empty most of the time, so formation of enzyme–substrate complexes depends strongly on how often substrate molecules collide with the enzyme. In this region, each increase in $[S]$ raises the frequency of successful binding events, producing a near-linear rise in reaction rate. The enzyme is not saturated, so the limiting factor is substrate availability rather than enzyme capacity. This is why the reaction shows first-order dependence on substrate at low $[S]$. As $[S]$ increases further, this proportionality gradually weakens as saturation begins. Hence, the rate is directly proportional to $[S]$ at very low substrate concentration.
333. When an enzyme is saturated with substrate, the rate becomes limited primarily by:
ⓐ. Enzyme turnover capacity, not $[S]$
ⓑ. Diffusion of water molecules in solution
ⓒ. The number of peptide bonds in the enzyme
ⓓ. The number of DNA bases in the enzyme
Correct Answer: Enzyme turnover capacity, not $[S]$
Explanation: At saturation, essentially every enzyme active site is occupied most of the time, so additional substrate cannot increase the fraction of enzyme bound to substrate. The reaction rate is then determined by how fast the enzyme can convert bound substrate to product and release it, allowing the next catalytic cycle. This maximum catalytic throughput is captured by $V_{max}$ for a fixed amount of enzyme. In this state, substrate is no longer the limiting factor; the enzyme’s catalytic cycling is. Therefore, changing $[S]$ further produces little to no increase in rate. Hence, under saturation the rate is limited by enzyme turnover capacity, not $[S]$.
334. The Michaelis constant $K_m$ is defined as the substrate concentration at which:
ⓐ. The enzyme is completely denatured
ⓑ. The reaction rate becomes zero
ⓒ. The reaction rate is half of $V_{max}$
ⓓ. The reaction rate becomes double of $V_{max}$
Correct Answer: The reaction rate is half of $V_{max}$
Explanation: $K_m$ is a key parameter that connects substrate concentration to reaction rate for many enzyme-catalyzed reactions. By definition, it is the substrate concentration at which the reaction velocity reaches half of its maximum value, i.e., $v = \frac{1}{2}V_{max}$. This point is useful because it lies in the mid-range of the enzyme’s response to substrate, where changes in $[S]$ still significantly affect rate. $K_m$ is commonly used in exams to interpret enzyme efficiency and compare how different enzymes respond to substrate levels. The definition is specific and does not describe denaturation or zero rate. Therefore, $K_m$ is the $[S]$ at which the rate is half of $V_{max}$.
335. If Enzyme X has a lower $K_m$ than Enzyme Y for the same substrate, it usually indicates:
ⓐ. Enzyme X has lower specificity for the substrate
ⓑ. Enzyme X reaches $V_{max}$ only at very low temperature
ⓒ. Enzyme X cannot be saturated by substrate
ⓓ. Enzyme X binds the substrate effectively at lower $[S]$
Correct Answer: Enzyme X binds the substrate effectively at lower $[S]$
Explanation: $K_m$ reflects how much substrate is needed for an enzyme to reach half of $V_{max}$. A lower $K_m$ means that half-maximal velocity is achieved at a lower substrate concentration, which typically implies effective substrate binding under dilute conditions. In practical terms, the enzyme forms enzyme–substrate complexes readily even when $[S]$ is not high. This is often interpreted as higher apparent affinity in standard textbook-level questions. It does not mean the enzyme cannot saturate; saturation still occurs, but at lower $[S]$ relative to an enzyme with higher $K_m$. Therefore, a lower $K_m$ usually indicates effective binding at lower $[S]$.
336. If $[S]$ is increased from high to even higher values while the enzyme is already saturated, the rate will:
ⓐ. Remain nearly unchanged near $V_{max}$
ⓑ. Fall sharply due to reduced collisions
ⓒ. Drop to zero because substrate blocks the enzyme permanently
ⓓ. Become strictly proportional to $[S]$ again
Correct Answer: Remain nearly unchanged near $V_{max}$
Explanation: Once saturation is reached, active sites are occupied almost continuously, so the enzyme is operating close to its maximal catalytic capacity. Increasing $[S]$ further does not create additional active sites or increase the fraction of enzyme bound to substrate beyond “almost all.” The limiting step becomes the enzyme’s turnover cycle, so the velocity ensure stays near $V_{max}$. This is why the rate–substrate curve flattens at high substrate levels. Any additional substrate mainly remains unbound in solution without increasing catalytic throughput. Therefore, the rate remains nearly unchanged near $V_{max}$.
337. At $[S] = K_m$, the reaction velocity $v$ is:
ⓐ. Equal to $V_{max}$
ⓑ. Equal to $2V_{max}$
ⓒ. Equal to $\frac{1}{2}V_{max}$
ⓓ. Equal to zero
Correct Answer: Equal to $\frac{1}{2}V_{max}$
Explanation: The defining property of $K_m$ is that it corresponds to the substrate concentration at which the reaction rate is half of the maximum possible rate for that enzyme concentration. Therefore, when $[S] = K_m$, the velocity is exactly $v = \frac{1}{2}V_{max}$. This relationship is frequently used to interpret enzyme kinetics questions and to compare enzymes under the same conditions. It also marks the transition region where the rate changes from being strongly dependent on $[S]$ (low substrate) to becoming less sensitive as saturation approaches. The relationship is definitional and ensures a clear numerical link between $K_m$ and $V_{max}$. Hence, at $[S]=K_m$, $v$ equals $\frac{1}{2}V_{max}$.
338. Increasing substrate concentration can most directly reduce the effect of a competitive inhibitor because:
ⓐ. The inhibitor converts into product at high $[S]$
ⓑ. Substrate outcompetes inhibitor for the same active site
ⓒ. The enzyme permanently loses its active site at high $[S]$
ⓓ. The inhibitor becomes a coenzyme at high $[S]$
Correct Answer: Substrate outcompetes inhibitor for the same active site
Explanation: In competitive inhibition, the inhibitor and substrate compete for binding to the same active site region on the enzyme. If $[S]$ is increased, the probability that a substrate molecule occupies the active site instead of the inhibitor increases. This shifts binding toward enzyme–substrate complex formation, helping restore reaction rate toward the uninhibited maximum at sufficiently high substrate levels. The key idea is competition for the same binding site, not conversion of inhibitor into product. This concept connects substrate concentration directly to observed inhibition strength. Therefore, higher $[S]$ can reduce competitive inhibition because substrate outcompetes the inhibitor at the active site.
339. When $[S]$ is very high and the reaction rate has reached $V_{max}$, the factor that would most effectively increase the rate further is:
ⓐ. Decreasing the temperature strongly
ⓑ. Lowering the pH far from optimum
ⓒ. Removing all substrate from the mixture
ⓓ. Increasing enzyme concentration
Correct Answer: Increasing enzyme concentration
Explanation: At $V_{max}$, the enzyme is saturated, so the reaction rate is limited by the total number of active sites available and how fast each site can catalyze turnover. Adding more substrate will not help because the existing enzymes are already working near full capacity. Increasing enzyme concentration adds more active sites, allowing more enzyme–substrate complexes to form at the same time, which raises the maximum achievable rate. This is why $V_{max}$ is proportional to enzyme concentration under saturating substrate. Changing temperature or pH away from optimum generally reduces activity rather than increasing it. Therefore, increasing enzyme concentration is the most effective way to raise the rate beyond the current $V_{max}$ level.
340. The most appropriate shape for a typical enzyme velocity vs $[S]$ plot (with fixed enzyme) is:
ⓐ. A curve that rises quickly then levels off
ⓑ. A straight line increasing forever
ⓒ. A straight line decreasing forever
ⓓ. A flat line at zero for all $[S]$
Correct Answer: A curve that rises quickly then levels off
Explanation: At low $[S]$, the rate increases rapidly as more substrate binds to available active sites, producing a strong rise in velocity. As $[S]$ continues to increase, a growing fraction of enzyme molecules remain occupied, and the rate becomes less sensitive to additional substrate. Eventually, saturation is approached and the velocity nears $V_{max}$, creating a plateau. This explains the classic hyperbolic pattern associated with many enzyme systems under basic conditions. The curve reflects a transition from substrate-limited behavior to enzyme-limited behavior. Therefore, the plot rises quickly at first and then levels off.
341. In competitive inhibition, the inhibitor most commonly:
ⓐ. Binds at the active site and competes with substrate
ⓑ. Binds only to the enzyme–substrate complex
ⓒ. Permanently destroys the enzyme by breaking peptide bonds
ⓓ. Converts the substrate directly into product without enzyme
Correct Answer: Binds at the active site and competes with substrate
Explanation: Competitive inhibitors resemble the substrate sufficiently to fit into the enzyme’s active site. By occupying the active site, they prevent the substrate from binding at that moment, reducing the formation of enzyme–substrate complex. Because binding is competitive and reversible in most standard cases, increasing substrate concentration can reduce inhibitor impact by increasing the chance that substrate occupies the active site instead. The enzyme’s catalytic machinery is not destroyed; it is simply blocked temporarily. This is why competitive inhibition primarily affects substrate binding rather than the catalytic step itself. Therefore, competitive inhibition occurs by active-site competition with the substrate.
342. A defining kinetic feature of competitive inhibition is that it can be reduced by:
ⓐ. Decreasing enzyme concentration to zero
ⓑ. Increasing substrate concentration
ⓒ. Increasing inhibitor concentration only
ⓓ. Lowering temperature below freezing always
Correct Answer: Increasing substrate concentration
Explanation: In competitive inhibition, the substrate and inhibitor compete for the same active site. When substrate concentration rises, substrate molecules are more likely to bind the active site than the inhibitor at any given time. This shifts the balance toward more enzyme–substrate complexes, increasing reaction rate toward the uninhibited maximum at sufficiently high $[S]$. The inhibitor is not removing the enzyme permanently; it is creating competition for binding. This property distinguishes competitive inhibition from cases where inhibition cannot be overcome by extra substrate. Hence, increasing substrate concentration reduces the effect of competitive inhibition.
343. In competitive inhibition, the apparent $K_m$ is typically:
ⓐ. Decreased because substrate binds tighter
ⓑ. Converted into $V_{max}$ at all concentrations
ⓒ. Unchanged because inhibitor never affects binding
ⓓ. Increased because more substrate is needed to reach $\frac{1}{2}V_{max}$
Correct Answer: Increased because more substrate is needed to reach $\frac{1}{2}V_{max}$
Explanation: $K_m$ is the substrate concentration needed to reach half of $V_{max}$, so it reflects how readily the enzyme achieves a given rate at a given $[S]$. With competitive inhibition, the inhibitor occupies the active site part of the time, so a higher substrate concentration is required to achieve the same level of enzyme–substrate complex formation. As a result, the curve shifts right and the apparent $K_m$ increases. This change is about competition for binding, not about lowering the maximum capacity of the enzyme at saturating substrate. Therefore, competitive inhibition increases the apparent $K_m$ because more substrate is needed to reach $\frac{1}{2}V_{max}$.
344. In competitive inhibition, $V_{max}$ is generally:
ⓐ. Reduced because enzyme is permanently damaged
ⓑ. Increased because inhibitor activates enzyme
ⓒ. Unchanged because high $[S]$ can outcompete inhibitor
ⓓ. Forced to zero regardless of substrate level
Correct Answer: Unchanged because high $[S]$ can outcompete inhibitor
Explanation: $V_{max}$ represents the maximum rate when enzyme active sites are effectively saturated with substrate. In competitive inhibition, the inhibitor does not destroy the enzyme; it only competes for the same active site. At sufficiently high substrate concentration, substrate can occupy active sites most of the time, allowing the enzyme to reach its usual maximal turnover rate. Therefore, the maximum achievable rate remains essentially the same, even though more substrate is required to reach it. This is why competitive inhibition shifts the curve but does not lower the plateau. Hence, $V_{max}$ remains unchanged because high $[S]$ can outcompete the inhibitor.
345. Which description best matches a competitive inhibitor?
ⓐ. A molecule that binds at a site different from the active site
ⓑ. A molecule structurally similar to substrate that fits the active site
ⓒ. A molecule that binds only after product formation
ⓓ. A molecule that replaces the enzyme’s peptide bonds
Correct Answer: A molecule structurally similar to substrate that fits the active site
Explanation: Competitive inhibitors typically resemble the substrate in shape or key functional groups, allowing them to fit into the enzyme’s active site. Because they occupy the same binding region, they prevent substrate binding during that time, lowering the reaction rate at a given substrate concentration. This resemblance is why competition occurs directly at the active site rather than through a separate regulatory site. The interaction is often reversible, so substrate can displace the inhibitor when $[S]$ increases. The enzyme is not chemically converted into something else; it is simply blocked from binding the real substrate. Therefore, a competitive inhibitor is a substrate-like molecule that fits the active site.
346. If a competitive inhibitor is added, the substrate concentration required to reach $V_{max}$:
ⓐ. Becomes lower than before
ⓑ. Becomes roughly unchanged at all levels
ⓒ. Becomes zero because inhibitor makes enzyme faster
ⓓ. Becomes higher than before
Correct Answer: Becomes higher than before
Explanation: Competitive inhibitors reduce the fraction of enzyme active sites available for substrate binding at any given substrate concentration. To achieve near-saturation of enzyme with substrate in the presence of an inhibitor, a higher substrate concentration is needed to outcompete the inhibitor for active-site occupancy. The maximum rate itself can still be reached, but it requires pushing the binding equilibrium strongly in favor of substrate by increasing $[S]$. This is why the rate–$[S]$ curve shifts to the right under competitive inhibition. The change is about needing more substrate, not about creating a new maximum. Hence, the substrate concentration required to approach $V_{max}$ becomes higher.
347. A key hallmark of competitive inhibition is that the inhibitor:
ⓐ. Binds to active site and prevents substrate binding
ⓑ. Lowers $V_{max}$ by destroying enzyme structure
ⓒ. Works only at very high temperature always
ⓓ. Forms phosphodiester bonds with the substrate
Correct Answer: Binds to active site and prevents substrate binding
Explanation: Competitive inhibitors act by occupying the active site, blocking the substrate from binding at that moment. Because substrate and inhibitor cannot bind the active site simultaneously, the inhibitor directly reduces enzyme–substrate complex formation at a given $[S]$. The enzyme’s structure is not necessarily destroyed, which is why $V_{max}$ can remain unchanged at very high substrate concentration. This mechanism is fundamentally about binding competition, not about covalent backbone formation or temperature-only behavior. The effect is therefore strongest when substrate is low and decreases as substrate becomes abundant. Thus, competitive inhibition is characterized by active-site binding that prevents substrate binding.
348. In a scenario where substrate and inhibitor both fit the same active site, increasing inhibitor concentration (with fixed enzyme and substrate) will most directly:
ⓐ. Increase reaction rate by activating the enzyme
ⓑ. Leave the reaction rate unchanged always
ⓒ. Reduce reaction rate due to fewer enzyme–substrate complexes
ⓓ. Convert enzyme into a substrate-like molecule
Correct Answer: Reduce reaction rate due to fewer enzyme–substrate complexes
Explanation: When inhibitor and substrate compete for the same active site, higher inhibitor concentration increases the chance that the inhibitor occupies the active site rather than the substrate. This lowers the frequency of productive enzyme–substrate complex formation and decreases the observed reaction velocity at that fixed substrate concentration. The enzyme is not being used up; it is being blocked more often. Because competition is occurring at the binding step, the immediate effect is reduced catalysis due to reduced substrate access to the active site. This is the central cause-effect chain in competitive inhibition questions. Therefore, increasing inhibitor concentration reduces the rate by lowering enzyme–substrate complex formation.
349. Which change is most consistent with competitive inhibition when comparing kinetic parameters?
ⓐ. $K_m$ increases while $V_{max}$ remains unchanged
ⓑ. $K_m$ decreases while $V_{max}$ increases
ⓒ. $K_m$ and $V_{max}$ both decrease together
ⓓ. $K_m$ and $V_{max}$ both increase together
Correct Answer: $K_m$ increases while $V_{max}$ remains unchanged
Explanation: Competitive inhibition primarily interferes with substrate binding because inhibitor competes for the active site. This means more substrate is needed to achieve the same velocity, so the apparent $K_m$ increases. However, at sufficiently high substrate concentration, substrate can outcompete the inhibitor and the enzyme can still reach its normal maximum catalytic capacity, so $V_{max}$ remains unchanged. This parameter pattern is the standard diagnostic signature of competitive inhibition in exam-level kinetics. It reflects a right-shift of the curve without lowering the plateau. Hence, the consistent change is increased $K_m$ with unchanged $V_{max}$.
350. A student claims “Competitive inhibition permanently inactivates enzymes.” The most accurate correction is:
ⓐ. Competitive inhibition converts enzymes into cofactors
ⓑ. Competitive inhibition always breaks enzyme peptide bonds
ⓒ. Competitive inhibition always lowers $V_{max}$ to zero
ⓓ. Competitive inhibition usually involves reversible active-site binding
Correct Answer: Competitive inhibition usually involves reversible active-site binding
Explanation: Competitive inhibition typically occurs when an inhibitor binds the active site through non-covalent interactions, competing with substrate. Because the binding is commonly reversible, the inhibitor can dissociate and allow substrate binding, especially when substrate concentration increases. This is why competitive inhibition can often be overcome by raising $[S]$ and why $V_{max}$ can remain unchanged. Permanent inactivation is more consistent with irreversible inhibition, not competitive reversible binding. The competitive mechanism is therefore about temporary occupancy of the active site, not destruction of enzyme structure. Hence, the correct correction is that competitive inhibition usually involves reversible active-site binding.
351. In non-competitive inhibition (idea level), the inhibitor typically:
ⓐ. Binds only at the active site like substrate
ⓑ. Binds at a site other than the active site
ⓒ. Replaces the substrate permanently in solution
ⓓ. Acts only by increasing substrate concentration
Correct Answer: Binds at a site other than the active site
Explanation: Non-competitive inhibitors reduce enzyme activity by binding to a site distinct from the active site, often altering enzyme conformation or catalytic efficiency. Because the inhibitor does not need to occupy the substrate-binding pocket, it can inhibit even when substrate is present. This binding can change how well the enzyme performs catalysis, even if substrate can still bind. The core idea is “different site, functional disruption,” rather than direct competition for the same binding region. This is why increasing substrate concentration generally does not fully reverse the inhibition. Therefore, non-competitive inhibitors typically bind at a site other than the active site.
352. A key idea-level feature of non-competitive inhibition is that it is generally:
ⓐ. Overcome completely by increasing substrate concentration
ⓑ. Independent of inhibitor binding to the enzyme
ⓒ. Not fully reversed by increasing substrate concentration
ⓓ. Caused only by lowering temperature
Correct Answer: Not fully reversed by increasing substrate concentration
Explanation: In non-competitive inhibition, the inhibitor reduces catalytic activity without directly competing for the active site. Since the inhibitor binds elsewhere, adding more substrate does not force the inhibitor off the enzyme in the same way as competitive inhibition. Even if substrate binding continues, the enzyme’s catalytic function is impaired because the inhibitor alters the enzyme’s working form or catalytic step. This is the central conceptual difference tested between competitive and non-competitive inhibition. The inhibition therefore persists across substrate increases, though the exact pattern can vary in detailed kinetics. Hence, non-competitive inhibition is not fully reversed by increasing substrate concentration.
353. The most appropriate idea-level effect of non-competitive inhibition on $V_{max}$ is that it:
ⓐ. Decreases because effective active enzyme is reduced
ⓑ. Increases because inhibitor activates enzyme
ⓒ. Becomes infinite at high substrate concentration
ⓓ. Stays unchanged because inhibitor binds active site
Correct Answer: Decreases because effective active enzyme is reduced
Explanation: $V_{max}$ reflects the maximum rate when the enzyme system is working at full catalytic capacity under saturating substrate. Non-competitive inhibitors reduce the effective catalytic ability of enzyme molecules, so even with plenty of substrate present, the maximum achievable rate is lowered. This is because the inhibitor-bound enzyme cannot contribute fully to catalysis or becomes less efficient. Since the limitation is not substrate availability, the plateau itself drops. This idea-level result contrasts with competitive inhibition, where the plateau can remain the same. Therefore, non-competitive inhibition decreases $V_{max}$ by reducing effective active enzyme.
354. A common idea-level description of non-competitive inhibition is that the inhibitor:
ⓐ. Prevents substrate from ever binding to the enzyme
ⓑ. Always mimics the substrate shape perfectly
ⓒ. Acts only when substrate concentration is zero
ⓓ. Makes the enzyme catalytically less effective after binding elsewhere
Correct Answer: Makes the enzyme catalytically less effective after binding elsewhere
Explanation: Non-competitive inhibitors bind at a site distinct from the active site and reduce enzyme activity by altering conformation or catalytic function. Substrate may still bind, but the enzyme’s ability to convert substrate to product is decreased. This is why the key concept is reduced catalytic efficiency rather than blocked binding. The inhibitor does not have to resemble the substrate because it is not competing for the same binding pocket. The defining feature is functional disruption through allosteric or non-active-site binding. Hence, non-competitive inhibition makes the enzyme less effective after binding elsewhere.
355. If an inhibitor binds to both free enzyme and enzyme–substrate complex at a non-active site, the inhibition is best considered:
ⓐ. Competitive (active-site only)
ⓑ. Non-competitive (idea-level allosteric binding)
ⓒ. Substrate activation
ⓓ. Denaturation only
Correct Answer: Non-competitive (idea-level allosteric binding)
Explanation: Non-competitive inhibition at the idea level refers to inhibitors that do not compete with substrate for the active site and can bind regardless of whether substrate is present. Binding to both free enzyme and the enzyme–substrate complex at another site fits this concept because it reduces catalytic output without requiring active-site competition. The result is decreased effective enzyme activity and reduced maximum rate. This is consistent with the allosteric binding idea that the inhibitor affects function through a separate site. It differs from competitive inhibition, which is mainly active-site competition. Therefore, this pattern is best considered non-competitive at the idea level.
356. Compared with competitive inhibition, non-competitive inhibition is more likely to:
ⓐ. Leave $V_{max}$ unchanged at high $[S]$
ⓑ. Increase $V_{max}$ as inhibitor rises
ⓒ. Lower $V_{max}$ even when $[S]$ is very high
ⓓ. Require inhibitor to resemble substrate closely
Correct Answer: Lower $V_{max}$ even when $[S]$ is very high
Explanation: Competitive inhibition can often be reduced at high substrate concentration because substrate outcompetes the inhibitor for the active site, allowing $V_{max}$ to be approached. In non-competitive inhibition, the inhibitor binds elsewhere and reduces catalytic effectiveness, so raising substrate concentration cannot fully restore maximum catalytic capacity. As a result, the observed plateau rate drops, meaning $V_{max}$ is lowered even when substrate is abundant. The inhibitor’s binding does not depend on substrate resemblance because it is not competing for the same pocket. This is the core contrast commonly tested. Hence, non-competitive inhibition lowers $V_{max}$ even at very high $[S]$.
357. A student says “Non-competitive inhibitors must look like the substrate.” The best correction is:
ⓐ. They usually bind at a different site, so resemblance is not required
ⓑ. They always bind the active site, so resemblance is required
ⓒ. They act only by increasing substrate concentration
ⓓ. They only function at $pH\ 2$
Correct Answer: They usually bind at a different site, so resemblance is not required
Explanation: Substrate resemblance is important mainly for competitive inhibitors because they must fit into the active site. Non-competitive inhibitors commonly bind at a different site, often called an allosteric site, where shape similarity to the substrate is not necessary. Their inhibitory effect comes from altering enzyme conformation or catalytic performance rather than blocking substrate binding directly. This is why they can inhibit even when substrate concentration is high. The concept is about binding location and functional consequence, not mimicry. Therefore, non-competitive inhibitors usually bind at a different site, so resemblance is not required.
358. In idea-level terms, adding more substrate to a non-competitively inhibited reaction most often results in:
ⓐ. Full restoration of normal maximum rate
ⓑ. No change at all in reaction rate
ⓒ. Complete removal of inhibitor from enzyme
ⓓ. Limited improvement but not full recovery of maximum rate
Correct Answer: Limited improvement but not full recovery of maximum rate
Explanation: Increasing substrate concentration can increase the rate to some extent by forming more enzyme–substrate complexes among the uninhibited enzyme molecules. However, because the inhibitor reduces catalytic capacity through non-active-site binding, the maximum achievable rate remains lower than normal. The inhibitor does not get displaced from its site simply by adding substrate, so full recovery of $V_{max}$ is not expected. This distinguishes non-competitive inhibition from competitive inhibition, where high $[S]$ can largely overcome the inhibitor. The key idea is that the plateau stays depressed. Hence, more substrate may give limited improvement but not full recovery of maximum rate.
359. The simplest idea-level statement about $K_m$ in pure non-competitive inhibition is that it is:
ⓐ. Increased because substrate cannot bind at all
ⓑ. Decreased because inhibitor strengthens binding
ⓒ. Often unchanged because binding site is not directly competed
ⓓ. Converted into $V_{max}$ at all conditions
Correct Answer: Often unchanged because binding site is not directly competed
Explanation: Non-competitive inhibitors act by binding outside the active site and reducing catalytic effectiveness, not primarily by preventing substrate binding. Because substrate-binding ability may remain similar, the substrate concentration needed to reach half of the new maximum can remain conceptually similar in the simplest model. In exam-oriented “idea level” treatment, the signature feature emphasized is decreased $V_{max}$ with $K_m$ often unchanged. The focus is that the inhibitor does not compete for the active site, so it does not necessarily change substrate affinity. While real systems can vary, the textbook conceptual pattern is commonly tested. Therefore, $K_m$ is often unchanged at the idea level.
360. Non-competitive inhibition is most directly explained by the idea that the inhibitor:
ⓐ. Blocks substrate entry by occupying the same pocket
ⓑ. Binds elsewhere and reduces catalytic function
ⓒ. Increases enzyme concentration automatically
ⓓ. Converts substrate into inhibitor chemically
Correct Answer: Binds elsewhere and reduces catalytic function
Explanation: The defining concept of non-competitive inhibition is binding at a site other than the active site, leading to reduced enzymatic activity. This binding can distort the active site, reduce turnover efficiency, or alter enzyme dynamics needed for catalysis. Because it does not rely on occupying the substrate-binding pocket, it is not simply a competition problem. This explains why increasing substrate concentration does not fully remove the inhibitory effect. The central idea is functional impairment through non-active-site binding. Therefore, non-competitive inhibition is explained by inhibitor binding elsewhere and reducing catalytic function.
361. $K_m$ is best interpreted (in basic exam sense) as the substrate concentration at which:
ⓐ. $v = V_{max}$
ⓑ. $v = 2V_{max}$
ⓒ. $v = \frac{1}{2}V_{max}$
ⓓ. $v = 0$
Correct Answer: $v = \frac{1}{2}V_{max}$
Explanation: $K_m$ is defined as the substrate concentration that produces half of the maximum velocity for a given fixed enzyme amount, i.e., $v = \frac{1}{2}V_{max}$. This point lies in the mid-range of the velocity–substrate curve, where the enzyme is neither mostly empty nor fully saturated. Because it is a definition, it remains a reliable anchor for graph-based questions and quick calculations. Conceptually, $K_m$ indicates how much substrate is needed to drive the enzyme to a significant fraction of its maximum rate. Lower or higher $K_m$ values shift how quickly the curve rises with $[S]$. Hence, $K_m$ corresponds to half-maximal velocity.
362. For an enzyme-catalyzed reaction measured at fixed enzyme amount, $V_{max}$ means the:
ⓐ. Highest rate when enzyme is saturated
ⓑ. Lowest rate when substrate is absent
ⓒ. Rate at $[S] = 0$ only
ⓓ. Rate that ignores enzyme concentration
Correct Answer: Highest rate when enzyme is saturated
Explanation: $V_{max}$ is the maximum velocity observed when substrate concentration is high enough that nearly all enzyme active sites are occupied most of the time. In this saturated condition, adding more substrate does not increase the rate because the limiting factor becomes the enzyme’s catalytic turnover capacity. $V_{max}$ therefore reflects how fast the available enzyme molecules can process substrate under ideal saturation. It is not the rate at zero substrate and it is not independent of enzyme amount. If enzyme concentration increases, $V_{max}$ rises because more active sites are present. Thus, $V_{max}$ is the highest rate under saturation.
363. If enzyme concentration is doubled (substrate kept saturating), the most expected change is:
ⓐ. $K_m$ doubles and $V_{max}$ stays same
ⓑ. $K_m$ halves and $V_{max}$ stays same
ⓒ. $K_m$ stays same and $V_{max}$ doubles
ⓓ. $K_m$ becomes zero and $V_{max}$ halves
Correct Answer: $K_m$ stays same and $V_{max}$ doubles
Explanation: With saturating substrate, the reaction is enzyme-limited, so adding more enzyme increases the total number of active sites and increases the maximum achievable rate. Therefore, doubling enzyme concentration typically doubles $V_{max}$. In contrast, $K_m$ is a property tied to how substrate concentration relates to reaching half of the maximum rate for that enzyme’s binding/catalytic system, and it does not depend on how much enzyme is present. So changing enzyme amount shifts the plateau but does not shift the substrate concentration scale of half-saturation. This is a common trap in board questions. Hence, $V_{max}$ doubles while $K_m$ remains unchanged.
364. An enzyme with a lower $K_m$ (same substrate, similar conditions) generally reaches high velocity at:
ⓐ. Higher $[S]$ than before
ⓑ. Only after denaturation
ⓒ. Only at $[S] = 0$
ⓓ. Lower $[S]$ than before
Correct Answer: Lower $[S]$ than before
Explanation: A lower $K_m$ means the enzyme achieves $v = \frac{1}{2}V_{max}$ at a lower substrate concentration. In practical exam interpretation, this implies the enzyme becomes significantly active even when substrate is relatively scarce. The velocity–$[S]$ curve rises more quickly along the substrate axis, so less substrate is needed to reach a given fraction of the maximum rate. This does not require denaturation or zero substrate; it is about how strongly the system responds to substrate availability. Therefore, lower $K_m$ is associated with reaching high velocity at lower $[S]$.
365. In competitive inhibition, the typical parameter change is:
ⓐ. $K_m$ increases; $V_{max}$ unchanged
ⓑ. $K_m$ unchanged; $V_{max}$ decreases
ⓒ. $K_m$ decreases; $V_{max}$ unchanged
ⓓ. $K_m$ decreases; $V_{max}$ increases
Correct Answer: $K_m$ increases; $V_{max}$ unchanged
Explanation: Competitive inhibitors compete with substrate for the same active site, so more substrate is needed to achieve the same occupancy of active sites. This shifts the velocity–$[S]$ curve to the right, increasing the apparent $K_m$ because $\frac{1}{2}V_{max}$ is reached at a higher $[S]$. However, at sufficiently high substrate concentration, substrate can outcompete the inhibitor and the enzyme can still reach its usual maximal turnover rate. Thus, the plateau ($V_{max}$) remains essentially unchanged in the standard model. The key idea is altered binding competition, not reduced catalytic capacity. Hence, $K_m$ rises while $V_{max}$ stays the same.
366. In non-competitive inhibition (idea level), the most expected change is:
ⓐ. $K_m$ increases; $V_{max}$ unchanged
ⓑ. $K_m$ decreases; $V_{max}$ unchanged
ⓒ. $K_m$ unchanged; $V_{max}$ decreases
ⓓ. $K_m$ increases; $V_{max}$ increases
Correct Answer: $K_m$ unchanged; $V_{max}$ decreases
Explanation: In the basic non-competitive idea, the inhibitor binds at a site other than the active site and reduces catalytic effectiveness of the enzyme. Because substrate binding is not the main step being directly competed, the substrate concentration scale for half-saturation is often treated as unchanged in simplified questions, so $K_m$ remains the same. The major visible effect is a lower maximum achievable rate because some enzyme molecules are functionally less effective even at saturating $[S]$. Therefore, the plateau drops, meaning $V_{max}$ decreases. This distinguishes it from competitive inhibition where the plateau is maintained. Hence, $V_{max}$ decreases while $K_m$ is unchanged.
367. When $[S] \gg K_m$, the reaction velocity is best expected to be close to:
ⓐ. $0$
ⓑ. $\frac{1}{2}V_{max}$
ⓒ. $V_{max}$
ⓓ. $2V_{max}$
Correct Answer: $V_{max}$
Explanation: When substrate concentration is much greater than $K_m$, active sites are occupied most of the time and the enzyme approaches saturation. In this region, adding more substrate produces little additional increase in rate because the limiting step becomes the enzyme’s turnover capacity rather than substrate availability. As a result, the velocity approaches the plateau value, $V_{max}$. This is why the velocity–$[S]$ curve flattens at high substrate levels. The condition $[S] \gg K_m$ is a standard clue for “near maximal rate” in exam problems. Therefore, the rate is close to $V_{max}$.
368. At $[S] = K_m$, the enzyme active sites are (idea level) roughly:
ⓐ. About 50% occupied on average
ⓑ. About 0% occupied on average
ⓒ. About 90% occupied on average
ⓓ. About 100% occupied on average
Correct Answer: About 50% occupied on average
Explanation: In the basic Michaelis–Menten interpretation, $K_m$ corresponds to the substrate concentration at which $v = \frac{1}{2}V_{max}$. Idea-level reasoning links half-maximal velocity with about half of the enzyme active sites being occupied by substrate on average. This makes $K_m$ a convenient marker for the “half-saturation” point of the enzyme. It is not a guarantee of exact occupancy in every system, but it is the standard conceptual mapping used in board and competitive questions. It helps students visualize what $K_m$ means in binding terms. Hence, at $[S] = K_m$, sites are roughly 50% occupied.
369. If $V_{max}$ is high but $K_m$ is also high for the same substrate, the best inference is:
ⓐ. Enzyme is fast at saturation but needs higher $[S]$
ⓑ. Enzyme is slow at saturation and needs lower $[S]$
ⓒ. Enzyme cannot reach saturation at any $[S]$
ⓓ. Enzyme works only at $pH\ 2$
Correct Answer: Enzyme is fast at saturation but needs higher $[S]$
Explanation: A high $V_{max}$ means that once the enzyme is saturated, it can process substrate at a high maximum rate for the given enzyme amount. A high $K_m$ means more substrate is needed to reach half of that maximum rate, so the enzyme becomes strongly active only at comparatively higher substrate concentrations. Together, these indicate an enzyme that can be very effective when substrate is abundant but is less effective under low substrate availability. This combined interpretation is commonly tested in “compare enzymes” questions. It does not imply impossibility of saturation or a fixed pH requirement. Therefore, it suggests fast saturation performance but higher $[S]$ requirement.
370. In a velocity vs $[S]$ curve, the most direct experimental indicator of $V_{max}$ is:
ⓐ. The initial steep slope at low $[S]$
ⓑ. The point where $v = \frac{1}{2}V_{max}$
ⓒ. The $[S]$ value where $v = 0$
ⓓ. The plateau level reached at high $[S]$
Correct Answer: The plateau level reached at high $[S]$
Explanation: $V_{max}$ is the maximum rate approached when substrate becomes sufficiently high that the enzyme is effectively saturated. On a standard velocity vs $[S]$ plot, this appears as the plateau region where further increases in $[S]$ produce minimal change in velocity. That plateau height represents the enzyme-limited maximum under the given conditions. The initial slope reflects behavior at low substrate, and the half-maximum point is used to locate $K_m$, not $V_{max}$. Therefore, the clearest indicator of $V_{max}$ is the plateau level at high $[S]$.
371. In most living tissues, water content by fresh weight is typically closest to:
ⓐ. $70-90\%$
ⓑ. $10-20\%$
ⓒ. $30-40\%$
ⓓ. $95-99\%$
Correct Answer: $70-90\%$
Explanation: Most cells and tissues are predominantly aqueous, so the fresh (wet) weight includes a large fraction of water. Water acts as the solvent for biochemical reactions, enables diffusion of solutes, and supports macromolecular interactions in cytoplasm. Because fresh weight includes both water and dissolved/particulate solids, the water fraction is high but not near-complete in most tissues. This typical range is frequently used to contrast wet weight with dry weight composition. It also explains why drying a tissue causes a major loss of mass. Hence, the common fresh-weight water content is about $70-90\%$.
372. Which item is best described as a “trace element” in living systems?
ⓐ. Carbon
ⓑ. Oxygen
ⓒ. Hydrogen
ⓓ. Zinc
Correct Answer: Zinc
Explanation: Trace elements are required in very small amounts compared with major elements like carbon, hydrogen, oxygen, and nitrogen. Despite their low concentration, they are essential because they often serve as cofactors for enzymes and stabilize protein structure or active-site chemistry. Zinc is a classic example because it supports catalytic function in several enzymes and contributes to proper biomolecular interactions. The term “trace” refers to quantity needed, not importance for life. Even small deficiency can disrupt metabolic pathways due to loss of enzyme efficiency. Therefore, zinc is best classified as a trace element.
373. The Benedict’s test is most directly used to detect:
ⓐ. Starch granules
ⓑ. Reducing sugars
ⓒ. Amino acids
ⓓ. Fatty acids
Correct Answer: Reducing sugars
Explanation: Benedict’s test detects sugars that can act as reducing agents due to a free aldehyde or keto group capable of reducing metal ions under the test conditions. Such sugars include glucose and other monosaccharides, and some disaccharides after considering their free anomeric carbon. The positive result reflects the sugar’s reducing property rather than merely being a carbohydrate. This is why sucrose, despite being a sugar, is typically negative unless first hydrolyzed. The test is commonly used to connect carbohydrate structure with chemical reactivity. Hence, Benedict’s test is used to detect reducing sugars.
374. Which disaccharide is typically classified as non-reducing because both anomeric carbons are involved in the linkage?
ⓐ. Sucrose
ⓑ. Maltose
ⓒ. Lactose
ⓓ. Cellobiose
Correct Answer: Sucrose
Explanation: A sugar is non-reducing when it lacks a free anomeric carbon that can open into a reactive aldehyde or keto form. In sucrose, the glycosidic bond connects the anomeric carbon of glucose to the anomeric carbon of fructose, so neither unit retains a free reducing end. As a result, sucrose does not readily reduce Benedict’s or similar reagents under standard conditions. This property directly arises from the bonding pattern, not from sweetness or solubility. The concept is tested to link glycosidic linkage position with reducing behavior. Therefore, sucrose is the typical non-reducing disaccharide.
375. The glycosidic linkage most characteristic of cellulose is:
ⓐ. $\alpha(1-4)$
ⓑ. $\alpha(1-6)$
ⓒ. $\beta(1-4)$
ⓓ. $\beta(1-6)$
Correct Answer: $\beta(1-4)$
Explanation: Cellulose is a structural polysaccharide built from repeating glucose units joined mainly by $\beta(1-4)$ glycosidic bonds. This linkage produces straight, unbranched chains that align closely and form extensive hydrogen bonding between chains. The resulting microfibrils provide high tensile strength, which explains cellulose’s major role in plant cell wall structure. The key exam point is that linkage type determines chain geometry and therefore function as structural or storage. This also explains why cellulose is not readily digested by many animals without specialized enzymes. Hence, cellulose is characterized by $\beta(1-4)$ linkages.
376. Chitin is best described as a polymer mainly of:
ⓐ. Ribose units
ⓑ. Adenine units
ⓒ. Fatty acid units
ⓓ. $N$-acetylglucosamine units
Correct Answer: $N$-acetylglucosamine units
Explanation: Chitin is a structural polysaccharide found in arthropod exoskeletons and fungal cell walls. Its monomer is $N$-acetylglucosamine, a modified glucose derivative, which forms long chains that contribute rigidity and strength. The polymer architecture supports protective and structural roles rather than energy storage. This “modified sugar polymer” idea is commonly tested to distinguish chitin from cellulose and from proteins. The presence of the acetylated amino group changes chemical properties and strengthens interactions in the matrix. Therefore, chitin is mainly a polymer of $N$-acetylglucosamine units.
377. “Essential amino acids” are those amino acids that:
ⓐ. Cannot form peptide bonds
ⓑ. Must be obtained from the diet
ⓒ. Always carry a negative charge
ⓓ. Are never used in proteins
Correct Answer: Must be obtained from the diet
Explanation: Essential amino acids are required for protein synthesis but cannot be synthesized in adequate amounts by the human body. Because the biosynthetic pathways for these amino acids are absent or insufficient, they must be supplied through dietary intake to maintain normal growth and tissue repair. This concept is central to nutrition-based biomolecules questions and helps explain protein-quality evaluation. When essential amino acids are deficient, protein synthesis can be limited even if total protein intake seems adequate. The term “essential” therefore reflects a physiological requirement for external supply. Hence, essential amino acids must be obtained from the diet.
378. At approximately neutral pH, most amino acids exist predominantly as:
ⓐ. Only uncharged molecules
ⓑ. Only negatively charged ions
ⓒ. Zwitterions with both charges
ⓓ. Only positively charged ions
Correct Answer: Zwitterions with both charges
Explanation: Amino acids contain both an acidic carboxyl group and a basic amino group, so they can gain and lose protons depending on pH. Near neutral pH, the carboxyl group is typically deprotonated (negative), while the amino group is protonated (positive), creating a zwitterion. This internal charge balance affects solubility, buffering behavior, and electrophoretic movement in biochemical separation. The zwitterionic form is a fundamental reason amino acids behave as amphoteric molecules. This concept is frequently tested as a base idea before moving to isoelectric point. Therefore, amino acids commonly exist as zwitterions at neutral pH.
379. In double-stranded DNA, if the number of adenine bases is 30, the number of thymine bases is:
ⓐ. 30
ⓑ. 15
ⓒ. 60
ⓓ. 90
Correct Answer: 30
Explanation: In double-stranded DNA, base pairing follows strict complementarity: adenine pairs with thymine, and guanine pairs with cytosine. Because each adenine on one strand pairs with exactly one thymine on the opposite strand, their counts are equal in a complete double-stranded molecule. This equality reflects pairing rules rather than total base content or strand length alone. The concept is used to connect structure (specific pairing) with composition calculations in exam questions. It also underlies why DNA maintains uniform helix width through purine–pyrimidine pairing. Hence, 30 adenines correspond to 30 thymines.
380. The RNA type most directly involved in bringing amino acids to the ribosome is:
ⓐ. rRNA
ⓑ. mRNA
ⓒ. snRNA
ⓓ. tRNA
Correct Answer: tRNA
Explanation: Transfer RNA (tRNA) acts as the adaptor molecule that carries specific amino acids to the ribosome during protein synthesis. Each tRNA has an anticodon region that recognizes complementary codons on mRNA, ensuring the correct amino acid is added to the growing polypeptide chain. This role links the nucleotide sequence of mRNA to the amino acid sequence of protein, which is the core logic of translation. The amino acid attachment occurs at the acceptor end of tRNA, enabling repeated delivery during elongation. This specificity is essential for faithful decoding of genetic information. Therefore, tRNA is the RNA that brings amino acids to the ribosome.
381. At the isoelectric point ($pI$), an amino acid in an electric field will most likely:
ⓐ. Move strongly toward the cathode
ⓑ. Move strongly toward the anode
ⓒ. Show no net migration overall
ⓓ. Split into two different amino acids
Correct Answer: Show no net migration overall
Explanation: At $pI$, an amino acid exists predominantly as a zwitterion with equal positive and negative charges, giving it zero net charge. In an electric field, net movement depends on net charge; when the net charge is zero, there is no overall attraction toward either electrode. Although internal charges are present, their effects cancel out in terms of directional migration. This is why amino acids show minimum electrophoretic mobility at their $pI$ under standard conditions. The concept is used to separate amino acids by electrophoresis by adjusting buffer pH. Therefore, at $pI$ an amino acid shows no net migration overall.
382. The peptide bond is considered rigid mainly because it has:
ⓐ. Partial double-bond character due to resonance
ⓑ. Complete ionic character between amino acids
ⓒ. A glycosidic linkage between two sugars
ⓓ. A phosphodiester linkage between nucleotides
Correct Answer: Partial double-bond character due to resonance
Explanation: The peptide bond involves resonance between the carbonyl group and the amide nitrogen, which delocalizes electrons across the bond. This delocalization gives the peptide bond partial double-bond character, making it shorter and less free to rotate than a typical single bond. Because rotation is restricted, the peptide linkage tends to be planar, which strongly influences protein backbone geometry. This rigidity is a key reason proteins have predictable secondary structures like helices and sheets. The concept is tested to connect chemical bonding with 3-D structure formation. Hence, peptide bonds are rigid mainly due to partial double-bond character from resonance.
383. Glycogen branching is primarily formed through:
ⓐ. Only $\alpha(1-4)$ bonds in a straight chain
ⓑ. Only $\beta(1-4)$ bonds in a straight chain
ⓒ. Only $\beta(1-6)$ bonds at branch points
ⓓ. $\alpha(1-6)$ bonds at branch points
Correct Answer: $\alpha(1-6)$ bonds at branch points
Explanation: Glycogen consists of glucose units mainly connected by $\alpha(1-4)$ glycosidic bonds in the linear segments, but branching occurs when a glucose residue forms an additional linkage. The specific bond that creates the branch point is the $\alpha(1-6)$ glycosidic bond, which connects a new chain to the main chain. This branching increases the number of terminal ends, enabling rapid addition and removal of glucose units during energy demand. The linkage type directly determines the architecture and functional advantage of glycogen as a storage polysaccharide. Therefore, glycogen branching is formed through $\alpha(1-6)$ bonds at branch points.
384. DNA with higher $G \equiv C$ content generally shows higher melting temperature mainly because:
ⓐ. $G \equiv C$ pairs have larger base size than $A=T$
ⓑ. $G \equiv C$ pairs have stronger bonding interactions
ⓒ. $G \equiv C$ pairs reduce the helical diameter
ⓓ. $G \equiv C$ pairs prevent replication entirely
Correct Answer: $G \equiv C$ pairs have stronger bonding interactions
Explanation: Double-stranded DNA stability depends on the strength of base-pair interactions and stacking within the helix. $G \equiv C$ base pairs form stronger bonding interactions than $A=T$ pairs, so more energy (higher temperature) is needed to separate the strands. As the fraction of $G \equiv C$ increases, the overall helix becomes more resistant to heat-driven strand separation. This is why DNA rich in $G \equiv C$ typically requires higher temperature to “melt” into single strands. The concept links base composition to physical stability and is common in competitive-style reasoning questions. Hence, higher $G \equiv C$ content increases melting temperature because the bonding interactions are stronger.
385. A transition-state analog inhibitor is expected to be highly effective mainly because it:
ⓐ. Binds the active site more tightly than the substrate
ⓑ. Converts the enzyme into a cofactor permanently
ⓒ. Increases $V_{max}$ by activating the enzyme
ⓓ. Breaks peptide bonds inside the enzyme
Correct Answer: Binds the active site more tightly than the substrate
Explanation: Enzymes are designed to stabilize the transition state of a reaction more than the substrate itself, which is a major reason they lower activation energy. A transition-state analog resembles the transition state’s shape and charge distribution, so it fits the active site exceptionally well. Because the active site has highest complementarity for the transition state, the analog can bind very strongly and occupy the catalytic pocket effectively. This strong binding blocks substrate access and sharply reduces reaction rate even at low inhibitor levels. The potency comes from exploiting the enzyme’s strongest binding preference, not from destroying the enzyme. Therefore, a transition-state analog inhibitor is effective because it binds the active site more tightly than the substrate.
386. In triacylglycerols (fats and oils), fatty acids are linked to glycerol mainly by:
ⓐ. Peptide bonds
ⓑ. Glycosidic bonds
ⓒ. Phosphodiester bonds
ⓓ. Ester bonds
Correct Answer: Ester bonds
Explanation: Triacylglycerols are formed when the three hydroxyl groups of glycerol react with the carboxyl groups of three fatty acids. This reaction produces ester bonds, creating a neutral, highly hydrophobic storage molecule suited for long-term energy storage. The ester linkage is the defining chemical connection in fats and oils and explains their nonpolar nature and water insolubility. This also clarifies why these lipids are not considered “true polymers” like proteins or nucleic acids, which have repeating monomers linked by peptide or phosphodiester bonds. The bond type is therefore central to lipid structure questions. Hence, fatty acids are linked to glycerol by ester bonds.