IB Biology Quiz: Apply Enzymes And Metabolism
20 questions · exam conditions
0:00
Apply Enzymes And MetabolismQuestion 1 of 20

In an experiment investigating an enzyme-catalyzed reaction, the rate is measured at various substrate concentrations while enzyme concentration is kept constant. A point is reached where increasing the substrate concentration further does not increase the reaction rate. Which statement best explains this plateau?

The enzyme has been fully consumed by the reaction and must be synthesized again.
The end-product of the reaction has accumulated and is acting as a competitive inhibitor.
The active sites of all available enzyme molecules are saturated with substrate.
The reaction has reached equilibrium, and the forward and reverse reactions are proceeding at equal rates.
← Back to quizzes

IB Biology Quiz

IB Biology Quiz: Apply Enzymes And Metabolism

Practice Apply Enzymes And Metabolism in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Enzymes And Metabolism, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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.

All questions

Question 1

In an experiment investigating an enzyme-catalyzed reaction, the rate is measured at various substrate concentrations while enzyme concentration is kept constant. A point is reached where increasing the substrate concentration further does not increase the reaction rate. Which statement best explains this plateau?

  1. The enzyme has been fully consumed by the reaction and must be synthesized again.
  2. The end-product of the reaction has accumulated and is acting as a competitive inhibitor.
  3. The active sites of all available enzyme molecules are saturated with substrate. (correct answer)
  4. The reaction has reached equilibrium, and the forward and reverse reactions are proceeding at equal rates.
Explanation: When the reaction rate stops increasing despite adding more substrate, it indicates that the enzyme concentration has become the limiting factor. At this point, nearly all enzyme active sites are occupied with substrate molecules at any given moment. The enzyme is working at its maximum possible rate (Vmax) under those conditions, and the rate is limited by how quickly the enzyme can process the substrate and release the product.

Question 2

The enzyme phosphofructokinase (PFK) is a key regulator of glycolysis. High concentrations of ATP bind to a site on PFK distinct from its active site, which leads to a decrease in the enzyme's activity. Which term correctly describes the role of ATP in this context?

  1. A competitive inhibitor, because it reduces the overall rate of glycolysis.
  2. An allosteric inhibitor, because it binds at a regulatory site and changes the enzyme's conformation. (correct answer)
  3. A coenzyme, because ATP is directly involved in the energy transfers of the pathway.
  4. An irreversible inhibitor, because high ATP levels signal that the cell has sufficient energy.
Explanation: Since ATP binds to a site other than the active site (an allosteric site) and causes a change in activity, it is acting as an allosteric regulator. Because it decreases activity, it is an allosteric inhibitor. This is a form of non-competitive inhibition and a key example of feedback regulation in metabolism. It is not competitive because it does not bind to the active site.

Question 3

Pepsin is a protease that functions optimally at the highly acidic pH of 2 found in the stomach. If pepsin passes into the small intestine, where the pH is approximately 8, its activity ceases. What is the most likely reason for this loss of activity?

  1. The alkaline pH causes the peptide bonds in the pepsin's primary structure to break, destroying the enzyme.
  2. The change in H+ concentration alters the ionic bonds in the enzyme's tertiary structure, changing the shape of the active site. (correct answer)
  3. The substrate molecules are denatured by the alkaline pH and can no longer fit into pepsin's active site.
  4. Hydroxide ions (OH⁻) act as competitive inhibitors, binding directly to the active site and blocking the substrate.
Explanation: Enzyme structure, and therefore function, is highly dependent on pH. Extreme changes in pH disrupt the pattern of ionic and hydrogen bonds that hold the enzyme in its specific three-dimensional tertiary structure. This change in conformation alters the active site, so the substrate can no longer bind effectively, leading to a loss of activity. This process is called denaturation. Breaking peptide bonds (primary structure) requires more extreme conditions.

Question 4

Methanol is toxic because it is converted to formaldehyde by the enzyme alcohol dehydrogenase. A standard emergency treatment for methanol poisoning is the administration of a large quantity of ethanol. Which statement best explains the biochemical basis for this treatment?

  1. Ethanol is a non-competitive inhibitor of alcohol dehydrogenase, changing its shape so it cannot bind methanol.
  2. Ethanol provides the necessary energy to reverse the conversion of methanol to formaldehyde, detoxifying the system.
  3. Ethanol causes the rapid denaturation of alcohol dehydrogenase, preventing any further reaction.
  4. Ethanol acts as a competitive inhibitor, competing for the active site of alcohol dehydrogenase and slowing formaldehyde production. (correct answer)
Explanation: Ethanol and methanol are structurally similar. Because of this, ethanol can also bind to the active site of alcohol dehydrogenase. When administered in high doses, ethanol outcompetes methanol for the active sites. This slows down the rate at which methanol is converted to toxic formaldehyde, giving the body time to excrete the methanol harmlessly. This is a classic medical application of competitive inhibition.

Question 5

Phenylketonuria (PKU) is a metabolic disorder resulting from a mutation in the gene for the enzyme phenylalanine hydroxylase. This enzyme normally converts the amino acid phenylalanine into tyrosine. In an individual with untreated PKU, what are the expected consequences for amino acid levels?

  1. The concentrations of both phenylalanine and tyrosine will be abnormally high.
  2. The concentration of phenylalanine will be abnormally high, and the concentration of tyrosine will be abnormally low. (correct answer)
  3. The concentration of phenylalanine will be abnormally low, and the concentration of tyrosine will be abnormally high.
  4. The concentrations of both phenylalanine and tyrosine will be abnormally low.
Explanation: If the enzyme that converts phenylalanine (the substrate) to tyrosine (the product) is non-functional, the metabolic pathway is blocked. This will cause the substrate, phenylalanine, to accumulate to toxic levels. Simultaneously, the product, tyrosine, will not be synthesized from this pathway, leading to abnormally low levels. This requires a basic understanding of substrate-product relationships in a metabolic pathway.

Question 6

The synthesis of the amino acid isoleucine from threonine is a metabolic pathway controlled by end-product inhibition. Isoleucine, the final product, acts as an allosteric inhibitor of the first enzyme in the pathway, threonine deaminase. If a bacterial culture is provided with a high concentration of external isoleucine, what would be the expected immediate outcome within the cells?

  1. The rate of threonine synthesis will increase to compensate for the inhibition.
  2. The activity of threonine deaminase will decrease, leading to an accumulation of threonine. (correct answer)
  3. The concentration of all intermediates in the pathway will increase significantly.
  4. The genes coding for the pathway's enzymes will be permanently repressed.
Explanation: High levels of isoleucine will bind to the allosteric site of threonine deaminase, inhibiting its function. This will stop the conversion of threonine into the next intermediate. Consequently, the enzyme's activity decreases, and its substrate, threonine, will accumulate because it is no longer being consumed by this pathway. This is a classic example of negative feedback.

Question 7

Heavy metal ions such as mercury (Hg²⁺) are potent enzyme inhibitors. They often bind to the sulfhydryl (-SH) groups of cysteine residues within the enzyme's structure. What is the most direct consequence of this binding?

  1. It mimics the substrate, acting as a competitive inhibitor that blocks the active site.
  2. It provides a cofactor needed for the reaction, causing the reaction to proceed too quickly.
  3. It disrupts disulfide bridges and other bonds crucial for maintaining the enzyme's tertiary structure. (correct answer)
  4. It alters the DNA sequence that codes for the enzyme, causing a mutated protein to be produced.
Explanation: Cysteine residues and their sulfhydryl groups are often involved in forming disulfide bridges, which are vital for stabilizing the tertiary and quaternary structures of proteins. When heavy metal ions bind to these groups, they disrupt these critical bonds, causing the enzyme to lose its specific 3D shape (denature) and, consequently, its function. This is a form of non-competitive, often irreversible, inhibition.

Question 8

A metabolic pathway consists of four sequential reactions, A → B → C → D, each catalyzed by a specific enzyme. The final product, D, acts as a non-competitive inhibitor of the first enzyme (E₁), which catalyzes A → B. If the cell has an excess of product D, what will be the effect on the concentrations of the intermediates?

  1. Concentrations of A, B, and C will all increase.
  2. Concentration of A will increase, while concentrations of B and C will decrease. (correct answer)
  3. Concentration of A will decrease, while concentrations of B and C will increase.
  4. Concentrations of A, B, and C will all decrease.
Explanation: Excess D will inhibit E₁, the enzyme converting A to B. This creates a bottleneck at the beginning of the pathway. As a result, the substrate for this enzyme, A, will accumulate as it is not being consumed. Since the production of B is blocked, the subsequent intermediates, B and C, will not be formed and their concentrations will decrease as they are used up by the later steps of the pathway or other cellular processes.

Question 9

Consider a metabolic pathway S → I₁ → P, catalyzed by enzymes E₁ and E₂. If the Vmax of E₁ is much greater than the Vmax of E₂, which of the following is the most likely consequence under conditions of high initial substrate (S) concentration?

  1. The concentration of the intermediate I₁ will remain very low throughout the process.
  2. The overall rate of production of P will be determined by the activity of E₁.
  3. The concentration of the intermediate I₁ will increase until its production rate equals its consumption rate. (correct answer)
  4. The enzyme E₂ will experience end-product inhibition from the accumulation of P.
Explanation: If E₁ works much faster than E₂, it will rapidly convert S into the intermediate I₁. Since E₂ consumes I₁ slowly, I₁ will accumulate. The concentration of I₁ will continue to rise, increasing the rate of the reaction catalyzed by E₂ (as per Michaelis-Menten kinetics) until the rate of I₁ consumption by E₂ matches the rate of I₁ production by E₁. The step catalyzed by E₂ is the rate-limiting step of the pathway.

Question 10

An enzyme-catalysed reaction is proceeding at 75% of its Vmax. According to the Michaelis-Menten equation, what is the ratio of the substrate concentration [S] to the Michaelis constant Km?

  1. 1:1
  2. 2:1
  3. 3:1 (correct answer)
  4. 4:1
Explanation: The Michaelis-Menten equation is V = (Vmax * [S]) / (Km + [S]). We are given that V = 0.75 * Vmax. Substituting this in gives: 0.75 * Vmax = (Vmax * [S]) / (Km + [S]). We can cancel Vmax from both sides: 0.75 = [S] / (Km + [S]). Rearranging the equation: 0.75 * (Km + [S]) = [S], which gives 0.75 * Km + 0.75 * [S] = [S]. Subtracting 0.75 * [S] from both sides gives 0.75 * Km = 0.25 * [S]. To find the ratio [S]/Km, we rearrange to [S]/Km = 0.75 / 0.25 = 3. Therefore, the ratio of [S] to Km is 3:1.

Question 11

The enzyme lysozyme breaks down bacterial cell walls. Its active site can bind a chain of six sugar units. However, its catalytic activity is highest when it binds a full chain of six units, and it shows very low activity if it binds a chain of only two or three units. This is an example of:

  1. Competitive inhibition by shorter sugar chains.
  2. Enzyme saturation, where only the longer chain is able to fully saturate the active site.
  3. Allosteric activation, where the first few sugar units act as activators for the later units.
  4. The induced-fit model, where optimal binding of the full substrate induces the ideal catalytic conformation. (correct answer)
Explanation: This observation strongly supports the induced-fit model. The binding of the full-length, six-unit substrate likely induces the precise conformational change in the lysozyme active site that is required for efficient catalysis. Shorter chains can bind, but they fail to induce this optimal structure, resulting in low activity. This demonstrates that the interaction and fit between the entire substrate and enzyme are critical for function, a key tenet of the induced-fit model.

Question 12

A student conducts an enzyme assay at a fixed, non-saturating substrate concentration and a constant temperature and pH. If the student repeats the experiment but doubles the concentration of the enzyme, what is the most probable outcome for the initial reaction rate?

  1. The initial rate will remain the same, as substrate concentration is the limiting factor.
  2. The initial rate will increase four-fold, because reaction rate is proportional to the square of enzyme concentration.
  3. The initial rate will increase, but by less than double, due to substrate limitation.
  4. The initial rate will approximately double, as there are twice as many available active sites. (correct answer)
Explanation: When the substrate concentration is not saturating, the reaction rate is limited by both the substrate concentration and the enzyme concentration. If the enzyme concentration is doubled, there will be twice as many active sites available to bind with the substrate molecules. This will lead to a doubling of the rate of effective collisions and thus a doubling of the initial reaction rate. The rate is directly proportional to the enzyme concentration in these conditions.

Question 13

An investigation shows that a newly discovered drug reduces the activity of a specific bacterial enzyme. When the reaction is run in the presence of the drug at a very high substrate concentration, the maximum reaction rate approaches the rate observed without the drug. What can be deduced about the mechanism of the drug?

  1. The drug is a competitive inhibitor that binds to the active site. (correct answer)
  2. The drug is a non-competitive inhibitor that binds to an allosteric site.
  3. The drug is an irreversible inhibitor that permanently denatures the enzyme.
  4. The drug acts as an allosteric activator, but only at low substrate concentrations.
Explanation: The key observation is that high substrate concentrations can overcome the inhibition. This is characteristic of competitive inhibition, where the inhibitor and substrate compete for the same active site. By increasing the substrate concentration, the probability of the substrate binding to the active site increases, allowing the reaction rate to approach its normal Vmax. Non-competitive and irreversible inhibition cannot be overcome by adding more substrate. An activator would increase the rate.

Question 14

In biotechnology, enzymes are often immobilized by attaching them to a solid support or trapping them in beads. For example, lactase is immobilized to produce lactose-free milk. What is a primary biochemical advantage of immobilization in industrial processes?

  1. Immobilization allows the enzyme to be easily recovered from the product and reused in subsequent batches. (correct answer)
  2. Immobilization significantly increases the enzyme's Vmax compared to its free-floating state.
  3. Immobilized enzymes are completely resistant to changes in temperature and pH that would normally cause denaturation.
  4. The process of immobilization converts competitive inhibitors present in the milk into useful cofactors for the enzyme.
Explanation: The major practical and economic advantage of immobilizing enzymes is that they are not free in the solution with the product. This allows for easy separation of the enzyme from the product stream, preventing contamination of the product with the enzyme. It also means the expensive enzyme can be easily recovered and reused for many cycles, making the process more cost-effective.

Question 15

An enzyme is incubated with an inhibitor, leading to a significant loss of activity. The mixture is then subjected to dialysis, a process that separates small molecules from large molecules. After dialysis, the enzyme's full activity is restored. What can be concluded about the nature of the inhibitor?

  1. The inhibitor was an irreversible inhibitor that was removed by dialysis.
  2. The inhibitor was a reversible inhibitor that dissociated from the enzyme and was removed. (correct answer)
  3. The inhibitor was a protein that was denatured and removed during the dialysis process.
  4. The inhibitor was consumed during the initial reaction, and dialysis removed the reaction products.
Explanation: Dialysis removes small molecules (like most inhibitors) while retaining large molecules (like enzymes). The restoration of activity indicates that the inhibitor was removed from the solution, allowing the enzyme to function again. This means the bond between the inhibitor and the enzyme was weak and non-covalent, allowing them to dissociate. This is the definition of reversible inhibition (either competitive or non-competitive). Irreversible inhibitors form strong, covalent bonds and would not be removed from the enzyme by dialysis.

Question 16

An enzyme's activity is studied under two conditions. Without any inhibitor, its Vmax is 100 µmol s⁻¹ and its Km is 2 mM. In the presence of a competitive inhibitor, its apparent Km becomes 8 mM, while its Vmax remains 100 µmol s⁻¹. At what substrate concentration will the initial reaction velocity be the same in both conditions?

  1. The velocities can never be the same at any finite substrate concentration. (correct answer)
  2. When the substrate concentration is equal to the average of the two Km values (5 mM).
  3. When the substrate concentration is equal to Vmax (100 mM).
  4. The velocities are only the same as the substrate concentration approaches infinity.
Explanation: The Michaelis-Menten equation is V = (Vmax * [S]) / (Km + [S]). To find where the velocities are equal, we set V(no inhibitor) = V(inhibitor): (100 * [S]) / (2 + [S]) = (100 * [S]) / (8 + [S]). We can cancel (100 * [S]) from both sides (assuming [S] > 0), leaving 1 / (2 + [S]) = 1 / (8 + [S]). This simplifies to 2 + [S] = 8 + [S], which gives 2 = 8. This is a contradiction, meaning there is no finite, positive substrate concentration where the velocities are identical. The inhibited reaction is always slower than the uninhibited one, although they converge at [S]=0 (V=0) and as [S] approaches infinity (V approaches Vmax).

Question 17

An enzyme extracted from a thermophilic bacterium living in hot springs has an optimal temperature of 85°C. When this enzyme is incubated at 37°C (human body temperature), the reaction rate is very low. What is the best explanation for this observation?

  1. The enzyme is irreversibly denatured at 37°C because its structure is not stable at this lower temperature.
  2. The substrate molecules have insufficient kinetic energy to bind effectively to the active site at 37°C.
  3. The enzyme has insufficient flexibility at 37°C, resulting in fewer effective collisions with the substrate molecules. (correct answer)
  4. A competitive inhibitor, normally inactive at 85°C, binds to the active site at 37°C, blocking the reaction.
Explanation: While low temperatures reduce kinetic energy (as in B), the structure of thermophilic enzymes is specifically adapted to be stable and functional at high temperatures. This often means they are more rigid than other enzymes. At temperatures far below their optimum, they lack the necessary flexibility for the conformational changes (like induced fit) required for efficient catalysis, leading to a very low rate of reaction. It is not denatured (A), as denaturation occurs at temperatures above the optimum.

Question 18

Niacin (vitamin B3) is a dietary requirement for humans as it is a precursor for the synthesis of the coenzyme NAD⁺. In cellular respiration, what is the fundamental role of NAD⁺?

  1. It acts as the final substrate in the Krebs cycle, being converted into ATP.
  2. It is an allosteric inhibitor that regulates the speed of glycolysis.
  3. It functions as an electron carrier, accepting high-energy electrons during catabolic reactions. (correct answer)
  4. It is a structural component of the mitochondrial inner membrane.
Explanation: Coenzymes are non-protein organic molecules that assist enzymes in their catalytic function. NAD⁺ is a classic example; its role in cellular respiration is to accept high-energy electrons (becoming reduced to NADH) from glucose breakdown in glycolysis and the Krebs cycle. It then transports these electrons to the electron transport chain, where their energy is used to synthesize ATP. It is a carrier molecule, not a substrate, inhibitor, or structural component.

Question 19

Why does increasing the concentration of the substrate often fail to overcome the effects of a non-competitive inhibitor?

  1. The non-competitive inhibitor has a much higher affinity for the active site than the natural substrate.
  2. The inhibitor is chemically transformed by the enzyme into a product that denatures other enzyme molecules.
  3. The inhibitor permanently and covalently binds to the active site, making it impossible for the substrate to bind.
  4. The inhibitor binds to the enzyme at an allosteric site, altering the active site's conformation and catalytic efficiency. (correct answer)
Explanation: Non-competitive inhibitors do not bind to the active site, so they do not compete with the substrate. They bind to an allosteric site elsewhere on the enzyme. This binding causes a conformational change that alters the shape of the active site, making it less effective at catalyzing the reaction. Since the inhibitor and substrate are not competing for the same site, adding more substrate does not displace the inhibitor and cannot restore the enzyme's maximum catalytic rate.

Question 20

The induced-fit model of enzyme action has largely replaced the lock-and-key model. Which statement provides the best distinction between the two models?

  1. In the lock-and-key model the active site is rigid, while in the induced-fit model the active site is flexible and can change shape upon substrate binding. (correct answer)
  2. The induced-fit model involves binding at an allosteric site, whereas the lock-and-key model involves binding only at the active site.
  3. The lock-and-key model describes reversible binding of a substrate, while the induced-fit model describes irreversible binding.
  4. In the induced-fit model the enzyme is denatured by the substrate, while in the lock-and-key model the enzyme remains unchanged.
Explanation: The core difference lies in the nature of the active site. The older lock-and-key model proposed a rigid active site with a fixed shape that perfectly matched the substrate. The induced-fit model proposes that the active site is flexible and that the binding of the substrate induces a conformational change in the enzyme, resulting in a tighter, more precise fit that facilitates the reaction.