IB Biology C1.1: Enzymes and metabolism worked solutions

A. Glucose units are joined into a larger molecule, so the reaction is anabolic, and joining by loss of water is condensation. B and D name hydrolysis, which splits molecules, and C and D call a building reaction catabolic.

C. All enzymes lower the activation energy of the reaction they catalyse. They are not used up (A is wrong), each is specific to its substrate (B is wrong) and each has its own optimum pH (D is wrong).

B. In induced fit the active site changes shape as the substrate binds. A describes a rigid lock-and-key mould, C is untrue because the substrate is strained, and D places the binding away from the active site.

C. The two peaks differ by 90-55=35 kJ, and the reactant level is the same for both. A is the overall energy change, B is the activation energy with the enzyme and D is the activation energy without it.

D. Heat breaks the weak bonds that hold the tertiary structure, so the active site changes shape and the substrate no longer fits. Collisions become more frequent with heating (A), substrate does not run out as temperature rises (B), and the enzyme is not a reactant that is used up (C).

A. At pH 8 the charges on the amino acids change, the shape of the active site changes and the substrate no longer fits, so the rate is low. B, C and D ignore the effect of pH on the active site.

B. At the plateau all the active sites are occupied, so only more enzyme raises the rate. More substrate (A) does nothing, a lower temperature (C) slows collisions and an added product (D) would not help.

A. If extra substrate overcomes the inhibitor, it competes for the active site, so it is a competitive inhibitor. A non-competitive inhibitor (B) and an irreversible inhibitor (D) are not outcompeted, and C does not explain a rate that matches the normal one.

C. Salivary amylase is secreted into the mouth, so it acts extracellularly. Glycolysis (A), the Krebs cycle (B) and DNA replication (D) take place inside cells.

A. Rate =8\div10=0.8 cm^3 s^{-1}. B inverts the ratio, C forgets the time and D gives the mean over 30 s (18\div30=0.6) instead of the first 10 s.

(a) Metabolism is the sum of all the enzyme-catalysed reactions in a cell (1). (b) Any three of: energy is released or taken in small manageable amounts, not all at once (1); intermediates can be used to branch off to other products (1); each step has its own enzyme, so each step can be controlled separately (1); an enzyme can be specific to a small change in shape (1).

(a) The active site has a particular shape and pattern of charge (1) that fits only lactose, not sucrose (1). (b) Heat breaks the weak bonds, such as hydrogen bonds, holding the tertiary structure (1); the active site changes shape, so lactose can no longer bind and no enzyme-substrate complex forms (1); the peptide bonds are not broken, so the polypeptide is intact (1).

(a) Any two of: the enzyme can be reused (1); the product is not contaminated by enzyme (1); the enzyme is more stable over a wider range of temperature and pH (1). (b) Substrate must diffuse to the beads, or some active sites are blocked by the support (1). (c) A slow flow gives time for the substrate to enter the active site and be converted (1).

(a) Independent variable: pH of the buffer (1); dependent variable: volume of oxygen collected in 60 s (1). (b) Two controls with reasons, such as the same temperature, because temperature changes the rate and can denature the enzyme (1), and the same concentration and volume of hydrogen peroxide, because the substrate level sets the rate (1). (c) Repeat at each pH at least three times and calculate a mean (1). (d) A control with boiled extract shows that oxygen comes from active catalase and not from the peroxide decomposing by itself (1).

(a) 40 ^\circC, or between 30 and 50 ^\circC (1). (b) (30-14)/14\times100=114\,\% (1). (c) The rate falls from 52 at 40 ^\circC to 20 at 50 ^\circC and to 0 at 60 ^\circC (1). (d) At the higher temperatures weak bonds in the protein break (1) and the active site changes shape, so the substrate no longer fits: denaturation (1).

Any six of: at low temperature particles have little kinetic energy (1); as temperature rises they move faster (1) and collide more often and with more energy (1); more enzyme-substrate complexes form per second, so the rate rises (1) to a maximum at the optimum (1); above the optimum weak bonds break (1); the active site changes shape and the substrate no longer fits, so the enzyme is denatured (1); the rate falls towards zero (1).

(a) Rate =10/20=0.50 cm^3 s^{-1} (1). (b) Rate =(26-22)/(100-60)=0.10 cm^3 s^{-1} (1). (c) The substrate is used up, so its concentration falls (1) and there are fewer collisions with the active sites, so the rate falls (1). (d) The initial rate is measured when the substrate concentration is known and highest and no product has built up (1).

Any six of: both lower the rate of the enzyme (1); both can bind reversibly (1); competitive inhibitor binds at the active site (1); non-competitive inhibitor binds at another, allosteric, site (1); competitive inhibitor resembles the substrate (1); non-competitive inhibitor changes the shape of the active site (1); more substrate overcomes a competitive inhibitor (1); more substrate does not overcome a non-competitive one (1); the top rate is unchanged by a competitive inhibitor and lowered by a non-competitive one (1).

(a) End-product inhibition, which is negative feedback (1). (b) Z binds at an allosteric site away from the active site (1); the shape of the active site changes, so W no longer fits (1); binding is reversible, so when Z falls the enzyme works again (1). (c) It prevents the cell from making more Z than it needs, which would waste resources (1).

(a) P: the rate climbs towards the normal value as substrate rises, 11 against 33 at 5 but 83 against 95 at 200 (1), so extra substrate outcompetes it and P is competitive (1). Q: the rate is about half the normal rate at every concentration, so extra substrate does not overcome it and Q is non-competitive (1). (b) (42-83)/83\times100=-49\,\% (1). (c) The rate with P would approach the normal rate (1) because at very high concentration the substrate outnumbers the inhibitor at the active site (1). (d) A non-competitive inhibitor (1), because its effect does not fall when substrate rises, so a surge of substrate cannot overcome it (1).

(a) D bonds irreversibly to the active site (1), so extra substrate cannot displace it, unlike a reversible competitive inhibitor (1). (b) Without T the wall is not built or repaired, so it is weak (1); water enters the cell by osmosis from the dilute medium and the weak wall cannot resist, so the cell bursts (1). (c) Any two: a mutation changes the active site of T so D no longer binds (1); the bacterium makes an enzyme that breaks D down (1); a change reduces the entry of D (1). (d) Human cells have no cell wall and no enzyme T, so there is nothing for D to bind (1).

(a) Catabolic, because fatty acids are oxidised and broken down (1). (b) Reactions are not 100 % efficient (1), so part of the energy is released as heat and not captured in ATP (1). (c) The heat holds the body temperature near the optimum for its enzymes (1). (d) In a cold environment without this heat the body would cool (1); molecules would move more slowly, there would be fewer collisions and the rate of metabolic reactions would fall (1).

(a) At 30 ^\circC: 1000/120=8.3 (1); at 40 ^\circC: 1000/70=14.3 (1). (b) The fastest rate is at 40 ^\circC (1); only every 10 ^\circC was tested, so the true optimum is between 30 and 50 ^\circC and more temperatures should be tested (1). (c) Mean \pm SD at 40 is 65 to 75 s and at 50 is 81 to 109 s (1); the ranges do not overlap, so the film clears more slowly at 50 and the difference is probably real (1). (d) At 60 ^\circC the time is 400 s, so the rate is very low (1) because heat has broken the weak bonds, changed the shape of the active site and denatured the protease (1).

(a) The base substitution changes a codon, so a different amino acid joins the polypeptide (1). The new side chain forms different hydrogen, ionic or disulfide bonds (1), so the polypeptide folds differently and the tertiary structure changes (1), which changes the shape of the active site so the substrate no longer fits (1). (b) Amino acids of the active site bind the substrate and take part in catalysis directly (1), so a change there is more likely to stop the enzyme than one at the surface (1).

(a) Both are intracellular enzymes: those of glycolysis are free in the cytoplasm (1) and those of the Krebs cycle are in the mitochondrial matrix (1). (b) Each step has its own enzyme, so energy is released in small amounts (1) and each step can be regulated separately (1). (c) ATP binds an allosteric site of an early enzyme and slows the pathway (1); when ATP is used its concentration falls, it detaches, and the pathway speeds up again, which is negative feedback (1).

(a) Without cross-links the wall is weak and cannot resist the pressure of the cell contents (1) because transpeptidase is permanently blocked (1). (b) The dilute medium is hypotonic to the cytoplasm (1), so water enters by osmosis and the weak wall cannot resist the pressure, so the cell bursts (1). (c) An isotonic medium has the same solute concentration as the cytoplasm, so there is no net movement of water into the cell (1) and the bacteria would not burst, although penicillin still blocks wall building (1).