Questions on enzyme action and specificity

Question 1

Explain why a given enzyme catalyses only one reaction, using the terms active site, complementary and enzyme-substrate complex in your answer.

Accepted Answer

B1 the active site is a region formed by the tertiary structure of the enzyme, with a particular shape B1 only a substrate whose shape is complementary to the active site can bind to it B1 when it binds, an enzyme-substrate complex forms and the reaction is catalysed B1 a molecule that is not complementary cannot enter the site, so no complex forms and that reaction is not catalysed

Question 2

Compare the lock-and-key model of enzyme action with the induced fit model.

Accepted Answer

B1 in lock and key the active site is already exactly complementary to the substrate, whereas in induced fit it is only close to complementary before binding B1 in lock and key nothing moves, whereas in induced fit the active site changes shape as the substrate binds B1 both models account for specificity, but only induced fit accounts for the rise in rate, because the closing of the site strains the substrate's bonds B1 induced fit is supported by crystal structures of enzymes with and without their substrate, such as hexokinase closing around glucose, which lock and key is not

Question 3

Explain four ways in which a closed active site can lower the activation energy of a reaction it catalyses.

Accepted Answer

B1 binding holds the substrate or substrates in the correct orientation for the reaction, reducing the energy wasted in collisions that would otherwise occur the wrong way round B1 as the site closes around the substrate it strains the bonds that are about to break, so they are partly broken before the reaction proper begins B1 the active site can exclude water from the reaction, creating a local chemical environment a beaker of solution could not reproduce B1 charged R groups lining the site can donate or accept protons directly, taking part in the catalysis chemically rather than only holding the substrate

Question 4

A biochemist adds extra enzyme to a reaction mixture already at equilibrium, and claims that, because the reaction now proceeds faster in both directions, more product will eventually be formed than before. Evaluate this claim.

Accepted Answer

B1 the claim is incorrect: adding more enzyme increases the rate at which equilibrium is reached in both directions, but does not move the position of equilibrium itself B1 the enzyme lowers the activation energy of the same route for the forward and the reverse reaction by the same amount, so both are speeded up in the same proportion B1 the ratio of product to reactant at equilibrium is fixed by the energy difference between them, which the enzyme does not change B1 judgement: more enzyme will bring the mixture to the same equilibrium sooner, which may be commercially useful, but it will not shift the balance towards more product

Question 5

An enzyme in a 2.0 cm³ sample converts 4.5 × 10⁻⁴ mol of substrate to product in 30 seconds. Calculate the rate of reaction in mol s⁻¹, and calculate how many moles would be converted in 2 minutes at that same rate.

Accepted Answer

M1 rate = amount converted ÷ time taken: 4.5 × 10⁻⁴ ÷ 30 A1 1.5 × 10⁻⁵ mol s⁻¹ M1 moles converted in 2 minutes = rate × 120 s A1 1.8 × 10⁻³ mol

Question 6

A single amino acid is changed within an enzyme's active site itself, rather than forty residues away. Explain why a change at the active site is more likely to abolish activity than a similarly sized change elsewhere in the protein, and explain one way the enzyme could still work despite the change.

Accepted Answer

B1 the active site's amino acids form the surface directly responsible for binding the substrate, and in induced fit, for straining it B1 a change there is more likely to alter the shape or the charge of that surface directly, rather than only indirectly through a knock-on effect on folding elsewhere B1 a change elsewhere in the protein may or may not disturb a bond that holds the fold together, so its effect on the site is less certain B1 the enzyme could still work if the new amino acid's R group is similar enough in size and properties to the original that the site remains complementary to the substrate

Question 7

Explain why enzymes are globular proteins, and why a fibrous protein such as collagen could not act as an enzyme.

Accepted Answer

B1 a globular protein folds with hydrophilic side chains facing outwards, so it dissolves in cytoplasm or plasma where the reaction happens B1 its amino acid sequence is irregular, so the surface it presents is irregular and can carry an active site of a particular shape B1 a fibrous protein has a repetitive sequence and forms long insoluble strands, so it is neither soluble nor able to present such a pocket

Question 8

A molecule built to resemble the strained, half-reacted form of a substrate binds to its enzyme far more tightly than the substrate itself does. Suggest what this tells us about the shape of the active site and which model of enzyme action it supports, and suggest why such a molecule would slow the enzyme down.

Accepted Answer

B1 the active site must be complementary to the strained transition state rather than to the relaxed substrate B1 this supports induced fit, in which the site moulds around the substrate and strains the bonds that are about to break B1 the molecule occupies the active site, so substrate cannot bind there and fewer enzyme-substrate complexes form

Question 9

A base substitution changes one amino acid in an enzyme, forty residues away from the active site along the chain, and the enzyme stops working. Explain how a change that far from the active site can destroy the enzyme's activity.

Accepted Answer

B1 the new amino acid has a different R group, so a bond that held the fold at that point, a hydrogen bond, an ionic bond or a disulfide bridge, may no longer form B1 those bonds hold the tertiary structure, and folding brings distant parts of the chain into contact, so a bond lost at one residue can move residues that line the active site B1 the active site is then a different shape, the substrate is no longer complementary to it, and no enzyme-substrate complex can form

Question 10

Describe three lines of evidence that led to the lock-and-key model being replaced by the induced fit model.

Accepted Answer

B1 X-ray crystallography caught enzymes in two shapes: hexokinase closes around a glucose molecule, which a rigid site could not do B1 molecules built to resemble the strained, half-reacted transition state bind far more tightly than the substrate itself, so the site must be shaped to grip the strained form B1 some enzymes accept a family of related substrates, as lipases hydrolyse triglycerides with a range of fatty acid tails, and a lock that accepts several keys is a poor lock

Question 11

Describe what happens to the shape of an induced-fit active site, from before the substrate binds to the formation of the enzyme-substrate complex.

Accepted Answer

B1 before binding, the active site is close to complementary to the substrate but not an exact fit B1 as the substrate binds, the active site moulds around it, changing shape to become more closely complementary B1 once bound, the enzyme-substrate complex is formed, with the substrate's bonds strained by the closing of the site

Question 12

Sucrase hydrolyses sucrose but has no effect on lactose, a different disaccharide. Suggest, in terms of active site shape, why sucrase does not catalyse the hydrolysis of lactose.

Accepted Answer

B1 the active site of sucrase has a particular shape, formed by its tertiary structure, that is complementary to sucrose B1 lactose has a different molecular shape, so it is not complementary to the active site and cannot bind there to form an enzyme-substrate complex B1 without an enzyme-substrate complex forming, the reaction is not catalysed, so lactose is hydrolysed only at its uncatalysed rate

Question 13

State what an enzyme does to the activation energy of a reaction, and state one quantity that an enzyme does not change.

Accepted Answer

B1 an enzyme lowers the activation energy of the reaction B1 it does not change the energy of the reactants or of the products, and so does not move the position of equilibrium

Question 14

A single molecule of catalase can decompose 6 × 10⁶ molecules of hydrogen peroxide per second. Calculate how many hydrogen peroxide molecules one catalase molecule decomposes in 5 minutes, giving your answer in standard form.

Accepted Answer

M1 convert the time to seconds: 5 × 60 = 300 s, and multiply the rate by it A1 6 × 10⁶ × 300 = 1.8 × 10⁹ molecules

Question 15

Name the model of enzyme action in which the active site is already exactly complementary to the substrate before binding, and name the model in which the active site changes shape as the substrate binds.

Accepted Answer

B1 lock and key B1 induced fit, named by Koshland